Beginner exercise on the Implicit function theorem

Question

My first exercise on the Implicit function theorem:

Show that the non-linear equation

$x^4 + e^y + sin(z) + z^5 = 1$

has a "local resolution function" $(x,y) \rightarrow z(x,y)$ (is this the right term?) at the point $(x',y',z') := (0,0,0)$. Furthermore calculate the "first order approximate function" of $z(.,.)$ at the point $(0,0)$

What I did so far

1. Put the equation into a function: $f(x,y,z) = x^4+e^y+sin(z)+z^5-1$
2. Test if $f(x',y',z') = 0$

$f(0,0,0) = 0^4+e^0+sin(0)+0^5 -1$

$= 0 + 1 + 0 + 0 -1 = 0$

Therefore this is solveable (?)

3. To show if the "local resolution function" exists, $Dzf(x,y,z)$ must be invertible for $(x', y', z')$

$Df(x,y,z) = \begin{bmatrix} 4x^3 & e^y & cos(z)+5z^4 \end{bmatrix}$

$Df(0,0,0) = \begin{bmatrix} 0 & 1 & 1 \end{bmatrix}$

$Dzf(x,y,z) = \begin{bmatrix} cos(z)+5z^4 \end{bmatrix}$

$Dzf(x', y', z') = Dzf(0,0,0) = \begin{bmatrix} cos(0)+5*0^4 \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix}$

$\det(Dzf(0,0,0)) = 1 \neq 0$

Therefore $Dzf$ is invertible and thus the "local resolution function" exists.

4. "first order approximate function":

$\vec{x} = (x,y)$

$y = z$

$g(\vec{x}) = y' + Dy(\vec{x'})*(\vec{x}-\vec{x'})$

$Dy(\vec{x}) = -Dyf(\vec{x}, \vec{y})^{-1} * D\vec{x}f(\vec{x}, \vec{y})$

$=> g(\vec{x}) = y' - Dyf(\vec{x}, \vec{y})^{-1} * D\vec{x}f(\vec{x}, \vec{y})*(\vec{x}-\vec{x'})$

$= 0 - \begin{bmatrix} 1 \end{bmatrix} * \begin{bmatrix} 0 & 1 \end{bmatrix} * (\vec{x} - (0, 0))$

$= -\begin{bmatrix} 0 & 1 \end{bmatrix} * \vec{x}$

$= -y$

since $\vec{x} = (x, y)$

I have the feeling I'm doing this horribly wrong, and I lack the knowledge to prove if that makes any sense.

Also Wolfram Alpha or similiar Questions seem to do something totally different on this kind of exercise, I wonder whether this is just another kind of task, or this is another way of doing it?

Answer 1

I'm going to answer by guesswork on what the German might be, generally agreeing with @Git Gud. 0. Your "that shows this is solvable" really should be "that shows that $(0,0,0)$ really is a point on the level-surface.

1. "Local resolution" probably means "there's a function z = f(x, y)" such that

$$ x^4 + e^y + sin( f(x, y) ) + f(x, y)^5 = 1 $$ for all $(x, y)$ near $(0, 0)$, and with $f(0, 0) = 0$. If the selected point had been, say, $(2, 3, z_0)$, then I'd have to show $f$ is defined near $(2, 3)$, and $f(2, 3) = z_0$, as well as the equation being satisfied. All those zeros in the selected point actually make it hard to see what's going on!

The way to show this is to look at the function $$ G(x, y, z) = x^4 + e^y + sin(z) + z^5 - 1 $$ at the chosen point, and evaluate its gradient there: $$ \nabla g(x, y, z) = (4x^3, e^y, cos(z) + 5z^4) \\ \nabla g(0, 0, 0) = (0, 1, 1) $$ Because this gradient is nonzero, the level-surface for $G$ near $(0,0,0)$ is a smooth surface (because of the inverse or implicit function theorem). Because the gradient at $(0,0,0)$ has a nonzero $z$-component, the projection of the level surface onto the $z = 0$ plane is locally a one-to-one map onto its image. That guarantees the existence of the function $f$ (locally).

Your steps ("3") are exactly these, and look great to me.

2. Local approximating function. Looking at the gradient, you can see that to remain on the level surface, a change in $y$ must be cancelled by an equal change in $z$, while a change in $x$ has no effect on $z$. (all this being "infinitesimally" or "up to first order"). So the local approximation of $G$ is

$$ \hat{G}(x, y, z) = y - z $$ The local approximation of $f$ (if that's what the question is looking for) is $$ \hat{f}(x, y) = -y. $$

In short: everything you did looks just fine. Nice work!