exp_q(x) keeps every qth term in the power series for exp(x)

The function exp_q(x) is defined by taking the power series for exp(x) and keeping only the terms whose index is a multiple of q. For example, exp₂(x) keeps only the even-numbered terms in the exponential power series and so equals cosh(x).

$\exp_2(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots = \cosh(x)$

In general,

$\exp_q(x) = \sum_{n=0}^\infty [q \mid n] \frac{x^n}{n!} = \sum_{n=0}^\infty \frac{x^{nq}}{(nq)!}$

The first sum uses Iverson’s bracket notation: a Boolean expression in brackets denotes the function that returns 1 when the expression is true and zero when it is false. Here the bracket equals 1 when q divides n and is zero otherwise.

Closed forms

Let ω = exp(2πi / q). Then

$\exp_q(x) = \frac{1}{q}\sum_{k=0}^{q-1} \exp(\omega^k x)$

This lets us find closed-form expressions for exp_q(x). For example, when q = 4, ω = i and

$\exp_4(x) = \frac{1}{2}\left( \cosh(x) + \cos(x) \right)$

Here’s a proof of the identity above:

$\begin{align*} \frac{1}{q} \sum_{k=0}^{q-1} \exp(\omega^k x) &= \frac{1}{q} \sum_{k=0}^{q-1} \sum_{n=0}^\infty \frac{\omega^{kn}x^n}{n!} \\ &= \sum_{n=0}^\infty \left( \frac{1}{q} \sum_{k=0}^{q-1} \omega^{kn}\right) \frac{x^n}{n!} \\ &= \sum_{n=0}^\infty [q \mid n] \frac{x^n}{n!} \\ &= \exp_q(x) \end{align*}$

In the proof we used the identity

$\frac{1}{q} \sum_{k=0}^{q-1} \omega^{kn} = [q \mid n]$

which is important in deriving the properties of the discrete Fourier transform.

Differential equations

The first time I saw the function exp_q(x) was in differential equations, though I didn’t know at the time the function had a name.

When a course in differential equations gets to power series solutions, a common example or homework problem is to solve

$y^{(k)}(x) = y(x)$

for k = 3 or 4, i.e. to find a function that equals its third or fourth derivative.

If the initial conditions are

and

$y^\prime(0) = y^{\prime\prime}(0) = \cdots = y^{(k-1)}(0) = 0$

the unique solution to

$y^{(k)}(x) = y(x)$

is y(x) = exp_k(x).

Mathematica and Mittag-Leffler

Mathematica does not have a built-in function implementing exp_q(x), but it does have an implementation of the Mittag-Leffler function, and so thanks to a relation between this function and exp_q(x) you can implement the latter as

expq[x_, q_] := MittagLefflerE[q, x^q]

Combinatorics

The first time I saw the notation exp_q(x) was in combinatorics. I had intended to include an application from that book here, but I make that the topic for the next post.

Closed forms

Differential equations

Mathematica and Mittag-Leffler

Combinatorics

Leave a Reply