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Looking for closed-forms of $\int_0^{\pi/4}\ln^2(\sin x)\,dx$ and $\int_0^{\pi/4}\ln^2(\cos x)\,dx$ A few days ago, I posted the following problems
Prove that
\begin{equation}
\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}\\[20pt]
-\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3)
\end{equation}
and the OP receives some good answers even I then could answer it.
My next question is finding the closed-forms for
\begin{align}
\int_0^{\pi/4}\ln^2(\sin x)\,dx\tag1\\[20pt]
\int_0^{\pi/4}\ln^2(\cos x)\,dx\tag2\\[20pt]
\int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt\tag3
\end{align}
I have a strong feeling that the closed-forms exist because we have nice closed-forms for
\begin{equation}
\int_0^{\pi/4}\ln(\sin x)\ dx=-\frac12\left(C+\frac\pi2\ln2\right)\\
\text{and}\\
\int_0^{\pi/4}\ln(\cos x)\ dx=\frac12\left(C-\frac\pi2\ln2\right).
\end{equation}
The complete proofs can be found here.
As shown by Mr. Lucian in his answer below, the three integrals are closely related, so finding the closed-form one of them will also find the other closed-forms. How to find the closed-forms of the integrals? Could anyone here please help me to find the closed-form, only one of them, preferably with elementary ways (high school methods)? If possible, please avoiding contour integration and double summation. Any help would be greatly appreciated. Thank you.
|
Following the same approach as in this answer,
$$ \begin{align} &\int_{0}^{\pi/4} \log^{2} (2 \sin x) \ dx = \int_{0}^{\pi/4} \log^{2}(2) \ dx + 2 \log 2 \int_{0}^{\pi/4}\log(\sin x) \ dx + \int_{0}^{\pi /4}\log^{2}(\sin x) \ dx \\ &= \frac{\pi}{4} \log^{2}(2) - \log (2) \left(G + \frac{\pi}{2} \log (2) \right) + \int_{0}^{\pi/4} \log^{2}(\sin x) \ dx \\ &= \int_{0}^{\pi /4} \left(x- \frac{\pi}{2} \right)^{2} \ dx + \text{Re} \int_{0}^{\pi/4} \log^{2}(1-e^{2ix}) \ dx \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \text{Im} \int_{{\color{red}{1}}}^{i} \frac{\log^{2}(1-z)}{z} \ dz \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \text{Im} \left(\log^{2}(1-i) \log(i) + 2 \log(1-i) \text{Li}_{2}(1-i) - 2 \text{Li}_{3}(1-i) \right) \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \left(\frac{\pi}{8} \log^{2}(2) - \frac{\pi^{3}}{32} + \log(2) \ \text{Im} \ \text{Li}_{2}(1-i) - \frac{\pi}{2} \text{Re} \ \text{Li}_{2}(1-i)- 2 \ \text{Im} \ \text{Li}_{3}(1-i)\right) . \end{align}$$
Therefore,
$$ \begin{align}\int_{0}^{\pi/4} \log^{2}(\sin x) \ dx &= \frac{\pi^{3}}{48} + G \log(2)+ \frac{5 \pi}{16}\log^{2}(2) + \frac{\log(2)}{2} \text{Im} \ \text{Li}_{2}(1-i) - \frac{\pi}{4} \text{Re} \ \text{Li}_{2}(1-i) \\ &- \text{Im} \ \text{Li}_{3}(1-i) \approx 2.0290341368 . \end{align}$$
The answer could be further simplified using the dilogarithm reflection formula $$\text{Li}_{2}(x) {\color{red}{+}} \text{Li}_{2}(1-x) = \frac{\pi^{2}}{6} - \log(x) \log(1-x) $$
and the fact that $$ \text{Li}_{2}(i) = - \frac{\pi^{2}}{48} + i G.$$
EDIT:
Specifically, $$\text{Li}_{2}(1-i) = \frac{\pi^{2}}{16} - i G - \frac{i \pi}{4} \log(2). $$
So $$\int_{0}^{\pi /4} \log^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{ \log(2)}{2} + \frac{3 \pi}{16} \log^{2}(2) - \text{Im} \ \text{Li}_{3}(1-i).$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/917154",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "35",
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|
The diophantine equation $a^2+ab-b^2=0$ I first tried with brute force with $-1000 \leq a,b \leq 1000$ but found no solution.
But then a simple argument showed me that there was no solution. Not only in the integers, but even for the rationals. What was that argument?
|
And why is it impossible to solve the equation? $a^2+ab-b^2=0$
$$a=\frac{-b\pm\sqrt{b^2+4b^2}}{2}=b\frac{-1\pm\sqrt{5}}{2}$$
you can see the whole decision no.
If $a^3+ab-b^2=0$ would have acted similarly. $b^2-ab-a^3=0$
$$b=\frac{a\pm\sqrt{a^2+4a^3}}{2}=a\frac{1\pm\sqrt{4a+1}}{2}=a\frac{1\pm{k}}{2}$$
$a$ - to choose such that the root was intact. $a=\frac{k^2-1}{4}$
If $a^2+pab-b^2=0$
$$a=\frac{-pb\pm\sqrt{p^2b^2+4b^2}}{2}=b\frac{-p\pm\sqrt{p^2+4}}{2}$$
There are solutions when: $p=0 $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920480",
"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int\frac{dx}{x\sqrt{x^2+1}}$ I would like to ask for some help regarding the following indefinite integral, tried integration by parts and trigonometric substitution which both brought me to $\int\frac{\sec\theta}{\tan\theta}d\theta$, and from this point it is messy to integrate by parts, any help would be appreciated.
$$\int\frac{dx}{x\sqrt{x^2+1}}$$
|
$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}$
$\displaystyle z^{2}=x^{2}+1\Rightarrow 2zdz=2xdx\Rightarrow xzdz=x^{2}dx=
(z^{2}-1)dx$
$\displaystyle\Rightarrow \frac{dz}{z^{2}-1}=\frac{dx}{xz}$
$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\int\frac{dz}{z^{2}-1}=\frac{1}{2}\int(\frac{1}{z-1}-\frac{1}{z+1})dz$
$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\frac{1}{2}ln\left| \frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1} \right|+c$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Let $p \geq 2$. Prove that if $2^p-1$ is prime then $p$ must also be prime. Would the following be a valid proof?
Let $r$ and $s$ be positive integers, then the polynomial $x^{rs}-1=(x^r -1)(x^{s(r-1)}+x^{s(r-2)}+\cdots+x^r+1)$.
So if $p$ is composite (say $rs$ with $1<s<p$), then $2^p-1$ is also composite (because it is divisible by $2^s-1$)
|
By contrapositive, if $p$ is not prime, then $2^p-1$ is not prime. Since $p$ is not prime, is composite, so $p=nr$ with $1<r<p, 1<n<p$. Now $2^p-1=(2^r)^n-1$, use the change $x=2^r$ we get $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+ \cdots+x+1).$$
Clearly $(x^{n-1}+x^{n-2}+x^{n-3}+ \cdots+x+1)>1.$ Since $x=2^r \wedge r>1, 2^r>1 \Rightarrow x-1>1.$ Thus, $x^n-1=2^{rn}-1=2^p-1$ is the product of terms greater than 1. Finally, $2^p-1$ is not prime
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920991",
"timestamp": "2023-03-29T00:00:00",
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|
Find the sum $1+\cos (x)+\cos (2x)+\cos (3x)+....+\cos (n-1)x$ By considering the geometric series $1+z+z^{2}+...+z^{n-1}$ where $z=\cos(\theta)+i\sin(\theta)$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+...+\cos(n-1)\theta$ = ${1-\cos(\theta)+\cos(n-1)\theta-\cos(n\theta)}\over {2-2\cos(\theta)}$
I've tried expressing $\cos(n\theta)$ as ${e^{in\theta}+e^{-in\theta}} \over {2}$ but I don't think that will lead anywhere. Does it help that $1+z+z^{2}+z^{3}+...+z^{n-1}$=$e^{0i\theta}+e^{i\theta}+e^{2i\theta}+e^{3i\theta}+...+e^{(n-1)i\theta}$?
So the sum $\sum_{r=0}^{n-1} e^{ir\theta}$=${e^{ni\theta}-1} \over {e^{i\theta}-1}$
Thank you in advance :)
|
$$M=cos(\theta)+cos(2\theta)+cos(3\theta) +....+cos((n-1)\theta)=\\\frac{2sin(\frac{\theta}{2})}{2sin(\frac{\theta}{2})}M=\\\frac{2sin(\frac{\theta}{2})}{2sin(\frac{\theta}{2})}cos(\theta)+cos(2\theta)+cos(3\theta) +....+cos((n-1)\theta)=\\
\frac{(sin(\frac{3\theta}{2})-sin(\frac{1\theta}{2}))+(sin(\frac{5\theta}{2})-sin(\frac{3\theta}{2}))+(sin(\frac{7\theta}{2})-sin(\frac{5\theta}{2}))++...+(sin(\frac{(\frac{n+1}{2}\theta}{2})-sin(\frac{n-1}{2}\frac{1\theta}{2}))}{2sin(\frac{\theta}{2})}\\so\\M=
\frac{-sin(\frac{1\theta}{2})+sin(\frac{n+1}{2}\theta)}{2sin(\frac{\theta}{2})}\\then\\find \\M+1\\M+1=1+\frac{-sin(\frac{1\theta}{2})+sin(\frac{n+1}{2}\theta)}{2sin(\frac{\theta}{2})}
$$
|
{
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|
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y)=(y-1)^2$$
this equality shows that $x$ and $z$ have same sign(both of them are positive or negative).like this we can conclude that $xy=(x-1)^2$ and $yz=(x-1)^2$.So all variables are positive or negative.If all of them are negative then $x+y+z \neq 2$ so all of $x,y,z$ are non negative. I don't know what to do for showing that none of $x,y,z$ will be be bigger than $\frac{4}{3}$.
|
Note we also have $x^2+y^2+z^2=2$ (expand $(x+y+z)^2=4$).
Note $y^2+z^2 \geq 2yz=2-2x(y+z)=2-2x(2-x)$, so
$x^2+y^2+z^2 \geq 3x^2-4x+2$
But $3x^2-4x+2 >2$ for $x>\frac{4}{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have a quadratic equation with the form:
$$-\frac{b}{2}x^2 + bx + b = 0$$
Applying $x_1$ for our equation I get $b=0$. Why?
The final answer is: $x^2-2x-2$.
|
The quadratic (not quartic) equation of a real number $x$ referres to the monic quadratic polynomial that annihilates $x$. So, the equation $ax^2+bx+c=0$ should only be verified on $x = x_1$ and $x= x_2$.
As expected, making $x = x_1$ you should attain a zero regardless of $b$, you should do that calculation again.
The correct answer should be the one that makes the polynomial monic, i.e., the term $x^2$ with coefficient 1, so $$\frac{b}{2} = -1 \Leftrightarrow b= -2$$ As you observed in the correct answer.
|
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|
Finding maxima values of a multivariate function
Find the maximal value of the function for $a=24.3$, $b=41.5$:
$$f(x,y)=xy\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}$$
Using the second derivative test for partial derivatives, I find the critical point in terms of $a$ and $b$ by taking partial derivatives of $x$ and $y$ and equating them to $0$.
$$f_y=0$$
$$f_x=0$$
Getting
$$y \left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right ) -\frac{yx^2}{a^2}=0$$
and
$$x \left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right ) - \frac{xy^2}{b^2}=0$$
Then i combine both equations together to get
$$\left ( 1-\frac{y^2}{b^2} \right ) =\frac{2x^2}{a^2}$$
and
$$\left ( 1-\frac{x^2}{a^2} \right ) =\frac{2y^2}{b^2}$$
Solving both equations to get the critical points in terms of a and b. I got
$$(0,0)$$
$$\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}}\right)$$
Hence to get the maximum value I substitute the second critical point back into the original function. And I let $a=24.3$, $b=41.5$ to get the maximal point. However, I don't seem to get the right answer.
Is my method correct?
|
Another way is to consider equivalently the maximum of
$$\frac{f^2}{a^2b^2}=\frac{x^2}{a^2} \frac{y^2}{b^2} \left(1-\frac{x^2}{a^2} -\frac{y^2}{b^2} \right)$$
which is the product of three positive terms with constant sum, so each term must be equal to (in this case) $\frac13$ at maximum.
|
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|
Known exact values of the $\operatorname{Li}_3$ function We know some exact values of the trilogarithm $\operatorname{Li}_3$ function.
Known real analytic values for $\operatorname{Li}_3$:
*
*$\operatorname{Li}_3(-1)=-\frac{3}{4} \zeta(3)$
*$\operatorname{Li}_3(0)=0$
*$\operatorname{Li}_3\left( \frac{1}{2} \right) = \frac{7}{8} \zeta(3) - \frac{1}{12} \pi^2 \ln 2 + \frac{1}{6} \ln^3 2$
*$\operatorname{Li}_3(1) = \zeta(3),$ where $\zeta(3)$ is the Apéry's constant
*$\operatorname{Li}_3\left(\phi^{-2}\right)=\frac{4}{5} \zeta(3) + \frac{2}{3} (\ln \phi)^3 - \frac{2}{15} \pi^2 \ln \phi,$ where $\phi$ is the golden ratio.
Using identities for the list above we could also get:
*
*$\operatorname{Li}_3(2) = \frac{\pi^2}{4} \ln 2 + \frac{7}{8} \zeta(3) - \frac{\pi}{2} \ln^2(2) \cdot i,$ or we could write into this alternate form.
*$\operatorname{Li}_3\left(\phi^2\right) = \frac{4}{5} \zeta(3) - \frac{2}{3} \ln^3 \phi + \frac{8\pi^2}{15}\ln \phi - 2\pi\ln^2(\phi) \cdot i,$ or there is an alternate form here.
We know even less about complex argumented values:
*
*$\operatorname{Li}_3(i)=-\frac{3}{32}\zeta(3) +\frac{\pi^3}{32} i$
*$\operatorname{Li}_3(-i)=-\frac{3}{32}\zeta(3) -\frac{\pi^3}{32} i.$
There are some partial result for complex cases:
*
*$\Im\left[ \operatorname{Li}_3 \left( \frac{1 \pm i}{\sqrt{2}} \right) \right] = \pm \frac{7\pi^3}{256}$, and there is an expression for the real part in term of derivatives of digamma function,
*other related specific values around the unit circle like this or this, etc.
Lucian said the following in this question:
*
*$\Re\left[\text{Li}_3\left(\dfrac{1+i}2\right)\right]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).$
While working on Lucian's problem I was able to specify this one
*
*$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$
Than I got the idea to write a question, maybe someone could give us some more specific values of the trilogarithm function.
|
It appears that
$$
\Im\left[\mathrm{Li}_3\left((-1)^{\frac{1}{2^n}}\right) \right]=\frac{(2^n-1)(2^{n+1}-1)}{3\cdot2^{3n+2}}\pi^3
$$
but that's by experimentation.
|
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|
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
Prove $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$
Things I have done: after trying many ways and failing, I reached the fact that$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2<\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{100}{101}\right)=\frac{1}{101}$
So $\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true. it remains to show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}$.
I'm thinking of applying my approach on proving this part. something like this. $$\frac{1}{225}<\frac{1}{x}=\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right) \times B<\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2$$
And another thing I'm curious about it,is there a way to approximate value of $\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}$ ?
|
Note that $$\frac{n-1}{n}\gt\frac{n-2}{n-1}$$
for $n\gt 1$. So, we have $$\begin{align}\left(\frac 12\times \frac 34\times \cdots\times \frac{99}{100}\right)^2&\gt \left(\frac 12\times \frac 34\times\frac 56\times \cdots\times \frac{99}{100}\right)\left(\color{red}{\frac 12}\times \frac 23\times \frac{4}{5}\times\cdots\times \frac{98}{99}\right)\\&=\frac{1}{200}\\&\gt\frac{1}{225}\\&=\frac{1}{15^2}\end{align}$$
|
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|
finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to integrate. I tried to split them into two $\displaystyle\int\frac{2}{x+2
}dx$ and $\displaystyle\int\frac{1-2x}{x^2+1}dx$ but I don't know how to do after.
|
Use
$$\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C.$$
Note that $$\int\frac{2}{x+2}dx=2\int\frac{(x+2)'}{x+2}dx$$
$$\int\frac{1-2x}{x^2+1}dx=-\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx=-\int\frac{(x^2+1)'}{x^2+1}dx+\int\frac{1}{x^2+1}dx.$$
Here, set $x=\tan\theta$ for $$\int\frac{1}{x^2+1}dx.$$
|
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|
Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $ I'm looking for a closed form of this integral.
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$
where $\operatorname{Li}_2$ is the dilogarithm function.
A numerical approximation of it is
$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$
As Lucian said $I$ has the following equivalent forms:
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$
According to Mathematica it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us this form.
With Maple using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's answer, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits.
I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.
I hope it exists. Similar integrals are the following.
$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$
Related techniques are in this or in this paper. This one also could be useful.
|
My attempt. This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.
First, from Landen's identity of the following form
$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \quad z \notin [1, \infty)$$
we observe that
\begin{align*}
I
&= -\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \left\{ \mathrm{Li}_2\left(-\frac{x}{1-x}\right) - \frac{1}{2}\log^{2}(1-x) \right\} \, dx \\
&= -\int_{-\infty}^{0} \frac{2\mathrm{Li}_2 (t) + \log^{2}(1-t)}{2\sqrt{1-2t}(1-t)} \, dt
\tag{1}
\end{align*}
By noting that
$$ \frac{d}{dt} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) = \frac{1}{2\sqrt{1-2t}(1-t)}, $$
integration by parts and the substitution $x = (1-2t)^{-1/2}$ shows that (1) is equal to
\begin{align*}
I
&= \int_{-\infty}^{0} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) \frac{2\log(1-t)}{t(t-1)} \, dt \\
&= \int_{0}^{1} \frac{8x \arctan x}{1 - x^4} \log \left( \frac{1+x^2}{2x^2} \right) \, dx \tag{2}
\end{align*}
The following observation
$$ \Re \log \left(\frac{1+ix}{\sqrt{2}} \right) = \frac{1}{2}\log \left( \frac{1+x^2}{2} \right)
\quad \text{and} \quad
\Im \log \left(\frac{1+ix}{\sqrt{2}} \right) = \arctan x $$
somehow seems to suggest complex-analytic approach, but I have not been successful with such approaches so far. Next, from the following simple formula
$$ \log \left( \frac{1+x^2}{2x^2} \right) \, dx = \int_{0}^{1} \frac{d}{dy} \log \left( \frac{y^2+x^2}{y^2 + 1} \right) \, dy $$
the integral (2) can be further decomposed into the following form
$$ I = \int_{0}^{1}\int_{0}^{1} \frac{16xy \arctan x}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy. \tag{3} $$
Simple calculation shows that
$$ \int_{0}^{1}\int_{0}^{1} \frac{16xy}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy = 2\zeta(2), $$
so I suspect that the situation in (3) is not that bad.
|
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|
Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
|
$$ \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + \dots \right)^2 = \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + O(x^4) \right)^2 =$$
$$= 1+ \dfrac{1}{4}x^2 + \dfrac{1}{64}x^4 + \dfrac{1}{256} x^6+1-\dfrac{1}{4}x^2 - \dfrac{1}{8}x^3 + \dfrac{1}{8}x^3 +\dfrac{1}{16}x^4 - \dfrac{1}{64}x^5 + O(x^4) =$$
$$=1+x + O(x^4)$$
|
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|
Rubiks Cube Landing on Red A Rubik’s cube in which each side is painted one of six colors (white, orange, red, blue, green and yellow).
