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Question Number 125994 by Dwaipayan Shikari last updated on 16/Dec/20
((Σ_(n=0) ^∞ e^(−n^2 ) )/(Σ_(n=0) ^∞ e^(−2n^2 ) ))
$$\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } }{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}{n}^{\mathrm{2}} } } \\ $$
Answered by Olaf last updated on 16/Dec/20
Jacobi Theta function : ϑ(z,τ) = Σ_(n=−∞) ^(n=+∞) exp(πin^2 τ+2πinz) ϑ(0,(i/π)) = Σ_(n=−∞) ^(n=+∞) e^(−n^2 ) = 2Σ_(n=0) ^∞ e^(−n^2 ) −1 (1) ϑ(0,((2i)/π)) = Σ_(n=−∞) ^(n=+∞) e^(−2n^2 ) = 2Σ_(n=0) ^∞ e^(−2n^2 ) −1 (2) (1) and (2) : ((Σ_(n=0) ^∞ e^(−n^2 ) )/(Σ_(n=0) ^∞ e^(−2n^2 ) )) = ((ϑ(0,(i/π))+1)/(ϑ(0,((2i)/π))+1))
$$\mathrm{Jacobi}\:\mathrm{Theta}\:\mathrm{function}\:: \\ $$$$\vartheta\left({z},\tau\right)\:=\:\underset{{n}=−\infty} {\overset{{n}=+\infty} {\sum}}\mathrm{exp}\left(\pi{in}^{\mathrm{2}} \tau+\mathrm{2}\pi{inz}\right) \\ $$$$\vartheta\left(\mathrm{0},\frac{{i}}{\pi}\right)\:=\:\underset{{n}=−\infty} {\overset{{n}=+\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } \:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } −\mathrm{1}\:\left(\mathrm{1}\right) \\ $$$$\vartheta\left(\mathrm{0},\frac{\mathrm{2}{i}}{\pi}\right)\:=\:\underset{{n}=−\infty} {\overset{{n}=+\infty} {\sum}}{e}^{−\mathrm{2}{n}^{\mathrm{2}} } \:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}{n}^{\mathrm{2}} } −\mathrm{1}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\::\:\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } }{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}{n}^{\mathrm{2}} } }\:=\:\frac{\vartheta\left(\mathrm{0},\frac{{i}}{\pi}\right)+\mathrm{1}}{\vartheta\left(\mathrm{0},\frac{\mathrm{2}{i}}{\pi}\right)+\mathrm{1}} \\ $$
Commented by Dwaipayan Shikari last updated on 16/Dec/20
Thanking you! But is there any other way to approximate?
$${Thanking}\:{you}!\:{But}\:{is}\:{there}\:{any}\:{other}\:{way}\:{to}\:{approximate}? \\ $$
Commented by Olaf last updated on 16/Dec/20
There is no asymptotic formula but a computation is possible. See here : https://hlabrande.fr/maths/pubs/theta.pdf
$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{asymptotic}\:\mathrm{formula} \\ $$$$\mathrm{but}\:\mathrm{a}\:\mathrm{computation}\:\mathrm{is}\:\mathrm{possible}. \\ $$$$\mathrm{See}\:\mathrm{here}\:: \\ $$$$\mathrm{https}://\mathrm{hlabrande}.\mathrm{fr}/\mathrm{maths}/\mathrm{pubs}/\mathrm{theta}.\mathrm{pdf} \\ $$
Commented by Dwaipayan Shikari last updated on 16/Dec/20
Σ_(n≥0) ^∞ e^(−n^2 ) ∼∫_0 ^∞ e^(−x^2 ) dx+lim_(x→0) (e^(−x^2 ) /2)+i∫_0 ^∞ ((e^(−(ix)^2 ) −e^(−(−ix)^2 ) )/(e^(2πx) −1))dx = ∫_0 ^∞ e^(−x^2 ) dx+(1/2)+0 ∼(((√π)+1)/2) Can we use this ? Abel plana formula Σ_(z≥0) ^∞ f(z)dz=∫_0 ^∞ f(z)dz+((f(0))/2)+i∫_0 ^∞ ((f(it)−f(−it))/(e^(2πt) −1))dt
$$\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } \sim\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}}+{i}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\left({ix}\right)^{\mathrm{2}} } −{e}^{−\left(−{ix}\right)^{\mathrm{2}} } }{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx} \\ $$$$=\:\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{0}\:\sim\frac{\sqrt{\pi}+\mathrm{1}}{\mathrm{2}} \\ $$$${Can}\:{we}\:{use}\:{this}\:? \\ $$$${Abel}\:{plana}\:{formula} \\ $$$$\underset{{z}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}{f}\left({z}\right){dz}=\int_{\mathrm{0}} ^{\infty} {f}\left({z}\right){dz}+\frac{{f}\left(\mathrm{0}\right)}{\mathrm{2}}+{i}\int_{\mathrm{0}} ^{\infty} \frac{{f}\left({it}\right)−{f}\left(−{it}\right)}{{e}^{\mathrm{2}\pi{t}} −\mathrm{1}}{dt} \\ $$
Commented by Dwaipayan Shikari last updated on 16/Dec/20
Thanking you
$${Thanking}\:{you} \\ $$

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