Jacobi
Find the value of 1 + 2 ψ ( 4 1 ) 1 + 2 ψ ( 4 ) , where ψ ( x ) = n = 1 ∑ ∞ e − n 2 π x .
The answer is 0.5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
@Mark Hennings Sir, could you please suggest some good books for learning these topics??? I only browse through Wolfram pages and get to know the basics of it, but I haven't really had any rigorous treatment......
Log in to reply
@Brian Lie Any help, please??
Log in to reply
You need to start by reading a good book on Analysis. There are books on the Jacobi Theta functions, but they rely on a vast array of results that come from standard analysis texts, and so will be hard to read until you have got your head around the basics. Randolph's lift from Wolfram Mathworld comments that this result follows from the Poisson sum formula - this is covered in good Analysis texts. The Mathworld page references Apostol's book on analysis - old, but none the worse for that!
Please, see Jacob Theta Function :
ψ = x → ∑ n = 1 ∞ e π ( − n 2 ) x
2 ψ ( 4 1 ) + 1 2 ψ ( 4 ) + 1 = ϑ 3 ( 0 , e − 4 π ) ϑ 3 ( 0 , e − 4 π ) = 2 1
The function ψ famously satisfies the identity 1 + 2 ψ ( x − 1 ) 1 + 2 ψ ( x ) = x 1 x > 0 and so the answer is 2 1 .