User:DavidCBryant/Sandbox

Now I have my own private sandbox! DavidCBryant 16:27, 23 November 2006 (UTC)

Adding a temporary page User:DavidCBryant/Generalized continued fraction.

Theta functions in terms of the nome
Instead of expressing the Theta functions in terms of z and &tau;, we may express them in terms of arguments w and the nome q, where w = exp(&pi;iz) and q = exp(&pi;i&tau;). In this form, the functions become



\begin{align} \vartheta(w; q)& = \sum_{n=-\infty}^\infty (w^2)^n q^{n^2}\quad& \vartheta_{01}(w; q)& = \sum_{n=-\infty}^\infty (-1)^n (w^2)^n q^{n^2}\\ \vartheta_{10}(w; q)& = \sum_{n=-\infty}^\infty (w^2)^{\left(n+\frac{1}{2}\right)} q^{\left(n + \frac{1}{2}\right)^2}\quad& \vartheta_{11}(w; q)& = i \sum_{n=-\infty}^\infty (-1)^n (w^2)^{\left(n+\frac{1}{2}\right)} q^{\left(n + \frac{1}{2}\right)^2} \end{align} $$

So we see that the Theta functions can also be defined in terms of w and q, without reference to the exponential function. These formulas can, therefore, be used to define the Theta functions over other fields where the exponential function might not be everywhere defined, such as fields of p-adic numbers.

Secondary Hack Area
I'm trying to set up a nice-looking Padé table for the exponential function here.

Tertiary Hack Area

 * x &asymp; y.



\begin{align} h_0& = b_0& k_0& = 1\\ h_1& = b_1 b_0 + a_1& k_1& = b_1\\ h_{i+1}& = b_{i+1} h_i + a_{i+1} h_{i-1}& k_{i+1}& = b_{i+1} k_i + a_{i+1} k_{i-1}\, \end{align} $$



f(z) = \sum_{n=1}^\infty \left(z^2 + n\right)^{-2}.\, $$


 * $$x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots\,}}}} $$


 * $$ z = \sqrt{a}\,x + i \sqrt{b} \,y . $$


 * $$ \int_a^b x^2\,dx $$


 * $$\varphi =

\begin{cases} \arctan(\frac{y}{x}) & \mbox{if } x > 0\\ \arctan(\frac{y}{x}) + \pi & \mbox{if } x < 0 \mbox{ and } y \ge 0\\ \arctan(\frac{y}{x}) - \pi & \mbox{if } x < 0 \mbox{ and } y < 0\\ +\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y > 0\\ -\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y < 0\\ 0 & \mbox{if } x = 0 \mbox{ and } y = 0. \end{cases}$$