Suppose each side of the Rubik’s cube consists of only one color, if the Rubik’s cube is tossed 6 times what is the probability that the cube will land on the red side at least 4 times?
|
Add up the following:
*
*The probability that it lands on the red side exactly $4$ times is $\dbinom{6}{4}\cdot\dfrac{5^2}{6^6}=\dfrac{375}{46656}$
*The probability that it lands on the red side exactly $5$ times is $\dbinom{6}{5}\cdot\dfrac{5^1}{6^6}=\dfrac{30}{46656}$
*The probability that it lands on the red side exactly $6$ times is $\dbinom{6}{6}\cdot\dfrac{5^0}{6^6}=\dfrac{1}{46656}$
Another way to look at it:
*
*The total number of ways it can land is $6^6=46656$
*The number of ways it can land on the red side exactly $4$ times is $\dbinom{6}{4}\cdot5^2=375$
*The number of ways it can land on the red side exactly $5$ times is $\dbinom{6}{5}\cdot5^1=30$
*The number of ways it can land on the red side exactly $6$ times is $\dbinom{6}{6}\cdot5^0=1$
In either case, the probability of it landing on the red side at least $4$ times is $\dfrac{375+30+1}{46656}$
|
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|
Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here
$\int \dfrac{1}{4 x^4} \dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here please help
|
Hint: Let be $u = x^2$, then $du = 2x\,dx$, and so you got
$$ \int\,x^3\sqrt{x^2+4}\,dx = \dfrac{1}{2}\int\,u\sqrt{u+4}\,du $$
and the let be $t = u+4$, so $u = t-4$, $dt = du$, and
$$ \dfrac{1}{2}\int\,u\sqrt{u+4}\,du = \dfrac{1}{2}\int\,(t-4)\sqrt{t}\,dt $$
you can calculate this integral by yourself.
|
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|
Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer would be $0$, but apparently this is wrong.
The correct answer is $\displaystyle{{1 \over 2}\left(\,a - b\,\right)}$.
|
Substitute $x=1/t,$ then our expression becomes $$\lim_{t \to 0} \frac{\sqrt{1+at^2+t^4}-\sqrt{1+bt^2+t^4}}{t^2}$$
Note that here we can apply the L'Hospitals rule.
$$\frac{\frac{1}{2}(1+at^2+t^4)^{-1/2}(2at+4t^3)}{2t} \to \frac {a}{2}$$
as $t \to 0.$
Therefore it easy to see that required limit is $$\frac{a-b}{2}.$$
|
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|
How to prove this equation? $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$ Suppose $a, b$, and $c$ are nonzero real numbers which satisfy the equation: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$
Prove: if $n$ is an odd integer, then $a^n + b^n + c^n=(a+b+c)^n$
I was thinking of relating this to natural logs and derivatives somehow but then i'm not even sure how that would help.
|
Consider the expression
$$\frac{1}{x}+\frac{1}{b}+\frac{1}{c} = \frac{1}{x+b+c}.$$
This can be simplified to
$$(x+b+c)(x(b+c)+bc) - x(bc)=0.$$
This is a second degree polynomial in $x$. If we set $x=-b$, then this satisfies the given equation, hence $x+b$ is a factor of this polynomial. By symmetry $x+c$ is also a factor of this polynomial. Thus for some constant $K$ we get
$$(x+b+c)(x(b+c)+bc) - x(bc)= K(x+b)(x+c).$$
By direct comparison the constant $K$ turns out to be $b+c$. Thus
$$(x+b+c)(x(b+c)+bc) - x(bc)= (b+c)(x+b)(x+c).$$
Said differently we have
$$(a+b+c)(a(b+c)+bc) - a(bc)= (b+c)(a+b)(a+c)=0.$$
Thus
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}.$$
is equivalent to
$$(a+b)(b+c)(c+a)=0.$$
Thus at least one of the following must be true
$$a=-b \quad \text{ or } \quad b=-c \quad \text{ or } \quad c=-a.$$
Now we can use the fact that $n$ is odd to claim that
$$a^n + b^n + c^n=(a+b+c)^n.$$
|
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|
Equivalent of $ x(x+1)(x+2)\cdots(x+n)$? Assume $x>0$. Is there an equivalent for this quantity $$ x(x+1)(x+2)\cdots(x+n)$$ as $n$ tends to $+\infty$?
I've tried to write $$x(x+1)(x+2)\cdots(x+n)=x^{n+1}\left(1+\frac 1x\right)\left(1+\frac 2x\right)\cdots\left(1+\frac nx\right)$$ I don't know if I'm on the right track... Thanks for your help!
|
I showed many years ago that
the closest $n$-th power
of an integer
to $x(x+1)...(x+n-1)$
for integer $x$
is $(x+[(n-1)/2])^n$
whenever $x$ is large enough
compared to $n$.
A sufficient condition is,
with $u = x+[(n-1)/2]$,
$u \geq (n^2-n)/4$
for even $n$
and
$u \geq (n^2+3n-4)/4$
for odd $n$.
Some more precise results,
with $u = x+[(n-1)/2]$
(the proofs of which are not easy):
The smallest value of $u$
such that $u^n$
is closer to $x(x+1)...(x+n-1)$
than $(u-1)^n$
is
$ \dfrac{n^2-1}{12} + \dfrac{n-1}{4}
+ \dfrac{13}{20} + O\left(\dfrac1{n}\right)$.
The smallest value of $u$
such that $(u-1/2)^n$
is closer to $x(x+1_...(x+n-1)$
than $(u-3/2)^n$
is
$ \dfrac{n^2-1}{24} + \dfrac{n-1}{8}
+ \dfrac{13}{10} + O\left(\dfrac1{n}\right)$.
|
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|
The maximum possible value of $ (xv - yu)^2 $ over the surface ... The maximum possible value of $ (xv - yu)^2 $ over the surface given by the equations
$ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $ is :
I solved it and my answer comes out to be $9$ but the correct answer is $36$.
Here is what I did :
Put $ x = r_1\cos\theta$ and $y = r_1\sin\theta $
$ u = r_2\sin\theta $ and $ v = r_2\cos\theta$
Using $ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $
we get $ r_1 = \sqrt 2 $ and $ r_2 = \frac 3{\sqrt 2} $
Now $ (xv - yu)^2 = (r_1r_2)^2(\cos^2\theta - \sin^2\theta)^2 $
$= 9(1-2\sin^2\theta)^2$
Now this will be maximum when $ (1- 2\sin^2\theta)^2 $ is maximum which is at $ \sin^2\theta = 0,1 $ so the maximum value of the expression comes out to be $9$.
Where did I go wrong here? Or am I correct?
|
Another way :
$$(xv-yu)^2=(x^2+y^2)(u^2+v^2)-(xu+yv)^2=36-(xu+yv)^2\le 36.$$
The equality is attained when $(x,y,u,v)=\left(\frac{4\sqrt 2}{3},-\frac 23,1,2\sqrt 2\right).$
P.S. When $(x,y,u,v)=\left(\frac{4\sqrt 2}{3},-\frac 23,1,2\sqrt 2\right)$, the followings hold :
$$xu+yv=0,\ \ \ x^2+y^2=4,\ \ \ u^2+v^2=9.$$
|
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|
Finding the second derivative of a function (implicit differentiation) I am not sure if I got the correct answer or not. It was a homework from my textbook, and it does not have answers for even number questions...
Original Function: $x^2y-4x=5$
$\frac{dy}{dx}$ = $\frac{4-2xy}{x^2}$
It seems I got to finding the dy/dx part correctly according to wolfram alpha
Heres the steps to finding $\frac{dy^2}{d^2x}$
*
*$\frac{(x^2)(-2y-2x\frac{dy}{dx}) - ((-2xy+4)(2x))}{x^4}$
*$\frac{(x^2)(-2y-2x\frac{4-2xy}{x^2}) - ((-2xy+4)(2x))}{x^4}$
*$\frac{(-2x^2y+4x^2y-8x) - (-4x^2y+8x)}{x^4}$
*$\frac{6x^2y-16x}{x^4}$
*$\frac{6(5+4x)-16x}{x^4}$
*$\frac{30+24x-16x}{x^4}$
*$\frac{30+8x}{x^4}$ = $\frac{dy^2}{d^2x}$
Can anyone tell me if I did it correctly?
|
i have also
$y'=\frac{4-2xy}{x^2}$
and by the quotient rule
$y''=\frac{(-2y-2xy')x^2-(4-2xy)2x}{x^4}$
plugging $y''$ in the equation above we obtain
$y''=\frac{\left(-2y-2x\left(\frac{4-2xy}{x^2}\right)\right)x^2-(4-2xy)2x}{x^4}$
simplifying this we get
$y''=\frac{6xy-16}{x^3}$
|
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|
proportion of primes in a polynomial sequence It is conjectured (Bunyakovsky) that when $P(x)$ is a polynomial from $\mathbb{Z}[X]$, irreducible, with positive leading coefficient and so that the integers $P(n)$ , $n\gt0$ do not share a common factor different from $1$, then there are infinitely many primes in the sequence $P(n)$ , $n\gt0$ .
But what about the proportion (density) of those primes? It seems that it can vary a lot, even when the degree of the polynomial is not changed.
For instance, for degree $10$, let $$P(x)=1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+x^{10}$$
and $$Q(x)=-691+2073 x-287 x^2-3285 x^3+1420 x^4+2310 x^5-1190 x^6-1050 x^7+525 x^8+525 x^9+105 x^{10}$$
there are more than $60$ values of $n$ below $1000$, for which $P(n)$ is prime, whereas in the same range, $Q(n)$ is prime only when $n=129$ or $n=539$.
Can somebody find another such polynomial sequence, with an even lesser proportion of primes?
|
Chinese remainder theorem.
Pick $R(x)$ with positive coefficients so that $$\begin{align}
R(x)&\equiv x^9(x-2)\pmod 3\\
R(x)&\equiv x^7(x-1)(x-2)(x-3)\pmod 5\\
R(x)&\equiv x^5(x-1)(x-2)(x-4)(x-5)(x-6)\pmod 7\\
R(x)&\equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)\\&\equiv x^{10}-1\pmod {11}
\end{align}
$$
Then $R(n)$ will be divisible by one of $3,5,7,11$ unless $n\equiv 1\pmod 3,4\pmod 5,$ etc. This means the only $n$ we need to check is $n\equiv 1144\pmod {1155=3\cdot5\cdot7\cdot 11}$. So in particular, $R(n)$ is not prime for $n=0,1,2,3\dots,2298$.
Note, the above polynomials are not hard to compute, since, for example, $$\begin{align}
(x-1)(x-2)(x-4)(x-5)(x-6)&\equiv\frac{x^6-3^6}{x-3}
\\&\equiv x^5+3x^4+3^2x^3+3^3x^2+3^4x+3^5\\
&\equiv x^5+3x^4+3x^3+6x^2+4x+5\pmod 7
\end{align}$$
On the other hand, your first $P(x)$ is the polynomial for the primitive $22$nd roots of unity, which means that if $p\mid P(n)$ then $n$ is a primitive $22$nd root of unity, modulo $p$. That can only happen if $p\equiv 1\pmod {22}$, so, in particular, $P(n)\equiv 0\pmod {2,3,5,7}$ is not possible, and the only eliminated value modulo $11$ is $10$. This gives a far larger set of prospective value of $n$.
|
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|
Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
\frac{(\sin 2x)}{(\cos 2x)}=2\sin x
$$
$$
\frac{(2\sin x \cos x)}{((\cos x)^2-(\sin x)^2)}=2\sin x
$$
$$
2\sin x\cos x=2\sin x\cos^2 x-2\sin^3 x
$$
$$
\sin x\cos x=\sin x(\cos^2 x-\sin^2x)
$$
$$
\cos x=\cos^2 x-\sin^2 x
$$
$$
\cos x=\cos^2 x+\cos^2x-1
$$
$$
2\cos^2x-\cos x-1=0
$$
Solving this quadratic equation gives $\cos x=1$ or $\cos x=-\frac 12$
$$
\cos x=1=\cos \theta
$$
$$
x=n2\pi
$$
$$
\cos x=-\frac 12=\cos(\frac{2\pi}3)
$$
$$
x=\pm\frac{2\pi}3+n2\pi
$$
|
We have $2\sin x\cos2x=\sin2x=2\sin x\cos x\iff2\sin x(\cos2x-\cos x)=0$
$\sin x=0\implies x=m\pi\ \ \ \ (1)$ where $m$ is any integer
else $\cos2x-\cos x=0\iff\cos2x=\cos x\implies2x=2r\pi\pm x$ where $r$ is any integer
Taking the '+' sign, $x=r\pi$ which is same as $(1)$
Taking the '-' sign, $2x=2r\pi-x\iff x=\dfrac{2r\pi}3$
|
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|
how many solutions to non-linear simultaneous equations I'm doing a Lagrange multiplier optimization problem, and I wound up with the following simultaneous equations:
$2x + 1 -2\lambda x = 0$
$4y-2 \lambda y = 0$
$6z-2 \lambda z = 0$
$-x^2 - y^2 - z^2 + 1 = 0$
I am asked to find ALL solutions, and I managed to find 2 solutions (had to assume that y and z were both zero so equation 2 and 3 would make sense) :
$\lambda_1 = \frac{1}{2}, x_1 = -1$
$\lambda_2 = \frac{3}{2}, x_2 = 1$
But how do I know that these two are ALL the solutions? Is there any way to tell how many solutions I should be able to get, or maybe suggest how to get more solutions?
Thanks.
|
Since the 2nd and 3rd equation are contradictory when both y and z are not zero, I must try several scenarios where just y, just z, and both y and z are equal to 0. For the y=z=0 case:
Solve equation 1 for x:
$x = \frac{-1}{2-2\lambda}$
plug into equation 4, solve for lambda:
$\lambda = \frac{8 \pm \sqrt{64-48}}{8}$
For each of these two answers, plug back into the x equation for 2 solutions:
$
x_1 = -1,
y_1 = 0,
z_1 = 0,
\lambda_1 = \frac{1}{2}$
$
x_2 = 1 ,
y_2 = 0 ,
z_2 = 0 ,
\lambda_2 = \frac{3}{2}
$
Now take the case where z = 0,
Solve equation 2 for lambda:
$\lambda = 2$
Into equation 1:
$x = \frac{1}{2}$
both of these into equation 4:
$y = \pm \sqrt{\frac{3}{4}}$
Giving two more solutions of:
$x_3 = \frac{1}{2}, y_3 = \sqrt{\frac{3}{4}}, z_3 = 0, \lambda_3 = 2$
$x_4 = \frac{1}{2}, y_4 = -\sqrt{\frac{3}{4}}, z_4 = 0, \lambda_4 = 2$
Taking the case where y = 0:
Solve equation 3 for lambda:
$\lambda = 3$
Into equation 1:
$x = \frac{1}{4}$
both into equation 4:
$z = \pm \sqrt{\frac{15}{16}}$
Giving the final two solutions of:
$x_5 = \frac{1}{4}, y_5 = 0, z_5 = \sqrt{\frac{15}{16}}, \lambda_5 = 3$
$x_6 = \frac{1}{4}, y_6 = 0, z_6 = -\sqrt{\frac{15}{16}}, \lambda_6 = 3$
I can't think of any other possibility that would be mathematically legal. If I assume x=0, the first equation becomes $1=0$ which can't happen. Similarly, if I assume lambda = 0, then the 4th equation will eventually become impossible.
|
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|
Integrate by partial fraction decomposition $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$
Here's what I have so far...
$$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$
$$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$
$$\mathrm A(x^2+2x+5) + \mathrm B(x^2+x)+\mathrm C(x+1)=\\$$
$$(\mathrm A+\mathrm B)x^2 + (2\mathrm A + \mathrm B + \mathrm C)x + (5\mathrm A+\mathrm C)\\$$
$$\mathrm A=-3,\;\mathrm B=8,\;\mathrm C = 31$$
$$$$
$$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx = \int\bigg(-\frac{3}{x+1}+\frac{8x+31}{x^2+2x+5}\bigg)dx\Rightarrow$$
$$\int-\frac{3}{x+1}dx +\int\frac{8}{x^2+2x+5}dx+\int\frac{31}{x^2+2x+5}dx $$
Hopefully I've got it correct until this point (if not, someone point it out please!). I can do the first integration by moving the -3 out and using $u=x+1$ to get
$$-3 \ln(x+1)$$
but I'm stuck on the next two.
|
(Just working on the last integral, but not checking the partial fractions.) Complete the square:
$$\int \frac{dx}{x^2+2x+5} = \int \frac{dx}{(x+1)^2 + 4}$$
Now, substitute $x+1 = 2\tan(t)$. So, $dx = 2\sec^2 t\;dt$. Thus:
$$\begin{align}
\int \frac{dx}{(x+1)^2 + 4} &= \int\frac{2\sec^2(t)dt}{4(\tan^2t + 1)} \\
&= \frac{1}{2}\int\frac{\sec^2 t\; dt}{\sec^2t} \\
&= \frac{1}{2}\int 1\; dt\\
&= \cdots
\end{align}$$
|
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|
Compute $\sum_{n=0}^\infty \frac{(n+1)^2}{n!}$ This is what I've done so far:
$$\sum_{n=0}^\infty\frac{(n+1)^2}{n!} = 1 +4 +4.5 + \frac 83 + \frac{25}{24} +\frac{3}{10} + \frac {49}{720} +\dots$$
I know I need to manipulate $\frac{(n+1)^2}{n!}=\frac{(n+1)(n+1)}{n(n-1)(n-2)}=$
|
Using the standard Maclaurin series for $e^x$, we find that
$$xe^x=\sum_0^\infty
\frac{x^{n+1}}{n!}.$$
Differentiating we get
$$xe^x+e^x=\sum_0^\infty (n+1)\frac{x^n}{n!}.$$
Multiply by $x$, and differentiate again. We get
$$x^2e^x+3xe^x+e^x=\sum_0^\infty (n+1)^2\frac{x^n}{n!}.$$
Finally, set $x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/951107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving quadratic equations in the field $F_5$ Let $y = x^2 + 2x + 2 = 0$.
Solve the equation in the field $F_5$.
So I used the common $b^2 - 4ac$ formula and got that $x$ is either $-1/2$ or $-3/2$ but I'm not sure if this is in the field...
|
We have $a=1$, $b=2$, $c=2$, so $b^2-4ac=4-8=-4=1$. Moreover $2^{-1}=3$, so
$$
x=2^{-1}(-2\pm1)=3(-2\pm1)
$$
which means $x=3(-3}=-9=1$ or $x=3(-1)=-3=2$. The formula
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
indeed means
$$
x=(2a)^{-1}(-b\pm\sqrt{b^2-4ac})
$$
Check:
$$
1^2+2\cdot1+1=0,\qquad 2^2+2\cdot2+2=0
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/954562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \equiv n^4+2n^2 \pmod3 $$$$n^4+2n^2 = n^2(n^2+2)=n^2(n^2-1)=n^2(n+1)(n-1)\equiv 0 \pmod3$$
Divisble by 8: $$n^4+ 6n^3 + 11n^2+6n \equiv n(n^3-2n^2-5n-2) \pmod8 $$$$n(n^3-2n^2-5n-2) = n(n+1)(n^2-3n-2) = n(n+1)(n^2+5n+6) = n(n+1)(n+2)(n+3) = ??? \pmod8$$
So it seems that the polynomial has to be divisible by four, but I'm not sure how to show that it has to be divisible by two one more time to show that it's divisible by 8.
Any other approaches to solving the problem are welcome, as I'm preparing for an exam and would appreciate having multiple ways to tackle the problem. I think this problem was meant to be an exercise in proof by induction, but the other approach seemed more approachable.
|
$$n^4+6n^3+11n^2+6n=n(n^3+6n^2+11n+6)=n(n+1)(n^2+5n+6)$$
$$=n(n+1)(n+2)(n+3)$$ which is a Product of $4$ consecutive integers
Now see The product of n consecutive integers is divisible by n factorial
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Matrix Problem about commutative multiplication The problem is:
"Find all $2\times 2$ matrices A that have the property that for any $2\times 2$ matrix B, AB = BA."
Given hint: "The given equation must hold for all B. Try matrices B that have lots of zero entries."
