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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?
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
[tex]
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}.
[/tex]
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].
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]
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?
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).
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?