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Expressing multi-variable functions

  1. Jun 3, 2008 #1
    I have a simple question, let's say I have a function f = f(h(a,b), c, d). Can I express this as f = g(a, b, u(c,d))? Are the two expressions equivalent or is one different/more general than the other?
  2. jcsd
  3. Jun 3, 2008 #2
    Your question was not detailed enough that there could be clear answer, because you did not specify what kind of function u is supposed to be. Also... it seems that the h is misdirection there.

    The following claim should be true, perhaps it answers something:

    For arbitrary function [tex]f:\mathbb{R}^4\to\mathbb{R}[/tex], there exists functions [tex]g:\mathbb{R}^3\to\mathbb{R}[/tex] and [tex]u:\mathbb{R}^2\to\mathbb{R}[/tex], so that

    f(x_1,x_2,x_3,x_4) = g(x_1,x_2,u(x_3,x_4)),\quad\forall\; x_1,\ldots,x_4\in\mathbb{R}.

    The reason for this is that [tex]\mathbb{R}[/tex] and [tex]\mathbb{R}^2[/tex] have the same cardinality, so that there exists a bijection [tex]u:\mathbb{R}^2\to\mathbb{R}[/tex]. The g can then be defined with

    g(x_1,x_2,y) = f(x_1,x_2,u^{-1}(y)).

    You can make the question more difficult by assuming more about f and demanding g and u to satisfy some conditions.
  4. Jun 3, 2008 #3
    I don't think I agree with that.

    Let [tex] f(x_1,x_2,x_3,x_4)=x_1x_3+x_4[/tex]. Now, what function [tex]g(x_1,x_2,u(x_3,x_4))[/tex] is equal to [tex]f(x_1,x_2,x_3,x_4)[/tex]
  5. Jun 3, 2008 #4
    Do you know what cardinality and bijection mean?

    I don't know a nice formula that you could write into a calculator, but a function g defined by

    g(x_1,x_2,y) = x_1 (p_1\circ u^{-1})(y) + (p_2\circ u^{-1})(y),

    where [tex]u:\mathbb{R}^2\to\mathbb{R}[/tex] is some bijection, and [tex]p_1,p_2:\mathbb{R}^2\to\mathbb{R}[/tex] the projections, does the job.
  6. Jun 4, 2008 #5
    I read a little about cardinality...I think I understand. Can anyone direct me to where I can look to answer my original question? (i.e. what topics names I can look up in calculus or analysis)
  7. Jun 4, 2008 #6


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    In general, f(h(a,b), c, d) can be written as k(a,b,c, d) but cannot be written as g(a,b,u(c,d)) since the last assumes that c and d appear throughout f only in a specific form: u(c,d). k(a,b,c,d) is "more general" than either f(h(a,b),c,d) or g(a,b,u(c,d)) but it is impossible to say whether one of those two is "more general" than the other without specific h or u.
  8. Jun 4, 2008 #7
    Ok, here's where the original question came from...maybe this helps.

    Say I have a function [tex]F(a,b,c) = G(d,e)[/tex]. Assume the Implicit Function Theorem conditions are satisfied. So, I can solve for c as follows: [tex]c = H(G(d,e),a,b)[/tex]. Now, in this case, can I write c as [tex]c = N(d,e,M(a,b))[/tex]? Why or why not?
  9. Jun 4, 2008 #8
    Do you mean that this cannot be done when u has already been fixed, or something else has been assumed of the u, or is there a contradiction with my post?
  10. Jun 4, 2008 #9
    Anyone? Do the functions N and M exist so I can write [tex]c = N(d,e,M(a,b))[/tex]? By the way, N and M can be anything...I just wanted to know if they exist.
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