Product of functions to express any function

In summary, the conversation discusses the representation of one-dimensional and two-dimensional functions using a set of functions, including the ability to represent a two-dimensional function as a linear superposition of weighted products. The possibility of representing a function as a sum of products is discussed, with some members mentioning specific theorems that apply in certain cases. The conversation also touches on the topic of separable solutions and what this implies about the function.
  • #1
fog37
1,569
108
Hello Forum,

Let's say we have a complete set of functions ##u_{i} (x)## that can be used to represent anyone 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
 
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  • #2
fog37 said:
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 realize that sin(x-ct) is a sum of such products no?
fog37 said:
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.
 
  • #3
fog37 said:
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.
 
  • #4
Orodruin said:
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 said:
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.
 
  • #5
Orodruin said:
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.
 
  • #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?
 

FAQ: Product of functions to express any function

What is a product of functions?

A product of functions is a mathematical expression that combines two or more functions using the multiplication operation. It is represented as f(x)g(x), where f(x) and g(x) are the two functions being multiplied together.

Why is it useful to express any function as a product of functions?

Expressing a function as a product of functions can make it easier to analyze and understand. It can also allow for the use of different mathematical techniques and strategies to solve problems involving the function.

How do you find the product of two functions?

To find the product of two functions, you multiply the two functions together. This can be done by multiplying corresponding terms or using the distributive property.

What types of functions can be expressed as a product of functions?

Any type of function, including polynomial, exponential, logarithmic, and trigonometric functions, can be expressed as a product of functions.

What are some real-world applications of expressing functions as a product of functions?

Expressing functions as a product of functions is commonly used in engineering, physics, and economics, among other fields. It can be used to model and analyze various real-world phenomena, such as population growth, interest rates, and chemical reactions.

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