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# Modular Forms/Periodicity / Fourier Expansion Watch

1. This is probably a stupid question but answers from different perspectives greatly appreciated.

My notes and the book for my course Serre-see below- simply state that because

, i.e. because f is periodic a fourier expansion exists and then defines it to be an expansion of

- by answers from different perspective i mean, because i came across some notes online that where saying well-defined in 0 < Re (q) < 1 , and the fact that f(t)=f(t+1) is what extends it to be well-defined holomorphic everywhere
- whereas a simple answer, judging from my notes and the book Serre a couse in arithmetic which simply state due to its periodicity

Don't really understand, cheers.
2. bumpety-bump.
3. (Original post by xfootiecrazeesarax)
bumpety-bump.
It's not quite clear to me what you're asking! You seem to be making reference to Serre's book on arithmetic (but don't give a page reference) and to holomorphic functions and to periodicity and fourier transforms - but I can't extract what you're actually asking from that.

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Updated: November 15, 2016
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