I Product of functions to express any function

1. Jan 22, 2017

fog37

Hello Forum,

Let's say we have a complete set of functions $u_{i} (x)$ that can be used to represent any one dimensional function $f(x)$. We then find another and different set $v_{i} (x)$ that can do the same thing, i.e. represent any function $f(x)$ via a linear superposition.

I believe that any two-dimensional function $g(x,y)$ can then be represented as a linear superposition of weighted products $u_{i} (x) v_{j} (y)$: $$g(x,y=\Sigma a_{ij} u_{i} (x) v_{j} (y)$$

Is that correct? How do we call this process and when it is feasible?

I know that in Fourier theory a traveling field $f(x,t)$ can be expressed as a weighted sum of traveling plane waves which are not product function of time and space....

Thanks

2. Jan 23, 2017

Orodruin

Staff Emeritus
You realise that sin(x-ct) is a sum of such products no?
yes you can do that, it is just an expansion of a function in terms of a basis for the function space - not any different from expressing any vector in terms of a given basis.

3. Jan 23, 2017

Stephen Tashi

No. - In the sense that I know of no such mathematical result for arbitrary functions.

For continuous functions, there is the Kolmogorov-Arnold representation theorem https://en.wikipedia.org/wiki/Kolmogorov–Arnold_representation_theorem.

Of course, for a two variable function g(x,y) that can be expanded in a (convergent) two variable Mclaurin series, you get the result that $g(x,y) = \sum_{i,j} a_{i,j} u_i(x) v_j(y)$ where each$u_i(x)$ has the form $x^{N_i}$ and each $v_j (x)$ has the form $y^{N_j}$.

Perhaps some forum member can tell us a very general theorem that applies to your question if we restrict the set of functions to be infinitely smooth functions instead of literally "all" functions.

4. Jan 23, 2017

Orodruin

Staff Emeritus
To reconcile these apparently contradictory replies, I implicitly assumed that "any" referred to "any function in the function space spanned by the given functions". Of course, if taken literally, it is not a true statement.

5. Jan 23, 2017

Stephen Tashi

Further, we can contrast these two assumptions:
1) Assume that the set of products $u_i(x)v_j(y)$ is a basis for the set of functions containing $g(x,y)$.
or
2) Only assume that $u_i(x)$ and $v_j(y)$ are each a basis for some restricted set of functions of one variable (such as the infinitely differentiable functions in one variable) and that $g(x,y)$ is a member of some restricted set of functions in two variables (such as the set of infinitely differentiable functions in two variables).

Assumption 1 is , by definition, a "yes" answer to the question.

Assumption 2 is a not a "yes" answer by definition. It is plausible that for particular types of one-variable and two-variable functions the answer "yes" can be proven as a theorem. However, I don't know any theorems that state such general results. The case of a McLaurin expansion could probably be generalized to a case where $u_i$ and $v_i$ are polynomials.

6. Jan 23, 2017

fog37

Thanks everyone.

For example, let's consider the elastic drum problem and its modes of oscillations when fixed at its boundaries. We start by considering separable solutions of the form $g(x,y,t)= p(t) f(x) v(z)$....
What does the ability to make the solution function separable implicitly imply about the function?