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

Question

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 $x=0$.

Answer 1

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 $\Gamma(3/2)$.

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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.

Answer 2

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

Answer 3

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.

Answer 4

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 $i=\sqrt{-1}$

Answer 5

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):

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:

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