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

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

The Calculus Tree

We first start with a tree structure on $P,$ the set of real coefficient polynomials with domain and codomain $\mathbb{R}$, and call it the polynomial tree.  We will say $f\preceq g$ iff $g^{(n)}=\frac{d^ng}{dx^n}=f$ for some $n\in\mathbb{N}.$

Proposition 1.  This gives us a partial order on $P$.

Proof.  $f\preceq f$ since $f^{(0)}=f.$  If $f\preceq g$ and $g\preceq h,$ then $g^{(n)}=f$ and $h^{(m)}=g,$ thus $h^{(n+m)}=f;$ so $f\preceq h.$  And lastly if $f\preceq g$ and $g\preceq f,$ then $g^{(n)}=f$ and $f^{(m)}=g.$  Thus $f^{(m+n)}=f$ which implies $m+n=0$ and thus $m=n=0,$ so $f=g.$

The fact that differentiation yields a unique function gives us the tree structure:  the chain of derivatives (predecessors) is well ordered with minimal element $0.$  By the Weierstrass approximation theorem we can attempt to extend this to $C_0^\infty(\mathbb{R})$ the closure of the class of smooth functions with compact support.  We extend our indexing set to the ordinal $\omega+2$ and define the $\omega$th (and $(\omega+1)$th) derivative linearly by the rule

$\displaystyle \frac{d^\omega }{dx^\omega}\left(\frac{x^\omega}{\omega!}\right)=1$

$\displaystyle \frac{d^{\omega+1}f}{dx^{\omega+1}}=\frac{d}{dx}\left(\frac{d^{\omega}f}{dx^\omega}\right)=0.$

Recall integration is a set valued operation–sending a function to its set of antiderivatives which all differ by some constant.

Example 2.  Let $f(x)=e^x.$  We have

$\displaystyle e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$

Hence $f^{(n)}=f$ for finite $n,$ $f^{(\omega)}=1$ since the “last term” is $x^\omega/\omega!,$ and $f^{(\omega+1)}=0.$  If we begin to integrate, we will add polynomials trailing after $e^x.$  Continuing this denumerably we then obtain

$\displaystyle e^x\stackrel{\int\cdot\,dx^\omega}{\longmapsto}e^x+\sum_{n=1}^\infty c_m\frac{x^n}{n!}$

where $m=co(n)$ is the conumber of $n$ in the sense that it satisfies $N=n+m$ in the $n$th term of the $N$th partial sum.  The denumerable integration converges if the sequence of constants chosen satisfies

$\displaystyle\lim_{n\to\infty}\left|\frac{c_{m+1}}{(n+1)c_m}\right|=\lim_{m\to 1}\left|\frac{c_{m+1}}{(n+1)c_m}\right|<1.$

Suppose the sequence $c=(c_1,c_2,...)$ satisfies this condition and let us assume the sum converges to some $f,$ then we can write

$\displaystyle\int e^x\,dx^\omega=_ce^x+f=\sum_{n=0}^\infty\frac{x^n}{n!}+\sum_{n=1}^\infty c_m\frac{x^n}{n!}=1+\sum_{n=1}^\infty\left(\frac{c_m+1}{n!}\right)x^n.$

In particular we thus have $(e^x+f)^{(\omega)}=c_1+1$ since

$1=\lim_{n\to\infty} co(n):=co(\omega).$

We thus have $0\preceq e^x,$ $0\preceq f,$ $0\preceq e^x+f,$ and $0\preceq e^xf.$  We also define the degree of the functions which are not polynomials as $\omega.$  Let $p$ be a polynomial with degree $n$ and $f$ have infinite degree.  Then we clearly have that $fp$ and $f+p$ have degree $\omega.$

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 $\omega\neq\infty$).

Example 4.  The functions $e^x,\pm\sin(x),$ and $\pm\cos(x)$ are cyclic with orders $1,4,$ and $4$ respectively.  The only function with finite degree which is cyclic is $0$ with order $1.$

In fact being cyclic occurs iff

$\begin{array}{lcl}f^{(n)}-f=0&\Leftrightarrow&(D^n-1)f=0\\&\Leftrightarrow&(D-1)(D^{n-1}+D^{n-2}+\cdots+1)f=0\\&\Leftarrow&(D-\zeta_n^k)f=0\end{array}$

for at least some $k$ where $\zeta_n=e^{2\pi i/n}$ and $1\leq k\leq n.$  Thus we have

$\displaystyle f(x)=\sum_ic_ie^{\zeta_n^kx}.$

With this presentation it is clear that the order of the $i$th term in the above term is $n$ if $\gcd(k,n)\neq 1.$  Otherwise the order is $\gcd(k,n).$

Let us now define the equivalence relation where $f\sim g$ iff $f$ and $g$ are cyclic with same order and where $f^{(k)}=g$ such that $k\leq ord(f)=ord(g).$  Reflexivity is clear, and symmetry/transitivity involve simple $\mbox{mod~}ord(f)$ arithmetic.

Example 5.  Hence some equivalence classes are $\{\sin(x),\cos(x),-\sin(x),-\cos(x)\},$ $\{e^x\},$ and $\{p\}$ for any polynomial $p.$

Proposition 6$\preceq$ gives a tree structure on $C_0^\infty(\mathbb{R})/\sim,$ which we call the calculus tree.

Proof.  This is trivial considering the relation we modded out by was precisely loops in the poset–together with the fact that the poset has a unique minimal element.

Also see update.