- #1
eljose
- 492
- 0
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..:grumpy: :grumpy: )
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..:grumpy: :grumpy: )
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...