# Expressing multi-variable functions

• Physics_wiz
In summary: 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).
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?

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.

jostpuur said:
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 $$u:\mathbb{R}^2\to\mathbb{R}$$.

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 $$f(x_1,x_2,x_3,x_4)$$

Physics_wiz said:
I don't think I agree with that.

Do you know what cardinality and bijection mean?

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 $$f(x_1,x_2,x_3,x_4)$$

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.

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)

Physics_wiz said:
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.

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?

HallsofIvy said:
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 said:
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?

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.

## 1. What is a multi-variable function?

A multi-variable function is a mathematical concept that involves more than one independent variable. In other words, the output of the function is dependent on multiple input variables, rather than just one.

## 2. How is a multi-variable function expressed?

A multi-variable function is typically expressed using mathematical notation, with the input variables listed inside parentheses and the output variable after an equal sign. For example, the function f(x,y) = x + y would take two input variables (x and y) and output their sum.

## 3. What is the purpose of expressing a multi-variable function?

Expressing a multi-variable function allows us to understand and analyze the relationship between the input variables and the output variable. It also allows us to make predictions and solve problems involving multiple variables.

## 4. Are all multi-variable functions linear?

No, not all multi-variable functions are linear. A linear function is one in which the output variable changes at a constant rate in relation to the input variables. However, multi-variable functions can be non-linear, meaning the output variable changes at a non-constant rate in relation to the input variables.

## 5. Can a multi-variable function have more than two input variables?

Yes, a multi-variable function can have any number of input variables. The number of input variables is determined by the specific problem or situation being analyzed. For example, a function that calculates the volume of a box would have three input variables (length, width, and height).

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