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I The fractional derivative operator

  1. Feb 27, 2017 #1
    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. jcsd
  3. Feb 28, 2017 #2
    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.
     
  4. Feb 28, 2017 #3

    DrClaude

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    Staff: Mentor

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