I tried bashing by letting $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}$ and $B=\begin{pmatrix}w & x\\y & z\end{pmatrix}$, but as you can tell, this got messy quickly. Any help would be appreciated, thank you.
|
$$\begin{pmatrix}a & b\\c & d\end{pmatrix} \begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}a & 0\\c & 0\end{pmatrix}$$
$$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix} \begin{pmatrix}a & b\\c & d\end{pmatrix}=\begin{pmatrix}a & b\\0 & 0\end{pmatrix}$$
So, you get $b=c=0.$
$$\begin{pmatrix}a & b\\c & d\end{pmatrix} \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}=\begin{pmatrix}0 & a\\0 & c\end{pmatrix}$$
$$\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix} \begin{pmatrix}a & b\\c & d\end{pmatrix}=\begin{pmatrix}c & d\\0 & 0\end{pmatrix}$$
So, you get $c=0, a=d.$
That is,
$$A=\begin{pmatrix}a & 0\\0 & a\end{pmatrix}. $$ You only need to show that this matrices commute with any $2\times 2$ matrix.
|
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|
Find the equation of the locus of a point which has its sum of distance from (0,3) and (0,-3) equal to 8. I know the answer but due to various expressions under root, I'm unable to reach there.
Is there something which I'm missing to make the solution easier? By the way the answer given is $$\frac{x^2}{16} + \frac{y^2}{7}=1$$ which isn't equal to the one I'm getting, in fact, it is nowhere close to it.
My answer: (I just used the one plane distance formula here)$$30y^2+64^2+12(64)y=16^2•(x^2+y^2)+16^2(9+6y)$$
I see no point in expanding this.
|
We start from
$$\sqrt{x^2+(y+3)^2}+\sqrt{x^2+(y-3)^2}=8.$$
Bring $\sqrt{x^2+(y-3)^2}$ to the other side and square. We get
$$x^2+(y+3)^2=64-16\sqrt{x^2+(y-3)^2}+x^2+(y-3)^2.$$
A little algebra brings us to
$$12y-64=-16\sqrt{x^2+(y-3)^2}.$$
Minor cancellation gives
$$4\sqrt{x^2+(y-3)^2}=16-3y.$$
Square. We get some cancellation and arrive at
$$16x^2+7y^2=16^2-(16)(9)=(16)(7).$$
Finally, divide both sides by $(16)(7)$.
Remark: You may find it easier to deal with the slightly more general equation
$$\sqrt{x^2+(y+p)^2}+\sqrt{x^2+(y-p)^2}=k.$$
The steps are exactly the same, but we don't get the distraction of numbers. I certainly find it easier!
|
{
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|
Evaluation of $\int \frac{x\sin( \sqrt{ax^2+bx+c})}{ax^2+bx+c} \ dx\ $ How do we find $$\int \frac{x\sin( \sqrt{ax^2+bx+c})}{ax^2+bx+c} \ dx\ $$
NB: It is not mandatory that $ax^2+bx+c$ has only a single root
|
Hint:
$\int\dfrac{x\sin\sqrt{ax^2+bx+c}}{ax^2+bx+c}dx$
$=\int\dfrac{x}{ax^2+bx+c}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(ax^2+bx+c)^{n+\frac{1}{2}}}{(2n+1)!}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx(ax^2+bx+c)^{n-\frac{1}{2}}}{(2n+1)!}dx$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Compute $\displaystyle \lim_{x \rightarrow -3} \frac{\frac{1}{3} + \frac{1}{x}}{3+x}$ $$\lim_{x \rightarrow -3} \frac{\frac{1}{3} + \frac{1}{x}}{3+x}$$
Can I multiply the top and bottom by $3x$? If not, why not. I am having a hard time understanding when it is appropriate to multiply the numerator and denominator by the LCD.
|
\begin{align*}\lim_{x\rightarrow -3}\ \frac{\frac{1}{3} + \frac{1}{x}}{3+x} & = \lim_{x\rightarrow -3}\ \frac{x+3}{3x(3+x)} \\&= \lim_{x\rightarrow -3}\ \frac{1}{3x} = \frac{-1}{9}\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/963065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to find transformation matrix which converts matrix to simple standard form I have a matrix A$$ \left( \begin{array}{ccc}
0 & 1 \\
a^2 & 0\\
\end{array} \right) $$
Using eigen values, I convert it into simple standard form B:
$$\left( \begin{array}{ccc}
a & 0 \\
0 & -a\\
\end{array} \right) $$
How can I find the transformation matrix M which converts this to simple standard form. In other words, how to find M such that $MAM^{-1}=B$. I am actually interested in knowing the nature of the transformation. Also in general how does one find it?
|
*
*find it's eigenvalues using $|\lambda I -A| = 0$ equation, you must find them $a$ and $-a$.
Hint: $\lambda^2-a^2 = (\lambda-a)(\lambda+a)$
*find their corresponding eigenvectors $e_1$ and $e_2$, using the criteria of eigen-value-vector
$\lambda_1 e_1 = Ae_1 $
$$a\left(\!
\begin{array}{c}
x \\
y
\end{array}
\!\right)=\left(\!
\begin{array}{c}
0 & 1 \\
a^2 & 0
\end{array}
\!\right)\left(\!
\begin{array}{c}
x \\
y
\end{array}
\!\right)$$
\begin{cases}
ax=y \\
ay=a^2x
\end{cases}
so we notice that a set of solutions is of the form $\left(\!
\begin{array}{c}
x \\
ax
\end{array}
\!\right)$ let's pick the vector $e_1=\left(\!
\begin{array}{c}
1 \\
a
\end{array}
\!\right)$
(Do the same to find $e_2$)
according to my calculations $$e_1=\left(\!
\begin{array}{c}
1 \\
a
\end{array}
\!\right)$$
$$e_2=\left(\!
\begin{array}{c}
1 \\
-a
\end{array}
\!\right)$$
*construct the matrix $M$ with those eigenvectors
$$M=\left(\!
\begin{array}{c}
1 & 1 \\
a & -a
\end{array}
\!\right)$$
And you are done
$$M^{-1}=\left(\!
\begin{array}{c}
\frac{1}{2} & \frac{1}{2a} \\
\frac{1}{2} & \frac{-1}{2a}
\end{array}
\!\right)$$
Now :
$$M^{-1}AM=\left(\!
\begin{array}{c}
\frac{1}{2} & \frac{1}{2a} \\
\frac{1}{2} & \frac{-1}{2a}
\end{array}
\!\right)\left(\!
\begin{array}{c}
0 & 1 \\
a^2 & 0
\end{array}
\!\right)
\left(\!
\begin{array}{c}
1 & 1 \\
a & -a
\end{array}
\!\right)$$
$$M^{-1}AM=\left(\!
\begin{array}{c}
a & 0 \\
0 & -a
\end{array}
\!\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing the limit of an expression I have the following question. Determine the limit of the following expression.
$\lim_{x \rightarrow 0} \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4}$.
My attempt to this question is the following. Let the function $f(x) = \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4} = \frac{\cos(x \sqrt{2})}{x^4} - \frac{1}{(1+x^2) x^4}.$ Then $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos(x \sqrt{2})}{x^4} - \lim_{x \rightarrow 0} \frac{1}{(1+x^2) x^4} = 0 - Undefined$. So the limit does not exists.
But my professor's answer is $-\frac{5}{6}$. He used Taylor expansion. What am I doing wrong??
|
we get by simplifying your term
$\frac{(1+x^2)\cos(x\sqrt{2})-1)}{x^4(1+x^2)}$ and the taylor expansion for $x=0$ is $-\frac{5}{6}+O(x^2)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
|
More precisely, $\sqrt{a^2+4} = a + 2/a + O(1/a^3)$ as $a \to \infty$. In fact
it is easy to see that $$(a+2/a)^2 > a^2 + 4 > (a+2/a - 2/a^3)^2 \ \text{for} \ a > 1$$
so
$$a + \dfrac{2}{a} > \sqrt{a^2 + 4} > a + \dfrac{2}{a} - \dfrac{2}{a^3}$$
|
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|
Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\sqrt{6}-5)$$
Could you show me the steps to get the final result?
Thank you.
|
$$(\sqrt{3x}+\sqrt{2x})^2=17^2=289$$
$$(\sqrt{3x}+\sqrt{2x})^2 = 3x + 2x + 2\sqrt{6}x= (5+2\sqrt{6})x=289,$$
hence
$$x=\frac{289}{(5+2\sqrt{6})} = \frac{289 (5-2\sqrt{6})}{5^2-(2\sqrt{6})^2}
=289 (5-2\sqrt{6})$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the limit of ${x+\sqrt{x^2-4x+1}}$ at negative infinity I'm having trouble finding the following limit:
\begin{equation}
\lim_{x \to -\infty} {x+\sqrt{x^2-4x+1}}
\end{equation}
I tried to simplify it in many ways but couldn't get it to a form where I could evaluate the limit. How should I go about modifying this limit in order to evaluate it?
EDIT: Forgot the minus sign in front of the infinite, sorry.
|
$\displaystyle \lim_{x\to-\infty}x+\sqrt{x^2-4x+1}=\lim_{x\to-\infty}\frac{(x+\sqrt{x^2-4x+1})(x-\sqrt{x^2-4x+1})}{x-\sqrt{x^2-4x+1}}= \\ =\displaystyle \lim_{x\to-\infty}\frac{x^2-x^2+4x-1}{x-\sqrt{x^2-4x+1}}=\lim_{x\to-\infty}\frac{4-\frac{1}{x}}{1-\frac{|x|}{x}\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}=\frac{4}{2}=2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Algorithm to find the coefficient of GCD linear combination? One of the properties of the GCD of two integers is that it can be written as the linear combination of the two, is there an algorithm that can be used to find the coefficients of this linear combination?
|
The answer of @LonelyMathematician is perfect, but I confess that I have always found this way of describing the action just too confusing. Let’s step back a bit and look at what’s going on. In each step, you start with (dividend, divisor) and get a new pair, (newdividend, newdivisor), where the new dividend is the old divisor, and the new divisor is the remainder, equal to dividend - quotient times divisor. You see that this is a linear relationship expressible by a matrix equation:
\begin{align}
\begin{pmatrix}\text{newdividend}\\ \text{newdivisor}\end{pmatrix}
=
\begin{pmatrix}
0&1\\1&-\text{quotient}
\end{pmatrix}
\begin{pmatrix}\text{dividend}\\ \text{divisor}\end{pmatrix}\,.
\end{align}
Too wordy! Let’s do exactly the same computation as @LonelyMathematician, finding the greatest common divisor of $805$ and $154$. The first step is to say that
\begin{align}
\begin{pmatrix}154\\33\end{pmatrix}
=
\begin{pmatrix}0&1\\1&-5\end{pmatrix}
\begin{pmatrix}803\\154\end{pmatrix}\,.
\end{align}
We didn’t even need to write that out, except maybe to set our doubts to rest. The four matrices have the lower-right entries of $-5, -4, -1, -2$, these are the quotients of the divisions, but the matrices have to be written in the right order, and multiplied out.
\begin{align}
\begin{pmatrix}0&1\\1&-2\end{pmatrix}\begin{pmatrix}0&1\\1&-1\end{pmatrix}
\begin{pmatrix}0&1\\1&-4\end{pmatrix}\begin{pmatrix}0&1\\1&-5\end{pmatrix}=\begin{pmatrix}5&-26\\-14&73\end{pmatrix}\,,
\end{align}
and if we call this two-by-two matrix $A$, we see that
\begin{align}
\begin{pmatrix}11\\0\end{pmatrix}=A\begin{pmatrix}803\\154\end{pmatrix}\,,
\end{align}
in other words, the proper coefficients are found on the top row of $A$, namely $5$ and $-26$.
I think that you’ll agree that this method is very easily programmed.
|
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|
Proof that $\sum\limits_{i=1}^n \cos \sqrt{i}$ is unbounded. Please advice how to prove that $\sum\limits_{i=1}^n \cos \sqrt{i}$ is unbounded. By this I mean there exists no positive real $B$ such that for any natural $n$ $$-B <\sum\limits_{i=1}^n \cos \sqrt{i} < B$$
UPD: it looks like the sum is not bounded (it follows from Euler–Maclaurin formula), so it seems it is just necessary to fill out details.
|
Given any smooth function $f(x)$ defined over $(1 - \epsilon,\infty)$, we can rewrite
its partial sum over $\mathbb{Z}_{+}$ as a Riemann-Stieltjes integral:
$$\sum_{k=1}^n f(k) = \int_{1^{-}}^{n^{+}} f(x) d\lfloor x \rfloor \tag{*1}$$
Let $\;B_n(x)\;$ be the $n^{th}$ Bernoulli polynomial and $\;P_n(x) = B_n(\{x\})\;$, we know
$$\lfloor x \rfloor = x - \{ x \} = x - \frac12 - B_1(\{x\}) = x - \frac12 - P_n(x)$$
and $P_n$ satisfies the relation $P'_n(x) = n P_{n-1}(x)$ for $n > 1$.
Using them, we can transform RHS of $(*1)$ by repeat integration by parts.
$$\begin{align}
\sum_{k=1}^n f(k)
= & \int_{1}^{n} f(x) dx - \int_{1^{-}}^{n+} f(x) dP_1(x)\\
= & \int_{1}^{n} f(x) dx
- \bigg[ f(x) P_1(x) \bigg]_{1^{-}}^{n^{+}} + \frac12 \int_1^n f'(x) dP_2(x)\\
= & \int_{1}^n f(x) dx
- \bigg[ f(x) P_1(x) - \frac12 f'(x) P_2(x) \bigg]_{1^{-}}^{n^{+}} - \frac{1}{3!} \int_1^n f''(x) dP_3(x)\\
\vdots &\\
= & \int_{1}^n f(x) dx
+ \underbrace{\frac12 (f(1) + f(n) ) + \sum_{k=1}^{p}\frac{B_{2k}}{(2k)!}\bigg[f^{(2k-1)}(x)\bigg]_1^n}_{C_{n,p}}
+ R_{n,p}
\end{align}
$$
The last line is the famous Euler-Maclaurin formula and the error term $R_{n,p}$ has the form:
$$
R_{n,p}
= -\frac{1}{(2p)!} \int_1^n f^{(2p)}(x) P_{2p}(x) dx
= \frac{1}{(2p+1)!} \int_1^n f^{(2p+1)}(x) P_{2p+1}(x) dx
\tag{*2}
$$
For our problem, $f(x) = \cos\sqrt{x}$. We only need to keep the expansion up to $p = 1$. We have
$$\begin{align}
\int_1^n f(x) dx &= 2\sqrt{n} \sin\sqrt{n} +2 \cos\sqrt{n}- 2(\sin 1 + \cos 1)\\
C_{n,1} &= \frac12(\cos\sqrt{n} + \cos 1) -\frac{1}{24}\left(
\frac{\sin\sqrt{n}}{\sqrt{n}} - \sin 1\right)
\end{align}$$
For the error term $R_{n,1}$, we will use the second form in $(*2)$.
Let $K = \sup\limits_{0 \le x \le 1}|P_3(x)| = \frac{1}{12\sqrt{3}}$, we have
$$\begin{align}|R_{n,1}|
&= \frac{1}{3!}\left|\int_1^n \left(
\frac{\sin\sqrt{x}}{8 x^{3/2}}
+ \frac{3\cos\sqrt{x}}{8 x^2}
-\frac{3\sin\sqrt{x}}{8 x^{5/2}}
\right) P_3(x) dx\right|\\
&\le \frac{K}{6}\int_1^n \left(
\frac{1}{8 x^{3/2}}
+ \frac{3}{8 x^2}
+\frac{3}{8 x^{5/2}}
\right) dx\\
&\le \frac{K}{6}\int_1^\infty \left(
\frac{1}{8 x^{3/2}}
+ \frac{3}{8 x^2}
+\frac{3}{8 x^{5/2}}
\right) dx\\
&= \frac{7K}{48} = \frac{7}{576\sqrt{3}} \approx 0.0070164
\end{align}
$$
So for all $n$, we have
$$\sum_{k=1}^n \cos\sqrt{k} =
2\sqrt{n} \sin\sqrt{n} + \frac{5}{2} \cos\sqrt{n}- \frac{47 \sin 1 + 36 \cos 1}{24} + \epsilon_n$$
with the error term $|\epsilon_n| \le \frac{1}{24\sqrt{n}} + |R_{n,1}| \le 0.05$.
From this, we see for large $n$, the behaviour of the partial sums $\sum\limits_{k=1}^n \cos\sqrt{k}$ are dominated by the term $2\sqrt{n}\sin\sqrt{n}$. It simply oscillate between $\pm 2\sqrt{n}$ as $n$ increases and hence unbounded.
|
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|
One Step in an Integration I was doing the integration
$$
\int\frac{1}{(u^2+a^2)^2}du
$$
and I had a look at the lecturer's steps where I got stuck in the following step:
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
I guess it is integrating this by parts, but I could't see the trick. Could somebody help me with this please?
I had some work after I asked this question here. Here is my approach.
First note
$$
d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{2a^2u}{(u^2+a^2)^2}du,
$$
then we have
$$
\int\frac{1}{(u^2+a^2)^2}du=\int\left(-\frac{1}{2a^2u}\right)\left(-\frac{2a^2u}{(u^2+a^2)^2}\right)du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right).
$$
Then integrate by parts, we have
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}-\int\frac{a^2}{u^2(u^2+a^2)}du.
$$
But notice
$$
\frac{a^2}{u^2(u^2+a^2)}=\frac{1}{u^2}-\frac{1}{u^2+a^2},
$$
then we get
$$
\int\frac{a^2}{u^2(u^2+a^2)}du=-\frac{1}{u}-\int\frac{1}{u^2+a^2}du,
$$
then
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}+\frac{1}{u}+\int\frac{1}{u^2+a^2}du.
$$
But
$$
\frac{1}{u}-\frac{a^2}{u(u^2+a^2)}=\frac{u}{u^2+a^2},
$$
then
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
Is this approach okay? I am positive with this but not sure it is comprehensive.
|
Let $$I(a)= \int \frac{1}{a+x^2} = \frac{1}{\sqrt{a}}\arctan(\frac{x}{\sqrt{a}})$$
Then,
$$I'(a) = \int \frac{-1}{(a+x^2)^2} = d/da( \frac{1}{\sqrt{a}}\arctan(\frac{x}{\sqrt{a}})) $$
Find the expression in RHS and then put $a=u^2$. And you are done! :)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result when I test (using www.WolframAlpha.com) for various values of $a$ where $0<a<1$.
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$
D_2
\, =\,
\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
How could I go about proving:-
(1) $D_1$ = $D_2$,
(SOLVED, I think, by my two answers below, but using WolframAlpha to obtain integral solutions)
$$$$
(2) $D_1$ = $R$ or $D_2$ = $R$.
(MOVED to a separate question:
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$).
UPDATE 1
You can see how WolframAlpha
produces these results by inputting the following input texts:-
For Eqtn 1 with a=0.1 input: integrate (3*0.1(sinx)^2)/((1-0.1*cosx)^4) from x=0 to 2*pi
For Eqtn 2 with a=0.1 input: integrate (cosx)/((1-0.1*cosx)^3) from x=0 to 2*pi
for Result with a=0.1 input: evaluate 3 0.1 pi/(1-0.1^2)^(5/2)
UPDATE 2
WolframAlpha also computes expressions for the indefinite integrals as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
$$$$
$$$$
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression are equivalent to each other. This could be useful. For example we can define a difference function $f_3 = f_1-f_2$ whose indefinite integral $I_3 = I_1-I_2$ will exclude the common awkward third term.
Let us assume that $f_3$ is continuously integrable over the range $0,2\pi$ (we cannot be sure by inspection alone, but it can be shown, see my answer below). Then, if $D_1=D_2$ over the range $0,2\pi$ then $D_1-D_2=0$ and so $D_3$ (=$\int_0^{2\pi}f_3\,d\theta$) should have value zero. This is expanded on in my answer below.
|
(as per David H's suggestion to use Integration By Parts)
Hypothesis: The two definite integrals $D_1$ and $D_2$ have the same value, i.e. $D_1 = D_2$.
