The operate expq(x) is outlined by taking the ability collection for exp(x) and preserving solely the phrases whose index is a a number of of q. For instance, exp2(x) retains solely the even-numbered phrases within the exponential energy collection and so equals cosh(x).
Basically,
![exp_q(x) = sum_{n=0}^infty [q mid n] frac{x^n}{n!} = sum_{n=0}^infty frac{x^{nq}}{(nq)!}](https://www.johndcook.com/expq2.svg)
The primary sum makes use of Iverson’s bracket notation: a Boolean expression in brackets denotes the operate that returns 1 when the expression is true and nil when it’s false. Right here the bracket equals 1 when q divides n and is zero in any other case.
Closed varieties
Let ω = exp(2πi / q). Then
This lets us discover closed-form expressions for expq(x). For instance, when q = 4, ω = i and
Right here’s a proof of the id above:
![begin{align*} frac{1}{q} sum_{k=0}^{q-1} exp(omega^k x) &= frac{1}{q} sum_{k=0}^{q-1} sum_{n=0}^infty frac{omega^{kn}x^n}{n!} &= sum_{n=0}^infty left( frac{1}{q} sum_{k=0}^{q-1} omega^{kn}right) frac{x^n}{n!} &= sum_{n=0}^infty [q mid n] frac{x^n}{n!} &= exp_q(x) end{align*}](https://www.johndcook.com/expq5.svg)
Within the proof we used the id
![frac{1}{q} sum_{k=0}^{q-1} omega^{kn} = [q mid n]](https://www.johndcook.com/expq6.svg)
which is vital in deriving the properties of the discrete Fourier remodel.
Differential equations
The primary time I noticed the operate expq(x) was in differential equations, although I didn’t know on the time the operate had a reputation.
When a course in differential equations will get to energy collection options, a typical instance or homework downside is to unravel
for okay = 3 or 4, i.e. to discover a operate that equals its third or fourth by-product.
If the preliminary circumstances are
and
the distinctive answer to
is y(x) = expokay(x).
Mathematica and Mittag-Leffler
Mathematica doesn’t have a built-in operate implementing expq(x), however it does have an implementation of the Mittag-Leffler operate, and so due to a relation between this operate and expq(x) you’ll be able to implement the latter as
expq[x_, q_] := MittagLefflerE[q, x^q]
Combinatorics
The primary time I noticed the notation expq(x) was in combinatorics. I had meant to incorporate an software from that e book right here, however I make that the subject for the subsequent submit.
