Kapteyn series

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

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

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

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

• Schlömilch's series