The fractional derivative operator

  • #1
127
5
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:

Answers and Replies

  • #2
474
66
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.
 

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