Tag Archives: calculus

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 \omegath (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 nth term of the Nth 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 ith 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.