Fractional Order Derivative

Question

A friend and I were talking about derivatives, and he asked an interesting question. Since both of us have not taken our calculus courses yet, neither of us were sure of the answer. His question has two parts: 1. can you have a fractional order derivative i.e. could you have $\frac{d^ny(x)}{dx^n}$ where, for example, $n=\frac{1}{2}$? And, 2. can you have a variable as the order of a derivative i.e. $\frac{d^ny(x)}{dx^n}$ where n=some variable? I have doubts as to whether either are even possible, but I decided to post the qestion to find out for sure.

Answer 1

Oh boy, this is a fun one to answer. First, yes, these things exist and they fall under a thing called fractional calculus.

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To start, let's assume that the functions we are looking at are smooth so that all the normal derivatives exist.

Secondly, we'll also want $\frac{d^a}{dx^a}=D_x^a$, the $a$th derivative w.r.t. $x$.

Third, we want $D_x^aD_x^b=D_x^{a+b}$. Specifically, $D^{1/2}D^{1/2}=D^1$.

Now, one approach would be to use the limit definition of the derivative:

$$D_x^1f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$

As we've assumed all these derivatives exist, it suffices to replace these limits with one-sided limits:

$$^+D_x^1f(x)=\lim_{h\to0^+}\frac{f(x+h)-f(x)}h$$

Take the derivative again and you'll get

$$^+D_x^2f(x)=\lim_{h\to0^+}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$

One can then derive the following formula with induction:

$$^+D_x^nf(x)=\lim_{h\to0^+}\frac1{h^n}\sum_{k=0}^n(-1)^k\binom nkf(x+(n-k)h)$$

Since $\binom nk=0$ for every $n-k\in\mathbb N_{>0}$, we have

$$^+D_x^nf(x)=\lim_{h\to0^+}\frac1{h^n}\sum_{k=0}^\infty(-1)^k\binom nkf(x+(n-k)h)$$

Let $h=1/N$ and this becomes

$$^+D_x^nf(x)=\lim_{N\to\infty}N^n\sum_{k=0}^\infty(-1)^k\binom nkf\left(x+\frac{n-k}N\right)$$

This allows an extension to fractional derivatives, where we use generalized binomial coefficients:

$$^+D_x^af(x)=\lim_{N\to\infty}N^a\sum_{k=0}^\infty(-1)^k\binom akf\left(x+\frac{a-k}N\right)$$

However, the above sum has the annoying attitude to be divergent for negative values, hence, we may include an additional parameter to fix that:

$$_b^+D_x^af(x)=\lim_{N\to\infty}N^a\sum_{k=0}^{bN}(-1)^k\binom akf\left(x+\frac{a-k}N\right)$$

This is an extended version of the Grunwald-Letnikov derivative, and as I have previously shown, the above extends to integrals via Riemann sums:

$$_1^+D_x^{-1}f(x)=\lim_{N\to\infty}\frac1N\sum_{k=1}^Nf\left(x-\frac kN\right)=\int_{x-1}^xf(t)~\mathrm dt$$

Pretty cool, right?

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Now, the above limit is quite horrendous and truthfully hard to evaluate at any non-integer value. While the following definition for fractional derivatives aren't much simpler, they are usually more direct for solving. To begin, we instead look at fractional integrals.

Let us have $_uJ_x^a$ be the $a$th integral w.r.t. $x$ around the point $u$. That is,

$$_uJ_x^af(x) = \int_u^x \int_u^{\sigma_1} \cdots \int_u^{\sigma_{n-1}} f(\sigma_{a}) \, \mathrm{d}\sigma_{a} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1$$

Note that $u$ has an impact on the integral. It is, perhaps, a curious thing, that we have the following formula:

$$_uJ_x^nf(x)=\frac1{(n-1)!}\int_u^x(x-t)^{n-1}f(t)~\mathrm dt$$

which may be proven by induction. Here, $n!$ is the factorial, where define it through the Gamma function,

$$x!=\int_0^\infty t^xe^{-t}~\mathrm dt,\quad x>-1$$

Quite obviously, the above formula extends to non-integer values. By repeatedly differentiating, you will find with the help of the fundamental theorem of calculus that

$$D_x^a~_uJ_x^bf(x)\stackrel?=D_x^{a-b}f(x)$$

Unfortunately, the right side doesn't have a $u$, which might be a problem, so we need to include an additional parameter:

$$_uD_x^a~_uJ_x^bf(x)=~_uD_x^{a-b}f(x)$$

Also note that the order is necessary, since

$$f(x)=\frac d{dx}\int_u^xf(t)~\mathrm dt\ne\int_u^x\frac d{dt}f(t)~\mathrm dt=f(x)-f(0)$$

That is, we need to take the integral first, then the derivative. This leads to a natural fractional derivative:

$$_uD_x^af(x)=\begin{cases}\frac1{(-\{a\})!}\frac{d^{\lceil a\rceil}}{dx^{\lceil a\rceil}}\int_u^x(x-t)^{-\{a\}}f(t)~\mathrm dt,&a\ge0\\\frac1{(a-1)!}\int_u^x(x-t)^{a-1}f(t)~\mathrm dt,&a<0\end{cases}$$

Where $\{a\}$ is the fractional part of $a$. For example,

$\{\pi\}=0.1415926\dots$

$\{-2.7\}=0.3$

And $\lceil a\rceil$ is the ceiling of $a$. For example,

$\lceil\pi\rceil=4$

$\lceil-2.7\rceil=-2$

Using this definition, you may find that

$$_0D_x^ax^b=\frac{b!}{(b-a)!}x^{b-a}$$

as a mere example. Particularly, we have things like

$$_0D_x^{1/2}\sqrt x=\frac{\sqrt\pi}2$$

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Of course, there are many many other fractional derivatives out there (indeed, no two fractional derivatives are necessarily the same), so I suggest you browse the Wikipedia more if you are interested, and if you feel ready to do some crazy integrals, try some research.

