Converting an implicit 3-variable equation into explicit function

Question

I have the equation: $$xz(y+1)+e^y=1$$ and I would like to be able to write $y$ as a function of $x$ and $z$ i.e. $y=f(x,z)$. How would I go about this? I think this might have something to do with partial derivatives. Eventually, I want to determine:

$$\lim_{(x,z)\to (0,0)}\frac{f(x,z)}{x^2+z^2}$$

Alternatively, how do I compute this limit if I can't find a formula for $y=f(x,z)$ ?

Answer 1

Using implicit derivation you can compute the Taylor aproximation of $y$ around $(0,0)$.

$y(0,0)=0$,

$$ y_x=\frac{-z(y+1)}{e^y+xz} \;\; y_z=\frac{-x(y+1)}{e^y+xz} $$ So $y_x(0,0)=0$, $y_z(0,0)=0$.

You can also compute the second derivatives (Verify, I did not do the calculations carefully): $$ y_{xx}(0,0)=0 \;\; y_{xz}(0,0)=-1\;\; y_{zz}(0,0)=0 $$ Therefore $$ y(x,z)=-\frac{zx}{2}+ O(\left(\sqrt{x^2+z^2}\right)^3) $$ and then $$ \lim_{(x,z)\rightarrow (0,0)} \frac{y(x,z)}{x^2+z^2} \mbox{ does not exist.} $$.

Answer 2

If $xz \to 0^-$ (i.e. $xz$ is always negative approaching $0$), then you can find a large negative solution for $y$ such that $xz(y+1)$ is approximately $1$ (slightly less) and $e^y$ is almost $0$ (slightly more). Then your limit of $f(x,z)/(x^2 + z^2)$ for this solution for $y = f(x,z)$ would be $- \infty$ as $(x,z) \to 0$ along trajectories such that $xz \to 0^-$. Of course, this solution for $xz < 0$ cannot be extended to a continuous or smooth solution $f(x,z)$ because you must have $f(0,0) = 0$. So maybe you can avoid this gotcha by insisting that $f$ is continuous/differentiable, and then $f$ could potentially be uniquely defined to be a nice solution for which a limit exists.

Answer 3

The point $(0,0,0)$ satisfies the given equation $$F(x,y,z):=xz(y+1)+e^y-1=0,\ \tag{1}$$ and I take it that you are interested in the function $$f:\ (x,z)\mapsto y=f(x,z)$$ defined in a neighborhood of $(0,0)$, satisfying $f(0,0)=0$ and $F\bigl(x,f(x,z),z\bigr)\equiv0$.

For the moment we put $xz=:u$; then $(1)$ will be replaced by $$G(u,y):=u(y+1)+e^y-1=0\ .\tag{2}$$ Now $(0,0)$ is a solution of $(2)$; furthermore $$G_u(0,0)=1,\quad G_y(0,0)=1\ne0\ .$$ It follows that $(2)$ defines in the neighborhood of $(0,0)$ a $C^1$-function $y=\phi(u)$ with $\phi(0)=0$ and $$\phi'(0)=-{G_u(0,0)\over G_y(0,0)}=-1\ .$$ Therefore $$y=-u+o(|u|)\qquad(u\to0)\ ,$$ and returning to $x$ and $z$ we have $$f(x,z)=-xz+o(|xz|)\qquad\bigl((x,z)\to(0,0)\bigr)\ .$$ As $|xz|\leq {1\over2}(x^2+z^2)$ it follows that $${f(x,z)\over x^2+z^2}=-{x z\over x^2+z^2}+o(1)\qquad\bigl((x,z)\to(0,0)\bigr)\ .$$ From this we conclude that the limit in question does not exist: Approaching $(0,0)$ horizontally we get the limit $0$, and approaching $(0,0)$ from north-west we get the limit $-{1\over2}$.