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 $n$?


1 Answer 1


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 $|b_s^n(\alpha)|\le c_1 e^{-c_2n}$.


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.

  • $\begingroup$ @Alice The function is good because its Fourier transform at the integer values gives $b^n_s$. $\endgroup$
    – Andrew
    Commented Jul 13, 2011 at 21:36

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