Kapteyn series is a series expansion of analytic functions on a domain in terms of the Bessel function of the first kind. Kapteyn series are named after Willem Kapteyn, who first studied such series in 1893.^[1][2] Let ${\displaystyle f}$ be a function analytic on the domain

${\displaystyle D_{a}=\left\{z\in \mathbb {C} :\Omega (z)=\left|{\frac {z\exp {\sqrt {1-z^{2)))){1+{\sqrt {1-z^{2))))}\right|\leq a\right\))$

with ${\displaystyle a<1}$. Then ${\displaystyle f}$ can be expanded in the form

${\displaystyle f(z)=\alpha _{0}+2\sum _{n=1}^{\infty }\alpha _{n}J_{n}(nz)\quad (z\in D_{a}),}$

where

${\displaystyle \alpha _{n}={\frac {1}{2\pi i))\oint \Theta _{n}(z)f(z)dz.}$

The path of the integration is the boundary of ${\displaystyle D_{a))$. Here ${\displaystyle \Theta _{0}(z)=1/z}$, and for ${\displaystyle n>0}$, ${\displaystyle \Theta _{n}(z)}$ is defined by

${\displaystyle \Theta _{n}(z)={\frac {1}{4))\sum _{k=0}^{\left[{\frac {n}{2))\right]}{\frac {(n-2k)^{2}(n-k-1)!}{k!))\left({\frac {nz}{2))\right)^{2k-n))$

Kapteyn's series are important in physical problems. Among other applications, the solution ${\displaystyle E}$ of Kepler's equation ${\displaystyle M=E-e\sin E}$ can be expressed via a Kapteyn series:^[2][3]

${\displaystyle E=M+2\sum _{n=1}^{\infty }{\frac {\sin(nM)}{n))J_{n}(ne).}$

Relation between the Taylor coefficients and the ${\displaystyle \alpha _{n))$ coefficients of a function

Let us suppose that the Taylor series of ${\displaystyle f}$ reads as

${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}.}$

Then the ${\displaystyle \alpha _{n))$ coefficients in the Kapteyn expansion of ${\displaystyle f}$ can be determined as follows.^[4]: 571

${\displaystyle {\begin{aligned}\alpha _{0}&=a_{0},\\\alpha _{n}&={\frac {1}{4))\sum _{k=0}^{\left\lfloor {\frac {n}{2))\right\rfloor }{\frac {(n-2k)^{2}(n-k-1)!}{k!(n/2)^{(n-2k+1)))}a_{n-2k}\quad (n\geq 1).\end{aligned))}$

Examples

The Kapteyn series of the powers of ${\displaystyle z}$ are found by Kapteyn himself:^{[1]: 103,  [4]: 565}

${\displaystyle \left({\frac {z}{2))\right)^{n}=n^{2}\sum _{m=0}^{\infty }{\frac {(n+m-1)!}{(n+2m)^{n+1}\,m!))J_{n+2m}\{(n+2m)z\}\quad (z\in D_{1}).}$

For ${\displaystyle n=1}$ it follows (see also ^[4]: 567)

${\displaystyle z=2\sum _{k=0}^{\infty }{\frac {J_{2k+1}((2k+1)z)}{(2k+1)^{2))},}$

and for ${\displaystyle n=2}$ ^[4]: 566

${\displaystyle z^{2}=2\sum _{k=1}^{\infty }{\frac {J_{2k}(2kz)}{k^{2))}.}$

Furthermore, inside the region ${\displaystyle D_{1))$,^[4]: 559

${\displaystyle {\frac {1}{1-z))=1+2\sum _{k=1}^{\infty }J_{k}(kz).}$