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

$$\frac{\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-n^2}}{\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-2n^2}}$$

Answered by Olaf last updated on 16/Dec/20

$$\mathrm{Jacobi}\mathrm{Theta}\mathrm{function}:$$$$\vartheta (z,\tau )=\underset{n=-\mathrm{\infty }}{\sum ^{n=+\mathrm{\infty }}}\exp (\pi {in}^2\tau +2\pi inz)$$$$\vartheta (0,\frac{i}{\pi })=\underset{n=-\mathrm{\infty }}{\sum ^{n=+\mathrm{\infty }}}{e}^{-n^2}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-n^2}-1\left(1\right)$$$$\vartheta (0,\frac{2i}{\pi })=\underset{n=-\mathrm{\infty }}{\sum ^{n=+\mathrm{\infty }}}{e}^{-2n^2}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-2n^2}-1\left(2\right)$$$$\left(1\right)\mathrm{and}\left(2\right):\frac{\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-n^2}}{\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-2n^2}}=\frac{\vartheta (0,\frac{i}{\pi })+1}{\vartheta (0,\frac{2i}{\pi })+1}$$

Commented by Dwaipayan Shikari last updated on 16/Dec/20

$$Thankingyou!Butisthereanyotherwaytoapproximate?$$

Commented by Olaf last updated on 16/Dec/20

$$\mathrm{There}\mathrm{is}\mathrm{no}\mathrm{asymptotic}\mathrm{formula}$$$$\mathrm{but}\mathrm{a}\mathrm{computation}\mathrm{is}\mathrm{possible}.$$$$\mathrm{See}\mathrm{here}:$$$$\mathrm{https}://\mathrm{hlabrande}.\mathrm{fr}/\mathrm{maths}/\mathrm{pubs}/\mathrm{theta}.\mathrm{pdf}$$

Commented by Dwaipayan Shikari last updated on 16/Dec/20

$$\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{e}^{-n^2}\sim {\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx+\underset{x\to 0}{\lim }\frac{e^{-x^2}}{2}+i{\int }_0^{\mathrm{\infty }}\frac{e^{-{\left(ix\right)}^2}-e^{-{(-ix)}^2}}{e^{2\pi x}-1}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx+\frac{1}{2}+0\sim \frac{\sqrt{\pi }+1}{2}$$$$Canweusethis?$$$$Abelplanaformula$$$$\underset{z⩾0}{\sum ^{\mathrm{\infty }}}f\left(z\right)dz={\int }_0^{\mathrm{\infty }}f\left(z\right)dz+\frac{f\left(0\right)}{2}+i{\int }_0^{\mathrm{\infty }}\frac{f\left(it\right)-f(-it)}{e^{2\pi t}-1}dt$$

Commented by Dwaipayan Shikari last updated on 16/Dec/20

$$Thankingyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com