Answer 2

Let me add this "destructive" answer as an addition. We take a conceptual approach to the "half" derivative $Hf$ of $f$. What properties should a "half derivative" $Hf$ of $f$ have?

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(1) Clearly, we want $HHf = f'$ for all twice half-differentiable functions $f$, so applying the half derivative twice should give the ordinary derivative.

However, there are many ways of defining a suitable half derivative with property (1), so this doesn't give a unique definition of $H$. We need more properties. The next thing which comes to my mind is

(2) For the ordinary $n$-th derivative, very basic differentiation rules state that $$(f+g)^{(n)} = f^{(n)} + g^{(n)}\quad\text{and}\quad(\lambda f)^{(n)} = \lambda f^{(n)}.$$

It's only natural to ask for the same for the half derivative. So $H$ should be linear, that is $$H(f+g) = Hf + Hg\quad\text{and}\quad H(\lambda f) = \lambda Hf$$ for all half-differentiable functions $f,g$ and all $\lambda\in\mathbb{R}$.

(In other words, the set of all half-differentiable functions forms a vector space and $H$ is a linear operator on it.)

So do (1) and (2) give us a unique definition of a half derivative? Unfortunately, we will see that the answer is "no", as these properties turn out to be contradictory if we additionally demand only the following tiny bit

(3) All constant functions as well as the identity function $\operatorname{id} : x\mapsto x$ are three times half-differentiable.

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Let us derive the contradiction from (1), (2) and (3). For the zero function $\mathbf{0} : x \mapsto 0$ we have by additivity (part of the linearity property) $$ H\mathbf{0} = H(\mathbf{0} + \mathbf{0}) = H\mathbf{0} + H\mathbf{0}, $$ so subtracting $H\mathbf{0}$ on both sides gives $$ H\mathbf{0} = \mathbf{0}. $$

Now for the constant function $\mathbf{1} : x\mapsto 1$ we have $$ (H\mathbf{1})' = HH(H\mathbf{1}) = H(HH\mathbf{1}) = H(\mathbf{1}') = H\mathbf{0} = \mathbf{0},$$ so $H\mathbf{1}$ is a constant function, say $H\mathbf{1} = (x \mapsto a)$ with $a\in\mathbb{R}$. Now by linearity $$ \mathbf{0} = \mathbf{1}' = HH\mathbf{1} = H(x\mapsto a) = H(a\cdot\mathbf{1}) = a\cdot H\mathbf{1} = a\cdot (x\mapsto a) = (x\mapsto a^2),$$ showing that $a = 0$ and therefore $$H\mathbf{1} = \mathbf{0}.$$ Therefore, by linearity any constant function $x\mapsto b$ with $b\in\mathbb{R}$ has the half derivative $$ H(x\mapsto b) = H(b\cdot \mathbf{1}) = b\cdot H\mathbf{1} = b\mathbf{0} = \mathbf{0} $$ Now $$(H\operatorname{id})' = HH(H\operatorname{id}) = H(HH\operatorname{id}) = H(\operatorname{id}') = H\mathbf{1} = \mathbf{0},$$ so $H\operatorname{id} = (x \mapsto c)$ is constant and therefore its half derivative is $\mathbf{0}$. This gives the final contradiction $$ \mathbf{0} = H(H\operatorname{id}) = HH\operatorname{id} = \operatorname{id}' = \mathbf{1}. $$

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Conclusion. For me, there is no convincing concept of a half derivative. Of course, people can define fractional derivatives in this way or another and derive conclusions, but for it being really a "fractional derivative", something is missing (as (1), (2) and (3) cannot all be true), and there is no way out by the above contradiction.

Answer 3

Yes. Fractional derivatives exist.

See for example, this Wikipedia entry on Fractional Calculus. As I mention in a comment above; there are eight "related" posts about fractional calculus.

Indeed, the order of a derivative need not be in $\mathbb N$. Rational orders of a derivative exist, as you were suspecting in your post.

And even irrational numbers can be an orders of a derivative.

Answer 4

Another way to define fractional derivatives

Recall that the exponential function $e^{ax}$ is very easy to derive: $$\frac{d^n}{dx^n} e^{ax} = a^n e^{ax}$$

Assume that $f$ can be written as a sum or an integral of exponentials (I will here only go for the integral case): $$f(x) = \int a(\xi) e^{c \xi x} \, d\xi$$

Taking ordinary derivatives and assuming that the derivatives can be moved inside the integral then gives $$f^{(n)}(x) = \int a(\xi) (c\xi)^n e^{c \xi x} \, d\xi$$

This formula directly generalizes to fractional $n$.

Example

Probably you haven't heard about it yet, but there is something called the Fourier transform. It can be thought of as splitting a function into frequencies (using $e^{i \omega t}$): $$f(t) = \int \hat f(\omega) e^{i \omega t} \, d\omega$$ The derivatives then become $$f^{(n)}(t) = \int (i\omega)^n \hat f(\omega) e^{i \omega t} \, d\omega$$ There are some technicalities with defining $i^n$ for rational $n$, but the problems can be overcome.