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 nth differential root of unity. This is my own term, I don’t know if any actual mathematicians use it.

Like an nth 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 nth 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 nth 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 nth root of unity for every integer r such that 0 \le r < n, because any linear combination of differential nth roots of unity is a differential nth 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 nth 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 nth root of unity, because it is the horizontal translation of Z_r a units to the left, so its nth 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 sth 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?

Advertisements

One response to “Differential roots of unity

  1. Pingback: Notes on gradients | The House Carpenter

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s