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I'm new to this "fractional derivative" concept and try, using wikipedia, to solve a problem with the half-derivative of the zeta at zero, in this instance with the help of the zeta's Laurent-expansion.

Part of this fiddling is now to find the half-derivative $$ {d^{1/2}\over dx^{1/2}}{1 \over 1-x}$$

First I would like to understand, whether there is a short/closed form for this ot whether I have to express the fraction as a power series first and then to differentiate termwise.

Next I would like to know the value at $x=0$.

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  • $\begingroup$ A related problem. You need to use the Riemann-Liouville fractional derivative definition. $\endgroup$ May 15, 2013 at 1:02

5 Answers 5

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Use what you know about whole number derivatives. Inductively, you can prove $$\frac{d^n}{dx^n}\frac1{1-x}=\frac{n!}{(1-x)^{n+1}}$$ Now express $n!$ using the $\Gamma$ function ($\Gamma(n+1)$), and you can extend the definition to non-integral $n$: $$\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}=\frac{\Gamma(3/2)}{(1-x)^{3/2}}$$ At $x=0$, this is just $\Gamma(3/2)$.


To confirm that this method works, observe that you can also inductively prove $$\begin{align}\frac{d^n}{dx^n}\frac1{(1-x)^{3/2}}&=\frac{\frac{(2n+1)!}{4^n\cdot n!}}{(1-x)^{n+3/2}}\\&=\frac{\frac{\Gamma(2n+2)}{4^n\Gamma(n+1)}}{(1-x)^{n+3/2}}\end{align}$$ and extend to nonintegral $n$, so that $$\begin{align}\frac{d^{1/2}}{dx^{1/2}}\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}&=\frac{d^{1/2}}{dx^{1/2}}\frac{\Gamma(3/2)}{(1-x)^{3/2}}\\&=\frac{\Gamma(3/2)\frac{\Gamma(3)}{2\Gamma(3/2)}}{(1-x)^{2}}\\&=\frac{\frac{2!}{2}}{(1-x)^{2}}\\&=\frac{1}{(1-x)^2}\\&=\frac{d}{dx}\frac{1}{1-x}\end{align}$$ and all is as it should be.

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  • $\begingroup$ Hmm. Very nice. The first move (the generalizing to n'th derivative and then assuming n fractional) was really simple... phew... In another comment (to @Angela) I'll express also a problem which blocked me from proceeding $\endgroup$ May 14, 2013 at 21:43
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    $\begingroup$ +1; your last bit is one of the good ways to check a semiderivative; semidifferentiating a function twice should certainly yield the function's usual first derivative. $\endgroup$ May 15, 2013 at 2:39
  • $\begingroup$ (acc) This made it very simple and answered both partial questions quickly. Thanks! $\endgroup$ May 15, 2013 at 7:56
  • $\begingroup$ This is nice and all, but it doesn't work too well in the negative direction. $\endgroup$ Oct 23, 2016 at 12:43
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$$\frac{d^k}{dx^k} \left(\frac{1}{1-x} \right)=\frac{d^k}{dx^k} (1+x+x^2+...)=\Gamma(k+1) \sum_{n=k}^{\infty} {n\choose k}x^{n-k}\\ \sum_{n=k}^{\infty} {n\choose k}x^{n-k}=\frac{1}{(1-x)^{k+1}}\\ \frac{d^k}{dx^k} \frac{1}{1-x}=\frac{\Gamma(k+1)}{(1-x)^{k+1}}$$

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  • $\begingroup$ This is very nice, too. I pondered first to use this myself but was then blocked by some remarks (wikipedia?) on the funny effect, that fractional derivatives of a constant are not zero but introduce the parameter x in the formula. The power series for my function now has such a constant term. But I cannot locate in your formulae where it occurs and where it vanishes/cancels when it is half differentiated and then collected by the binomial sum. Can you explain this a bit more? $\endgroup$ May 14, 2013 at 21:47
  • $\begingroup$ That remark about the constant term I got actually not from wikipedia but from another resource: mathpages.com/home/kmath616/kmath616.htm which is a nice article, btw. but made me inconfident that I could solve the half derivative for the expanded complete powerseries on my own. $\endgroup$ May 14, 2013 at 21:58
  • $\begingroup$ I made a minor edit (format) to your answer. Hope you do not mind. $\endgroup$
    – user17762
    May 15, 2013 at 4:33
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See Reference for a complete solution of the problem of differentiation and integration of real order of rational polynomials.

