Jacobi and Ramanujan to the rescue!

Calculus Level 5

r = 1 1 e π r 2 = 1 2 ( π 1 a Γ ( b a ) 1 ) \large \sum_{r=1}^{\infty} \frac{1}{e^{\pi r^{2}}} = \frac{1}{2} \left( \frac{\pi^{\frac{1}{a}}}{\Gamma(\frac{b}{a})}-1\right)

Given a a and b b are coprime positive integers, find a + b a+b .


The answer is 7.

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1 solution

Tapas Mazumdar
Mar 25, 2017

One of the Jacobi theta functions gives us the identity

ϑ ( z ; τ ) = 1 + 2 n = 1 ( e π i τ ) n 2 cos ( 2 π n z ) \vartheta (z ; \tau) = 1 + \displaystyle 2 \sum_{n=1}^{\infty} {\left( e^{\pi i \tau} \right)}^{n^2} \cos (2 \pi n z)

Setting n = r n=r , z = 0 z = 0 and τ = i \tau = i , we get

ϑ ( 0 ; i ) = 1 + 2 r = 1 ( e π ) r 2 r = 1 1 e π r 2 = 1 2 [ ϑ ( 0 ; i ) 1 ] \vartheta (0 ; i) = 1 + \displaystyle 2 \sum_{r=1}^{\infty} {\left( e^{-\pi} \right)}^{r^2} \implies \sum_{r=1}^{\infty} \dfrac{1}{e^{\pi r^2}} = \dfrac 12 \left[ \vartheta (0 ; i) - 1 \right]

The explicit value of ϑ ( 0 ; i ) \vartheta (0 ; i) is given by

φ ( e π ) = ϑ ( 0 ; i ) = π 4 Γ ( 3 4 ) \varphi \left( e^{-\pi} \right) = \vartheta (0 ; i) = \dfrac{\sqrt[4]{\pi}}{\Gamma \left(\frac 34 \right)}

Hence

r = 1 1 e π r 2 = 1 2 [ π 1 4 Γ ( 3 4 ) 1 ] \displaystyle \sum_{r=1}^{\infty} \dfrac{1}{e^{\pi r^2}} = \dfrac 12 \left[ \dfrac{\pi^{\frac 14}}{\Gamma \left(\frac 34 \right)} - 1 \right]

So, a + b = 4 + 3 = 7 a+b = 4+3 = \boxed{7} .

thats great ......how did you learn all this.......... stuff thats awesome!!!!!

A Former Brilliant Member - 3 years, 7 months ago

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Whenever I stumble upon such problem, I try to learn the concepts in my free time via Brilliant wiki, wikipedia and wolframmathworld. Though they are stuff mainly from pure mathematics and I'm preparing for JEE so I do not believe I have any better practice in this field, however, enough to tackle such direct problems with some thorough self-reading.

Tapas Mazumdar - 3 years, 7 months ago

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That's great!!!!

A Former Brilliant Member - 3 years, 7 months ago

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