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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

    HallsofIvy

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    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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