Register to reply

Expressing multi-variable functions

by Physics_wiz
Tags: expressing, functions, multivariable
Share this thread:
Physics_wiz
#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?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
jostpuur
#2
Jun3-08, 09:39 PM
P: 2,066
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
#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
#4
Jun3-08, 10:55 PM
P: 2,066
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
#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
#6
Jun4-08, 07:24 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,569
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
#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
#8
Jun4-08, 10:27 AM
P: 2,066
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
#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.


Register to reply

Related Discussions
Multi variable limits Calculus & Beyond Homework 11
Multi-Variable math help Calculus & Beyond Homework 9
Multi variable derivative Calculus & Beyond Homework 6
A multi-variable problem Set Theory, Logic, Probability, Statistics 6
Best way of evaluating limits of multi variable functions Calculus 41