# Half order derivative of ${1 \over 1-x }$

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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A related problem. You need to use the Riemann-Liouville fractional derivative definition. – Mhenni Benghorbal May 15 '13 at 1:02

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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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 – Gottfried Helms May 14 '13 at 21:43
+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. – J. M. May 15 '13 at 2:39
(acc) This made it very simple and answered both partial questions quickly. Thanks! – Gottfried Helms May 15 '13 at 7:56

$$\frac{d^k}{dx^k} \left(\frac{1}{1-x} \right)=\frac{d^k}{dx^k} (1+x+x^2+...)=\Gamma(k+1) \sum_{n=k}^{\infty} {n\choose k}x^{n-k}\\ \sum_{n=k}^{\infty} {n\choose k}x^{n-k}=\frac{1}{(1-x)^{k+1}}\\ \frac{d^k}{dx^k} \frac{1}{1-x}=\frac{\Gamma(k+1)}{(1-x)^{k+1}}$$

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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? – Gottfried Helms May 14 '13 at 21:47
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. – Gottfried Helms May 14 '13 at 21:58
I made a minor edit (format) to your answer. Hope you do not mind. – user17762 May 15 '13 at 4:33

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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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? – Gottfried Helms May 15 '13 at 6:38
@GottfriedHelms: I did not understand your question well? Can you elaborate more? Have you tried the power series method using the Caputo definition? – Mhenni Benghorbal May 15 '13 at 6:43
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) – Gottfried Helms May 15 '13 at 6:54
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... – Gottfried Helms May 15 '13 at 6:59
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! – Gottfried Helms May 15 '13 at 7:36