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I was doing some mathematics doodling today and I wrote down

$$\frac{d^{n}}{dx^{n}}x^n = n!.$$

Looking at this made me wonder if when $n = 1/2$, if $D^{1/2}_0\sqrt{x} = \Gamma(1/2)$. Where $D^{1/2}_{0}$ is the Riemann-Liouville derivative with order 1/2 and lower terminal $0$. Indeed, it appears that

$$D^{1/2}_0 \sqrt{x} = \frac{1}{\Gamma(3/2)} \frac{d}{dx}\int_{0}^{x}\frac{\sqrt{t}}{\sqrt{x-t}}dt= \frac{2}{\Gamma(1/2)} \frac{d}{dx} \frac{x\pi}{2} = \sqrt{\pi}$$

as expected. The Wikipedia entry (which I've verified) suggests that "$D^{1/2}_{0}D^{1/2}_{0} = \frac{d}{dx}$", that is, two half derivatives is equivalent to the usual derivative and verifies it in the case of $f(x) = x$ by computing $D_0^{1/2}D_0^{1/2}x = 1$.

However, for a constant $c$,

$$D_0^{1/2}c = \frac{1}{\Gamma(3/2)} \frac{d}{dx} \int_{0}^{x} \frac{c}{\sqrt{x-t}}dt = \frac{1}{\Gamma(3/2)}\frac{d}{dx} 2c\sqrt{x} = \frac{c}{\sqrt{\pi x}}.$$

Now using these facts, I was curious what function satisfied $D_{0}^{1/2}f(x) = 1$ and noticed that

$$D_{0}^{1/2}D_{0}^{1/2}\sqrt{x} = D_{0}^{1/2} \Gamma(1/2) = D_{0}^{1/2}\sqrt{\pi} = \frac{\sqrt{\pi}}{\sqrt{\pi x}} = \frac{1}{\sqrt{x}} \neq \frac{d}{dx}\sqrt{x} =\frac{1}{2\sqrt{x}}.$$

At this point I am confused and I would like to better understand what is going on.

Wikipedia aside, I thought I was 100% certain I remember reading that the primary advantage of using the Riemann-Liouville definition was that the composition of fractional derivative operators "added up" in the way you would like, but this seems to contradict this. If it doesn't, then what was the purpose of Wikipedia's repeated half differentiation of $x$?

I think I must have made a simple arithmetic error, but I cannot find where the missing factor of 2 would come from to fix this.

JessicaK
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  • The convolution or Fourier transform definition of the fractional derivative really mean that $D^{1/2}( x^a \color{red}{1_{x > 0}}) = \frac{\Gamma(a+1)}{\Gamma(a-1/2+1)} x^{a-1/2}\color{red}{1_{x > 0}}$. That's why the fractional derivative of $x^0 1_{x > 0}$ doesn't have to be constant, since $x^0 1_{x > 0}$ is not constant itself. Note $D^1 1_{x >0} = \delta(x)$, so the formula I wrote is correct only when the fractional derivative is represented by a function. – reuns Dec 02 '17 at 04:42
  • @reuns I'm not sure I understand your point by introducing a different definition than the one asked in the question. It doesn't look like your definition of a fractional derivative is consistent with Riemann-Liouville. Could you please elaborate? – JessicaK Dec 03 '17 at 01:35
  • @reuns My point is that my post specifically mentions this is a question about the Riemann-Liouville fractional derivative and your comment begins by using the Fourier transform to define the derivative instead. To me, this makes your comment completely unrelated to the question asked. – JessicaK Dec 03 '17 at 04:19
  • Of course I'm talking of what you wrote. https://en.wikipedia.org/wiki/Riemann%E2%80%93Liouville_integral#Definition is the convolution definition of the fractional integral : $I^\alpha f = \frac{x^{\alpha-1} 1_{x > 0}}{\Gamma(\alpha)} \ast f$. You need to prove $I^\alpha I^\beta f = I^{\alpha+\beta} f$ (at least for $f(x) = x^b 1_{x >0}$). Then $D^\alpha$ is defined as its inverse operator. – reuns Dec 03 '17 at 04:23
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    @reuns I'm sorry, I don't understand why we have an indicator function $1_x>0$ when both the Wikipedia entry you linked and I.Podlubnys introductory text says that $f$ needs to be locally integrable, which $\sqrt(x)$ is on the non-negative reals. I have no problem with the fact that the half derivative of a constant is nonzero and I can see how $D^{1/2} D^{1/2} x^{0} \neq D^{1} x^{0}$ but it's not clear to me why that is the case either. – JessicaK Dec 03 '17 at 04:51
  • $F \ast G(x) = \int_{-\infty}^\infty F(t) G(x-t)dt$. You can't change the definition without loosing the commutativity, associativity, or the fact $\frac{d}{dx}$ is the convolution inverse of $1_{x >0}$ (ie. $\delta' \ast 1_{x >0} = \delta$). Then $\frac{1}{\Gamma(\alpha)} \int_a^x f(t) (x-t)^{\alpha-1}dt = \frac{x^{\alpha-1} 1_{x > 0}}{\Gamma(\alpha)} \ast f 1_{x > a}$. In your question you took $a=0, f(x) = x^{1/2}$. It is true that $D^{1/2} D^{1/2} (x^01_{x >0}) = D^1 (x^0 1_{x >0})=\delta(x)$, that's the point of all my comments. – reuns Dec 03 '17 at 06:43

