Expressing multi-variable functions


by Physics_wiz
Tags: expressing, functions, multivariable
Physics_wiz
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#1
Jun3-08, 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?
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jostpuur
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#2
Jun3-08, 09:39 PM
P: 1,988
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.
Physics_wiz
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#3
Jun3-08, 10:37 PM
P: 231
Quote Quote by jostpuur View Post
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]

jostpuur
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#4
Jun3-08, 10:55 PM
P: 1,988

Expressing multi-variable functions


Quote Quote by Physics_wiz View Post
I don't think I agree with that.
Do you know what cardinality and bijection mean?

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

[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.
Physics_wiz
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#5
Jun4-08, 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)
HallsofIvy
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#6
Jun4-08, 07:24 AM
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Quote Quote by Physics_wiz View Post
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). 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.
Physics_wiz
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#7
Jun4-08, 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?
jostpuur
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#8
Jun4-08, 10:27 AM
P: 1,988
Quote Quote by HallsofIvy View Post
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).
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?
Physics_wiz
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#9
Jun4-08, 07:10 PM
P: 231
Quote Quote by Physics_wiz View Post
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?
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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