Formal approaches to fractional derivative: There are several definitions for Fractional derivative. The most widely known one is the Riemann-Liouville fractional derivative

$$ f^{(q)}(x) = \frac{1}{\Gamma(k-q)} \frac{d^k}{dx^k} \int_{a}^{x}\, (x-t)^{k-q-1}\,f(t)\,dt\>, \quad (k-1 < q < k )\,, $$

and according to this definition you will have the following answer

$$ f^{\left(\frac{1}{2}\right)}(x)={\frac {{_2F_1\left(1,1;\,\frac{1}{2};\,x\right)}}{\sqrt {x}\sqrt {\pi }}}, $$

where $ _2F_1 $ is the hypergeometric function. The limit of the above function as $x\to 0^+$ goes to infinity. However, there is another definition known as the Caputo definition and is defined as

$$ f^{(q)}(x) = \frac{1}{\Gamma(k-q)} \int_{a}^{x}\, (x-t)^{k-q-1}\,\frac{d^k}{dt^k}f(t)\,dt\>, \quad (k-1 < q < k )\,, $$

and this definition will give you a different answer, namely

$$ f^{\left(\frac{1}{2}\right)}(x) = {\frac {1}{\sqrt {\pi } \left( x-1 \right) ^{3/2}}} \left( {\it \rm arctanh} \left( {\frac {\sqrt {x}}{ \sqrt {x-1}}} \right) -\sqrt {x}\sqrt {x-1}\right), $$

and the limit in this case goes to $0$.

Note: The simplest approach to your problem is summarized in the steps

1) Compute the Taylor series of the function,

2) Use the Formula

$$ \frac{d^q}{dx^q} x^m = \frac{\Gamma(m+1)}{\Gamma(m-q+1 )} x^{m-q}\,, $$

which corresponds to the Riemann-Liouville definition of the function $x^m$, or the formula (2.29), page 15 which corresponds to Caputo definition.

See here to see the power series technique.

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  • $\begingroup$ Last open question: What shall I do with the fractional powers of x, when I want look at the fractional derivatives at x=zero? (That's my current goal) I've seen the nice closed form of ${1\over 1-x}$ in the answer above which can be evaluated at $x=0$. But expanded into the power series and looking at the monomials I've always that fractional powers of zero: How does this cancel? $\endgroup$ May 15, 2013 at 6:38
  • $\begingroup$ @GottfriedHelms: I did not understand your question well? Can you elaborate more? Have you tried the power series method using the Caputo definition? $\endgroup$ May 15, 2013 at 6:43
  • $\begingroup$ Mhenni, Not yet. It's much that I have to chew mentally now, that handling with integrals is really unfamiliar to me (I've left out that exercises in my old courses in the 70ies) and I'm doing tiny steps into it since some weeks. I should also mention: phewww.w.. that all came up because I wanted simply check a certain real number which I could not find in wolframalpha, oeis and where I had a vague idea that it possibly might be related to the half derivative of the zeta at zero... Now I'm getting involved into a full featured research question on its own... (but well, it's interesting) $\endgroup$ May 15, 2013 at 6:54
  • $\begingroup$ ahhh.. and I think now: roots of zero with positive rational order should be evaluated to zero anyway, because from $x=0^{1/2}$ follows $x\cdot x=0 \to x=0$. I don't know where I had my brain... $\endgroup$ May 15, 2013 at 6:59
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    $\begingroup$ I've just downloaded your thesis, thank you very much. Moreover, the exercises in it seem to give useful redundancies to avoid simple misunderstandings. I'll read deeper into it from today now - thanks again! $\endgroup$ May 15, 2013 at 7:36
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While @alex.jordan's answer was nice and all, it fails for $-n\in\mathbb N$, where the logarithm should come into play. To account for such behavior, one could use the following formula:

$$\frac{d^n}{dx^n}\ln(x)=\frac{\ln(x)-\gamma-\psi^{(0)}(1-n)}{x^n\Gamma(1-n)}$$

which may also be proven by induction or derived straight from formulas. Here we have the Euler-Mascheroni constant and the digamma function.

