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}

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