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How do you DO analytic continuation?

  1. Sep 18, 2009 #1
    they talk about the existence of analytic continuation, but how do you find (the power series/product), calculate, compute the analytic continuation? how do you actually do analytic continuation on a function?
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 19, 2009 #2

    HallsofIvy

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    That depends strongly on the specific function involved.
     
  4. Sep 19, 2009 #3
    i mean, when you do not have a function whose Taylor series converge to it everywhere, how do you find the analytic continuation?
     
  5. Sep 20, 2009 #4

    lurflurf

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    One could write ten books about this. Often analytic continuation is not practical. The method ones uses differers greatly with the specific problem.
    In very simple cases the following works
    1) find the series expansion of f and a number of derivatives
    f(z+h)=f(z)+f'(z)h+f''(x)h^2+...
    f'(z+h)=f'(z)+f''(z)h+f'''(x)h^2+...
    f''(z+h)=f''(z)+f'''(z)h+f''''(x)h^2+...
    f'''(z+h)=f'''(z)+f'(z)h+f'''''(x)h^2+...
    ...
    in operator form
    f(z+h)=exp(hD)f(z)
    f'(z+h)=exp(hD)f'(z)
    [D^n]f(z+h)=exp(hD)[D^n]f(z)

    find a series for f about z=a
    given f and derivatives at z=a
    use the series to find f and derivatives at z=b
    find a new series for f about z=b
    now you can find f and derivatives at z=c where c can be found by expansion about b, but not expansion about a
    hopefully this is enough but if not repeat more times

    example
    f=1/(1-x)
    find an expansion of f about z=0 (radius=1)
    find an expansion about z=-sqrt(2)/2 (radius=1+sqrt(2)/2)
    compute f for some z such that |z+sqrt(2)/2|<1+sqrt(2)/2 (ie z=-2)
    have fun!
     
    Last edited: Sep 20, 2009
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