Why is&nbsp;∑limn=1∞e−(n10)2&nbsp;almost equal to&nbsp;5π-12&nbsp;(agreeing up to 427 digits)?

Step 1Defineψ(q)≡−12+12ϑ3(0,e−πq)=∑n≥1e−πn2qwhere ϑ3(z,q) is Jacobi's third theta function. An application of the Poisson summation formula leads to the identityϑ3(0,e−πq)=q,ϑ3(0,e−πq)whose proof is outlined in Jacobi (1828). Substitutingq=(πk2)−1 yields∑n≥1e−n2k2=−12+12ϑ3(0,e−1k2)=−12+12⋅πk2ϑ3(0,e−π2k2)so that∑n≥1e−n2k2=−12+k2π(1+2ψ(πk2))=−12+k2π+kπ∑n≥1e−(πkn)2.When k=10 the error has an order of magnitude oflog10(10πe−100π2)=1+log10π−100π2log10e=bf−427which corresponds to your observation.

We have an answer to "Why is∑limn=1∞e−(n10)2almost equal to5π-12(agreeing up to 427 digits)?"

Kymani Patrick

Answered question

2022-03-14

Why is $\sum {lim}_{n = 1}^{\infty} e^{- {(\frac{n}{10})}^{2}}$ almost equal to $5 \sqrt{π} - \frac{1}{2}$ (agreeing up to 427 digits)?

Answer & Explanation

bicicletagyp

Beginner2022-03-15Added 4 answers

Step 1
Define

ψ (q) \equiv - \frac{12}{+} {\frac{12}{ϑ}}_{3} (0, e^{- π q}) = \sum_{n \geq 1} e^{- π n^{2} q}

where

ϑ_{3} (z, q)

is Jacobi's third theta function. An application of the Poisson summation formula leads to the identity

ϑ_{3} (0, e^{- \frac{π}{q}}) = \sqrt{q}, ϑ_{3} (0, e^{- π q})

whose proof is outlined in Jacobi (1828). Substituting

q = {(π k^{2})}^{- 1}

yields

\sum_{n \geq 1} e^{- \frac{n^{2}}{k^{2}}}

= - \frac{12}{+} {\frac{12}{ϑ}}_{3} (0, e^{- \frac{1}{k^{2}}})

= - \frac{12}{+} \frac{12}{\cdot} \sqrt{π k^{2}} ϑ_{3} (0, e^{- π^{2} k^{2}})

so that

\sum_{n \geq 1} e^{- \frac{n^{2}}{k^{2}}}

= - \frac{12}{+} \frac{k}{2} \sqrt{π} (1 + 2 ψ (π k^{2}))

= - \frac{12}{+} \frac{k}{2} \sqrt{π} + k \sqrt{π} \sum_{n \geq 1} e^{- {(π k n)}^{2}} .

When

k = 10

the error has an order of magnitude of

\log_{10} (10 \sqrt{π} e^{- 100 π^{2}})

= 1 + \log_{10} \sqrt{π} - 100 π^{2} \log_{10} e

= b f - 427

which corresponds to your observation.

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