# I Product of functions to express any function

#### 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

#### Orodruin

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I know that in Fourier theory a traveling field f(x,t)f(x,t)f(x,t) can be expressed as a weighted sum of traveling plane waves which are not product function of time and space....
You realise that sin(x-ct) is a sum of such products no?
Is that correct? How do we call this process and when it is feasible
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.

#### Stephen Tashi

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?
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.

#### Orodruin

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

#### Stephen Tashi

To reconcile these apparently contradictory replies, I implicitly assumed that "any" referred to "any function in the function space spanned by the given functions".
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.

#### 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?