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## Main Question or Discussion Point

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...