Changing the argument to $1-x$, we have

$$\frac{d^n}{dx^n}\ln(1-x)=\frac{\ln(1-x)-\gamma-\psi^{(0)}(1-n)}{(x-1)^n\Gamma(1-n)}$$

where we applied chain rule and put it in the denominator.

$$\frac{d^{n+1}}{dx^{n+1}}\ln(1-x)=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-n)}{(x-1)^{n+1}\Gamma(-n)}$$

$$\frac{d^n}{dx^n}\frac{-1}{1-x}=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-n)}{(x-1)^{n+1}\Gamma(-n)}$$

$$\frac{d^n}{dx^n}\frac1{1-x}=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-n)}{-(x-1)^{n+1}\Gamma(-n)}$$

I will remark the only problem with this is with $n\in\mathbb N$, where will get the indeterminate form $\frac\infty\infty$, which can be treated as a limit to get the desired values. Also note that by applying L'Hospital's rule with respect to $n$ removes the logarithm for $n\in\mathbb N$, which explains why we usually do not see it occur.

Lastly, the half derivative is given as

$$\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}=\frac{\ln(1-x)-\gamma-\psi^{(0)}(-1/2)}{-(x-1)^{3/2}\Gamma(-1/2)}$$

$$=\frac{\ln(1-x)+\ln(4)}{2\sqrt\pi(x-1)^{3/2}}$$

And at $x=0$,

$$=\frac{-i\ln(2)}{\sqrt\pi}$$

where $i=\sqrt{-1}$

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Introduction

As is usually the case with Fractional Derivatives, there is also not "the Fractional Derivative". They differ depending on the method of derivation used and the type of Differential Operator. I'll use $\operatorname{D_{x}^{\alpha}}$ as a differential operator given by $\operatorname{D_{x}^{\alpha}} = \frac{\operatorname{d}^{\alpha}}{\operatorname{d}x^{\alpha}}$. Note: I'll also use $\Gamma\left( \cdot \right)$ as the Complete Gamma Function.

Calculation

"Euler Method"

We can do this even mor generaly for $\operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{c} \right]$.

Leonhard Euler came up with a nice method for some Fractional Derivatives (see Euler's Approach). He has sorted the derivatives of the function by order and is trying to find a pattern in it. He then tried to transfer this pattern from integer orders to fractional ones, which I always affectionately call the "Euler method":

However, the whole thing here also depends on the sign of $c$. In the following i'll assume that the $c$ that i'll is given by $c \in \mathbb{R}^{+}$.

$$ \begin{align*} \operatorname{D_{x}^{0}}\left[ \left( a \cdot x + b \right)^{c} \right] &= \left( a \cdot x + b \right)^{c}\\ \operatorname{D_{x}^{1}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a \cdot c \cdot \left( a \cdot x + b \right)^{c - 1}\\ \operatorname{D_{x}^{2}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{2} \cdot c \cdot \left( c - 1 \right) \cdot \left( a \cdot x + b \right)^{c - 2}\\ \operatorname{D_{x}^{3}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{3} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdot \left( a \cdot x + b \right)^{c - 3}\\ \operatorname{D_{x}^{4}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{4} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdot \left( c - 3 \right) \cdot \left( a \cdot x + b \right)^{c - 4}\\ \operatorname{D_{x}^{5}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{5} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdot \left( c - 3 \right) \cdot \left( c - 4 \right) \cdot \left( a \cdot x + b \right)^{c - 5}\\ &\cdots\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{n} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdots \left( c - n + 1 \right) \cdot \left( a \cdot x + b \right)^{c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{n} \cdot \frac{c!}{\left( c - n \right)!} \cdot \left( a \cdot x + b \right)^{c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{\alpha} \cdot \frac{\Gamma\left( c + 1 \right)}{\Gamma\left( c - \alpha + 1 \right)} \cdot \left( a \cdot x + b \right)^{c - \alpha},\, &\text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $$

$$\fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{\alpha} \cdot \frac{\Gamma\left( c + 1 \right)}{\Gamma\left( c - \alpha + 1 \right)} \cdot \left( a \cdot x + b \right)^{c - \alpha},\, \text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $} \tag{1}$$ wich can be simplefied via using the Falling Factorial $\left( x \right)_{y} = x^{\underline{y}}$.

