# I The fractional derivative operator

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1. Feb 27, 2017

### Kumar8434

I've been thinking about it since yesterday and have noticed this pattern:

We have, the first order derivative of a function $f(x)$ is:
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} .......(1)$$
The second order derivative of the same function is:
$$f''(x)=\lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x)}{h}$$
By putting $x=x+h$ in (1), we can have $f'(x+h)$.
So,$$f''(x)=\lim_{h\rightarrow 0}\frac{\lim_{h\rightarrow 0}\frac{f(x+h+h)-f(x+h)}{h}-\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}{h}$$
Or , $$f''(x)=\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}.....(2)$$
You can check by L'Hospital's rule that this limit evaluates to $f''(x)$.
Now, the third order derivative of the same function is:
$$f'''(x)=\lim_{h\rightarrow 0}\frac{f''(x+h)-f''(x)}{h}$$
By putting $x=x+h$ in (2), we can get $f''(x+h)$.

So, $$f'''(x)=\lim_{h\rightarrow 0}\frac{\lim_{h\rightarrow 0}\frac{f(x+3h)-2f(x+2h)+f(x+h)}{h^2}-\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}}{h}$$
which gives $$f'''(x)=\lim_{h\rightarrow 0} \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}......(3)$$
Again, by repeating the same process, we can get that:
$$f''''(x)=\lim_{h\rightarrow 0}\frac{f(x+4h)-4f(x+3h)+6f(x+2h)-4f(x+h)+f(x)}{h^4}....(4)$$
So, we notice that the coefficient of $f(x+(n-r)h)$ in the expression of $f^{n}(x)$ ($n^{th}$ derivative of $f(x)$) is actually $(-1)^{r}\cdot {^n}C_r$, same as the coefficient of $x^{n-r}$ in the expansion of $(x-1)^n$.
It can be proved that:
$$f^n(x)=\lim_{h\rightarrow 0}\frac{\sum_{r=0}^n(-1)^{r}\cdot ^{n}C_r\cdot f(x+(n-r)h)}{h^n}$$
where $f^n(x)$ is the $n^{th}$ order derivative of the function $f(x)$.

Now, to generalize this to fractional order derivatives, we just have to generalize the coefficients, which must be similar to the generalization of the expansion of $(x-1)^n$ to fractional exponents.

I'm not very good with binomial theorem, but I guess that it should be:
$$f^n(x)=\lim_{h\rightarrow 0}\frac{f(x+nh)-n\cdot f(x+(n-1)h)+\frac{n(n-1)}{2!}\cdot f(x+(n-2)h)-....}{h^n}$$
, where $n$ can be fractional. Have I done anything wrong?

Last edited: Feb 28, 2017
2. Feb 28, 2017

### SlowThinker

You can't have a noninteger number of addends.
Try to figure out how it would look for (x^0.5)' or (x^1.5)'. I suspect you won't get very far.

3. Feb 28, 2017