# Differential roots of unity

Lately I’ve been thinking about how to prove the existence and uniqueness of solutions to differential equations. This is a very hard problem in general; it’s solved in the case where the equation is first-order by the Picard-Lindelöf theorem, but that theorem involves more analysis than I’m familiar with at the moment. Instead, I’ve just been looking at a very particular case where proving existence and uniqueness turns out to be very easy.

Suppose $n$ is a positive integer and $f$ is an entire complex-valued function such that for every complex number $x$, $D^n f = f$, i.e. $f$ is a solution of the homogeneous linear differential equation with constant coefficients whose characteristic equation is $\lambda^n = 1$. You might therefore call $f$ an $n$th differential root of unity. This is my own term, I don’t know if any actual mathematicians use it.

Like an $n$th root of unity, $f$ has the property that for every pair of non-negative integers $a$ and $b$ such that $a \equiv b \mod n$, $D^a f = D^b f$, because by the definition of congruence, there is an integer $q$ such that $a = q n + b$, so

$\begin{array}{rcl} D^a f &=& D^{q n + b} f \\ &=& D^b (D^{q n} f) \\ &=& D^b f \end{array}$

if you accept that $D^{q n} f = f$, which is obvious and can be proven by induction. Because of this property, we can define $D^r f$ for every integer $r$ modulo $n$ as the common value of $D^r f$ for integers $r$ in the congruence class associated with $r$.

The key insight which allows all the $n$th differential roots of unity to be found is that the Taylor series of $f$ at 0 can be written as a sum of $n$ series, where each series in the sum consists of the terms whose indices in the original series are in a particular congruence class modulo $n$. So for every complex number $x$,

$\begin{array}{rcl} f(x) &=& \sum_{k = 0}^\infty \frac {D^k f(0) x^k} {k!} \\ &=& \sum_{r = 0}^{n - 1} \sum_{k = 0}^\infty \frac {D^{n k + r} f(0) x^{n k + r}} {(n k + r)!} \\ &=& \sum_{r = 0}^{n - 1} \sum_{k = 0}^\infty \frac {D^r f(0) x^{n k + r}} {(n k + r)!} \\ &=& \sum_{r = 0}^{n - 1} D^r f(0) \sum_{k = 0}^\infty \frac {x^{n k + r}} {(n k + r)!}. \end{array}$

This is assuming that for every integer $r$ such that $0 \le r < n$, the series $\sum_{k = 0}^\infty \frac {x^{n k + r}} {(n k + r)!}$ (which will be denoted by $Z_r(x)$) converges, but this can be easily proven by the ratio test:

$\begin{array}{rcl} \lim_{k \to \infty} \frac {x^{n (k + 1) + r} / ((n (k + 1) + r)!)} {x^{n k + r} / ((n k + r)!)} &=& \lim_{k \to \infty} \frac {x^{n (k + 1) + r} / x^{n k + r}} {((n (k + 1) + r)!) / ((n k + r)!)} \\ &=& \lim_{k \to \infty} \frac {x^n} {(n k + r + 1) (n k + r + 2) \dotsm (n k + r + n)} \\ &=& 0, \end{array}$

since $(n k + r + 1) (n k + r + 2) (n k + r + n)$ approaches $\infty$ as $k$ approaches $\infty$.

So, $f$ is a linear combination of $Z_0$, … and $Z_{n - 1}$. Therefore, the set of all such linear combinations definitely includes all the differential $n$th roots of unity; however, it may also include other functions. In order to ensure that this is not the case, we only need to show that $Z_r$ is a differential $n$th root of unity for every integer $r$ such that $0 \le r < n$, because any linear combination of differential $n$th roots of unity is a differential $n$th root of unity too (since the associated differential equation is linear). Well, for every complex number $x$, using the rule for differentiation of power series, we have

$\begin{array}{rcl} D^n Z_r(x) &=& \frac {\mathrm d^n} {\mathrm d x^n} \sum_{k = 0}^\infty \frac {x^{n k + r}} {(n k + r)!} \\ &=& \sum_{k = 1}^\infty \frac {(n k + r) (n k + r - 1) \dotsm (n k + r - (n - 1)) x^{n k + r - n}} {(n k + r)!} \\ &=& \sum_{k = 1}^\infty \frac {x^{n (k - 1) + r}} {(n (k - 1) + r)!} \\ &=& \sum_{k = 0}^\infty \frac {x^{n k + r}} {(n k + r)!} \\ &=& Z_r(x), \end{array}$

