Derivation that e^pi is approximately pi+20

== Email to Eric Weisstein about https://mathworld.wolfram.com/AlmostInteger.html, 26 November 2023 ==

Hello Eric,

Credit for the idea to differentiate the Jacobian identity at τ=i goes to Aaron Doman, a.k.a. @MathFromAlphaToOmega, who mentioned it in a comment on a video by popular math YouTuber @Mathologer (Burkard Polster). You may want to contact Aaron to ask if they are fine with being credited for it ([contact details removed]). I am okay with being named if you wish to, but all I did was notice that it contradicted what's said on MathWorld, flesh out the details to confirm that it's correct, and of course get in touch with you.

Starting from the definition of the Jacobi theta function ϑ(0,τ) = Σe^(iπτn²), where the sum goes from n=-∞ to +∞, and the Jacobian identity ϑ(0,-1/τ) = √(τ/i)*ϑ(0,τ) (compare https://en.wikipedia.org/wiki/Theta_function#Jacobi_identities), differentiating on both sides gives ϑ'(0,-1/τ) = √(-iτ)/(2τ)*ϑ(0,τ)+√(τ/i)*ϑ'(0,τ) = Σ[2πτn²-1)/(2√(-iτ)]e^(iπτn²). At the specific value τ=i, this evaluates to Σ(iπn²-i/2)e^(-πn²).

On the other hand, one can start with just the left side, getting ϑ'(0,-1/τ) = (iπ/τ²)Σn²e^(-iπn²/τ), which at τ=i is equal to Σ(-iπn²)e^(-πn²).

Equating the two different ways to write ϑ'(0,-1/τ) gives Σ[(iπn²-i/2)+iπn²]e^(-πn²) = 0, which after dividing by i simplifies to
Σ(2πn²-1/2)e^(-πn²) = 0.

Split the sum into three parts (-∞ to -1, 0, and +1 to +∞), and notice that the sums from -∞ to -1 and +1 to +∞ are identical:
2Σ(2πn²-1/2)e^(-πn²) = 1/2, where the sum goes from n=1 to +∞. Multiplying by 2 gives the interesting sum Σ(8πn²-2)e^(-πn²) = 1 (with the sum from n=1 to +∞; compare https://www.wolframalpha.com/input?i=sum+%288pi*n%C2%B2-2%29e%5E%28-pi*n%C2%B2%29+from+n%3D1+to+infinity). I haven't found anyone who has calculated it explicitly, but I am sure that sum is well known. I think the idea to combine it with the approximation π ≈ 22/7 is novel.

This converges very fast. Using just the first term (n=1) gives e^π ≈ 8π-2 = 7π-2+π ≈ 7(22/7)-2+π = 20+π.

Funnily, the approximation π ≈ 22/7 makes the formula an order of magnitude more precise (7π-2 = 19.991..., e^π-π = 19.9991...). I guess it's by coincidence that much of the error cancels out. Even if it didn't, I think 19.991 would still be close enough to 20 to be interesting.

Ironic that the MathWorld article prominently features π ≈ 22/7 in the sentence just above the first mention of e^π-π ≈ 20.

Kind regards,
Daniel Bamberger