Evaluating the Grünwald–Letnikov derivative of the power function

Question

I recently came across the Grünwald–Letnikov derivative and I wanted to use it to evaluate fractional derivatives of various functions. Specifically I used this expression of the derivative $$D^qf(x)=\lim_{h \to 0} \sum_{0 \leq m < \infty}\frac{(-1)^m {q \choose m}f(x+(q-m)h)}{h^q}$$ I know that this derivative has the property that $D^{\alpha} (D^{\beta}f(x)) = D^{\alpha+\beta}f(x))$. And so when evaluating the $q^{\text{th}}$ fractional derivative of $e^{a x}$, I got the expected result of $a^qe^{a x}$. However when I try to evaluate $D^q(x^p)$ where $0<q<1$ and $p \in \mathbb{N}$, I seem to get that $D^q(x^p)=0$. Here is my process

$$D^q(x^p)=\lim_{h \to 0} \sum_{0 \leq m < \infty}\frac{(-1)^m {q \choose m}(x+(q-m)h)^p}{h^q}$$ $$ = \lim_{h \to 0} \sum_{0 \leq m < \infty}\sum_{0\leq n \leq p}\frac{(-1)^m {q \choose m}{p \choose n}x^{p-n}(q-m)^{n}h^{n}}{h^q}$$ $$ = \lim_{h \to 0} \sum_{0 \leq m < \infty}\frac{(-1)^m {q \choose m}x^{p}}{h^q} + \lim_{h \to 0} \sum_{0 \leq m < \infty} \sum_{1\leq n \leq p}\frac{(-1)^m {q \choose m}{p \choose n}x^{p-n}(q-m)^{n}h^{n}}{h^q}$$ Now we can use the binomial theorem to say that $$\sum_{0 \leq m < \infty}\frac{(-1)^m {q \choose m}x^{p}}{h^q} =\frac{x^p}{h^q}\sum_{0 \leq m < \infty} {q \choose m}(-1)^m1^{q-m} = \frac{x^p}{h^q} (1-1)^q = 0$$ And for the term $$\lim_{h \to 0} \sum_{0 \leq m < \infty} \sum_{1\leq n \leq p}\frac{(-1)^m {q \choose m}{p \choose n}x^{p-n}(q-m)^{n}h^{n}}{h^q}$$ Since $n \geq 1$ and $q < 1$, then for all $n$ we get that $\lim_{h \to 0} \frac{h^n}{h^q}=0$ and so the entire above expression becomes $0$.

Thus $D^q(x^p)=0$. Now, this clearly does not satisfy the property of compsing fractional derivatives because we get that $D^{\frac{1}{2}}(D^{\frac{1}{2}} (x^p)) = 0$ instead of the expected $D^{\frac{1}{2}}(D^{\frac{1}{2}} (x^p)) = D(x^p) = px^{p-1}$.

I think I'm getting this result because this expression actually has two limits, $h$ and the upper bound of the sum with $m$ and so really our expression should be

$$ D^q(x^p) = \lim_{h \to 0} \lim_{k \to \infty} \sum_{0 \leq m < k}\sum_{0\leq n \leq p}\frac{(-1)^m {q \choose m}{p \choose n}x^{p-n}(q-m)^{n}h^{n}}{h^q}$$ $$ = \lim_{h \to 0} \lim_{k \to \infty} (\sum_{0 \leq m < k}(-1)^m {q \choose m})\frac{x^{p}}{h^q} + \lim_{h \to 0} \lim_{k \to \infty} \sum_{1\leq n \leq p}{p \choose n}x^{p-n}\frac{h^{n}}{h^q} \sum_{0 \leq m < k} (-1)^m {q \choose m}(q-m)^{n}$$

Now we can see that $$\lim_{k \to \infty} \sum_{0 \leq m < k}(-1)^m {q \choose m} = 0$$ $$\lim_{h \to 0} \frac{1}{h^q} = \infty$$ $$\lim_{h \to 0} \frac{h^n}{h^q} = 0$$

And I am fairly confident that $$\lim_{k \to \infty} \sum_{0 \leq m < k} (-1)^m {q \choose m}(q-m)^{n} = \pm \infty$$ And so if we try to evaluate one limit before another, we'll get $ D^q(x^p) = 0+\infty$ and trying to do both gets us $D^q(x^p) = 0 \cdot \infty + \infty \cdot 0$. And so clearly we can't evaluate this limit from an arbitrary direction. That begs the question of how can we exactly evaluate this limit to get an answer?

