eljose
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Could someone explain "Hardy-Littlewood" circle method?..in fact according to Wikipedia they take:
[tex]f(z)=\sum_{n=0}^{\infty} a(n) z^n[/tex]
So the inverse transform to get the a(n) is:
[tex]2i \pi a(n)= \oint dzf(z)z^{-(n+1)}[/tex]
This is what i understand (Don't make me look at wikipedia because the explanation is similar and there's no example.. )
The main objective of the method is supposed to get an "asymptotyc" expression for the a(n) [tex]a(n)\sim g(n)[/tex] where the function g is known but how is this done?..thanx...
[tex]f(z)=\sum_{n=0}^{\infty} a(n) z^n[/tex]
So the inverse transform to get the a(n) is:
[tex]2i \pi a(n)= \oint dzf(z)z^{-(n+1)}[/tex]
This is what i understand (Don't make me look at wikipedia because the explanation is similar and there's no example.. )
The main objective of the method is supposed to get an "asymptotyc" expression for the a(n) [tex]a(n)\sim g(n)[/tex] where the function g is known but how is this done?..thanx...