Is it theoretically possible to find g(x) from this equation?

[Zomboy]

Is it theoretically possible to find "g(x)" from this equation?

So through my workings on this question I came up with this equation:

(x^2)yz - (y^2)(x^2) - x + g(x) = g(y,z)

* where g(x) is some function of x and g(y,z) is some function of y and z

I'd like to derive g(x) from this although I get the feeling that it simply can't be done. Can say a function of (x,y,z) + a function of (x) ever give you a function of (y,z)? Why can't you do this?

 

[micromass]

Your notation is bad. When you write g(x), it implies that is a function in one variable. But g(y,z) is a function of two variables. This is impossible.

 

[HallsofIvy]

Are you saying the g(x) and g(y,z) are different functions (in which case, you shouldn't use the same letter!) or that g(y,z) is "the same function as g(x) except using two variables" (which is meaningless).

 

[Mark44]

Furthermore, (x^2)yz - (y^2)(x^2) - x is a function of three variables, x, y, and z.

 

[Norwegian]

Notation is bad, but we should be able to interpret g(x) and g(y,z) as two different functions, the point being that the first depends only on x, while the latter depends on y and z.

Such a g(x) does btw not exist in your particular case. For example, y=z=0 gives g(x)=x+C, and y=1,z=0, gives g(x)=x²+x+D, where C and D are some constants.

Can say a function of (x,y,z) + a function of (x) ever give you a function of (y,z)?

Yes, trivially, for example (x+y+z) + (-x) = (y+z).

 

[I like Serena]

Well, if you meant that g(x) and g(y,z) are different functions, then you would get:
(x^2)yz - (y^2)(x^2) - x + g(x) = h(y,z)
Its partial derivative with respect to x is:
2xyz-2xy^2-1+g'(x)=0

Since this is an expression with y and z in it, this means that there is no such g'(x) that depends only on x.

So we have to assume that you actually meant g(x,y,z) with depends on all of them.
In that case, you get:
$$2xyz-2xy^2-1+{\partial g \over \partial x}(x,y,z)={\partial g \over \partial x}(x,y,z)$$

But then, this would only be true if $2xyz-2xy^2-1=0$.

So the question becomes: what did you actually mean?
The problem that you state has no sensible solution however it was meant.