Using integration by parts
$$
\int uv' \,d\theta= uv-\int u'v \,d\theta
$$
let us define
$$
u = \frac{1}{(1-a\cos\theta)^3}
\qquad
v' = \cos\theta
$$
So ($u'$ obtained from WolframAlpha)
$$
u'=\frac{-3a\sin\theta}{(1-a\cos\theta)^4}
\qquad
v=\sin\theta
$$
Then
$$
\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta= uv - \int u'v\, d\theta
$$
$$=\frac{-\sin\theta}{(1-a\cos\theta)^3}
-\int \frac{-3a\sin^2\theta}{(1-a\cos\theta)^4}
$$
So
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Now consider the definite integrals over the range $0,2\pi$
$$\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\left[\frac{-\sin\theta}{(1-a\cos\theta)^3}\right]_0^{2\pi}
+\int_0^{2\pi} \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta
$$
The term in square brackets clearly goes to zero over the range $0,2\pi$ because $\sin(0)=\sin(2\pi)=0$ and therefore we have $D_1=D_2$.
|
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|
Proving a palindromic integer with an even number of digits is divisible by 11 I'm in an introductory course for discrete math so I'm a novice at English proofs. I'm not sure if my reasoning here is valid or if I'm using modular arithmetic correctly. Specifically the line I marked with $(**)$. I would appreciate any feedback. Sorry if any part of the proof seems obvious we are generally expected to spell everything out.
Objective: Prove every palindromic integer with an even number of digits is divisible by $11$.
Proof:
Consider a palindromic integer $p$ in the form of $x_{1} x_{2} …. x_{n-1}x_{n}x_{n}x_{n-1}...x_{2}x_{1}$ where $p$ has $2n$ digits.
This can be expanded as:
$$x_{1} + x_{2}\cdot10 + … + x_{n}\cdot10^{n} + x_{n}\cdot10^{n+1} + … + x_{2}\cdot10^{2n} + x_{1}\cdot10^{2n+1}$$
$(**)$ $10 \equiv -1 \pmod{11}$ so if we take $\pmod{11}$ of the expression we can replace $10\equiv -1$.
$$x_{1} + x_{2}\cdot(-1) + … + x_{n}\cdot(-1)^{n} + x_{n}\cdot(-1)^{n+1} + … + x_{2}\cdot(-1)_{2n} + x_{1}\cdot(-1)^{2n+1}$$
Since we know $2n$ is even (and therefore $2n + 1$ is odd), and $(-1)^{a} = 1$ when $a$ is even and $= -1$ when $a$ is odd we can rewrite the expression as:
$$x_{1} - x_{2} + … + x_{n} - x_{n} + … + x_{2} - x_{1} = 0$$
Therefore since $p \equiv 0 \pmod{11}$, we have that $11$ divides $p$.
|
You have some off-by-one errors, but you have the right idea. Note that:
\begin{align*}
p
&= (x_1 + x_2 10 + \cdots + x_n 10^{n-1}) + (x_{n} 10^n + x_{n-1} 10^{n+1} + \cdots + x_1 10^{2n - 1}) \\
&= \sum_{k=1}^n x_k 10^{k-1} + \sum_{k=1}^n x_k 10^{2n-k} \\
&= \sum_{k=1}^n x_k10^{k-1}(1 + 10^{2n - 2k - 1}) \\
&\equiv \sum_{k=1}^n x_k10^{k-1}(1 + (-1)^{2n - 2k - 1}) \pmod {11} \qquad\qquad\text{since $10 \equiv -1 \pmod{11}$} \\
&\equiv \sum_{k=1}^n x_k10^{k-1}(1 -1) \pmod {11} ~~~~~~~~~~~~~~~~~~~\qquad\qquad\text{since $2n - 2k - 1$ is odd} \\
&\equiv 0 \pmod {11}
\end{align*}
|
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|
Evaluating Limit - $(1-\cos(x^2))/(x^3\sin(x))$ How would you go about evaluating the following limit as $x$ approaches $0$?
$$\lim_{x\to 0}\frac{1-\cos(x^2)}{x^3\sin(x)}$$
|
$$\frac{1-\cos(x^2)}{x^3\sin(x)}=\frac{2\sin^2\left(\frac{x^2}{2}\right)}{x^3\sin(x)}=\frac{1}{2}\left(\frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}}\right)^2\frac{x}{\sin(x)}$$
then use $\displaystyle \lim_{x\to 0}\frac{x}{\sin(x)}=1$, you get your result.
|
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|
Solving $y^2 - yx - y + x = 0$ for $y$? I solved this equation for $y$ by inspection and confirmed it with Wolfram Alpha -
$y^2 - yx - y + x = 0$
I got the values $y = 1$ and $y = x$
However I was wondering is there a formal method for solving it? I expressed it as a polynomial -
$y^2 + (-1 - x)y + x = 0$
and used the quadratic formula but it just left me with an awkward expression involving powers of $x$ and a square root...no $y = 1$ and $y = x$ which is what I was looking for...
So how would I go about solving this formally?
|
$$y^2-(x+1)y+x=0$$
First method - using the quadratic formula:
$$y_{1,2}=\frac{x+1\pm\sqrt{(x+1)^2-4x}}{2}=\frac{x+1\pm\sqrt{(x-1)^2}}{2}=\frac{(x+1)\pm(x-1)}{2}$$
Thus the solutions are $y=x$ or $y=1$.
Second method - taking common factors:
$$y^2-yx-y+x=y(y-x)-1(y-x)=(y-1)(y-x)=0$$
Thus the solutions are $y=x$ or $y=1$.
|
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|
Find Max Of $M=|z^{3}-z+2|$ Give a complex number $z$: $|z|=1$
Find Max $M=|z^{3}-z+2|$
Could someone help me solve this ?
|
Let $z=x+yi$, hence $$\displaystyle \small \left|z^3-z+2\right|=\left|(x+yi)^3-(x+yi)+2\right|=\left|x^3+3x^2yi-3xy^2-y^3i-x-yi+2\right|= \\ \small =\left|(x^3-3xy^2-x+2)+i(3x^2y-y^3-y)\right|=\left|(x^3-3xy^2-x+2)+iy(3x^2-y^2-1)\right|$$
It is given that $|z|=1 \ \mathrm{thus\,} \ x^2+y^2=1 \rightarrow y^2=1-x^2$.
Setting it to the equation we get $$\displaystyle \small M=\left|z^3-z+2\right|=\left|(x^3-3x(1-x^2)-x+2)+iy(4x^2-2)\right|=\left|(4x^3-4x+2)+iy(4x^2-2)\right|$$
Now we can evaluate and get $$\displaystyle \small M=\sqrt{(4x^3-4x+2)^2+y^2(4x^2-2)^2}=\sqrt{(4x^3-4x+2)^2+(1-x^2)(4x^2-2)^2}$$
Can you proceed?
|
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|
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
Can someone please explain to me how to solve this?
because I tried everything I know and it didn't work.
P.S: I'm in 8th grade so no quadratic formula.
|
Expand $\left(a + \frac{1}{a}\right)^4$ by using foil, repeatedly using the famous result $(a+b)^2 = a^2 + 2ab + b^2$ (which is best in this case) or as I've done, you can use the binomial theorem assisted by pascal's triangle:
$$\left(a + \frac1{a}\right)^4 = \left(\sqrt{3}\right)^4\\
\\ a^4 + 4\cdot a^3\frac{1}{a} + 6\cdot a^2\left(\frac{1}{a}\right)^2 + 4\cdot a\left(\frac{1}{a}\right)^3 + \left(\frac{1}{a}\right)^4 = 3^{4/2}\\
a^4 + 4a^2 + 6 + \frac{4}{a^2} + \frac{1}{a^4} = 3^2\\
a^4 + \frac{1}{a^4} + 4\left(a^2 + \frac{1}{a^2}\right) = 9 - 6\\
a^4 + \frac{1}{a^4} + 4\left(\Big(a + \frac{1}{a}\Big)^2 - 2\, a\frac{1}{a}\right) = 3\\
a^4 + \frac{1}{a^4} + 4\left((\sqrt{3})^2 - 2\right) = 3\\
\therefore \quad a^4 + \frac{1}{a^4} = 3 - 4\cdot(3-2) = -1
$$
What you see above is a gut response to this question, if you've paid attention, you will notice my usage of $a^2 + b^2 = (a+b)^2 - 2ab$ and the important substitution of $a + \frac{1}{a} = \sqrt{3}$
There are loads of ways to do algebra. The results matter more than the method. Intuition matters more than anything else.
|
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|
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
|
Divide both sides of your inequality by (a+b+c), so you now have
$$bc+ac+ab\geq\frac {9abc}{(a+b+c)}$$
Now divide both sides by a so you get
$$\frac{(bc)}{a} + c +b \ge\frac {9bc}{(a+b+c)}$$
Next divide by b
$$\frac c a + \frac a b + 1 \ge \frac {9c}{(a+b+c)}$$
Finally, divide by c
$$\frac 1 a + \frac 1 b + \frac 1 c \ge \frac 9{(a+b+c)}$$
(I have used > instead of greater than or equal to because I can't get that symbol on my keyboard.)
|
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|
Prove this limit of $x^4$ using $\epsilon - \delta$ $\lim_{x\to a} x^4 = a^4$ using epsilon and delta.
We know $|x^4 - a^4| < \epsilon \space \text{such that} \space |x - a| < \delta$
So we have to find a $\delta$ for, which it works.
$|x^4 - a^4| = |x^2 + a^2||x-a||x+a|$
$|x^2 + a^2||x-a||x+a| < \epsilon \space \text{such that} \space |x - a| < \delta$
Let's require,
$|x - a| < 1 \implies |x + a| < 2|a| + 1$
$1 > |x - a|$
$|x - a| \ge |x| - |a|$
Therefore,
$|x| - |a| < |x - a| < 1$
$|x| < 1 + |a|$
$x^2 < (1 + |a|)^2$
$\implies x^2 + a^2 < (1 + |a|)^2 + a^2 \implies |x^2 + a^2| < (1 + |a|)^2 + a^2$
$ \implies |x + a| < 2|a| + 1$
All together, this implies, finally:
$ |x^2 + a^2| \cdot |x+a| < ((1 + |a|)^2 + a^2)\cdot (2|a| + 1)$
This requires,
$|x - a| < \text{min}(1, \frac{1}{((1 + |a|)^2 + a^2)\cdot (2|a| + 1)})$
Thus, $\delta = \frac{1}{((1 + |a|)^2 + a^2)\cdot (2|a| + 1)})$
Thanks, please suggest if this is correct!
|
Aside from a couple of typos, it looks pretty good! Here's a cleaned up version of your proof.
Given any $\epsilon > 0$, let $\delta = \min\left(1, \dfrac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)}\right) > 0$. Then if $0 < |x - a| < \delta$, notice that:
\begin{align*}
|x^4 - a^4|
&= |x - a| \cdot |x + a| \cdot |x^2 + a^2| \\
&= |x - a| \cdot |(x - a) + 2a| \cdot |(x - a)^2 + 2a(x - a) + 2a^2| \\
&\leq |x - a| \cdot (|x - a| + 2|a|) \cdot (|x - a|^2 + 2|a||x - a| + 2a^2) &\text{triangle inequality}\\
&< |x - a| \cdot (1 + 2|a|) \cdot (1 + 2|a| + 2a^2) &\text{} |x - a| < \delta \leq 1 \\
&< \frac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)} \cdot (1 + 2|a|)(1 + 2|a| + 2a^2)
&(\star)\\
&= \epsilon
\end{align*}
where the second-to-last step $(\star)$ follows because:
$$
|x - a| < \delta \leq \frac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)}
$$
|
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|
Determinant of a block matrix times the determinant of the inverse of a block Let $A$ be an $n\times n$ invertible matrix. Let $a$ be a number in $\mathbb{F}$, let $\alpha$ be a row $n$-tuple of numbers from $\mathbb{F}$ and let $\beta$ be a column $n$-tuple of numbers from $\mathbb{F}$. Show that $$
|A|^{-1}\left|\begin{array}{cc}
a & \alpha \\
\beta & A
\end{array}\right|=a-\alpha A^{-1}\beta
$$
So far, I have the following:
$$
\begin{pmatrix}
1 & 0\\
-\beta a^{-1} & I
\end{pmatrix}
\begin{pmatrix}
a & \alpha\\
\beta & A
\end{pmatrix}=\begin{pmatrix}
a & \alpha\\
0 & A-a^{-1}\beta\alpha
\end{pmatrix}\\
\left|\begin{array}{cc}
a & \alpha \\
\beta & A
\end{array}\right|=\left|\begin{array}{cc}
a & \alpha\\
0 & A-a^{-1}\beta\alpha
\end{array}\right|=|a||A-a^{-1}\beta\alpha|=|aA-\beta\alpha|\\
|A^{-1}||aA-\beta\alpha|=|a-A^{-1}\beta\alpha|
$$
I don't know where to go from here...
|
You can use another decomposition.
$$
\begin{pmatrix}
1 & -\alpha A^{-1} \\
0 & A^{-1}
\end{pmatrix}
\begin{pmatrix}
a & \alpha \\
\beta & A
\end{pmatrix}
=
\begin{pmatrix}
a - \alpha A^{-1} \beta & 0 \\
A^{-1} \beta & I
\end{pmatrix}
$$
Taking determinants on both sides gives
$$
|A|^{-1}
\begin{vmatrix}
a & \alpha \\
\beta & A
\end{vmatrix}
=
a - \alpha A^{-1} \beta
,
$$
as desired.
|
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|
Can we express the following ordinary generating function? I wish to express the following power series
$$ \sum_{k \ge 0} \binom{n-k}{m} x^k$$
where $n,m$ are positive integer such that $0< m \le n$
|
In fact, the polynomial part $$\sum_{k=0}^n \binom{n-k}{m} x^k= \sum_{k=0}^{n-m} \binom{n-k}{m} x^k + \sum_{k=n-m+1}^n \binom{n-k}{m} x^k$$
$$=\sum_{k=0}^{n-m} \binom{n-k}{m} x^k$$
$$ =x^{n} \sum_{k=0}^{n-m} \binom{n-k}{m} x^{k-n} $$ $$=x^{n-m} [\sum_{i=0}^{\infty} \binom{m+i}{m} \frac{1}{x^i} - \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}]$$
$$ =x^{n-m} \frac{1}{(1- (1/x))^{m+1}} - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i} $$
$$ = \frac{x^{n+1}}{(x- 1)^{m+1}} - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}.$$
Hence, $$\sum_{k=0}^{\infty} \binom{n-k}{m} x^k = - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}.$$
|
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"url": "https://math.stackexchange.com/questions/994619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Convergence of series with ^(n+1)
I want to check, if this series is convergent or not:
$$\sum\limits_{n=1}^\infty \frac{n^n(n!)}{(2n)!}$$
I tried with the ratio test:
$$\frac{\frac{(n+1)^{n+1}(n+1)!}{(2n+2)!}}{\frac{n^n*n!}{(2n)!} } =\frac{(n+1)^{n+1}(n+1)!(2n)!}{(2n+2)!(n!)n^n}=\frac{(n+1)^{n+1}(n+1)}{n^n(2n+2)(2n+1)}$$
But how should I proceed?
Thank you for your help
queenD
|
\begin{align}
\dfrac{n^n n!}{(2n)!} &= \dfrac{n^n}{(n+1)(n+2) \cdots 2n} \\
&= \prod_{k=1}^n \dfrac{n}{n+k} \leq \prod_{k= \lfloor n/2\rfloor + 1}^n \dfrac{n}{n+k} \\
&\leq \left(\dfrac{n}{n + n/2}\right)^{n- \lfloor n/2\rfloor } = \left(\frac{2}{3}\right)^{n- \lfloor n/2\rfloor }\\ &\leq \left(\dfrac{2}{3}\right)^{n - n/2}=\left(\sqrt{\dfrac{2}{3}}\right)^n
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
|
Troubled by the quotient rule? Then it might be helpful to rewrite your question:
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{1 + 4\cos x}{2\sqrt{x+ 4\sin x}}\right) = \frac{1}{2} \left(\frac{\mathrm d}{\mathrm dx}(1 + 4\cos x)(x + 4\sin x)^{-1/2}\right)$$
Now, just use the product rule accompanied by the chain rule.
Try it out!
Here's how I would proceed:
Let $u = x + 4\sin x \implies u' = 1 + 4\cos x$
Substituting, your question now becomes:
$$\frac{1}{2}\left(\frac{\mathrm d}{\mathrm dx} (u') \cdot (u^{-1/2})\right)\\
= \frac{1}{2}\left(u''\cdot u^{-1/2}\ +\ u'\cdot \Big(-\frac{1}{2}u^{-3/2}\cdot u'\Big)\right)\\
= \frac{u''}{2\cdot u^{1/2}} - \frac{u'}{4\cdot u^{3/2}}\\
= \frac{2u\cdot u'' \ - \ u'}{4\cdot u^{3/2}}$$
where $u'' = \frac{\mathrm d}{\mathrm dx}\left( 1 + 4\cos x\right) = - 4\sin x$
Now, you would get the same result if you had directly applied the quotient rule (which is a special case of the product rule). You can infact derive the quotient rule by applying the product rule to $\frac{\mathrm d}{\mathrm dx} f(x)\cdot\Big(g(x)\Big)^{-1 }$ . I leave this as an exercise to you but in order to convince yourself further, redo the question directly using the quotient rule :
$$ \frac{ \mathrm d}{\mathrm dx} \left(\frac{u'}{2\sqrt{u}}\right) = \frac{1}{2}\frac{\mathrm d}{\mathrm dx}\left(\frac{u'}{u^{1/2}}\right) = \frac{1}{2}\left(\frac{u''\sqrt u \ - \ u'\frac{1}{2 \sqrt{u}}}{u}\right) \equiv \frac{2u\cdot u'' \ - \ u'}{4\cdot u^{3/2}}$$
Alternatively, set $ y = \frac{u'}{2\sqrt u} \implies 2y\sqrt u = u'$ and then differentiate and solve to get $y'$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Equation $(a+b)^a=a^b$ How can we find the positive integer solutions to $(a+b)^a=a^b$?
Since $a+b>a$, it is necessary that $a<b$, otherwise the left-hand side is less than the right-hand side. So let $b=a+x$. The equation turns into $(2a+x)^a=a^{a+x}$, or equivalently, $\left(2+\dfrac{x}{a}\right)^a=a^x$.
|
Your equation means that $(2+\frac xa)^a$ is an integer. This is only possible if $2+\frac xa$ is itself an integer. In particular, $x$ is divisible by $a$. Therefore we may write $b=ka$ for some integer $k\geq2$. Using this in the original equation, we obtain $$((k+1)a)^a = a^{ka}$$ or equivalently $$(k+1)^a = a^{(k-1)a}$$ which further simplifies to $$k+1=a^{k-1}.$$ This implies $k+1\geq 2^{k-1}$, leaving us with the possibilities $k=2,3$. These give us the solutions $(k,a)\in\{(2,3),(3,2)\}$, which gives us the solutions of the original equation: $$a=2,b=6,\\a=3,b=6.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1011317",
"timestamp": "2023-03-29T00:00:00",
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|
Find $a$, $b$ and $c$ in $\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3$
Find the values of the positive constants $a$, $b$ and $c$ given that when $x$ is sufficiently small for terms in $x^4$, and higher powers of $x$, to be neglected then:
$$
\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3 \space\space\text{(assume $|bx| < 2$)}
$$
Expanding $e^{ax}$ first, we get:
$$
1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}
$$
and expanding $(2+bx)^{-1}$ we get:
$$
\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}
$$
I tried dividing the expressions and comparing the coefficients, but didn't really get anywhere. Is there an easier way to do this?
|
Hint
As Integrator said, multiply.
Another simple way : consider the function $$e^{a x}=(2+b x)(\frac{1}{2}+\frac{x^2}{4}-cx^3)$$ Use the Taylor series for the lhs (just as you did) and expand the rhs. Now, identify as many terms as you can. You will then get the following equations $$ a-\frac{b}{2}=0$$ $$a^2-1=0$$ $$\frac{a^3}{6}-\frac{b}{4}+2 c=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1013253",
"timestamp": "2023-03-29T00:00:00",
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|
solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ solve the differential equation.
