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Functional independence

  1. May 16, 2005 #1
    Given a set of m real functions of n variables, what is a necessary and sufficient condition for the functions to be functionally independent ?

    A set a functions [tex] f_i(x_1,...x_n)\quad i=1,...m[/tex] are functionally independent, if the only function [tex]\phi(u_1,...u_m)[/tex] such that [tex]\phi(f_1,....f_m)=0[/tex] is [tex]\phi=0[/tex].

    For example if [tex] f(x,y,z)=x^3+y^2+z\quad g(x,y,z)=z^2+y\quad h(x,y,z)=z^4-x^3-2z^2y-z[/tex]

    Then clearly f,g,h are lin. indep....

    But they are functionally dependent, in the sense that [tex] h(x,y,z)=g(x,y,z)^2-f(x,y,z) [/tex].

    One of my problem is that [tex]\phi[/tex] is acting on functions f_i into R (functional), on f_i into functions of (x_1,...x_n) (operator), or on R^m f_i(x1...xn) ???

    Because if [tex]\phi[/tex] is a functional, then it suffices for example that a function is some Fourier transform, derivative or iteration of the other functions, instead of just operations on real numbers....
    Last edited: May 16, 2005
  2. jcsd
  3. May 16, 2005 #2

    matt grime

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    I am not sure that there will be a necessary and sufficient condition on the functions (other than such a phi exists), at least not in terms of the m and n.

    I'm a little confused as to what you mean by functional - phi isn't a 'linear' functional, which is something defined on Banach spaces. Should I just take it to mean a function of a set of functions? Are we allowed to use infinite sums - ie is this an algebraic or analytic dependence?

    Obviously if the f_i are polynomials of the x_j then we can get some idea then from the degrees of the polys. If we allow functions such as sin and cos then we are in to all manner of difficulties.
  4. May 16, 2005 #3
    Well I'll take an example : [tex]f(x,y)=x^2+y^2, g(x,y)=x^4+(2x^2+1)y^2+y^4[/tex]

    Then clearly, if [tex]\phi(x,y)[/tex] is a function R^2->R, then [tex]\phi(f(x,y),g(x,y))=0\Rightarrow\phi=0[/tex]...Hence f,g are functionally independent.

    However, if [tex]\phi[/tex] is a functional : [tex](C(R^2,R),C(R^2,R))->R [/tex] then I can define [tex]\phi(f,g)(x,y)=(f\circ p\circ f-g)(x,y) [/tex].

    with [tex] p: R->R^2, p(x)=(x,\cdot) [/tex] something a bit weird I should look....

    Then f,g are functionally dependent, because

    [tex] (f\circ p\circ f)(x,y)=f(f(x,y),y)=f(x^2+y^2,y)=(x^2+y^2)^2+y^2=x^4+(2x^2+1)y^2+y^4=g(x,y)[/tex]
  5. May 16, 2005 #4


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    your definition is probably too strict. for example if your functions f1,.,,,fm are not surjective onto R^m, then it is easy to find a non zero function that maps their image to zero, hence almost any functions are dependent in your definition.

    you might want something more like assuming your functions define an open map. your examples suggest you just do not want them to map entirely into any algebraic hypersurface.

    so you might ask that the induced substitution map from polynomials g(X1,...,Xm) of m variables, to functions with the same domain as the f's. taking (f1,...,fm) to g(f1,..,fm), is injective, i.e. sends only g=0 to 0.

    i.e. you need to restrict the class of functions you substitute into to get a reasonable notion.

    then once you do this, if you also restrict the f's to be say polynomials as well, then I would say the answer is they are (algebraically) independent if and only if they define independent transcendentals over R.

    Myself I would ask for simpler sufficient criteria suitable to the job at hand, like maybe they have injective derivative somewhere.
    Last edited: May 16, 2005
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