User:DavidCBryant/Sandbox

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

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

Contents

[edit] Hack area

This user is interested in Chemistry.
Caltech This user is or was a student at the California Institute of Technology

[edit] Theta functions in terms of the nome

Instead of expressing the Theta functions in terms of z and τ, we may express them in terms of arguments w and the nome q, where w = exp(πiz) and q = exp(πiτ). 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.

[edit] Secondary Hack Area

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

A portion of the Padé table for the exponential function ez
0 1 2 3
0 \frac{1}{1} \frac{1}{1 - z} \frac{1}{1 - z + {\scriptstyle\frac{1}{2}}z^2} \frac{1}{1 - z + {\scriptstyle\frac{1}{2}}z^2 - {\scriptstyle\frac{1}{6}}z^3}
1 \frac{1 + z}{1} \frac{1 + {\scriptstyle\frac{1}{2}}z}{1 - {\scriptstyle\frac{1}{2}}z} \frac{1 + {\scriptstyle\frac{1}{3}}z} {1 - {\scriptstyle\frac{2}{3}}z + {\scriptstyle\frac{1}{6}}z^2} \frac{1 + {\scriptstyle\frac{1}{4}}z} {1 - {\scriptstyle\frac{3}{4}}z + {\scriptstyle\frac{1}{4}}z^2 - {\scriptstyle\frac{1}{24}}z^3}
2 \frac{1 + z + {\scriptstyle\frac{1}{2}}z^2}{1} \frac{1 + {\scriptstyle\frac{2}{3}}z + {\scriptstyle\frac{1}{6}}z^2} {1 - {\scriptstyle\frac{1}{3}}z} \frac{1 + {\scriptstyle\frac{1}{2}}z + {\scriptstyle\frac{1}{12}}z^2} {1 - {\scriptstyle\frac{1}{2}}z + {\scriptstyle\frac{1}{12}}z^2} \frac{1 + {\scriptstyle\frac{2}{5}}z + {\scriptstyle\frac{1}{20}}z^2} {1 - {\scriptstyle\frac{3}{5}}z + {\scriptstyle\frac{3}{20}}z^2 - {\scriptstyle\frac{1}{60}}z^3}
3 \frac{1 + z + {\scriptstyle\frac{1}{2}}z^2 + {\scriptstyle\frac{1}{6}}z^3}{1} \frac{1 + {\scriptstyle\frac{3}{4}}z + {\scriptstyle\frac{1}{4}}z^2 + {\scriptstyle\frac{1}{24}}z^3} {1 - {\scriptstyle\frac{1}{4}}z} \frac{1 + {\scriptstyle\frac{3}{5}}z + {\scriptstyle\frac{3}{20}}z^2 + {\scriptstyle\frac{1}{60}}z^3} {1 - {\scriptstyle\frac{2}{5}}z + {\scriptstyle\frac{1}{20}}z^2} \frac{1 + {\scriptstyle\frac{1}{2}}z + {\scriptstyle\frac{1}{10}}z^2 + {\scriptstyle\frac{1}{120}}z^3} {1 - {\scriptstyle\frac{1}{2}}z + {\scriptstyle\frac{1}{10}}z^2 - {\scriptstyle\frac{1}{120}}z^3}

[edit] Tertiary Hack Area

x ≈ 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}