1 Answers1

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Lets keep track of some facts-

  1. $\Gamma(3/2) = \frac{\sqrtπ}{2}$ and $\Gamma(1/2) = \sqrt{π}$.

  2. $\frac d{dx} \sqrt{x} = \frac1{2\sqrt x}$ (These two you got right.)

  3. $ D^{\alpha} f:= \frac d{dx}J^{1-\alpha}f = \frac{1}{\Gamma(1-\alpha)}\frac d{dx}\int_0^x \frac {f(t)}{(t-x)^{\alpha}} \, dt$ (here you have the wrong Gamma constant for $\alpha = 1/2$., e.g. see Wikipedia.)

Forgive me for dropping the 0 subscript, but here we see where the missing factor of 2 is. With this we see (where you can verify the integrals via e.g. Wolfram Alpha) $$ D^{1/2}\sqrt x = \frac{1}{\Gamma(1/2)}\frac d{dx} \frac{xπ}2 = \frac{\sqrt{π}}{2}$$ which by the way is on the Wikipedia page, and similarly (here you got the right answer somehow) $$ D^{1/2} 1 = \frac{1}{\Gamma(1/2)}\frac d{dx} 2\sqrt x = \frac1{\sqrt{πx}}$$ The integral formula makes it clear that $D^\alpha f$ is $\mathbb R$-linear in $f$. Hence we can piece these together, $$ D^{1/2}(D^{1/2} \sqrt{x}) = D^{1/2}\frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{2}D^{1/2} 1 = \frac{\sqrt{π}}{2} \frac1{\sqrt{πx}} = \frac1{2\sqrt x} \color{green}{\overset{\checkmark}{=} \frac{d}{dx} \sqrt x}$$

There is a proof on Wikipedia that the fractional integral commutes $J^{\alpha+\beta} = J^{\alpha}J^\beta$, but I quote the following paper (which has proofs),

Although the Riemann-Liouville integral operator $D^{−α}, α ∈ \mathbb R^+ $ has the semigroup property[...] the RL derivative operator does not. However, we have the following interesting result.

If $x(t) ∈ C^1 [0,T] , α_i ∈ [0,1] \ (i\in 1,2 ) $ and $ α_1+ α_2 ∈ (0, 1 ]$, then $D^{α_1} · D^{α_2} x( t) = D^{α_1+ α_2} x (t) $ .

By the way, linearity implies that you know what solves $D^{1/2} f = 1$ since you computed $D^{1/2}\sqrt x $ is a constant.

Calvin Khor
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