Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

$${Thanking}\:{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com