Exercise 6.13

Let f be a function from ℤ to the complex numbers. Suppose that p is a prime and that f ( n + p ) = f ( n ) for all n ∈ ℤ . Let f ^ ( a ) = p − 1 ∑ ⁡ t f ( t ) ψ − a ( t ) . Prove that f ( t ) = ∑ ⁡ a f ^ ( a ) ψ a ( t ) . This result is directly analogous to a result in the theory of Fourier series.

Answers

Proof. Let f ^ ( a ) = p − 1 ∑ ⁡ t f ( t ) ψ − a ( t ) . Then

∑ a = 0 p − 1 f ^ ( a ) ψ a ( t ) = ∑ a = 0 p − 1 p − 1 ∑ s = 0 p − 1 f ( s ) ψ − a ( s ) ψ a ( t ) = p − 1 ∑ s = 0 p − 1 f ( s ) ∑ a = 0 p − 1 ψ − a ( s ) ψ a ( t ) = p − 1 ∑ s = 0 p − 1 f ( s ) ∑ a = 0 p − 1 ψ a ( t − s ) = ∑ s = 0 p − 1 f ( s ) δ ( s , t ) = f ( t ) □
User profile picture
2022-07-19 00:00
Comments