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Expressing multivariable functions 
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#1
Jun308, 09:17 PM

P: 231

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
Jun308, 09:39 PM

P: 2,023

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 [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]. The g can then be defined with [tex] g(x_1,x_2,y) = f(x_1,x_2,u^{1}(y)). [/tex] You can make the question more difficult by assuming more about f and demanding g and u to satisfy some conditions. 


#3
Jun308, 10:37 PM

P: 231

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] 


#4
Jun308, 10:55 PM

P: 2,023

Expressing multivariable functions
[tex] g(x_1,x_2,y) = x_1 (p_1\circ u^{1})(y) + (p_2\circ u^{1})(y), [/tex] 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. 


#5
Jun408, 12:13 AM

P: 231

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)



#6
Jun408, 07:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338




#7
Jun408, 08:42 AM

P: 231

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? 


#8
Jun408, 10:27 AM

P: 2,023




#9
Jun408, 07:10 PM

P: 231




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