When I try to evaluate it in python, I can never seem to get the derivative to converge to a single function when I try to decrease $h$ and increase $k$. Decreasing $h$ seems to inflate the values of the function and increasing $k$ seems to deflate them. This is expected, but no matter how big $k$ is or how small $h$ is, changing the other will radically alter the function.

Also it seems like even if we can find a way to evaluate these limits, the fractional derivative of $x^p$ is going to be a $p-1^{\text{th}}$ degree polynomial, which is counter to my expectation that the fractional derivative of a power function should give us a fractional power of $x$ times some constant. Am I correct in my guess that the result will be a polynomial or is there something else going on?

(Side Note: One way I could potentially see of evaluating this is to let $p$ just be a real number greater than zero, so that we get

$$ D^q(x^p) = \lim_{h \to 0} \lim_{k \to \infty} \sum_{0 \leq m < k}\sum_{0\leq n \leq \infty}\frac{(-1)^m {q \choose m}{p \choose n}x^{n}(q-m)^{p-n}h^{p-n}}{h^q}$$ $$ = \sum_{0\leq n \leq \infty} (\lim_{h \to 0} \lim_{k \to \infty} \sum_{0 \leq m < k}\frac{(-1)^m {q \choose m}{p \choose n}(q-m)^{p-n}h^{p-n}}{h^q})x^{n}$$

And so if we can find a closed expression for $$\lim_{h \to 0} \lim_{k \to \infty} \sum_{0 \leq m < k}\frac{(-1)^m {q \choose m}{p \choose n}(q-m)^{p-n}h^{p-n}}{h^q}$$ We could potentially get a Taylor Series for $D^q(x^p)$.

Update: After further attempts at a numerical evaluation this derivative, I noticed something strange. I tried evaluating the fractional derivatives for various power and exponential functions and when I evaluated the limit for some $k$ and $h$, if I scaled $k$ by some factor $a$ and divided $h$ by $a$ and evaluated the limit again, I got the same exact result. And so really whatever I evaluate the limit to be in this case is a matter of what "initial" values I selected for $k$ and $h$. And so I'm led to believe that depending on what direction you go in for evaluating $k$ and $h$, you'll get completely different convergences unless this limit converges very slowly and I haven't noticed.

Answer 1

As probably seen by yourself already, your mistake stems from $$\lim_{m\to\infty}m^{q+1}(-1)^m\binom{q}{m}=\lim_{m\to\infty}\frac{m^{q+1}\Gamma(m-q)}{\Gamma(-q)\Gamma(m+1)}=\frac{1}{\Gamma(-q)}\neq0,$$ causing the defining sum for GLD to diverge for (integer values of) $p>0$. The actual reason is the behavior of $x\mapsto x^p$ for "large negative" $x$, as opposed to that of $x\mapsto e^{ax}$ with $a>0$.

But what if we consider the GLD (at $x>0$) of $$f_p(x)=\begin{cases}x^p,&x>0\\0,&x\leqslant 0\end{cases}$$ (and, by the way, let's allow $0<q<p$ be arbitrary real numbers)?

Here the idea (that comes from Laplace transform inversion) is the equality $$f_p(x)=\frac{\Gamma(p+1)}{2\pi i}\int_{c-i\infty}^{c+i\infty}s^{-p-1}e^{xs}\,ds\qquad(c>0)$$ (this can also be proven directly; for $x>0$ use this, and for $x\leqslant 0$ consider $\int_{c-iR}^{c+iR}$ and close the path of integration by a semicircular contour on the right; this introduces a term which vanishes after $R\to\infty$, and the resulting integral equals $0$ by Cauchy), which produces $$h^{-q}\sum_{m=0}^\infty(-1)^m\binom{q}{m}f_p\big(x+(q-m)h\big)=\frac{\Gamma(p+1)}{2\pi i}\int_{c-i\infty}^{c+i\infty}s^{-p-1}e^{(x+qh)s}\left(\frac{1-e^{-hs}}{h}\right)^q\,ds.$$ And the limit $h\to 0^+$ is easily justified (by DCT) to be taken under the integral sign, giving $$\lim_{h\to 0^+}(\ldots)=\frac{\Gamma(p+1)}{2\pi i}\int_{c-i\infty}^{c+i\infty}s^{q-p-1}e^{xs}\,ds=\frac{\Gamma(p+1)}{\Gamma(p-q+1)}x^{p-q},$$ expectedly generalising the case of integer values of $q$.