Series multisection

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

Consider the geometric series

${\displaystyle \sum _{n=0}^{\infty }z^{n}={\frac {1}{1-z}}\quad {\text{ for }}|z|<1.}$

By setting ${\displaystyle z\rightarrow z^{q}}$ in the above series, its multisections are easily seen to be

${\displaystyle \sum _{m=0}^{\infty }z^{qm+p}={\frac {z^{p}}{1-z^{q}}}\quad {\text{ for }}|z|<1.}$

Remembering that the sum of the multisections must equal the original series, we recover the familiar identity

${\displaystyle \sum _{p=0}^{q-1}z^{p}={\frac {1-z^{q}}{1-z}}.}$

The exponential function

${\displaystyle e^{z}=\sum _{n=0}^{\infty }{z^{n} \over n!}}$

by means of the above formula for analytic functions separates into

${\displaystyle \sum _{m=0}^{\infty }{z^{qm+p} \over (qm+p)!}={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}\omega ^{-kp}e^{\omega ^{k}z}.}$

The bisections are trivially the hyperbolic functions:

${\displaystyle \sum _{m=0}^{\infty }{z^{2m} \over (2m)!}={\frac {1}{2}}\left(e^{z}+e^{-z}\right)=\cosh {z}}$

${\displaystyle \sum _{m=0}^{\infty }{z^{2m+1} \over (2m+1)!}={\frac {1}{2}}\left(e^{z}-e^{-z}\right)=\sinh {z}.}$

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

${\displaystyle \sum _{m=0}^{\infty }{z^{qm+p} \over (qm+p)!}={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}e^{z\cos(2\pi k/q)}\cos {\left(z\sin {\left({\frac {2\pi k}{q}}\right)}-{\frac {2\pi kp}{q}}\right)}.}$

These can be seen as solutions to the linear differential equation ${\displaystyle f^{(q)}(z)=f(z)}$ with boundary conditions ${\displaystyle f^{(k)}(0)=\delta _{k,p}}$, using Kronecker delta notation. In particular, the trisections are

${\displaystyle \sum _{m=0}^{\infty }{z^{3m} \over (3m)!}={\frac {1}{3}}\left(e^{z}+2e^{-z/2}\cos {\frac {{\sqrt {3}}z}{2}}\right)}$

${\displaystyle \sum _{m=0}^{\infty }{z^{3m+1} \over (3m+1)!}={\frac {1}{3}}\left(e^{z}-e^{-z/2}\left(\cos {\frac {{\sqrt {3}}z}{2}}-{\sqrt {3}}\sin {\frac {{\sqrt {3}}z}{2}}\right)\right)}$

${\displaystyle \sum _{m=0}^{\infty }{z^{3m+2} \over (3m+2)!}={\frac {1}{3}}\left(e^{z}-e^{-z/2}\left(\cos {\frac {{\sqrt {3}}z}{2}}+{\sqrt {3}}\sin {\frac {{\sqrt {3}}z}{2}}\right)\right),}$

and the quadrisections are

${\displaystyle \sum _{m=0}^{\infty }{z^{4m} \over (4m)!}={\frac {1}{2}}\left(\cosh {z}+\cos {z}\right)}$

${\displaystyle \sum _{m=0}^{\infty }{z^{4m+1} \over (4m+1)!}={\frac {1}{2}}\left(\sinh {z}+\sin {z}\right)}$

${\displaystyle \sum _{m=0}^{\infty }{z^{4m+2} \over (4m+2)!}={\frac {1}{2}}\left(\cosh {z}-\cos {z}\right)}$

${\displaystyle \sum _{m=0}^{\infty }{z^{4m+3} \over (4m+3)!}={\frac {1}{2}}\left(\sinh {z}-\sin {z}\right).}$

Multisection of a binomial expansion

${\displaystyle (1+x)^{n}={n \choose 0}x^{0}+{n \choose 1}x+{n \choose 2}x^{2}+\cdots }$

at x = 1 gives the following identity for the sum of binomial coefficients with step q:

${\displaystyle {n \choose p}+{n \choose p+q}+{n \choose p+2q}+\cdots ={\frac {1}{q}}\cdot \sum _{k=0}^{q-1}\left(2\cos {\frac {\pi k}{q}}\right)^{n}\cdot \cos {\frac {\pi (n-2p)k}{q}}.}$