Changing variables algebraically

Question

Suppose one has two functions, $y(x)$ and $z(x)$, and one seeks to obtain $y(z)$ by substituting $x(z)$ into $y(x)$. Can this be done in a single step? Or must $z(x)$ first be inverted independently? For the sake of illustration, suppose the functions consist of transcendental functions combined by the elementary operations of addition, subtraction, multiplication, division, and exponentiation. For instance,

$y(x) = a\ln x + \displaystyle\frac{\exp(bx)}{c+x}$

$z(x) = \ln x \cdot \left(1 - \cos x \exp(a^2x) \right)^{r}$

Answer 1

To visualize y[z] without having to go through inversion, you can use ParametricPlot:

 Manipulate[
 ParametricPlot[{y[x, a, b, c], z[x, a, r]}, {x, 0, 1}, 
 AxesLabel -> {z, y}, AspectRatio -> 1, 
 PlotRange -> {{-10, 10}, Automatic}], 
 {{a, 1}, 0, 5, .1}, {{b, 1},  0, 5, .1}, {{c, 1}, 0, 5, .1}, 
 Delimiter, {{r, 1}, .5, 2, .1}, 
 ControlPlacement -> Left]

Answer 2

Well, I guess you could try using InverseFunction. However, this will only give explicit algebraic expressions when the function that is to be inverted is fairly simple.

Your example functions are:

y[x_, a_, b_, c_] := a Log[x] + Exp[b x]/(c + x)
z[x_, a_, r_] := Log[x] (1 - Cos[x] Exp[a^2 x])^r

And you can evaluate and plot $y(x(z))$ where $z \mapsto x(z)$ is the inverse function of $x \mapsto z(x)$

y[InverseFunction[z[#, a, r]&][x], a, b, c]

(* a Log[InverseFunction[z[#1, a, r] &][x]] 
   + E^(b InverseFunction[z[#1, a, r]&][x])/
  (c + InverseFunction[z[#1, a, r] &][x]) *)

Plot[Evaluate[% /. {a -> 1, b -> 1, c -> 1, r -> 1}], {x, 1, 2}]

Note that InverseFunction does not always behave like you think it might.

Answer 3

I am aware of two functions that can eliminate variable algebraically: Eliminate and Solve. They are described in this guide:

• Eliminating Variables

These functions work with polynomial equations (or equations that can be reduced to polynomials in some way).

Reduce is more generic, but it doesn't give any means of eliminating $x$ without solving for it first. The syntax to use would be

Reduce[y == f[x] && z == g[x], {y, x}]

It can be guided by giving additional assumption about the variables in the original set of equations/inequalities (for example $x \in \mathbb{R}$ or $x > 0$) or specifying a domain.

That said, a more convenient way to verify the $y = f(z)$ solution you get is to just substitute the original expressions for $y$ and $z$ in terms of $x$ into $y = f(z)$ and verify that the relation hold.