Therefore the general solution to the differential equation $D^n f = f$ is

$f = \sum_{r = 0}^{n - 1} A_r Z_r,$

where for every integer $r$ such that $0 \le r < n$, $A_r$ is an arbitrary complex number. Furthermore, $A_r$ will always have the property that $D^r f(0) = A_r$. We can use this property to discover some properties of the functions $Z_0$, $Z_1$, … and $Z_{n - 1}$. For convenience, let us define $A_r$ and $Z_r$ for every integer $r$ modulo $n$ as $A_r'$ and $Z_r'$, where $r'$ is the unique integer such that $0 \le r < n$ which is in the congruence class associated with $r$. This makes these properties simpler to express.

(1) For every integer $r$ modulo $n$, we have

$Z_r = \sum_{s \in \mathbb Z_n} A_s Z_s$

and for every integer $s$ modulo $n$, $D^s Z_r(0) = A_s$. But clearly $A_s = 1$ if $r = s$, otherwise $A_s = 0$; therefore, $D^s Z_r(0) = 1$ if $r = s$, otherwise $D^s Z_r(0) = 0$. In particular, $Z_0(0) = 1$ but $Z_r(0) = 0$ if $r$ is not 0.

(2) For every pair of integers $r$ and $s$ modulo $n$, $D^r Z_s$ is a differential $n$th root of unity, because $D^n D^r Z_s = D^{n + r} Z_s = D^r Z_s$ (since for every integer $r'$ in the congruence class associated with $r$, $n + r' \equiv r' \mod n$). Therefore we have

$D^r Z_s = \sum_{t \in \mathbb Z_n} A_t Z_t$

and for every integer $t$ modulo $n$, $D^t D^r Z_s(0) = A_t$. Now, $D^t D^r Z_s(0) = D^{t + r} Z_s(0)$, which we know from (1) is 1 if $t + r = s$ (i.e. $t = s - r$), otherwise 0. Therefore $D^r Z_s = Z_{s - r}$.

(3) For every integer $r$ modulo $n$ and every complex number $a$, $x \in \mathbb C \mapsto Z_r(a + x)$ is a differential $n$th root of unity, because it is the horizontal translation of $Z_r$ $a$ units to the left, so its $n$th derivative is simply the horizontal translation of $D^n Z_r = Z_r$ $a$ units to the left. Therefore, for every complex number $b$,

$Z_r(a + b) = \sum_{s \in \mathbb Z_n} A_s Z_s(b)$

and for every integer $s$ modulo $n$, $A_s$ is the $s$th derivative of $x \in \mathbb C \mapsto Z_r(a + x)$ at 0, which is simply $D_s Z_r(a)$, i.e. $Z_{r - s}(a)$ (by (2)). Therefore

$\begin{array}{rcl} Z_r(a + b) &=& \sum_{s \in \mathbb Z_n} Z_{r - s}(a) Z_s(b) \\ &=& \sum_{s, t \in \mathbb Z_n, s + t = r} Z_s(a) Z_t(b) \end{array}$

which can be generalised by induction to

$Z_r \left( \sum_{i = 1}^m a_i \right) = \sum_{s_1, \dotsc, s_m \in \mathbb Z_n, s_1 + \dotsb + s_m = r} Z_{s_1}(a_1) \dotsm Z_{s_m}(a_m),$

from which it follows in particular that for every positive integer $m$,

$Z_r(m a) = \sum_{s_1, \dotsc, s_m \in \mathbb Z_n, s_1 + \dotsb + s_m = r} Z_{s_1}(a) \dotsm Z_{s_m}(a).$

I’m going to stop here, but there are lots of questions to ask. First of all, when $n = 1$ this formula shows that $Z_0(m a) = (Z_0(a))^m$, and it can be generalised to hold when $m$ is rational, which allows $Z_0$ to be used to extend the exponential functions to $\mathbb C$. Can it also be generalised like this when $1 < n$? Also, we know that when $n = 2$, $Z_0(x) = \cosh x$ and $Z_1(x) = \sinh x$, and $\cosh x + \sinh x = e^x = Z_0(x)$ when $n = 1$; do any similar relations hold in general between the differential roots of unity?