How do you DO analytic continuation?

[lolgarithms]

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?

 

[HallsofIvy]

That depends strongly on the specific function involved.

 

[lolgarithms]

i mean, when you do not have a function whose Taylor series converge to it everywhere, how do you find the analytic continuation?

 

[lurflurf]

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!