Limit Definition for Half-Derivative

Question

The derivative of a function $f$ is defined as $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ Let $$d_1(f,h)=\frac{f(x+h)-f(x)}{h}$$ and, in fact, let all $d_n$ be defined by $$\lim_{h\to 0} d_n(f,h)=f^{(n)}(x)$$ In order to obtain $d_2$, we can plug $d_1$ into itself to get $$\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}$$ $$\frac{f(x+2h)-f(x+h)-f(x+h)-f(x)}{h^2}$$ $$\frac{f(x+2h)-2f(x+h)-f(x)}{h^2}$$ and, in general, we can use induction to prove that, for natural $n$, $$d_n(f,h)=\frac{1}{h^n}\sum_{k=0}^n (-1)^k \binom{n}{k}f(x+(n-k)h)$$ However, I am interested in finding $d_\frac{1}{2}$. It should satisfy $$d_\frac{1}{2}(d_\frac{1}{2}(f,h),h)=d_1(f,h)=\frac{f(x+h)-f(x)}{h}$$ So that when it is composed with itself as shown, $d_1$ is the result.

It can possibly be obtained by figuring out how to extend the expression $$\frac{1}{h^n}\sum_{k=0}^n (-1)^k \binom{n}{k}f(x+(n-k)h)$$ to non-integer $n$. Does anybody have any ideas about how to do this?

Answer 1

One can extend this in a similar manner that the binomial expansion theorem is extended. Note that $\binom nk=0$ if $k>n$ and $n$ is a natural number. Thus, we have the Grunwald-Letnikov derivative,

$$f^{(\alpha)}(x)=\lim_{h\to0}\frac1{h^\alpha}\sum_{k=0}^\infty(-1)^k\binom\alpha kf(x+(\alpha-k)h)\tag{$\alpha\ge0$}$$

An interesting point to note is that for $\alpha=-1$, we get

$$f^{(-1)}(x)=\lim_{h\to0}h\sum_{k=0}^\infty f(x-(1+k)h)$$

Or as you may better recognize it,

$$f^{(-1)}(x)=\lim_{n\to\infty}\frac1n\sum_{k=1}^\infty f\left(x-\frac kn\right)$$

Which is extraodinarily similar to

$$\int_{x-1}^xf(t)~\mathrm dt=\lim_{n\to\infty}\frac1n\sum_{k=1}^nf\left(x-\frac kn\right)$$

Which I suppose would mean that a better all-enveloping fractional derivative could be given by

$$f^{(\alpha)}(x)=\lim_{h\to0}\frac1{h^\alpha}\sum_{k=0}^{\lfloor1/|h|\rfloor}(-1)^k\binom\alpha kf(x+(\alpha-k)h)\tag{$\alpha\in\mathbb C$}$$

However, for non integer $\alpha$, this isn't well defined as $h\to0^-$, and so we assume we can replace this with the limit from the positive side:

$$f^{(\alpha)}(x)=\lim_{n\to\infty}n^\alpha\sum_{k=0}^{\lfloor n\rfloor}(-1)^k\binom\alpha kf\left(x+\frac{\alpha-k}n\right)\tag{$\alpha\in\mathbb C$}$$

Which now gives us sort of anti-derivatives as well. Mainly,

$$\int_0^nf(x)~\mathrm dx=\sum_{k=1}^nf^{(-1)}(k)$$

---

Of course, one might note nothing unique about this particular extension. One could include more parameters:

$$_\beta^+\mathbb D_x^\alpha f(x)=\lim_{n\to\infty}n^\alpha\sum_{k=0}^{\beta n}(-1)^k\binom\alpha kf\left(x+\frac{\alpha-k}n\right)$$

And if we used $\frac d{dx}f(x)=\lim_{h\to0^+}\frac{f(x)-f(x-h)}h$,

$$_\beta^-\mathbb D_x^\alpha f(x)=\lim_{n\to\infty}n^\alpha\sum_{k=0}^{\beta n}(-1)^k\binom\alpha kf\left(x+\frac{k-\alpha}n\right)$$