$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$
The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try.
Thanks in advance.
|
This question is a bit of a nightmare for an entrance exam. Anyway, first observe that $$d(x^2 + y^2) = 2xdx + 2ydy$$ This suggests that using a variable $u = x^2 + y^2$ is useful, as we also have the RHS (right hand side)
$$RHS = \sqrt{\frac{a^2 - u}{u}}$$
The denominator on the LHS (left hand side) is slightly tricker; the minus sign suggests a derivative of $1/x$. Let's try $v = y/x$, then $$dv = \frac{1}{x} dy - \frac{y}{x^2} dx$$
Hence
$x^2 dv = xdy - ydx$ and we can write the LHS
$$LHS = \frac{x dx + y dy}{x dy - y dx} = \frac{1}{2x^2} \frac{du}{dv}$$
If we can write the $x^2$ terms of $u$ and $v$ we will have an ODE in just those variables: $$ \frac{1}{v^2 + 1} = \frac{x^2}{x^2 + y^2} = \frac{x^2}{u} \ \ \hbox{ hence } \ x^2 = \frac{u}{v^2 + 1}$$ Thus we can write
$$LHS = \frac{v^2 + 1}{2u} \frac{du}{dv} = RHS = \sqrt{\frac{a^2 - u}{u}}$$
or
$$\frac{du}{dv} = 2\sqrt{u(a^2 - u)} . \frac{1}{v^2 + 1}$$
This equation is separable
$$\int \frac{du}{\sqrt{u(a^2 - u)}} = 2\int \frac{dv}{v^2 + 1}$$
...after a bit of work,
$$\arctan\left( \frac{\sqrt{u}}{\sqrt{a^2 - u}} \right) = \arctan v + C$$
or back in the original variables:
$$\arctan\left( \frac{\sqrt{x^2 + y^2}}{\sqrt{a^2 - x^2 - y^2}} \right) = \arctan\left(\frac{y}{x} \right) + C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1015021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Fibonacci divisibility Prove that the following holds: $3|F_n$ if and only if $4|n$
Base case for $n=1$:
$F_1$=1, so $F_1$ is not divisible by 3 and 1 is not divisble by 4.
So the proposition holds for $k=1$
Continue for n=2, n=3 and n=4.
Inductive step
Assume $3|F(4k)$. We now have to prove that the proposition also holds for $4(k+1)=4k+4$.
$F(4k+4)=F(4k+3)+F(4k+2)=2F(4k+2)+F(4k+1)=3F(4k+1)+2F(4k)=5F(4k)+3F(4k-1)$
$5F(4k)$ is divisible by 3 by our hypothesis and it is obvious that $3F(4k-1)$ is divisible by 3.
$F(4k+3)=3F(4k)+2F(4k-1)$
$3F(4k)$ is divisble by 3 by our hypothesis and since $n$ already is a multiple of 4 $(n-1)$ can't possibly be a multiple of 4 to. So by our hypothesis $2F(4k-1)$ is not divisible by 3 and thus $F(4k+3)$ is not divisible by 3.
I continue the proof for $F(4k+2)$ and $F(4k+1)$ and conclude that by induction $3|F_n$ holds if $4|n$ holds, but the proposition involves a bi-implication, so I need to prove that $4|n$ holds if $3|F_n$ holds. However, I don't know where to begin.
I also think my proof is too lengthy. So I would like to know if there is a simpler, more elegant way to prove the proposition (induction is not required).
|
Along the lines off mookid:
Lemma: In mod 3 arithmetic, for all $k \in \Bbb{N}$,
$$
\begin{array}{c}
F_{8k-7} \equiv 1 \\
F_{8k-6} \equiv 1 \\
F_{8k-5} \equiv 2 \\
F_{8k-4} \equiv 0 \\
F_{8k-3} \equiv 2 \\
F_{8k-2} \equiv 2 \\
F_{8k-1} \equiv 1 \\
F_{8k} \equiv 0 \\
\end{array}
$$
Proof: The basis is trivially established by inspecting $F_1$ through $F_8$. Assume
the Lemma for $k$. Then
$$
\begin{array}{c}
F_{8(k+1)-7} = F_{8k+1} \equiv F_{8k} + F_{8k-1} = 0+1 = 1 \\
F_{8(k+1)-6} = F_{8k+2} \equiv F_{8k+1} + F_{8k} = 1+0 = 1 \\
F_{8(k+1)-5} = F_{8k+3} \equiv F_{8k+2} + F_{8k+1} = 1+1 = 2 \\
F_{8(k+1)-4} = F_{8k+4} \equiv F_{8k+3} + F_{8k+2} = 2+1 = 0 \\
F_{8(k+1)-3} = F_{8k+5} \equiv F_{8k+4} + F_{8k+3} = 0+2 = 2 \\
F_{8(k+1)-2} = F_{8k+6} \equiv F_{8k+5} + F_{8k+4} = 2+0 = 2 \\
F_{8(k+1)-1} = F_{8k+7} \equiv F_{8k+6} + F_{8k+5} = 2+2 = 1 \\
F_{8(k+1)} = F_{8k+8} \equiv F_{8k+7} + F_{8k+6} = 2+1 = 0
\end{array}
$$
This establishes induction so the lemma is true.
Finally, the theorem to be proved can be read from the lemma, by noting that only $F_{8k-4}$ and $F_{8k}$ are zero mod $3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1015578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true
and for even $n$ and positives $a_i$ the following inequality is true.
$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
|
A proof for even $n$.
Let $a_i>0$, $a_{n+1}=a_1$ and $n$ is an even natural number. Prove that:
$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
Proof.
By C-S and AM-GM $\sum\limits_{i=1}^n\frac{a_i}{a_i+a_{i+1}}=\sum\limits_{k=1}^{\frac{n}{2}}\frac{a_{2k-1}}{a_{2k-1}+a_{2k}}+\sum\limits_{k=1}^{\frac{n}{2}}\frac{a_{2k}}{a_{2k}+a_{2k+1}}\geq\frac{\left(\sum\limits_{k=1}^{\frac{n}{2}}\sqrt{a_{2k-1}}\right)^2}{a_1+a_2+...+a_n}+\frac{\left(\sum\limits_{k=1}^{\frac{n}{2}}\sqrt{a_{2k}}\right)^2}{a_1+a_2+...+a_n}\geq$
$\geq\frac{a_1+a_2+...+a_n+\frac{n^2-2n}{2}\sqrt[n]{a_1a_2...a_n}}{a_1+a_2+...+a_n}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find this inequality is $a^3+b^3+c^3+3xabc+y(a+b+c)\ge z(ab+bc+ac)$
let $x,y,z\ge 0$ ,Assmue that
$$a^3+b^3+c^3+3xabc+y(a+b+c)\ge z(ab+bc+ac),\forall a,b,c\ge 0$$ if and only if
$$z^2\le \min{(16y,4y(1+x))}$$
My idea： $$\Longrightarrow $$
let $a=b=0,c>0$,then we have
$$c^3+yc\ge 0\Longrightarrow c(c^2+y)\ge 0$$
let
$a=b>0,c=0$, then we have
$$2a^3++2ay\ge za^2\Longrightarrow 2a^2+2y\ge za$$
then I can't,
Thank you
|
The necessary condition:
$a=b=c$ gives $a^2(1+x)-za+y \ge 0, \: \forall a \in \mathbb R \implies z^2 \le 4(1+x)y$
Further, $a=b, c = 0$ gives $2a^2-za+2y \ge 0, \: \forall a \in \mathbb R \implies z^2 \le 16y$, so together, we have $$z^2 \le 4y\;\min(4, 1+x).$$
The sufficient condition:
An easy way would be to use the symmetric function theorem$^\dagger$, i.e. for a symmetric polynomial of degree three, $P(x_1, x_2, \dots, x_n)$, we have
$$\forall x_i \ge 0, \;\; P(x_1, x_2, \dots, x_n) \ge 0 \iff \forall k \in \{1,2,\dots,n \},\;\; P(\underbrace{1, 1,\dots 1}_{k \text{ ones}}, 0, ...0) \ge 0$$
So in the case of our three variable cubic, we only need to show $P(1, 0, 0), P(1, 1, 0), P(1, 1, 1) \ge 0$. i.e. $1+y \ge 0, \; 2(1+y)\ge z, \; 1+x+y\ge z, $.
Clearly the first inequality holds. $z^2 \le 16y\le 4(1+y)^2$ gives you the second inequality, and $z^2\le 4(1+x)y \le (1+x+y)^2$ gives you the last.
$^\dagger$ I have forgotten a good reference for this theorem, except that it can be proved using mixing variables or $uvw$ method. Will add the reference if I can find it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $P(x)$ is a poly. of least degree and local Max. at $x=1$ and Local Min. at $x=3,$ Then $P'(0)$ is If $P(x)$ is a polynomial of least degree which has a local Maxima at $x=1$ and Local Minima at
$x=3.$ If $P(1)=6$ and $P(3)=2$. Then $P'(0)=$
$\bf{My\; Try::}$ Given function has one Maxima and one Minima So $P(x)$ must have least
degree $3$ polynomial. So Let $P(x)=Ax^3+Bx^2+Cx+D$ and $P'(x)=3Ax^2+2Bx+C.$
Now Given $P(1)=6\Rightarrow A+B+C+D = 6....................(1)$
and Given $P(3)=2\Rightarrow 27A+9B+3C+D=2..............(2)$
and Given $P'(1)=0\Rightarrow 3A+2B+C = 0.........................(3)$
and Given $P'(3)=0\Rightarrow 9A+6B+C = 0.........................(4)$
Now Subtract $(4)-(3)$, We Get $6A+4B=0\Rightarrow 3A+2B=0$
Similarly Sub $(2)-(1)\;,$ We Get $26A+8B+2C=-4\Rightarrow 13A+4B+C=-2$
Is there is any other method my which we can solve the above question in less complex way.
plz explain me, Thanks
|
Thanks Friends ,I have got it
My Solution:: Let $P'(x) = A\cdot (x-1)\cdot (x-3)\;,$ Then Integrate both side w. r. to $x\;,$ We Get
$\displaystyle P(x) = A\left[\frac{x^3}{3}-2x^2+3x\right]+\mathcal C\;,$ Now for calculation of $A$ and $\mathcal C\;,$ Put $x=1$ and $x=3.$
So $\displaystyle P(1) = \frac{4A}{3}+\mathcal C\Rightarrow 4A+3\mathcal C = 18$ and $\displaystyle P(3) = \mathcal C\Rightarrow \mathcal C = 2.$ Now Solve These
Equation we get $\displaystyle A=3$ and $\displaystyle \mathcal C = 2$. So $P'(0) = 3A = 9\Rightarrow P'(0)=9$
|
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"url": "https://math.stackexchange.com/questions/1021189",
"timestamp": "2023-03-29T00:00:00",
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|
Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ Finding the closed form of:
$$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$
where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$
It appears when we try to determine the summation $\displaystyle \sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, using the generating function:
\begin{align}
\sum_{n=1}^{\infty} \frac{H_n}{n^3} \, x^{n} &= - \frac{1}{2} \, \sum_{n=1}^{\infty}\frac{1}{n^2} \, \sum_{k=1}^{n} \frac{(1-x)^k}{k^2} - \frac{\zeta(2)}{2} \, \operatorname{Li}_2(x) + \frac{7 \, \zeta(4)}{8} - \frac{1}{4} \, \operatorname{Li}_2^2(1-x) + \frac{\zeta^2(2)}{4} + \operatorname{Li}_4(x) \\
& \hspace{5mm} + \frac{1}{4} \, \log^2 x \, \log^2(1-x) + \frac{1}{2}\log x \, \log (1-x) \, \operatorname{Li}_2(1-x) + \zeta(3) \, \log x - \log x \,
\operatorname{Li}_2(1-x)
\end{align}
when we write, $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$
Combined with Cleo's closed form here, I know what the closed form should be, but how do I derive the result ?
|
Different approach:
By Cauchy product we have
$$\operatorname{Li}^2_2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$
take $x=1/2$ and rearrange the terms to get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}=\frac12\operatorname{Li}^2_2\left(\frac12\right)+3\operatorname{Li}_4\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^32^n}$$
Plugging $
\sum_{n=1}^\infty \frac{H_n}{n^32^n}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$ and $\operatorname{Li_2}\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$
gives
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42
\end{align*}
|
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|
Prove $\lim_{x\to-2}\frac{x+8}{x+3}=6 $ using $\epsilon-\delta$
Prove:
$$\lim_{x\to-2}\frac{x+8}{x+3}=6 $$
I started with:
$ |\frac{x+8}{x+3} - 6| < \epsilon => |\frac{x+8-6x-18}{x+3} | < \epsilon => |\frac{-5x-10}{x+3} | < \epsilon =>| \frac{-5(x+2)}{x+3} | < \epsilon $
From the definition of limit we know that:
$ |x+2| <\delta $
Just don't know how to continue now... any suggestions?
|
First let us restrict $x$ such that $\vert x+2 \vert < 1/2$, i.e., $-5/2<x<-3/2$.
This means we have $1/2<x+3<3/2$, i.e., $\dfrac23 < \dfrac1{x+3} < 2$
Hence, we have
$$\left \vert \dfrac{-5(x+2)}{x+3}\right \vert = 5 \left \vert \dfrac{x+2}{x+3}\right \vert < 10 \vert x+2 \vert$$
Hence, choose your $\delta = \min(1/2,\epsilon/10)$ to ensure that $\left \vert \dfrac{-5(x+2)}{x+3}\right \vert$ is less than $\epsilon$.
|
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|
The inequality $ abc\geq(a-b+c)(a+b-c)(b+c-a)$ holds for $a,b,c\geq 0$
What is the proof that:
$$\forall \ a,b,c\geq 0:\quad a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
I tried :
we can write that expression $(a-b+c)(a+b-c)(b+c-a)$ as $$-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
then
$$a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
$$\Longleftrightarrow $$
$$a b c \geq -a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
Now consider $f(a)=abc-(-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3)$ for $b,c$ fixed
if i follow the metode of Quang Hoang
let $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes
$$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$
for $x,y,z \geq 0$
we use AM-GM inequality
$$(y+z)(z+x)(x+y)\ge 2(yz)^{1/2}(2zx)^{1/2}(2xy)^{1/2}.\tag{1}$$
but other case of $x,y,z$ is not clear.
*
*i will be appreciated if someone could explain it with more detail
|
Let $a\geq b\geq c$.
Hence, $$abc-(a+b-c)(a+c-b)(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$
$$=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=$$
$$=(a-b)(a(a-c)-b(b-c))\geq0$$
because $a\geq b\geq0$ and $a-c\geq b-c\geq0$.
Done!
|
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|
Finding the zeroes in a function I have come across a problem on my trigonometry homework where we need to find the zeros of a function without the use of a calculator.
The Equation:
Given one of the zeros is $x=5$
$f(x) = x^3+2x^2-23x-60$
Is there a more efficient way to go about this problem than simply factor it out?
|
Yes if one of the roots is $5$ it means $x-5$ is factor of the given polynomial.Dividing the polynomial by $x-5$ we get the quadratic equation $x^2+7x+12$. Now the roots of this quadratic equation are $\frac{-7+(49-4*12)^{\frac{1}{2}}}{2}$ and $\frac{-7-(49-4*12)^{\frac{1}{2}}}{2}$. Thus the remaining two roots are $-4$ and $-3$. In general if we have a quadratic equation $ax^2+bx+c$ it has roots $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
For dividing $x^3+2x^2-23x-60$ by $x-5$ take a look at Polynomial Long Division.
|
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|
The Cosine and an Enigmatic Parabola In the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$, the cosine function superficially resembles an inverted parabola of the form $-ax^2+1$:
I wanted to know more and computed the $L^2$ norm (surrounding square root omitted for readability):
$$
\int_{-\pi/2}^{\pi/2} \,(\cos(x)-(ax^2+1))^2\,dx \quad=\quad \frac{\pi^5 a^2}{80}+\frac{1}{6}(48-6\pi^2+\pi^3) a+\frac{3\pi}{2}-4
$$
Minimizing the expression on the right hand side with respect to $a$ yields the enigmatic solution
$$a=\frac{-20\pi^3+120\pi^2-960}{3\pi^5}=-0.431...$$
which indeed shows a nice agreement:
Thus, the parabola $ax^2+1$ with this value for $a$ is in some sense the best parabolic approximation to the cosine in the largest interval in which the cosine is concave like the parabola is.
By now you have probably guessed my question: What is going on? I don't know what I expected the optimum value of $a$ to look like, but this is definitely not it. Where do these "magic" numbers $20$, $120$, $960$ and $3$ come from? Is there any way to understand this result other than computing the integral?
|
These rather unexpected and large coefficients arise, in the calculation of the integral and of the successive minimization, only if we search an expression for $a$ where the numerator has no fractional terms and no "collected" factors: in fact, in this case we necessarily have to perform some multiplications that lead to relatively large coefficients. However, if we allow to have fractional terms and collected factors in the numerator, the coefficients become relatively simple.
To understand where these large coefficients come from, it is convenient to analyze the solution step by step and to keep the numbers as "simple" as possible. The initial definite integral is obtained by noting that, considering each term separately, we have
$$\displaystyle \int_{-\pi/2}^{\pi/2} \,\cos^2(x)=\frac{\pi}{2}$$
$$\displaystyle \int_{-\pi/2}^{\pi/2} \,(-2 \cos(x)(ax^2+1))\,dx=(8-\pi^2)a-4$$
$$\displaystyle \int_{-\pi/2}^{\pi/2} \,(ax^2+1)^2\,dx= \frac{\pi^5}{2^4 \cdot 5}a^2+\frac{\pi^3}{6}a +\pi$$
Summing these three espressions, we get that the whole integral is equal to
$$\displaystyle \frac{\pi^5}{2^4 \cdot 5}a^2+(\frac{\pi^3}{6}-\pi^2+8)a+\frac{3\pi}{2}-4 $$
The derivative is
$$\displaystyle \frac{\pi^5}{2^3 \cdot 5}a+( \frac{\pi^3}{6}-\pi^2+8) $$
which equalized to zero and solved for $a$ yields the solution
$$\displaystyle a=\frac{2^3 \cdot 5 ( - \frac{\pi^3}{6}+\pi^2-8)}{ \pi^5}$$
where the coefficients are rather simple and not particularly striking.
However, if we want to eliminate the fractional term in the numerator, we have to collect the quantity $\displaystyle \frac{1}{6}$. This leads to
$$\displaystyle a=\frac{2^3 \cdot 5 ( -\pi^3+6\pi^2-6\cdot 8)}{ 6\pi^5}= \frac{2^2 \cdot 5 ( - \pi^3+6\pi^2-6\cdot 8)}{ 3\pi^5}$$
and multiplying the constant factor $2^2 \cdot 5=20$ for each term in the numerator we get
$$\displaystyle a= \frac{-20 \pi^3+ 20 \cdot 6\pi^2- 20 \cdot 6\cdot 8}{ 3\pi^5} \\= \frac{ -20 \pi^3+120 \pi^2-960}{ 3\pi^5} $$
which is the form reported in the OP.
|
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|
$\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$ Can one help finding this limit
$$\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$$
L'Hospital's rule is permited.
(Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$)
|
Hint. Recall that, for $x$ near $0$, you have
$$\tan x = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\mathcal{O}(x^9)$$
$$ \sin x = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{17x^7}{720}+\mathcal{O}(x^9)$$ to obtain (using only relevant terms) $$\tan(\sin x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+\mathcal{O}(x^9)$$ $$\sin (\tan x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+\mathcal{O}(x^9)$$
$$x^{3}-\sin^{2}x\tan x = -\frac{x^7}{15}+\mathcal{O}(x^9)$$
Then the desired limit is equal to $-2$.
|
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|
Derive Formula for Sine Half Angle I got stack with another problem (From the book: The Forgotten Art of Spherical Trigonometry):
Not having OC as 1 always creates me complicated formulas to define the rest. Once again looking for the smart track that I am missing.
|
For ease of notation, let $\beta = \alpha/2$. Note that in the figure, $\triangle ABO \cong \triangle CBO$. Hence $$AO = CO = \cos \beta, \quad AB = CB = \sin \beta.$$ Since $DO/CO = \cos \alpha$, it follows that $DO = \cos \alpha \cos \beta$. Similarly, $CD/CO = \sin \alpha$, hence $CD = \sin \alpha \cos \beta$. But $ABED$ is a rectangle, thus $$BE = AD = AO - DO = \cos \beta - \cos \alpha \cos \beta = \cos \beta (1 - \cos \alpha),$$ and $$CE = CD - ED = CD - BA = \sin \alpha \cos \beta - \sin \beta.$$ Then by the Pythagorean theorem, $$\begin{align*} \sin^2 \beta &= BC^2 = BE^2 + CE^2 \\ &= \cos^2 \beta (1 - \cos \alpha)^2 + (\sin \alpha \cos \beta - \sin \beta)^2 \\ &= \cos^2 \beta (1 - \cos \alpha)^2 + \cos^2 \beta \sin^2 \alpha - 2 \sin \alpha \cos \beta \sin \beta + \sin^2 \beta, \end{align*}$$ and cancelling $\sin^2 \beta$ from both sides, rearranging, and factoring like terms gives $$\begin{align*} 2 \sin \alpha \cos \beta \sin \beta &= \cos^2 \beta \left( (1 - \cos \alpha)^2 + \sin^2 \alpha\right) \\ &= 2 \cos^2 \beta (1 - \cos \alpha).\end{align*}$$ Now cancelling again and isolating $\beta$ gives $$\cot \beta = \frac{\sin \alpha}{1 - \cos \alpha}.$$ All that is left now is to square both sides to get $$\cot^2 \beta = \frac{\sin^2 \alpha}{(1 - \cos \alpha)^2} = \frac{1 - \cos^2 \alpha}{(1 - \cos \alpha)^2} = \frac{1 + \cos \alpha}{1 - \cos \alpha},$$ so that $$\csc^2 \beta = 1 + \cot^2 \beta = \frac{2}{1 - \cos \alpha},$$ therefore, $\sin^2 \beta = \dfrac{1 - \cos \alpha}{2},$ or $$\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}.$$ Compared with Blue's diagram and answer, I think it should be fairly clear which approach is more elegant.
|
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|
Compute: $\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}$ Help me please to compute the limit of:
$
\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdot \left ( 1-\frac{2}{4} \right ) ^{\frac{4}{n}}\cdot \left ( 1-\frac{2}{5} \right ) ^{\frac{5}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}
$
What I did:
$
L=\frac{a_{n+1}}{a_n} = \frac{ \left ( 1-\frac{2}{n+3} \right ) ^{\frac{n+3}{n+1}}}{ \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}}=\frac{2/e}{2/e}=1
$
But it's not. Since $L=1$, I need use something else...
|
Let
$$f_n = \left ( 1-\dfrac{2}{3} \right ) ^{\dfrac{3}{n}}\cdot \left ( 1-\dfrac{2}{4} \right ) ^{\dfrac{4}{n}}\cdot \left ( 1-\dfrac{2}{5} \right ) ^{\dfrac{5}{n}}\cdot \cdot \cdot \left ( 1-\dfrac{2}{n+2} \right ) ^{\dfrac{n+2}{n}} $$
We have
$$\ln(f_n) = \sum_{k=1}^n \dfrac{k+2}{n}\ln\left(1-\dfrac2{k+2}\right)$$
Hence,
\begin{align}
n \ln(f_n) & = \sum_{k=1}^n (k+2)\left(\ln(k)-\ln(k+2)\right)=\sum_{k=1}^n (k+2)\ln(k) - \sum_{k=1}^n(k+2)\ln(k+2)\\
& = \sum_{k=1}^n (k+2)\ln(k) - \sum_{k=3}^{n+2}k\ln(k)\\
& = 3\ln(1) + 4\ln(2) + \sum_{k=3}^n (k+2) \ln(k) - \sum_{k=3}^n k \ln(k) - (n+1)\ln(n+1) - (n+2)\ln(n+2)\\
& = 2 \ln(2) + 2 \sum_{k=2}^n \ln(k) - (n+1)\ln(n+1) - (n+2)\ln(n+2)\\
& = 2 \ln(2) + 2 \ln(n!) - (n+1)\ln(n+1) - (n+2)\ln(n+2)\\
& = 2 \ln(2) + \ln(2\pi n) + 2n \ln(n) - 2n - (n+1)\ln(n+1) - (n+2)\ln(n+2) + \mathcal{O}(1/n)\\
\ln(f_n) & = -2 + \mathcal{O}(\ln(n)/n)
\end{align}
Hence, $f_n \to e^{-2}$.
|
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|
Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ How can I prove that
$$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$$
|
The sum is equal to
$$\frac1{16} \sum_{n=0}^{\infty} \frac{(2 n+1)!}{2^{3 n} n !(n+1)!} = \frac1{16} \sum_{n=0}^{\infty} \frac{2 n+1}{n+1} \frac1{2^{3 n}} \binom{2 n}{n} $$
The sum may be written as
$$\frac1{8} \sum_{n=0}^{\infty} \frac1{2^{3 n}} \binom{2 n}{n} - \frac1{16}\sum_{n=0}^{\infty} \frac1{n+1} \frac1{2^{3 n}} \binom{2 n}{n}$$
Note that
$$\sum_{n=0}^{\infty} \frac1{2^{3 n}} \binom{2 n}{n} x^{n} = \left (1-\frac{x}{2}\right )^{-1/2} $$
so that
$$\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \frac1{2^{3 n}} \binom{2 n}{n} = \int dx \left (1-\frac{x}{2}\right )^{-1/2} = -4 \left (1-\frac{x}{2}\right )^{1/2}+C$$
This sum is zero when $x=0$, so $C=4$, and therefore the sum in question is
$$\frac{\sqrt{2}}{8} - \frac{2-\sqrt{2}}{8} = \frac{\sqrt{2}-1}{4}$$
which is not the answer provided, but which checks against Mathematica.
|
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|
Proving this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ by induction Where $n \in \mathbb{N}$
and
$$
F_n = \begin{cases}
0 & \text{ if } n = 0 \\
1 & \text{ if } n = 1 \\
F_{n-1} + F_{n-2} & \text{ if } n > 1
\end{cases}
$$
This is basically describing the famous Fibonacci sequence.
If we try the base case with 1, it works (for $0$ I am not sure...)
When $n = 1$
$2 \cdot0 - 1^2 = (-1)^1 = -1$
For the inductive hypothesis, we assume this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true.
Now, for the inductive step, we try to prove for $n+1$, so for $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$.
Since $n$ is always a natural number, and it will be always or even or odd, the $-1$ raised to $n$ will be always either $-1$ (when $n$ is odd) or $1$ (when $n$ is even).
Thus, $F_{n+1} \cdot F_{n-1} - F^2_n$ = -($F_{n+2} \cdot F_{n} - F^2_{n+1}$).
Or simply:
$$
(-1)^n = -(-1)^{n+1}
$$
$$
1 \cdot (-1)^n = -(-1)\cdot(-1)^n
$$
$$
1 = -(-1)
$$
Which is true.
I don't know if this is sufficient... I can arrive at the second step and say that we know this $(-1)^n$ is equals to $F_{n+1} \cdot F_{n-1} - F^2_n$, but I am not sure...
|
Your proof is completely incorrect. You need to prove why $$F_{n+2} F_{n} - F_{n+1}^2 = -(F_{n+1} F_{n-1} - F_{n}^2)$$ to prove that $F_{n+2} F_{n} - F_{n+1}^2 = (-1)^{n+1}$.
You can check the base case. For the inductive step, we have
\begin{align}
F_{n+2} F_{n} - F_{n+1}^2 & = (F_{n+1}+F_{n})F_{n} - F_{n+1}^2 & \left(\because F_{n+2} = F_{n+1} + F_{n} \right)\\
& = F_{n+1} F_{n} + F_{n}^2 - F_{n+1}^2\\
& = F_{n+1} F_{n} - F_{n+1}^2 + F_{n}^2\\
& = F_{n+1}(F_{n}-F_{n+1}) + F_{n}^2\\
& = -F_{n+1} F_{n-1} + F_{n}^2 & \left(\because F_{n+1} = F_{n} + F_{n-1}\right)\\
& = -\left(F_{n+1}F_{n-1} - F_{n}^2\right)
\end{align}
Now you are done.
|
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|
How does the recursion relation work in the solution to this differential equation (using series)? Sorry for the vague title but it would not let me post the first step and last step of this equation (too many characters!).
How does $$\dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} = \dfrac{a_0\Gamma(\frac{2}{3})}{3^nn!3^n\Gamma(n+\frac{2}{3})}?$$
This is a text book example I'm having trouble following. Solve the differential equation
$y'' = xy$ using series. This makes sense in general. But when we get down to solving for the coefficients of the series, it confuses me.
Based of the relations $$a_nn(n-1) = 0 \text{ for } n=0,1,2$$ and $$a_nn(n-1) = a_{n-3} \text{ for } n = 3,4,\ldots$$
we have
$$\begin{align}
a_{3n} &= \dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} \\
&= \dfrac{a_0}{3^n n! 3^n(n-\frac{1}{3})(n-1-\frac{1}{3})(n-2-\frac{1}{3})\cdots (\frac{5}{3}) (\frac{2}{3})}\\
&= \dfrac{a_0\Gamma(\frac{2}{3})}{3^nn!3^n\Gamma(n+\frac{2}{3})}\\
\end{align}$$
Particularly, that middle step there is throwing me off.
The book also provided
$$\begin{align}
a_{3n+1} &= \dfrac{a_1}{(3n+1)(3n)(3n-2)(3n-3)\cdots 10 \cdot 9 \cdot 7 \cdot 6 \cdot 4 \cdot 3} \\
&= \dfrac{a_1}{3^n n! 3^n(n+\frac{1}{3})(n-1+\frac{1}{3})(n-2+\frac{1}{3})\cdots (\frac{7}{3}) (\frac{4}{3})}\\
&= \dfrac{a_1\Gamma(\frac{4}{3})}{3^nn!3^n\Gamma(n+\frac{4}{3})}\\
\end{align}$$
and
$$a_{3n+2} = 0,$$
but that last one makes sense.
As a side note, why do they put $3^n$ twice in the denominator? Why not $3^{2n}$?
|
Use the fact that $\Gamma{(x+1)} = x \Gamma{(x)} $:
$$\begin{align}\Gamma{\left (n+\frac{2}{3} \right )} &= \left (n-\frac{1}{3} \right ) \Gamma{\left (n-\frac{1}{3} \right )}\\ &= \left (n-\frac{1}{3} \right ) \left (n-\frac{4}{3} \right )\Gamma{\left (n-\frac{4}{3} \right )}\\ &=\cdots\\&=\left (n-\frac{1}{3} \right ) \left (n-\frac{4}{3} \right )\cdots \frac{2}{3}\Gamma{\left (\frac{2}{3} \right )}\end{align}$$
The two $3^n$'s come from the factorial and the gamma function, each.
|
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|
Help finding the residue of $1/(z^8+1)$ Help finding the residue of $1/(z^8+1)$
I'm integrating over $\{ Re^{it} | 0 \leq t \leq \pi \}$, and I found 4 simple poles at $z_0=e^{in\pi/8}$ where $n = 0,...,3$ and I'm trying to calculate $res(1/(z^8+1),z_0)$
calculating this: $$\lim_{z\to z_0} (z-z_0)f = \lim_{z\to z_0}\frac{z-z_0}{1+z^8},$$ now are there any algebra tricks I can do to simplify this?
|
You don't need to factor $z^8+1$. The function $f(z)=\frac{1}{z^{8}+1}$ has 8 simple poles $z_{k}$, $k\in\left\{ 0,1,2\ldots ,7\right\} $, which are the zeros of the equation $z^{8}+1=0\Leftrightarrow z^{8}=-1$, i.e. the complex numbers $z_{k}=e^{i\left( \pi +2k\pi \right) /8}=e^{i\left(2k+1 \right)\pi /8}$ (and not $z_{k}=e^{i k\pi /8}$ that you computed).
(...) are there any algebra tricks I can do to simplify this?
Since the numerator of $f(z)$ is $1$, the residue $\mathrm{Res}\left(
f(z);z_{k}\right) $ reduces to the inverse of the derivative of the
denominator at $z=z_{k}$
\begin{equation*}
\mathrm{Res }\left( f(z);z_{k}\right) =\frac{1}{\left. \frac{d}{dz}
\left( z^{8}+1\right) \right\vert _{z=z_{k}}}=\frac{1}{8z_{k}^{7}}=\frac{1}{
8e^{i7\left( 2k+1\right) \pi /8}},\tag{1}
\end{equation*}
because when $f$ is of the form $f(z)=\frac{p(z)}{q(z)}$, where both numerator and
denominator are analytical functions, the following equality holds
\begin{equation*}
\mathrm{Res}\left( f(z);z_{k}\right) =\frac{p(z_{k})}{q^{\prime }(z_k)}
,\tag{2}
\end{equation*}
provided that $z_{k}$ is a simple pole of $f(z)$ and $p(z_{k})\neq 0$, $
q(z_{k})=0$, $q'(z_{k})\neq 0$. (Theorem 2 in section 69 of Complex Variables
and Applications / J. Brown and R. Churchill, International ed. 2003,
McGraw-Hill).
ADDED 2. For instance $$\mathrm{Res}\left( f(z);z_{0}\right)=\frac{1}{8}e^{-i7\pi /8}=-\frac{1}{8}\cos \frac{\pi }{8}-i\frac{1}{8}\sin \frac{\pi }{8}=-\frac{1}{16}\sqrt{2+\sqrt{2}}-\frac{1}{16}i\sqrt{2-\sqrt{2}}.$$
ADDED. Grouping the residues with equal immaginary part and symmetrical real part, each of the two following sums becomes pure imaginary, and respectively a function of $\sin\frac{\pi}{8}$ and $\sin\frac{3\pi}{8}$:
\begin{eqnarray*}
\mathrm{Res }\left( f(z);z_{0}\right) +\mathrm{Res }\left( f(z);z_{3}\right)
&=&\frac{1}{8e^{i7\left( 1\right) \pi /8}}+\frac{1}{8e^{i7\left( 7\right)
\pi /8}} \\
&=&\frac{1}{8e^{i7\pi /8}}+\frac{1}{8e^{i\pi /8}}=-\frac{1}{8}\left( e^{i\pi
/8}+e^{i7\pi /8}\right) \\
&=&-\frac{1}{8}\left( i2\sin \frac{\pi }{8}\right) =-\frac{1}{8}i\sqrt{2-\sqrt{2}},
\end{eqnarray*}
and
\begin{eqnarray*}
\mathrm{Res }\left( f(z);z_{1}\right) +\mathrm{Res }\left( f(z);z_{2}\right)
&=&\frac{1}{8e^{i7\left( 3\right) \pi /8}}+\frac{1}{8e^{i7\left( 5\right)
\pi /8}} \\
&=&\frac{1}{8e^{i5\pi /8}}+\frac{1}{8e^{i3\pi /8}}=\cdots \\
&=&-\frac{1}{8}\left( i2\sin \frac{3\pi }{8}\right) =-\frac{1}{8}i\sqrt{2+
\sqrt{2}}.
\end{eqnarray*}
The improper real integral $I=\int_{-\infty }^{\infty }\frac{1}{x^{8}+1}dx$ evaluates to
\begin{eqnarray*}
I &=&2\pi i\sum_{k=0}^{3}\mathrm{Res }\left( f(z);z_{k}\right) =2\pi i\left( -\frac{1}{8}i\sqrt{2-\sqrt{2}}-\frac{1}{8}i\sqrt{2+\sqrt{2}}
\right) \\
&=&2\pi i\left( -\frac{1}{8}i\sqrt{4+2\sqrt{2}}\right) =\frac{\pi }{4}\sqrt{
4+2\sqrt{2}}.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1035389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$ How to find the sum of the following series:
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
Any hints.
|
The sum may be written as
$$\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1\cdot 4 \cdots (3k-2)}{5^k k!} $$
Now, based on a suggestion from @Robert Israel:
$$(1+t)^{-1/3} = 1 -\frac1{1!}\frac13 t + \frac{1}{2!}\left ( -\frac13 \right ) \left ( -\frac{4}{3} \right )t^2 - \cdots$$
so that
$$1-(1+t)^{-1/3} = \frac1{1!}\frac13 t - \frac{1}{2!}\left ( -\frac13 \right ) \left ( -\frac{4}{3} \right )t^2 + \cdots $$
The series on the RHS reproduces the series in question when $t=3/5$, so that the sum is
$$1-\left ( \frac{8}{5} \right )^{-1/3} = 1-\frac12 5^{1/3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1036288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that
$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$
How can I show this?
|
Using Rearrangement inequality for $a,a^2$ and $b,b^2$
$$a^3+b^3\gt a^2b+b^2a$$
Similarly,
$$b^3+c^3\gt b^2c+c^2b$$
and
$$c^3+a^3\gt c^2a+a^2c$$
Adding these three, we have
Using sing AM-GM, we have
$$2(a^3+b^3+c^3)> a^2b+a^2c+b^2c+b^2a+c^2a+c^2b$$
$$\dfrac{a^2b+a^2c+b^2c+b^2a+c^2a+c^2b}{6}\ge\sqrt[6]{a^6b^6c^6}$$
$$\implies a^2b+a^2c+b^2c+b^2a+c^2a+c^2b\geq 6abc $$
Finally,
$$2(a^3+b^3+c^3> a^2(b+c)+b^2(c+a)+c^2(a+b)\geq 6abc $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1037765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving $\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$ I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this?
$$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$$
Thanks in advance.
|
Here I derive the rhs by simplification.
Use $\displaystyle \sin x = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$ and $\displaystyle \cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$ to write
$$\begin{align}\frac{\cos x - \sin x +1}{\cos x + \sin x -1}&=\frac{\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
- \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} +1}
{\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
+ \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} -1}\\
&=\cot \frac{x}{2}\\
&=\frac{1+\cos x}{\sin x}
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1039642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
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|
calculation $f(x)$ from given expression $$f(x)+xf(-x)=x-2$$
what is $f(x$)?
I try to solve this problem, but I don't know how to remove $f(-x)$ or converting it to $f(x)$.
|
After solving such task as in the above answers, you should always test it: you put $\frac{x^2 +3x - 2}{1+x^2}$ instead of $f(x)$ into the given equation:
$$\frac{x^2 +3x - 2}{1+x^2} + x\frac{x^2 -3x - 2}{1+x^2} =$$
$$= \frac{x^2 + 3x - 2 +x^3 - 3x^2 - 2x}{1+x^2} =$$
$$= \frac{x^3 - 2x^2 + x - 2}{1+x^2} =$$
$$= \frac{(x-2) (x^2+1)}{1+x^2} = x-2$$
I forgot to do it once on a maths competition and it cost me a point.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1043543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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|
How to find $\lim\limits_{x \to 0} \frac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$? How to find $\;\;\lim\limits_{x \to 0} \dfrac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$ ?
|
We have
$$f(x) = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} \dfrac{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}$$
Hence,
$$f(x) = \dfrac{\sin(2x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{1+\tan(x) - 1 + \tan(x)} = \dfrac{2\sin(x)\cos(x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{2\tan(x)}$$
This gives us
$$f(x) = \cos^2(x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)$$
Now take the limit as $x \to 0$ to get $f(x) \to 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1043690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
What is the value of $a+b+c$?
What is the value of $a+b+c$?
if $$a^4+b^4+c^4=32$$
$$a^5+b^5+c^5=186$$
$$a^6+b^6+c^6=803$$
How to approach this kind of problem. Any help.