or

$$ \begin{align*} \operatorname{D_{x}^{0}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( a \cdot x + b \right)^{-c}\\ \operatorname{D_{x}^{1}}\left[ \left( a \cdot x + b \right)^{c} \right] &= -a \cdot c \cdot \left( a \cdot x + b \right)^{-c - 1}\\ \operatorname{D_{x}^{2}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{2} \cdot c \cdot \left( c + 1 \right) \cdot \left( a \cdot x + b \right)^{-c - 2}\\ \operatorname{D_{x}^{3}}\left[ \left( a \cdot x + b \right)^{c} \right] &= -a^{3} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdot \left( a \cdot x + b \right)^{-c - 3}\\ \operatorname{D_{x}^{4}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{4} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdot \left( c - 3 \right) \cdot \left( a \cdot x + b \right)^{-c - 4}\\ \operatorname{D_{x}^{5}}\left[ \left( a \cdot x + b \right)^{c} \right] &= -a^{5} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdot \left( c + 3 \right) \cdot \left( c + 4 \right) \cdot \left( a \cdot x + b \right)^{-c - 5}\\ &\cdots\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{c} \right] &= \left( -a \right)^{n} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdots \left( c + n - 1 \right) \cdot \left( a \cdot x + b \right)^{c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( -a \right)^{n} \cdot \frac{\left( c + n -1 \right)!}{\left( c - 1 \right)!} \cdot \left( a \cdot x + b \right)^{-c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( -a \right)^{\alpha} \cdot \frac{\Gamma\left( c + \alpha \right)}{\Gamma\left( c \right)} \cdot \left( a \cdot x + b \right)^{-c - \alpha},\, &\text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $$

$$\fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( -a \right)^{\alpha} \cdot \frac{\Gamma\left( c + \alpha \right)}{\Gamma\left( c \right)} \cdot \left( a \cdot x + b \right)^{-c - \alpha},\, \text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $} \tag{2}$$

You also can write this in terms of the Rising Factorial $x^{\left( y \right)} = x^{\overline{y}}$ ($\frac{\Gamma\left( c + \alpha \right)}{\Gamma\left( c \right)} = c^{\left( \alpha \right)}$). This is also sometimes called the Pochhammer Symbol.

That would mean that you'll get: $$ \begin{align*} \operatorname{D_{x}^{\frac{1}{2}}}\left[ \frac{1}{1 - x} \right] &= \left( 1 \right)^{0.5} \cdot \frac{\Gamma\left( 1 + 0.5 \right)}{\Gamma\left( 1 \right)} \cdot \left( 1 - 1 \cdot x \right)^{-1 - 0.5}\\ \operatorname{D_{x}^{\frac{1}{2}}}\left[ \frac{1}{1 - x} \right] &= \left( 1 \right)^{0.5} \cdot \frac{\Gamma\left( 1.5 \right)}{1} \cdot \left( 1 - x \right)^{-1.5}\\ \operatorname{D_{x}^{\frac{1}{2}}}\left[ \frac{1}{1 - x} \right] &= \pm\Gamma\left( 1.5 \right) \cdot \frac{1}{\left( 1 - x \right)^{1.5}}\\ \end{align*} $$

You can see a little Problem: $\left( x \right)^{0.5}$ has different branches at $x = 1$, which are two roots of unity of $1$ (aka $1$ and $-1$) aka you have to choose a branch. This get's even worst if see $\frac{1}{\left( 1 - x \right)^{-1.5}}$ aka you should norm wich branch you whant to take.

