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.

Laplace coefficients are Fourier coefficients used in Celestial mechanics calculations

$$ b^n_s (\alpha) \equiv {1 \over \pi} \int_0^{2\pi} {\cos n \phi \over (1 - 2 \alpha \cos \phi + \alpha^2)^s} d \phi $$

with $s = i + 1/2$ (a half integer) and $0<\alpha <1$.

The function $(1 - 2\alpha \cos \phi + \alpha^2)^{-s}$ is analytic (locally) so the Fourier coefficients decay rapidly. For large $n$ and $\alpha$ not small, I think that $b^n_{1/2} \sim c_1 e^{-c_2(1-\alpha)n}$ (with constants $c_1,c_2$). How do I show that this is true and how do I derive constants $c_1,c_2$? If this was a bad guess is there an exponential function that does approximate the coefficients at large $n$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Let $$ f(x,\alpha)=\frac1{c(x)(1+\alpha^2 x^2)(1-2\alpha \cos x+\alpha^2)^s}, $$ where $$c(x)=\sum_{n=-\infty}^\infty\frac1{1+\alpha^2 (x+2\pi n)^2}.$$ This function belongs to $L_2(\mathbb R)$, has an analytic continuation in the strip $|\Im z|<a=\log \alpha^{-1}$ and its Fourier transform at the integer points $\xi=n$ is equal to $b_s^n(\alpha)$. It is known that the wider is the strip the faster is the decay of the Fourier transform, namely as $e^{-b|\xi|}$ for $b\in(0,a)$, see the reference in the answer here. So a rough result implied from there seems to be this: for any $c_2<\log\alpha^{-1}\ $ there exists $c_1>0$ such that $|b_s^n(\alpha)|\le c_1 e^{-c_2n}$.

EDIT

But the same idea applied to the circle and not the line gives a simpler solution. Consider the function $$f(z)=\frac1{(1+\alpha^2-\alpha(z+1/z))^s}.$$ Then $$f(e^{i \phi})=\frac1{(1-2\alpha \cos \phi+\alpha^2)^s}.$$ The denominator of $f$ has roots $\alpha$, $1/\alpha$. So $f$ is analytic in the ring $R=\{\alpha<|z|<\alpha^{-1}\}\ $. Hence $$f(z)=\frac12\sum_{n=-\infty}^\infty b_s^n(\alpha)z^n,\quad z\in R.$$ From the Cauchy root test for convergence it follows that $\limsup_{n\to\infty} |b_s^n(\alpha)|^{1/n}= \alpha\ $. This can be written in exponential form as desired above.

share|improve this answer
    
@Alice The function is good because its Fourier transform at the integer values gives $b^n_s$. –  Andrew Jul 13 '11 at 21:36

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.