UPDATE: Thank you all for answers. Now I realize there is no integer solution. But is there any real number solution? I am curious to know.
|
The lowest possible values for fourth powers of integers are $$0^4 = 0 \qquad 1^4 = (-1)^4 = 1 \qquad 2^4 = (-2)^4 = 16$$
By inspection, it becomes very clear that two of $a,b,c$ must have absolute value $2$, and one must be zero, in order for your first equation to hold.
Without loss of generality, suppose $c=0$ and $a = \pm b = \pm 2$.
Now $a^5 + b^5$ must be $2^5 + 2^5$, $-2^5 + 2^5$, or $-2^5 -2^5$. None of these is equal to $186$, so there are no solutions in the integers.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0 ^{\pi}\left (\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$ How would you evaluate the integral
$$\int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$$
The answer from Wolfram is $0$.
Would you use a substitution or do it by parts? Would making the substitution $u=\dfrac x2$ help?
|
Using the triple angle formula for sine and the power reduction formula for the square of cosine, we have:
$$\begin{align}
\sin{\left(3\theta\right)}
&=3\sin{\theta}-4\sin^3{\theta}\\
&=\sin{\theta}\left(3-4\sin^2{\theta}\right)\\
&=\sin{\theta}\left(4\cos^2{\theta}-1\right)\\
&=\sin{\theta}\left[4\left(\frac{1+\cos{\left(2\theta\right)}}{2}\right)-1\right]\\
&=\sin{\theta}\left[1+2\cos{\left(2\theta\right)}\right]\\
\implies \sin{\left(3\theta\right)}\csc{\left(\theta\right)}&=1+2\cos{\left(2\theta\right)};~\theta\in\mathbb{R}\land\frac{\theta}{\pi}\notin\mathbb{Z}.\\
\end{align}$$
Letting $\theta=\frac{x}{2}$ then gives us the identity,
$$\sin{\left(\frac{3x}{2}\right)}\csc{\left(\frac{x}{2}\right)}=1+2\cos{x};~x\in(2k\pi,2(k+1)\pi),k\in\mathbb{Z}.$$
Substituting $u=\frac{\pi}{2}-x$ transforms the interval of integration into a symmetric interval about the origin, and so the integral of the odd component of the integrand automatically becomes zero:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\pi}\left(\frac{\pi}{2}-x\right)\sin{\left(\frac{3x}{2}\right)}\csc{\left(\frac{x}{2}\right)}\,\mathrm{d}x\\
&=\int_{0}^{\pi}\left(\frac{\pi}{2}-x\right)\left(1+2\cos{x}\right)\,\mathrm{d}x\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u\left(1+2\sin{u}\right)\,\mathrm{d}u\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u\,\mathrm{d}u+2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u\\
&=0+4\int_{0}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u=4\int_{0}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u.\\
\end{align}$$
At this stage, we know that the integral $\mathcal{I}$ cannot be zero because $u\sin{u}$ is positive on $u\in(0,\frac{\pi}{2}]$. Finally, evaluating integrals of products powers and trig functions is a common application of integration by parts, of which the last integral above is an easy example:
$$\begin{align}
\mathcal{I}
&=4\int_{0}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u\\
&=4\left[\left[-u\cos{u}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(-1)\cos{u}\,\mathrm{d}u\right]\\
&=4\left[-\frac{\pi}{2}\cdot0+0\cdot1+\int_{0}^{\frac{\pi}{2}}\cos{u}\,\mathrm{d}u\right]\\
&=4\left[0+\left[\sin{u}\right]_{0}^{\frac{\pi}{2}}\right]\\
&=4\left[1-0\right]=4.~\blacksquare\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Indefinite integral with sector of ellipse An ellipse is given by the following equation:
$$
152 x^2 - 300 x y + 150 y^2 - 42 x + 40 y + 3 = 0
$$
After solving for the midpoint we have:
$$
152 (x-1/2)^2 - 300 (x-1/2) (y-11/30) + 150 (y-11/30)^2 = 1/6
$$
Introducing polar coordinates:
$$
x = 1/2 + r \cos(\theta) \quad ; \quad y = 11/30 + r \sin(\theta)
$$
Giving:
$$
152\, r^2 \cos^2(\theta) - 300\, r^2\, \cos(\theta) \sin(\theta) +
150\, r^2 \sin^2(\theta) = 1/6 \quad \Longrightarrow \\
\frac{1}{2} r^2(\theta) = \frac{1/12}
{152 \cos^2(\theta) - 300 \cos(\theta) \sin(\theta) + 150 \sin^2(\theta)}
$$
The area of a sector of the ellipse is:
$$
\int_{\theta_1}^{\theta_2} \frac{1}{2} r^2(\theta) \, d\theta
$$
So it seems that we have to find the indefinite integral:
$$
\int \frac{1/12 \, d\theta}
{152 \cos^2(\theta) - 300 \cos(\theta) \sin(\theta) + 150 \sin^2(\theta)}
$$
And then I'm stuck. Because feeding this into MAPLE with
int(1/12/(152*cos(theta)^2-300*cos(theta)*sin(theta)+150*sin(theta)^2),theta);
quite to my surprise, gives a complex result:
$$
{\frac {1}{720}}\,i\sqrt {3}\ln \left( \left( \tan \left( 1/2\,\theta
\right) \right) ^{2}+ \left( {\frac {5}{38}}\,i\sqrt {3}+{\frac {75}
{38}} \right) \tan \left( 1/2\,\theta \right) -1 \right) \\ -{\frac {1}{720
}}\,i\sqrt {3}\ln \left( \left( \tan \left( 1/2\,\theta \right)
\right) ^{2}+ \left( -{\frac {5}{38}}\,i\sqrt {3}+{\frac {75}{38}}
\right) \tan \left( 1/2\,\theta \right) -1 \right)
$$
But I'm pretty sure that the area of an ellipse sector is a real number.
So the question is: does there exist a closed form for the abovementioned
integral that is real valued instead of complex? What is it? And why that complex result with MAPLE?
AfterMath (see accepted answer)
$$
A = (1/2,3/10) \quad ; \quad B = (1/2,1/3) \quad ; \quad C = (1/2,11/30) \\
P = (1/3,1/6) \quad ; \quad Q = (2/3,1/2) \\
R = (1/4,1/10) \quad ; \quad S = (3/4,3/5)
$$
Triangle edges are black; ellipse is $\color{red}{red}$ .Triangle areas:
$\Delta PAQ = 1/180$ , $\Delta PRB = \Delta QSB = 1/720$ .
It is conjectured that the ellipse sector area $\overline{CRBSC}$
is exactly $1/3$ of the total ellipse area;the latter being
$ = \pi\sqrt{3}/180$ . Can someone prove or disprove this conjecture?
|
The integral is elementary.
$$
I = \int \frac{d \theta}
{a \cos^2(\theta) - b \cos(\theta) \sin(\theta) + c \sin^2(\theta)}=
\int \frac{d \tan(\theta)}
{a - b \tan(\theta) + c \tan^2(\theta)}
$$
Let $u=\tan(\theta)$ , then:
$$
I = \int \frac{d u}{c u^2 - b u + a} =
\int \frac{d u/c}{\left[ u-b/(2c) \right]^2+a/c - \left[b/(2c)\right]^2}=\\
2 \int \frac{d (2c u-b)}{(2c u-b)^2+(4ac-b^2)}=
\frac{2}{\sqrt{4ac-b^2}} \int \frac{d\left[(2c u-b)/ \sqrt{4ac-b^2}\right]}
{1+\left[(2c u-b)/\sqrt{4ac-b^2}\right]^2}
$$
Now let $\Delta = \sqrt{4ac-b^2}$ and we're finished:
$$
I(\;\tan(\theta)\;) =
\frac{2}{\Delta} \arctan\left(\frac{2c\tan(\theta)-b}{\Delta}\right)
$$
Fill in the numbers for our sector: $a = 152\;,\;b = 300\;,\;c = 150$ ,
$\tan(\theta_2) = 16/15\;,\;\tan(\theta_1) = 14/15$ ; remember that an
$\arctan$ is only defined for arguments $\in \left[-\pi/2,+\pi/2\right]$
and that the whole area of the ellipse equals $\pi\sqrt{3}/180$ . Then:
$$
\mbox{area} = \frac{\pi\sqrt{3}}{180}/2 - \left[\;I(16/15)-I(14/15)\;\right]/12 =
\frac{\pi\sqrt{3}}{540}
$$
Thus establishing that the AfterMath conjecture is true. So finally we have all the ingredients to calculate the area of
the "ellipse with a hat" ( $\color{blue}{blue} + \color{red}{red} + \color{green}{green}$ ) :
$$
\color{blue}{\pi\sqrt{3}/180(1-1/3)+1/120}+\color{red}{2/720}+\color{green}{1/180} =
\frac{\pi\sqrt{3}}{270}+\frac{1}{60}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})
\\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$
|
\begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\
&= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\
&= 4\sin^2(\theta + \phi) \\
&= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\
&= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2) \\
&= 16\cos^2((\theta + \phi)/2) - 16\cos^4((\theta + \phi)/2)\end{align*}
\begin{align*}c^2 - b^2 &= \sec^2\theta + \sec^2\phi + 2\sec(\theta)\sec(\phi) - (\tan^2\theta + \tan^2\phi + 2\tan(\theta)\tan(\phi)) \\
&= 2 + 2\sec(\theta)\sec(\phi) - 2\tan(\theta)\tan(\phi)\end{align*}
\begin{align*}\cos(\theta)\cos(\phi)(c^2 - b^2) &= 2(\cos(\theta)\cos(\phi) + 1 - \sin(\theta)\sin(\phi)) \\
&= 2(1 + \cos(\theta + \phi)) \\
&= 4\cos^2((\theta + \phi)/2)\end{align*}
$cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2$
$cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2$
\begin{align*}8\cos^2(\theta)\cos^2(\phi)bc/a &= 8\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))(\sec(\theta) + \sec(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= 8(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= 8\sin(\theta + \phi)(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= \frac{8(2\sin((\theta + \phi)/2)\cos((\theta + \phi)/2))(2\cos((\theta + \phi)/2)\cos((\theta - \phi)/2))}{2\sin((\theta + \phi)/2)\cos((\theta - \phi)/2)} \\
&= 16\cos^2((\theta+\phi)/2)\end{align*}
The desired result follows.
|
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"url": "https://math.stackexchange.com/questions/1048023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
|
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
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\newcommand{\ds}[1]{\displaystyle{#1}}
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\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
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\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\Phi\pars{a,b}\equiv
\int_{0}^{1}{\dd x \over \pars{x^{2} + a^{2}}\root{x^{2} + b^{2}}}:\
{\large ?}}$.
With Euler-sub $\ds{\root{x^{2} + b^{2}} \equiv x + t}$ we found
$\ds{x = {b^{2} - t^{2} \over 2t}}$ such that:
\begin{align}
\Phi\pars{a,b}&=-4\int_{\verts{b}}^{\root{1 + b^{2}}\ -\ 1}{t\,\dd t\over
t^{4} + 2\pars{2a^{2} - b^{2}}t^{2} + b^{4}}
\end{align}
With $\ds{t\ \mapsto\ t^{1/2}}$:
\begin{align}
\Phi\pars{a,b}&
=-2\int_{b^{2}}^{2 + b^{2} - 2\root{1 + b^{2}}}
{\dd t \over t^{2} + 2\pars{2a^{2} - b^{2}}t + b^{4}}
\\[5mm]&=-2\int_{b^{2}}^{2 + b^{2} - 2\root{1 + b^{2}}}
{\dd t \over \pars{t + 2a^{2} - b^{2}}^{2} - \pars{2a^{2} - b^{2}}^{2}+ b^{4}}
\\[5mm]&=-2\int_{2a^{2}}^{2a^{2} + 2 - 2\root{1 + b^{2}}}
{\dd t \over t^{2} + 4a^{2}b^{2} - 4a^{4}}
\\[5mm]&=-\,{1 \over \verts{a}\root{b^{2} - a^{2}}}
\int_{\verts{a}/\root{b^{2} - a^{2}}}
^{\pars{a^{2} + 1 - \root{1 + b^{2}}}/\pars{\verts{a}\root{b^{2} - a^{2}}}}
{\dd t \over t^{2} + 1}
\\[5mm]&={1 \over \verts{a}\root{b^{2} - a^{2}}}\bracks{%
\arctan\pars{\verts{a} \over \root{b^{2} - a^{2}}}-
\arctan\pars{a^{2} + 1 - \root{1 + b^{2}} \over \verts{a}\root{b^{2} - a^{2}}}}
\end{align}
With the identity
$\ds{\quad\arctan\pars{x} - \arctan\pars{y}
=\arctan\pars{x - y \over 1 + xy}}$ we get
$$
\color{#66f}{\large\Phi\pars{a,b}}
=\color{#66f}{\large{1 \over a\root{b^{2} - a^{2}}}\,
\arctan\pars{\root{b^{2} - a^{2}} \over a\root{1 + b^{2}}}}
$$
|
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|
Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ My proof of the induction step goes as follows (supposing equality holds for all $k \in \{1,2,\dots n \})$: $$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^{n} i^3+(n+1)^3 \\ = \left( \sum_{i=1}^ni\right)^2+(n+1)^3 \\ = \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ = \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ = \frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4} \\ = \frac{n^4+6n^3+13n^2+12n+4}{4} \\ = \frac{(n^2+3n+2)^2}{4} \\ = \frac{[(n+1)(n+2)]^2}{4} \\ = \left(\frac{(n+1)(n+2)}{2}\right)^2 \\ = \left(\sum_{i=1}^{n+1}i\right)^2$$ I was a little disappointed in my proof because the algebra got really hairy. It took me a long while to see that I could twice unfoil the polynomial $n^4+6n^3+13n^2+12n+4$ and all in all the solution seems pretty inelegant. Does anyone have a smoother way to prove the induction step or bypass the algebra? I feel like their must be some concise way to get the same result.
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You could have made it a little easier here:
$$
\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} = (n+1)^2 \left( \frac{n^2}{4}+\frac{4(n+1)}{4} \right) = \frac{(n+1)^2(n^2+4n+4)}{4} = \frac{((n+1)(n+2))^2}{4}.
$$
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|
Convergence test for the series $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$
Determine convergence of the series
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$$
My proof: using comparison test I have
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} = \sqrt{n+\frac{1}{n}} - \sqrt{n} \to 0$$ for large $n$.
Therefore the series should be converges. Now that we need to show that $a_n \lt b_n$ for some $a_n$ s.t $\sum_{n=1}^{\infty} a_n$ converges. for all $n \in\mathbb N$,
$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} \lt \frac{1}{n^2}$ since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges then $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$ must be converges.
Is this right or I need to fix my $\frac{1}{n^2}$?
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Since $$\sqrt {n^2 + 1} < n + \frac {1} {2n}$$ we have $$\frac {\sqrt {n^2 + 1} - n} {\sqrt n} < \frac {1} {2n \sqrt n}.$$ So the series is dominated by $$\sum_{n = 1}^{\infty} \frac {1} {2n \sqrt n} = \frac {1} {2} \zeta \left(\frac {3} {2} \right).$$
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Integrating trigonometric functions. I need a little help with solving the following two integrals. I am not able to approach these two problems correctly. Little hints will suffice. Thanks for your help!
*
*$$\int\frac{\tan^2x\sec^2x}{1+ \tan^6x}\,dx$$
*$$\int (\sin x \cos x)^{1/3}\,dx$$
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$1.$
$\int\dfrac{\tan^2x\sec^2x}{1+\tan^6x}dx$
$=\int\dfrac{\tan^2x}{1+\tan^6x}d(\tan x)$
$=\int\dfrac{1}{3(1+\tan^6x)}d(\tan^3x)$
$=\dfrac{\tan^{-1}\tan^3x}{3}+C$
$2.$
$\int(\sin x\cos x)^\frac{1}{3}dx=\int\dfrac{\sin^\frac{1}{3}2x}{2^\frac{1}{3}}dx$
Let $u=\sin^\frac{1}{3}2x$ ,
Then $x=\dfrac{\sin^{-1}u^3}{2}$
$dx=\dfrac{3u^2}{2\sqrt{1-u^6}}du$
$\therefore\int\dfrac{\sin^\frac{1}{3}2x}{2^\frac{1}{3}}dx$
$=\int\dfrac{3u^3}{2^\frac{4}{3}\sqrt{1-u^6}}du$
$=\int\dfrac{3u^3}{2^\frac{4}{3}}\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{6n}}{4^n(n!)^2}du$
$=\int\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{6n+3}}{2^{2n+\frac{4}{3}}(n!)^2}du$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{6n+4}}{2^{2n+\frac{4}{3}}(n!)^2(6n+4)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{6n+4}}{2^{2n+\frac{7}{3}}(n!)^2(3n+2)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!\sin^{2n+\frac{4}{3}}2x}{2^{2n+\frac{7}{3}}(n!)^2(3n+2)}+C$
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The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)}{(x-a)}$
$0=(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$(x^2-a^2)^{\frac{1}{2}}(2a)=\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$2a\sqrt{x^2-a^2}=\frac{a(x+a)}{2\sqrt{x^2-a^2}}$
$2a(x+a)(x-a)=a(x+a)$
$2(x-a)=a$
$2x-2a=a$
$x=\frac{3}{2}a$
My book says x=2a. I'm not sure where I went wrong. Thanks in advance
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You differentiated $\sqrt{x^2-a^2}$ incorrectly. It looks like you forgot to apply the chain rule.
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|
Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
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Given $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = \frac{1}{2}\int\frac{\sec x(\sec x+\tan x)+\sec x(\sec x-\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
$$\displaystyle = \frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx+\frac{1}{2}\int\frac{\sec x\cdot (\sec x-\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
Now Use $\sec^2x-\tan^2 x= 1$ For Second Integral, We Get
$$\displaystyle = \frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx+\frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{13}{2}}}dx$$
Now For both Integral , Let $\sec x+\tan x=t\;,$ Then $\sec x(\sec x+\tan x)dx = dt$
We Get $$\displaystyle = \frac{1}{2}\int\frac{1}{t^{\frac{9}{2}}}dt+\frac{1}{2}\int\frac{1}{t^{\frac{13}{2}}}dt = -\frac{1}{2}\cdot \frac{2}{7}t^{-\frac{7}{2}}-\frac{1}{2}\cdot \frac{2}{11}t^{-\frac{11}{2}}+\mathcal{C}$$
So We Get $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = -\frac{1}{7}(\sec x+\tan x)^{-\frac{9}{2}}-\frac{1}{11}(\sec x+\tan x)^{-\frac{11}{2}}+\mathcal{C}$$
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|
Evaluation of $2$ limit problems Evaluation of $(a)\;\;\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
and $(b)\;\; \displaystyle \lim_{x\rightarrow 0^{-}}\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+...............+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+2x}.$
$\bf{My\; Try::(a):}$ Given $\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
Now Multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$.
So $\displaystyle \lim_{x\rightarrow \infty}\frac{\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}}{\left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}}\times \left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
So we Get $\displaystyle \lim_{x\rightarrow \infty} \frac{(a-2)x^3+(3-c)x^2+(b-3)x+(2-d)}{x^2\cdot \left(\sqrt{1+\frac{a}{x}+\frac{3}{x^2}+\frac{b}{x^3}+\frac{2}{x^4}}+\sqrt{1+\frac{2}{x}-\frac{c}{x^2}+\frac{3}{x^3}-\frac{d}{x^4}}\right)}$
Now If limit is finite , Then $(a-2)=0\Rightarrow a=2$
can we solve the first question is any other way, If yes the plz explain me,
and how can i solve $(2)$ one.
