Can we express all doubly periodic functions as one of doubly periodic function?

Singly Periodic Functions $e^{x},\cos(x),\sin(x),\tan(x), .. etc.$

Euler's identity is $$e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)$$

$$e^{-i\alpha}=\cos(\alpha)-i\sin(\alpha)$$

Thus, we can express all trigonometric functions via $e^{x}$ function

$$\cos(\alpha)=\frac{e^{i\alpha}+e^{-i\alpha}}{2}$$ $$\sin(\alpha)=\frac{e^{i\alpha}-e^{-i\alpha}}{2i}$$

$$\tan(\alpha)=\frac{e^{i\alpha}-e^{-i\alpha}}{i(e^{i\alpha}+e^{-i\alpha})}$$ $$\cot(\alpha)=\frac{i(e^{i\alpha}+e^{-i\alpha})}{e^{i\alpha}-e^{-i\alpha}}$$ $$.$$ $$.$$

And also we know that periodic functions or periodic signals into the sum of a (possibly infinite) set of simple oscillating functions are can be expressed via sines and cosines (or complex exponentials). wiki reference

$$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \, [a_n \cos(\tfrac{2\pi nx}{P}) + b_n \sin(\tfrac{2\pi nx}{P})]\\ = \sum_{n=-\infty}^\infty c_n\cdot e^{i \tfrac{2\pi nx}{P}}$$

where $f(x)$ is any periodic function and $P$ is period.

Elliptic functions are doubly periodic functions

The elliptic functions are inversions of the elliptic integrals. The two standard forms of these functions are known as Jacobi elliptic functions and Weierstrass elliptic functions.

Jacobi elliptic functions arise as solutions to differential equations of the form

$$y''(x)=A+By(x)+Cy^2(x)+Dy^3(x)$$

and Weierstrass elliptic functions arise as solutions to differential equations of the form

$$y''(x)=A+By(x)+Cy^2(x)$$

My questions:

Can we express all doubly periodic functions as one of doubly periodic function similarly as we do in singly periodic functions?

Is there any research on such base doubly periodic function that can be used to express all type of doubly periodic functions or signals in series? Please share references If you know any.

In other words,For example:

$y(x)=e^{x}$ is a solution of $$y'(x)=y(x)$$ and $e^{x}$ has been used as base function to express all singly Periodic Functions.

Maybe we can define a $f{(x)}$ as the solution of $$y''(x)=1+y(x)+y^2(x)+y^3(x)$$ .$f{(x)}$ can be used to express all of other doubly periodic functions . I do not know if it is possible or not?

• "$e^x$ has been used as base function to express all singly Periodic Functions". Hmmm. You might have in mind the function $x\mapsto e^{ix}$. – Did Sep 19 '13 at 7:58
• @Did Yes It is. – Mathlover Sep 19 '13 at 7:59
• For a given set of double periods, the collection of all meromorphic functions with that set of double periods is known to form a field. The field is generated by the corresponding Weierstrass's elliptic function $\wp$ and its derivatives $\wp'$. i.e. every double periodic meromorphic functions is a rational function of some $\wp$ and $\wp'$. – achille hui Sep 19 '13 at 10:27
• @achillehui You mean we can select a base function ($\wp, \wp'$) in elliptic functions? I saw in the link that "It can be shown that this field is $\Bbb{C}(\wp, \wp')$, so that all such functions are rational functions in the Weierstrass function and its derivative". Do you know how can I prove it? thanks – Mathlover Sep 19 '13 at 11:37
• Look at the references in the wiki page. In particular, Serge Lang's book "Elliptic functions" (GTM 112) has a proof of that in first chapter. – achille hui Sep 19 '13 at 11:42

The elliptic functions are doubly periodic indeed, but they cannot serve in the analysis of 2D-signals. First they are analytic (resp. meromorphic), which implies that they satisfy the CR-equations and so cannot be used to represent an arbitrary doubly periodic "picture", e.g., the function $(x,y)\mapsto\sin^2(2\pi x-6\pi y)$. Second they necessarily have poles.
For 2D signal analysis you can use double Fourier series instead. The basis functions are suitably normalized functions $$e_{jk}(x,y):=\exp\bigl(2\pi i(jx+ky)\bigr)\ ,$$ and you obtain the Fourier coefficients of an arbitrary doubly periodic function $$\ f:\ {\mathbb R}^2 /{\mathbb Z}^2\to{\mathbb C}$$ by integration, as in the one-dimensional case.