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Non integer calculus

  1. Jan 16, 2004 #1
    Seem to remember reading about RF experimenting with non-integer differentiation. I found it quite interesting to play with.
    I started with 'half' differentiation.

    e.g. f(x) = x^2 , D[1]f = 2x , D[2]f = 2
    but what about non-integer diff?
    e.g. D[0.5]f = px^1.5
    what is p?

    Clearly D[0.5]D[0.5]f = D[1]f and D[1](x^n) = nx^n-1
    So D[0.5](x^n) = p(n)x^(n-0.5) and
    D[0.5]( p(n)x^(n-0.5) ) = p(n)p(n-0.5)x^(n-1)
    which is equal to D[1](x^n) = nx^(n-1)
    so p(n)p(n-0.5) = n

    From this I figure the key to finding p is solving the formula...
    p(n)*p(n-0.5) = n .

    It looks abit like the gamma function
    G(n+1)/G(n) = n

    and considering the solution to the gamma function, I decide that solving my non-integer diff formula is beyond me.
    I've got a pretty good approximation though.

    p ~= ((x+0.5)^0.5 + x^0.5)/2 for x>1

    hmm. Is this 'genaralized' diff alreadly established math? Since calculus is so central to math,it seems to me that any generalization has broad applications.
  2. jcsd
  3. Jan 16, 2004 #2


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    Science Advisor

    Yes, it is established math. However, it is not done in that form because it can be shown that you can get the same results by using the standard calculus with different functions.
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