# Update on the Calculus Tree

Recall in the post The Calculus Tree we had defined

Definition 3.  A function $f$ is cyclic if $f^{(n)}=f$ for some nonzero finite number $n.$  The number $n$ is called the differential order of $f,$ and is denoted $ord(f).$  If a function is not cyclic we say $ord(f)=\infty$ (note in this definition $\omega\neq\infty$).

I said it followed that if $f$ was cyclic, then $(D-\zeta_n^k)f=0$ for some $\zeta_n=e^{2\pi i/n}.$  This was not justified (correspondingly I have fixed the arrows).  Rather, the condition $(D-\zeta_n^k)f=0$ is certainly sufficient for $f$ to be cyclic.

So real cyclic functions must have the form no more complicated than

$\displaystyle f(x)=\sum_i c_ie^{a_i x}\left(\cos(b_ix)+\sin(b_ix)\right)$

where $a_i^2+b_i^2=1$ and $(a_i+b_i)^k=1$ for all $i$ and some $k\geq 1.$

For example $f(x)=\cos(x)$ is cyclic (of differential order 4) with $c_1=c_2=1/2,$ $a_1=a_2=0,$ $b_1=1,$ and $b_2=-1$ yielding

$\displaystyle\begin{array}{lcl}f(x)&=&\frac{1}{2}\left(\cos(x)+\sin(x)+\cos(-x)+\sin(-x)\right)\\&=&\frac{1}{2}\left(\cos(x)+\sin(x)+\cos(x)-\sin(x)\right)\\&=&\cos(x).\end{array}$