Fourier series of $\exp(\sum_k a_k\cos(k\theta+\phi_k))$

Question

I need to know the Fourier series of exponential of general function, represented as

$c_n:=\int^{\pi}_{-\pi}\exp\left(\sum_{k=0}^{\infty}a_k\cos(k\theta+\phi_k)\right)\cos(n\theta+\psi_n)d\theta$.

($c_n$ and $\psi_n$ are what I want to know, and the others are given.)

This question gives a specific solution of this problem; the Fourier series of $e^{a\cos x}$:

$\int^{\pi}_{-\pi}e^{a\cos \theta}\cos(n\theta)d\theta=2\pi I_n(a)$,

where $I_n$ is modified Bessel function of the first kind.

This comes from gererating function of $I_n$:

$\exp(\frac{a}{2}(z+\frac{1}{z}))=\sum_{k=1}^{\infty}I_k(a)z^k$.

But I have failed to apply this relation to my generalized question because of the complexity.

Could anyone help me solve this question?

(I apologize my poor English.)

Answer 1

Basically you are asking for the Fourier transform ${\cal F}[e^f]$ of the exponential of a function $f(\theta)$, in terms of the Fourier transform ${\cal F}[f]$ of $f$ itself; the formal answer is $${\cal F}[e^f]=1+\sum_{n=1}^\infty \frac{1}{n!}(\underset{\underbrace{\text{n times}}}{{\cal F}[f]*\cdots *{\cal F}[f]})$$ where $\ast$ is a convolution. There is no further simplification for the general case.

Answer 2

Thank you for your answer.

Well, I have already known that relation, but I was motivated by your post.

$$\exp\left(\sum_k a_k\cos(k\theta+\phi_k)\right)=\sum_{n=0}^\infty\frac{1}{n!}\left(\sum_{k=0}^\infty a_k\cos(k\theta+\phi_k)\right)^n$$

Although this is difficult to calculate integral as it is, after some deformations (which maybe contain some mistakes), this will be:

$$=\sum_{n=0}^\infty\frac{1}{2^{n+1}n!}\sum_{p_0=0}^\infty\sum_{p_1=0}^\infty\cdots\sum_{p_n=0}^\infty\left(\prod_{l=0}^n a_{p_l}\right)\sum_{q_0=\pm1}\sum_{q_1=\pm1}\cdots\sum_{q_n=\pm1}e^{i\sum_{m=0}^nq_m\phi_{p_m}}\cdot e^{i\sum_{m=0}^nq_m p_m\theta}$$

This (some complicated & formal) notation can be integrated,

$$c_j:=\int_{-\pi}^{\pi}d\theta\exp\left(\sum_k a_k\cos(k\theta+\phi_k)\right)\cos j\theta$$ $$=\sum_{n=0}^\infty\frac{1}{2^{n+1}n!}\sum_{p_0=0}^\infty\sum_{p_1=0}^\infty\cdots\sum_{p_n=0}^\infty\left(\prod_{l=0}^n a_{p_l}\right)\sum_{q_0=\pm1}\sum_{q_1=\pm1}\cdots\sum_{q_n=\pm1}e^{i\sum_{m=0}^nq_m\phi_{p_m}}\cdot(\delta_{\sum_{m=0}^n q_m p_m+j}+\delta_{\sum_{m=0}^n q_m p_m-j})$$ where $\delta_r$ represents Kronecker delta.

It is difficult to make this more simple for me, but the interior of the summation seems to decrease rapidly with "well-behaved" $a_k$.

So this Fourier integral may be dominated by the smaller $n$ terms and I have managed to write down the cases $n=0$ and $n=1$.

Then the approximated Fourier cosine series for small $k$ : $$c_0\sim\frac{1}{2}a_0+\frac{1}{4}\left(a_0^2+\sum_{p=0}^\infty a_p^2\right)$$

$$c_1\sim\frac{1}{2}a_1\cos\phi_1+\frac{1}{2}\left(a_0a_1\cos(\phi_1)+\sum_{p=0}^{\infty}a_p a_{p+1}\cos(\phi_p-\phi_{p+1})\right)$$

As I am originally interested in lower-order terms, possibly this caluclation will be sufficient (not tested yet).