Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can we prove that there does not exist a function $f$, which satisfies this equation for all $R>0$: $$\int_0^{2 \pi} \frac{1}{\sqrt{1+R^2 \sin^2(x)}} f(R \cos(x))\, dx= 1.$$

share|improve this question
1  
What can you say about the parameter $R$? Is this expected to hold for a particular value, for a range of values, or for all values of $R$? –  mjqxxxx May 16 '12 at 15:28
1  
If to search for an analytical solution, calculations seem to indicate (if a didn't made any mistakes) that such a solution exists for small $x$, has convergence radius of 1 and converges for $x=\pm1$. So the question in this case reduces to can it be analytically continued to some region containing $\mathbb R$? –  Andrew May 17 '12 at 10:30
    
If your computation is not too long, do you care to post it? Or give a few hints, what you were using? Taylor expansion of $F$? I would be really grateful. –  plusepsilon.de May 17 '12 at 11:23

1 Answer 1

up vote 2 down vote accepted

Here's some considerations. Change of variables $z=R\cos x$ transforms the integral into $$ 2\int_{-R}^R\frac{f(z)dz}{\sqrt{(1+R^2-z^2)(R^2-z^2)}}. $$ Using Mma it is possible to calculate integrals $$ a_n=2\int_{-R}^R\frac{z^ndz}{\sqrt{(1+R^2-z^2)(R^2-z^2)}}, $$ for example, $a_0=\frac{4 K\left(\frac{R^2}{R^2+1}\right)}{\sqrt{R^2+1}}$, $a_2=4 \sqrt{R^2+1} \left(K\left(\frac{R^2}{R^2+1}\right)-E\left(\frac{R^2}{R^2+1}\right)\right)$, $a_{2n+1}=0\,$, where $K$ and $E$ are complete elliptic integrals of the first and second kind respectively. Maclaurin series expansion for $a_{2n}$ begins with $R^{2n}$ (it can be seem from the initial integral there $n$th term in expansion of $f$ has $R^n \cos^n x$.) This allows to obtain coefficients for $$ f(z)=\sum_{n=0}^\infty c_nz^n $$ from the equality $$ \sum_{n=0}^\infty c_n a_n(R)\equiv1 $$ one by one: $c_0$ from $c_0a_0(R)=1+O(R^2)$, then $c_2$ from $c_0a_0(R)+c_2a_2(R)=1+O(R^4)\,$ etc. The result is $$ f(z)=\frac1{2\pi}\bigg( 1+\frac{z^2}{2}-\frac{7 z^4}{24}+\frac{59 z^6}{240}-\frac{3013 z^8}{13440}+\frac{10147 z^{10}}{48384}-\frac{191833 z^{12}}{967680}+ $$ $$ \frac{52145347 z^{14}}{276756480}-\frac{11939071981 z^{16}}{66421555200}+\frac{55491350861 z^{18}}{322618982400}-\frac{28327222665017 z^{20}}{171633298636800}+\ldots\bigg)= $$ $$ 0.159155+0.079577 z^2-0.0464202 z^4+0.0391256 z^6-0.0356796 z^8+ 0.0333777 z^{10} $$ $$ -0.0315509 z^{12}+ 0.0299873 z^{14}-0.0286076 z^{16}+0.0273751 z^{18}-0.0262677 z^{20}+\ldots $$ So coefficients signs seems to be alternating beginning from the second. Absolute values are monotonously decreasing, but slowly, so if it goes further that way the convergence radius should be 1.

share|improve this answer
    
That is a really good approach, probably not the best one for obtaining a closed form for $f$. Thanks a lot. –  plusepsilon.de May 17 '12 at 14:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.