How to compute this "differential sequence" of functions?

Question

Given a real function $f$, define the following sequences of functions:

• $f_0(x) = f(x)$;
• $f_{i+1}'(x) = [f_i(x+1)-f_i(x-1)]/2$.

Some simple examples are:

• If $f(x)=x$, then $f_i(x)=x$ for all $i\geq 0$ (up to an additive constant; note that adding a constant to $f_i$ does not affect $f_{i+1}$).
• Similarly, if $f(x)=x^2$, then $f_i(x)=x^2$ for all $i\geq 0$.
• The "pattern" breaks at 3: if $f(x)=x^3$, then $f_i(x) = x^3 + i\cdot x$.

But for some functions, computing $f_i$ is much harder. I am particualrly interested in $f(x)=1/x$. I get $f_1(x) = (\ln(x+1)-\ln(x-1))/2$, but from this point on, the sequence becomes much harder to compute.

QUESTIONS:

• Is there a simple algorithm to compute $f_i(x)$ for a general $f$?
• If not, is there a simple expression for $f_i(x)$ for the case when $f(x)=1/x$, or at least an expression that approximates $f_i(x)$?
• Does this sequence I made up have a known name?

Answer 1

Some ideas. Let $X=C_0(\mathbb R)$ be the space of continuous functions $f$ from $\mathbb R$ to $\mathbb R$ such that $f(0)=0$. For each $f\in X$ and $x\in\mathbb R$ put $$T(f)(x)=\frac 12\int_0^x (f(t+1)-f(t-1))dt.$$ Then $T$ should be a linear operator from $X$ to $X$ such that $f_i=T^i(f)$ for each $f\in X$ and each nonnegative integer $i$ (here we assume that $T^0$ is the identity operator).

So it can be useful to calculate $T(f)$ for some functions $f\in X$. For instance, for each natural $n$, we have $$T(x^n)=\sum_{k=0}^{\lfloor (n-1)/2\rfloor} \frac 1{n-2k}\cdot {n \choose 2k+1} x^{n-2k}.$$

Maybe the operator $T$ will be continuous when we endow $X$ with the topology of uniform convergence on compact subsets of $\mathbb R$. Maybe then we shall be able to linearly extend the above formula to functions $f(x)\in X$ represented as a series of powers of $x$ with the infinite radius of convergence.

Answer 2

If $f$ is a continuous function on $\mathbb{R}$, then the convolution product $$Tf := f * \frac{1}{2}\mathbb{1}_{[-1,1]}$$ is well defined. Moreover, $$Tf(x) = \frac{1}{2}(F(x+1)-F(x-1))/2,$$ where $F$ is an antiderivative of $f$.

For every positive integer $n$, $T^nf := f * u^{*n}$, where $u^*n$ is the $n$-fold convolution product of $u:=\frac{1}{2}\mathbb{1}_{[-1,1]}$, equivalently the density of the sum of $n$ independent random variables which are unifom on $[-1,1]$.

The convolution product $u^{*n}$ can be computed explicitly since for every $n \ge 2$ and $s \in \mathbb{R}$ $$\mathbb{1}_{[0,1]}^{*n}(s) = \sum_{k=0}^n {n \choose k} (-1)^k \frac{(s-k)_+^{n-1}}{(n-1)!}.$$