# Series multisection

In mathematics, a multisection of a power series is a new power series composed of equally spaced terms extracted unaltered from the original series. Formally, if one is given a power series
    $\sum _{n=-\infty }^{\infty }a_{n}\cdot z^{n}$
then its multisection is a power series of the form
    $\sum _{m=-\infty }^{\infty }a_{qm+p}\cdot z^{qm+p}$
where p, q are integers, with 0 ≤ p < q. Series multisection represents one of the common transformations of generating functions.

## Multisection of analytic functions

A multisection of the series of an analytic function
    $f(z)=\sum _{n=0}^{\infty }a_{n}\cdot z^{n}$
has a closed-form expression in terms of the function $f(x)$:
    $\sum _{m=0}^{\infty }a_{qm+p}\cdot z^{qm+p}={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}\omega ^{-kp}\cdot f(\omega ^{k}\cdot z),$
where $\omega =e^{\frac {2\pi i}{q}}$ is a primitive q-th root of unity. This expression is often called a root of unity filter. This solution was first discovered by Thomas Simpson. This expression is especially useful in that it can convert an infinite sum into a finite sum. It is used, for example, in a key step of a standard proof of Gauss's digamma theorem, which gives a closed-form solution to the digamma function evaluated at rational values p/q.

## Examples
### Bisection

In general, the bisections of a series are the even and odd parts of the series.

### Geometric series

Consider the geometric series
    $\sum _{n=0}^{\infty }z^{n}={\frac {1}{1-z}}\quad {\text{ for }}|z|<1.$
By setting $z\rightarrow z^{q}$ in the above series, its multisections are easily seen to be
    $\sum _{m=0}^{\infty }z^{qm+p}={\frac {z^{p}}{1-z^{q}}}\quad {\text{ for }}|z|<1.$
Remembering that the sum of the multisections must equal the original series, we recover the familiar identity
    $\sum _{p=0}^{q-1}z^{p}={\frac {1-z^{q}}{1-z}}.$

### Exponential function

The exponential function
    $e^{z}=\sum _{n=0}^{\infty }{z^{n} \over n!}$
by means of the above formula for analytic functions separates into
    $\sum _{m=0}^{\infty }{z^{qm+p} \over (qm+p)!}={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}\omega ^{-kp}e^{\omega ^{k}z}.$
The bisections are trivially the hyperbolic functions:
    $\sum _{m=0}^{\infty }{z^{2m} \over (2m)!}={\frac {1}{2}}\left(e^{z}+e^{-z}\right)=\cosh {z}$
    $\sum _{m=0}^{\infty }{z^{2m+1} \over (2m+1)!}={\frac {1}{2}}\left(e^{z}-e^{-z}\right)=\sinh {z}.$
Higher order multisections are found by noting that all such series must be real-valued along the real line. By taking the real part and using standard trigonometric identities, the formulas may be written in explicitly real form as
    $\sum _{m=0}^{\infty }{z^{qm+p} \over (qm+p)!}={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}e^{z\cos(2\pi k/q)}\cos {\left(z\sin {\left({\frac {2\pi k}{q}}\right)}-{\frac {2\pi kp}{q}}\right)}.$
These can be seen as solutions to the linear differential equation $f^{(q)}(z)=f(z)$ with boundary conditions $f^{(k)}(0)=\delta _{k,p}$, using Kronecker delta notation. In particular, the trisections are
    $\sum _{m=0}^{\infty }{z^{3m} \over (3m)!}={\frac {1}{3}}\left(e^{z}+2e^{-z/2}\cos {\frac {{\sqrt {3}}z}{2}}\right)$
    $\sum _{m=0}^{\infty }{z^{3m+1} \over (3m+1)!}={\frac {1}{3}}\left(e^{z}-2e^{-z/2}\cos {\left({\frac {{\sqrt {3}}z}{2}}+{\frac {\pi }{3}}\right)}\right)$
    $\sum _{m=0}^{\infty }{z^{3m+2} \over (3m+2)!}={\frac {1}{3}}\left(e^{z}-2e^{-z/2}\cos {\left({\frac {{\sqrt {3}}z}{2}}-{\frac {\pi }{3}}\right)}\right),$
and the quadrisections are
    $\sum _{m=0}^{\infty }{z^{4m} \over (4m)!}={\frac {1}{2}}\left(\cosh {z}+\cos {z}\right)$
    $\sum _{m=0}^{\infty }{z^{4m+1} \over (4m+1)!}={\frac {1}{2}}\left(\sinh {z}+\sin {z}\right)$
    $\sum _{m=0}^{\infty }{z^{4m+2} \over (4m+2)!}={\frac {1}{2}}\left(\cosh {z}-\cos {z}\right)$
    $\sum _{m=0}^{\infty }{z^{4m+3} \over (4m+3)!}={\frac {1}{2}}\left(\sinh {z}-\sin {z}\right).$

### Binomial series

Multisection of a binomial expansion
    $(1+x)^{n}={n \choose 0}x^{0}+{n \choose 1}x+{n \choose 2}x^{2}+\cdots$
at x = 1 gives the following identity for the sum of binomial coefficients with step q:
    ${n \choose p}+{n \choose p+q}+{n \choose p+2q}+\cdots ={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}\left(2\cos {\frac {\pi k}{q}}\right)^{n}\cdot \cos {\frac {\pi (n-2p)k}{q}}.$