The plot of this would look like (for the main branch):

Gamma[1.5] (1/(1 - x)^1.5) by Wolfram|Alpha

And the function of the main branch would be: $$\fbox{$ \begin{align*} \operatorname{D_{x}^{\frac{1}{2}}}\left[ \frac{1}{1 - x} \right] &= \Gamma\left( 1.5 \right) \cdot \left| \frac{1}{\left( 1 - x \right)^{1.5}} \right|,\, \text{for}\, x \in \mathbb{R}_{< 1}\\ \end{align*} $} \tag{3}$$

If we insert $x = 0$ we get: $$ \begin{align*} &= \pm\Gamma\left( 1.5 \right) \cdot \frac{1}{\left( 1 - 0 \right)^{1.5}}\\ &= \pm\Gamma\left( 1.5 \right) \cdot \frac{1}{\left( 1 \right)^{1.5}}\\ &= \pm\Gamma\left( 1.5 \right) \cdot \frac{1}{\pm 1}\\ &= \pm\Gamma\left( 1.5 \right) \cdot \pm\frac{1}{1}\\ &= \pm\Gamma\left( 1.5 \right)\\ &= \pm\frac{1}{2} \cdot \sqrt{\pi}\\ \end{align*} $$

I got the exact value of the Complete Gamma Function form Wikipedia. Note: Both $\pm$ are independent of each other and $\frac{1}{2} \cdot \sqrt{\pi}$ would be the main branch.

Calculation Via Series Expansion

Another well-known and frequently used method to differentiate a function fractionally is simply to differentiate the series expansion of this function with addends from functions that are easier to differentiate fractionally.

$\frac{1}{1 - x}$ has a nice series expansion given by $$ \begin{align*} \frac{1}{1 - x} = 1 + \sum\limits_{k = 1}^{\infty}\left[ x^{k} \right]. \end{align*} $$

If we now apply the Formula $\left( 1 \right)$ where $a = 1 ~\wedge~ b = 0 ~\wedge~ c = k$ to it (this is a spezial case sometimes called "Euler's Fractional Derivative of Monomials", we would get this: $$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \operatorname{D_{x}^{\alpha}}\left[ 1 + \sum\limits_{k = 1}^{\infty}\left[ x^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \operatorname{D_{x}^{\alpha}}\left[ 1 \right] + \operatorname{D_{x}^{\alpha}}\left[ \sum\limits_{k = 1}^{\infty}\left[ x^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \operatorname{D_{x}^{\alpha}}\left[ 1 \right] + \sum\limits_{k = 1}^{\infty}\left[ \operatorname{D_{x}^{\alpha}}\left[ x^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \operatorname{D_{x}^{\alpha}}\left[ 1 \right] + \sum\limits_{k = 1}^{\infty}\left[ 1^{\alpha} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot \left( 1 \cdot x + 0 \right)^{k - \alpha} \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \operatorname{D_{x}^{\alpha}}\left[ 1 \right] + \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{align*} $$

Now only the Fractional Derivative of a constant is missing. The problem: the Fractional Derivative of a constant differs immensely depending on the definition of the differential operator. I mostly use the definition $\operatorname{D_{x}^{\alpha}}\left[ \text{constant} \right] = 0$. However, a better definition would be $\operatorname{D_{x}^{\alpha}}\left[ \text{constant} \right] = \begin{cases} 0,\, &\text{if}\, \alpha \geq 0\\ \sum\limits_{k = 1}^{\left\lfloor \left| \alpha \right| \right\rfloor}\left[ c_{k} \cdot x^{k} \right],\, &\text{if}\, \alpha < 0 \end{cases}$ where $\left\lfloor \cdot \right\rfloor$ is the Floor Function also called the Greatest Integer Function or Integer Value, since this also applies to $\alpha \in \mathbb{R}^{-}$ and can therefore be used for indefinite integrals. Using this would give us: $$ \begin{equation*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \operatorname{D_{x}^{\alpha}}\left[ 1 \right] + \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = 0 + \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{equation*} $$

$$\fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{align*} $} \tag{4}$$

A Plot of the it for $\alpha = \frac{1}{2}$ looks like:

Sum[Gamma(k + 1)/Gamma(k - 0.5 + 1) x^(k - 0.5),{k,1,[Infinity]}

This function has exactly one branch. if we plug $x = 0$ and $\alpha = \frac{1}{2}$ in, we will get: $$ \begin{align*} &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - 0.5 + 1 \right)} \cdot 0^{k - \alpha} \right]\\ &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k + 0.5 \right)} \cdot 0 \right]\\ &= \sum\limits_{k = 1}^{\infty}\left[ 0 \right]\\ &= 0\\ \end{align*} $$

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