# Expressing multi-variable functions

1. Jun 3, 2008

### Physics_wiz

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. Jun 3, 2008

### jostpuur

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 $$f:\mathbb{R}^4\to\mathbb{R}$$, there exists functions $$g:\mathbb{R}^3\to\mathbb{R}$$ and $$u:\mathbb{R}^2\to\mathbb{R}$$, 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 $$\mathbb{R}$$ and $$\mathbb{R}^2$$ have the same cardinality, so that there exists a bijection $$u:\mathbb{R}^2\to\mathbb{R}$$. 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.

3. Jun 3, 2008

### Physics_wiz

I don't think I agree with that.

Let $$f(x_1,x_2,x_3,x_4)=x_1x_3+x_4$$. Now, what function $$g(x_1,x_2,u(x_3,x_4))$$ is equal to $$f(x_1,x_2,x_3,x_4)$$

4. Jun 3, 2008

### jostpuur

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 $$u:\mathbb{R}^2\to\mathbb{R}$$ is some bijection, and $$p_1,p_2:\mathbb{R}^2\to\mathbb{R}$$ the projections, does the job.

5. Jun 4, 2008

### Physics_wiz

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. Jun 4, 2008

### HallsofIvy

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.

7. Jun 4, 2008

### Physics_wiz

Ok, here's where the original question came from...maybe this helps.

Say I have a function $$F(a,b,c) = G(d,e)$$. Assume the Implicit Function Theorem conditions are satisfied. So, I can solve for c as follows: $$c = H(G(d,e),a,b)$$. Now, in this case, can I write c as $$c = N(d,e,M(a,b))$$? Why or why not?

8. Jun 4, 2008

### jostpuur

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?

9. Jun 4, 2008

### Physics_wiz

Anyone? Do the functions N and M exist so I can write $$c = N(d,e,M(a,b))$$? By the way, N and M can be anything...I just wanted to know if they exist.