Examine whether or not the number $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^{n^2}}$ is transcedental.
Solution
Yes , it is. In more detail we can express the series in terms of a Jacobi function, that is:
$$\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{2^{n^2}} &=\frac{1}{2}\left [ \vartheta_3 \left ( \frac{1}{2} \right ) -1\right ] \\&= - \frac{1}{2} + \frac{1}{2}\prod_{n=1}^{\infty}\left ( 1- \frac{1}{2^{2n}} \right )\left ( 1+ \frac{1}{2^{2n-1}} \right )^2\end{aligned}$$
whereas
$$\vartheta _3 (z, q)= \sum_{n=-\infty}^{\infty}q^{n^2}e^{2\pi iz}$$
is one of the Jacobi theta functions. For $z=0$ we have that $\vartheta_3(q) = \sum \limits_{n=-\infty}^{\infty}q^{n^2}$ and we know from theory of theta functions that for $|q|<1$ the number $\vartheta_3(q)$ is transcedental.
Hence the result follows.
Solution
Yes , it is. In more detail we can express the series in terms of a Jacobi function, that is:
$$\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{2^{n^2}} &=\frac{1}{2}\left [ \vartheta_3 \left ( \frac{1}{2} \right ) -1\right ] \\&= - \frac{1}{2} + \frac{1}{2}\prod_{n=1}^{\infty}\left ( 1- \frac{1}{2^{2n}} \right )\left ( 1+ \frac{1}{2^{2n-1}} \right )^2\end{aligned}$$
whereas
$$\vartheta _3 (z, q)= \sum_{n=-\infty}^{\infty}q^{n^2}e^{2\pi iz}$$
is one of the Jacobi theta functions. For $z=0$ we have that $\vartheta_3(q) = \sum \limits_{n=-\infty}^{\infty}q^{n^2}$ and we know from theory of theta functions that for $|q|<1$ the number $\vartheta_3(q)$ is transcedental.
Hence the result follows.
No comments:
Post a Comment