Thanks
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Assuming $\;\lfloor x\rfloor\;$ is the floor function, we get that for $\;x\;$ very close to zero from the left,
$$\;\forall\,k\in\Bbb N\;,\;\;\lfloor x^k\rfloor=\begin{cases}-1&,\;\;k\;\;\text{is odd}\\{}\\\;\;\,0&,\;\;k\;\;\text{is even}\end{cases}\;$$ , and then:
$$\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+\ldots+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+2x}=\frac{\overbrace{-1+0-1+\ldots+0-1}^{\lfloor x\rfloor+\ldots+\lfloor x^{2n+1}\rfloor}+n+1}{1+0-1+2x}=$$
$$=\frac{-1(n+1)+n+1}{2x}=0\xrightarrow[x\to 0^-]{}0$$
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Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$ For any natural number $n$ ,let $S(n)$ denote the sum of the digits of $n$.Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$
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First, by the nice answer/hint provided by mfl, the only way that $S(S(n)) = 2$ is if $a+b+c = 2$ or $11$ or $20$.
Next, suppose that $a$, $b$, and $c$ are digits with $a+b+c=2$ or $11$ or $20$. Then since,
$$100a+10b+c+9 = 100a + 10(b+1) + (c-1),$$
the new digits are either $a$, $b+1$, and $c-1$ (when $c>0$) or $a$, $b$, and $c+9$ (when $c=0$). Now, if the new digits are $a$, $b+1$, and $c-1$, then the new digit sum is the same. If the new digits are $a$, $b$, and $c+9$, then the new digits sum is increased by $9$. Now, there are three cases:
*
*$a+b+c=2$, in which case $a+b+c+9=11$ so that $S(S(n))=2$,
*$a+b+c=11$, in which case $a+b+c+9=20$ so that again $S(S(n))=2$, or
*$a+b+c=20$. But in this case we can't have $c=0$ so that the new digits are $a$, $b-1$, and $c-1$. Thus, the new digit sum is the same as the original and again $S(S(n))=2$.
As a result $S(S(n))=2$ iff $S(S(n+9))=2$. There's now a simple pattern that leads to the count of 100 such integers. Here they are:
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A couple of definite integrals related to Stieltjes constants In a (great) paper "A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations" by Iaroslav V. Blagouchine, the following integral representation of the first Stieltjes constant $\gamma_1$ is given (on page 539):
$$\gamma_1=-\left[\gamma-\frac{\ln2}2\right]\ln2+i\int_0^\infty\frac{dx}{e^{\pi x}+1}\left\{\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\right\}.\tag1$$
It's possible to get rid of imaginary numbers in this formula, and rewrite it in terms of only real-valued functions:
$$2\int_0^\infty\frac{\arctan x}{1+x^2}\frac{dx}{e^{\pi x}+1}-\int_0^\infty\frac{x\ln(1+x^2)}{1+x^2}\frac{dx}{e^{\pi x}+1}=\gamma_1+\left[\gamma-\frac{\ln2}2\right]\ln2.\tag2$$
Question:
Is it possible to find closed forms separately for each integral on the left-hand side of $(2)$?
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Using contour integration, I am able to evaluate the first integral in terms of a slightly different infinite series which may or may not be able to be evaluated in closed form.
First notice that
$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1} &= \frac{1}{2} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \left(1- \tanh \left(\frac{\pi x}{2} \right) \right) \, dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \, dx- \frac{1}{4} \int_{-\infty}^{\infty} \frac{\arctan x}{1+x^{2}} \tanh \left(\frac{\pi x}{2} \right) \, dx \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \int_{-\infty}^{\infty} \frac{\log(1-ix)}{1+x^{2}} \tanh \left(\frac{\pi x}{2} \right) \, dx . \end{align} $$
Now let $ \displaystyle f(z) = \frac{\log(1-iz)}{1+z^{2}} \tanh \left(\frac{\pi z}{2} \right).$
Then since $ \displaystyle \int f(z) \ dz$ vanishes along the upper half of the circle $|z|=2N$ as $N \to \infty$ through the positive integers, we get
$$ \begin{align} &\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1} \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \ 2 \pi i \left(\text{Res}[f(z),i] + \sum_{n=1}^{\infty} \text{Res} [f(z), i(2n+1)] \right) \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \ 2 \pi i \left( \frac{\log(2)-1}{2 \pi} - \sum_{n=1}^{\infty}\frac{\log(2n+2)}{2 \pi(n^{2}+n)}\right) \\ &= \frac{\pi^{2}}{16} + \frac{\log (2)}{4} - \frac{1}{4} - \frac{\log (2)}{4} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} \\ &= \frac{\pi^{2}}{16} - \frac{1}{4} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} . \end{align}$$
EDIT:
In this answer, Olivier Oloa talks about the coefficients of the regular part of the Laurent series expansion at the origin of the poly-Hurwitz zeta function (which he refers to as the poly-Stieltjes constants).
It turns out that we can evaluate the above infinite series in terms of the poly-Stieltjes constants.
According to Theorem 2,
$$ \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} = \gamma_{1}(1,0) - \gamma_{1}(1,1)$$ which, if I understand correctly, reduces to $$\gamma_{1}(1,0) - \gamma_{1}.$$
As a side note, Wolfram Alpha does not return a very good approximation of the value of that infinite series. And if you ask for more digits, it will return a different result.
But it appears that the value of the infinite series is approximately $ 1.2577468869$.
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Find the equation which has key root $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ In my last question which was Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ ,I could find the coefficients(were very easy) of fourth-degree equation, so I went to study the case which has key root $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ and I found the equation as follow
$$x^8-k_{1}x^6+k_{2}x^4-k_{3}x^2+k_{4}=0$$
The roots of this quation take the form
$$x=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
$$x=\sqrt{a}+\sqrt{b}-\sqrt{c}$$
$$x=\sqrt{a}-\sqrt{b}-\sqrt{c}$$
$$x=\sqrt{a}-\sqrt{b}+\sqrt{c}$$
$$x=-\sqrt{a}+\sqrt{b}+\sqrt{c}$$
$$x=-\sqrt{a}-\sqrt{b}+\sqrt{c}$$
$$x=-\sqrt{a}+\sqrt{b}-\sqrt{c}$$
$$x=-\sqrt{a}-\sqrt{b}-\sqrt{c}$$
When $a$, $b$, and $c$ are real numbers
I could find only the $k1=4(a+b+c)$.
By which method can I find the others coefficients
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If $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ then $(x-\sqrt{a})^2=(\sqrt{b}+\sqrt{c})^2$, that is, $\xi=\sqrt{\alpha}+\sqrt{\beta}$ with $\alpha=4ax^2$, $\beta=4bc$ and $\xi=x^2+a-b-c$.
The result in your previous question yields $\xi^4-2(\alpha+\beta)\xi^2+(\alpha-\beta)^2=0$, that is, $$(x^2+a-b-c)^4-8(ax^2+bc)(x^2+a-b-c)^2+16(ax^2-bc)^2=0.$$ This polynomial is also $$\prod_{\pm}(x\pm\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}),$$ where the product is over the eight possible choices of the three $\pm$ signs.
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Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
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$$3\equiv1\pmod2\implies 1+2^n+3^n+4^n\equiv0\pmod2$$
If $n$ is odd, $1+4^n=1^n-(-4)^n$ is divisible by $1-(-4)=5$
Similarly, $2^n+3^n\equiv0\pmod5,$
For $n=4k+2,2^{4k+2}\equiv4\pmod5,3^{4k+2}\equiv-1,4^{4k+2}\equiv1$
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Solving the equation: $3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)$ Solving the equation:
$$3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)\tag{1}$$
I tried to write $(1)$ becomes
$$\sqrt{3}\sin \left(\frac{\pi}{3}-x\right)=\sin \left(\frac{\pi}{3}+2x \right)$$
Now, I have stuck :( , can you help me?
Thanks!
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Hint
$$\cos{x}(3-2\sin{x})=\sqrt{3}(2\sin{x}+1)(1-\sin{x})$$
$$\Longrightarrow \left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)^2(2\sin{x}+1)$$
(1):$$\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)=0\Longrightarrow x=\dfrac{\pi}{2}+2k\pi$$
(2):$$\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)(2\sin{x}+1)\tag{1}$$
let
$$\sin{\dfrac{x}{2}}+\cos{\dfrac{x}{2}}=t\Longrightarrow \sin{x}=t^2-1,\left(\sin{\dfrac{x}{2}}-\cos{\dfrac{x}{2}}\right)^2=2-t^2$$
use $(1)^2$
then we have
$$t^2\cdot(3-2t^2+2)^2=3(2-t^2)(2t^2-1)^2\Longrightarrow 16t^6-56t^4+52t^2-6=0$$
$$\Longrightarrow (2t^2-3)(2t^2-2t-1)(2t^2+2t-1)=0$$
so
$$\Longrightarrow t=\sqrt{\dfrac{3}{2}},-\sqrt{\dfrac{3}{2}},\dfrac{1}{2}(-1\pm \sqrt{3})-\dfrac{1}{2}(1+\sqrt{3})$$
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If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
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We'll show a bit more: $a^a b^b > \left( \frac{a+b}{2} \right)^{a+b} > a^b b^a.$
First we need a lemma: $P = (1+x)^{1+x} (1-x)^{1-x} > 1$ if $x < 1$.
Proof: $\ln P = (1+x)\ln (1+x) + (1-x) \ln (1-x) = x ( \ln (1+x) - \ln (1-x)) + \ln (1+x) + \ln (1-x) = 2x \left(x + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right) - 2 \left( \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \cdots \right) = 2 \left(\frac{x^2}{1 \cdot 2} + \frac{x^4}{3 \cdot 4} + \frac{x^6}{5 \cdot 6} + \cdots \right).$
Therefore, $\ln P$ is positive, and so $P > 1.$
In this lemma put $x = \frac{s}{t}$, where $t > s$. Then
$ \left( 1 + \frac{s}{t} \right)^{1+ \frac{s}{t}} \left(1 - \frac{s}{t}\right)^{1-\frac{s}{t}} > 1. $ Thus
$ \left( \frac{t+s}{t} \right)^{t+s} \left(\frac{t-s}{t} \right)^{t-s} > 1^t = 1.$
Thus $ (t + s)^{t+s}(t - s)^{t-s} > t^{2t}$. Letting $t + s = a$ and $t - s = b$, we see that $t = \frac{a+b}{2}$ and so $a^a b^b > \left( \frac{a+b}{2}\right)^{a+b}.$
In a similar way, letting $Q = (1+x)^{1-x} (1-x)^{1+x}$, we can show that $\ln Q < 0$ and so $Q < 1.$ Now continue as we did in the last proof.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1063223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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|
Linear transformation ker and image Let $\varphi\colon \mathbb{R}^4 \rightarrow \mathbb{R}^3$ be described by $\varphi(X)=AX$ where
$A=\begin{pmatrix}
3 & 2 & 1 & 3 \\
1 & 1 & 1 & 1 \\
2 & 1 & 0 & 1
\end{pmatrix} $
. Find base vectors of $\ker \varphi$ and $\operatorname{Im} \varphi$. In my opinion those vectors will be $[-1,1,0,0]$ and $[-1,0,1,0]$ for kernel and $[3,1,2]$ and $[3,1,1] $ for image. Am I correct?
|
Start with the definitions:
$$\ker\varphi:=\{X\in\mathbb R^4:\varphi(X)=0\},$$ that is, the set of all $(x_1,x_2,x_3,x_4)$ such that
$$
\begin{bmatrix}
3 & 2 & 1 & 3 \\
1 & 1 & 1 & 1 \\
2 & 1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\0
\end{bmatrix}.
$$
Can you find the solution set of this system? If so, you will be able to write down a basis for $\ker\varphi$. Note that row reducing yields
$$
\begin{bmatrix}
3 & 2 & 1 & 3 \\
1 & 1 & 1 & 1 \\
2 & 1 & 0 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
so
\begin{align}
x_1-x_3&=0,\\
x_2+2x_3&=0,\\
x_4&=0
\end{align}
and thus
$x_4=0$, $x_2=-2x_3$, $x_1=x_3$ and $x_3$ is free. Hence a basis for $\ker\varphi$ is
$$
\left\{\begin{bmatrix} 1\\-2\\1\\0\end{bmatrix}\right\}.$$
As for $\text{Im}\,\varphi$, again go to the definition
$$
\text{Im}\,\varphi:=\{Y\in\mathbb{R}^3:\varphi(X)=Y\text{ for some }x\in\mathbb{R}^4\}.$$ So look at all possible outputs of $\varphi$ acting on an $X$ after row reducing:
\begin{align}
\begin{bmatrix}
3 & 2 & 1 & 3 \\
1 & 1 & 1 & 1 \\
2 & 1 & 0 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
\end{align}
Since you have leading ones in columns 1, 2, and 4, use columns 1, 2, and 4 of the original matrix as the basis for $\text{Im}\,\varphi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1063505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
|
\begin{align}J&=\int_0^\infty \frac{(x^2-1)\ln x}{1+x^4}dx\\
&=\underbrace{\int_0^\infty \frac{x^2\ln x}{1+x^4}dx}_{u=\frac{1}{x}}-\int_0^\infty \frac{\ln x}{1+x^4}dx\\
&=-2\int_0^\infty \frac{\ln x}{1+x^4}dx\\
K&=\int_0^\infty \frac{\ln x}{1+x^4}dx\\
R&=\int_0^\infty\int_0^\infty \frac{\ln(xy)}{(1+x^4)(1+y^4)}dxdy\\
&\overset{u(x)=xy}=\int_0^\infty\int_0^\infty\frac{y^3\ln u}{(u^4+y^4)(1+y^4)}dudy=\frac{1}{4}\int_0^\infty\frac{\ln u}{u^4-1}\left[\ln\left(\frac{1+y^4}{u^4+y^4}\right)\right]_{y=0}^{y=\infty}du\\
&=\int_0^\infty\frac{\ln^2 u}{u^4-1}du=\frac{1}{2}\underbrace{\int_0^\infty\frac{\ln^2 u}{u^2-1}du}_{=0}-\frac{1}{2}\underbrace{\int_0^\infty\frac{\ln^2 u}{u^2+1}du}_{=\frac{\pi^3}{8}}=\boxed{-\frac{\pi^3}{16}}
\end{align}
On the other hand,
\begin{align}R&=2K\int_0^\infty \frac{1}{1+x^4}dx\\
S&=\int_0^\infty \frac{1}{1+x^4}dx\overset{u=\frac{1}{x}}=\int_0^\infty \frac{u^2}{1+u^4}du\\
2S&=\int_0^\infty \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}dx\overset{z=x-\frac{1}{x}}=\int_{-\infty}^{+\infty}\frac{1}{z^2+2}dz=\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_{-\infty}^{+\infty}\\
S&=\boxed{\frac{\pi}{2\sqrt{2}}}
\end{align}
Therefore,
\begin{align}K&=\frac{R}{2S}=\frac{-\frac{\pi^3}{16}}{2\times\frac{\pi}{2\sqrt{2}}}=-\frac{\pi^2}{8\sqrt{2}}\\
J&=-2K=\boxed{\frac{\pi^2}{4\sqrt{2}}}
\end{align}
NB:
I assume that,
\begin{align}\int_0^\infty \frac{\ln^2 x}{1+x^2}dx=\frac{\pi^3}{8}\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1067499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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|
How to compute the residue of $(z^2+2z+1)\sin\left(\frac{1}{1+z}\right)$ This was an example given in my notes but all it concluded was with something about an infinite principal part. How do we compute it? we have it equal to $ \left( z + 1 \right)^2 \cdot \sin \left( (z+1)^{-1} \right) = \left( z + 1 \right)^2 \cdot \left[ (z+1)^{-1} + \cdots \right] $. That's all I could get. What now?
|
$z=-1$ is an essential singularity of the function. So we have $$(z^2+2z+1)\sin \frac{1}{1+z}=(z+1)^2\sin \frac{1}{1+z} $$ therefore $$(z+1)^2\sin \frac{1}{1+z}=(z+1)^2 \left(\frac{1}{1+z}-\frac{1}{3!(1+z)^3}+\frac{1}{5!(1+z)^5}-\cdots\right)$$ this shows that Res$[(z+1)^2\sin \frac{1}{1+z}, -1]=-\frac{1}{3!}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1072071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$ Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$
Please guide me to solve this problem. I have differentiated it with respect to $x$ and make equal to zero, but couldn't get any point.
|
By Jensen $$\left(\frac{\sqrt[3]{1+x}+\sqrt[3]{1-x}}{2}\right)^3\leq\frac{1+x+1-x}{2}=1,$$
which gives $\sqrt[3]{1+x}+\sqrt[3]{1-x}\leq2$.
The equality occurs for $x=0$.
Id est, the answer is $2$.
We used the following inequality.
For non-negatives $a$ and $b$ we have:
$$\left(\frac{a+b}{2}\right)^3\leq\frac{a^3+b^3}{2},$$
which is $(a+b)(a-b)^2\geq0$.
Also we can use P_M.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1074216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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|
How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
|
One way is to look at
$$
1+2x^2+3x^4+4x^6+5x^8+\dots
$$
as
$$
1+2t+3t^2+4t^3+5t^4+\dots
$$
where $t=x^2$. The last series is the derivative of
$$
1+t+t^2+t^3+t^4+t^5+\dots=\frac1{1-t}
$$
Therefore,
$$
1+2t+3t^2+4t^3+5t^4+\dots=\frac1{(1-t)^2}
$$
and
$$
1+2x^2+3x^4+4x^6+5x^8+\dots=\frac1{(1-x^2)^2}
$$
Now, just multiply by $1+x$:
$$
\begin{align}
1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+\dots
&=\frac{1+x}{(1-x^2)^2}\\
&=\frac1{(1-x)(1-x^2)}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1075054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
*$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
*$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
*$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$
How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.
|
Note that if your equation includes $\sqrt{4x-1}$, this means that $4x-1\ge0$. Similarly, you are only looking for solutions among numbers that fulfill $x+2\ge0$, since the equation contains the expression $\sqrt{x+2}$.
So you can write $\sqrt{(4x-1)^2}=|4x-1|=4x-1$. But do not forget in the end include only those solutions where $4x-1\ge0$, i.e., $x\ge\frac14$. (And the same for the condition $x+2\ge0$.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1079486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
Surface Area (Integration) Find the surface area of the object by rotating $y = 4+3x^2$ about the $y$-axis, where $1 \leq x \leq 2$.
I've been using the formula:
Surface area:
Definite integral of $$2\pi y\Bigg(1+\Big(\dfrac{dx}{dy}\Big)^2\Bigg)dy$$
Limits are supposed to be in the form of $y$ for the above question.
I've been trying to do this question for the last 3 hours and I get stuck at $$\int\dfrac{\cos(x)}{\sin^5(x)}dx$$
Can anyone help me how to solve the above form?
Thanks in advance.
|
Since $x=\sqrt{\frac{y-4}{3}}, \;\;\frac{dx}{dy}=\frac{1}{2}\left(\frac{y-4}{3}\right)^{-\frac{1}{2}}\cdot\frac{1}{3}\text{ and }\left(\frac{dx}{dy}\right)^2=\frac{1}{36}\frac{1}{\frac{y-4}{3}}=\frac{1}{12(y-4)}$.
Then $S=\displaystyle\int_7^{16} 2\pi\sqrt{\frac{y-4}{3}}\sqrt{1+\frac{1}{12(y-4)}}dy=2\pi\int_7^{16}\sqrt{\frac{y-4}{3}}\sqrt{\frac{12y-47}{12(y-4)}}dy$,
so $\;\;S=\displaystyle\frac{\pi}{3}\int_7^{16}\sqrt{12y-47}dy$. $\;\;\;$ (Now let $u=12y-47$.)
Alternatively, $S=\displaystyle\int_1^2 2\pi x\sqrt{1+36x^2}dx$,
so letting $u=1+36x^2$ and $du=72xdx$ gives
$\;\;\displaystyle S=\frac{\pi}{36}\int_{37}^{145}\sqrt{u}du$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1080559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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|
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
|
Hint: $\sqrt{x^2 + 1} - x$ can be written as $\sqrt{x^2 + 1} - \sqrt{x^2}$.
Generally, if you have an expression involving $\sqrt a - \sqrt b$ then it
is often useful to "expand" the expression with $\sqrt a + \sqrt b$:
$$
x \, (\sqrt{x^2 + 1} - x) = \frac {x \,(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)} {\sqrt{x^2 + 1} + x}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.