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I Product of functions to express any function

  1. Jan 22, 2017 #1
    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. jcsd
  3. Jan 23, 2017 #2

    Orodruin

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    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.
     
  4. Jan 23, 2017 #3

    Stephen Tashi

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    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.
     
  5. Jan 23, 2017 #4

    Orodruin

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    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.
     
  6. Jan 23, 2017 #5

    Stephen Tashi

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    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.
     
  7. Jan 23, 2017 #6
    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?
     
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