I am having problems with the following exercise:


Let $\mathbf{h}: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ given by

$$ \mathbf{h}(x,y,z) = \begin{pmatrix} x^2 + (z-1)^2 -5 + e^{y-2} \\ ln\left(\frac{xz}{2}\right) \end{pmatrix} $$

(i) Show that $\mathbf{h}(2,2,1)=0$ and that $\mathbf{h} \in C^1(\mathbb{R}^2)$.

I have done this. I showed that the partial derivatives exist and that they are continous thus $\mathbf{h} \in C^1(\mathbb{R}^2)$.

I think I did the next exercise correct, but if I did something wrong please tell me to check it again or give a hint.

(ii) Show that one can apply the implicit function theorem in order to obtain some small enough $\epsilon >0 $ and a $C^1$ function $\mathbf{f}: (1-\epsilon, 1+\epsilon) \rightarrow \mathbb{R}^2$ such that

$$\mathbf{h}(\mathbf{f}(z), z) = (0,0), ~~~~\forall z \in (1-\epsilon, 1+\epsilon).$$

I have that $\mathbf{h}(2,2,1)=0$ so I need to have $\mathbf{f}(z)=(2,2)$. From the previous exercise I found out that the partialderivatives are

$$[D\mathbf{h}] = \begin{pmatrix} 2x & e^{y-2} & 2(z-1) \\ 1/x & 0 & 1/z \end{pmatrix}$$

and at $(2,2,1)$ I have that

$$\frac{\delta \mathbf{h}}{\delta x} (2,2,1) = \begin{pmatrix} 4 \\ 1/2 \end{pmatrix} \neq 0 $$

$$\frac{\delta \mathbf{h}}{\delta y} (2,2,1) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq 0 $$

thus I can use the Implicit function theorem to find $\mathbf{f}(z)$.

The next exercise is the one causing me problems.

(iii) Find the Jacobi Matrix D\mathbf{f}.

How do I do this? By looking at the Implicit function theorem I am thinking that I need to use the following:

$$\frac{\delta \mathbf{f}}{\delta x} (1) = -\frac{\frac{\delta \mathbf{h}}{\delta z} (2,2,1)}{\frac{\delta \mathbf{h}}{\delta x}(2,2,1)}$$ and $$\frac{\delta \mathbf{f}}{\delta y} (1) = -\frac{\frac{\delta \mathbf{h}}{\delta z} (2,2,1)}{\frac{\delta \mathbf{h}}{\delta y}(2,2,1)}$$

so that $$[D\mathbf{f}](1) = \begin{pmatrix} \frac{\delta \mathbf{f}}{\delta x} (1) \\ \frac{\delta \mathbf{f}}{\delta y} (1) \end{pmatrix}$$

But how do I calculate $\frac{\delta \mathbf{f}}{\delta x} (1)$ and $\frac{\delta \mathbf{f}}{\delta y} (1)$? Because if I use the formula above I get a vector divided with a vector and this does not make sense to me.

Best regards Husky


If we define $\phi:\mathbb{R} \to \mathbb{R}^3$ by $\phi(z) = (f(z),z)^T$, we have, with slight abuse of notation, $D\phi(z) = \begin{bmatrix} Df(z) \\ I\end{bmatrix} $.

We know that $h \circ \phi(z) = 0$ around $z=1$, so $D (h \circ \phi)(z) = Dh(\phi(z)) D \phi(z) = 0$. Rewriting gives $\begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} & {\partial h(\phi(z)) \over \partial z} \end{bmatrix} \begin{bmatrix} Df(z) \\ I\end{bmatrix} = 0$, or $\begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} \end{bmatrix} Df(z) = - {\partial h(\phi(z)) \over \partial z}$. Then $Df(z) = - \begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} \end{bmatrix}^{-1} {\partial h(\phi(z)) \over \partial z}$.

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  • $\begingroup$ I must honestly admit that I do not understand the above. Is there any other way to explain this? I do not understand why we define $\phi : \mathbb{R} \rightarrow \mathbb{R}^3$, since $\mathbf{h}$ goes from $\mathbb{R}^3 \rightarrow \mathbb{R}^2$ and $\mathbf{f}$ goes from $\mathbb{R} \rightarrow \mathbb{R}^2$ $\endgroup$ – Husky653 Jun 10 '15 at 17:11
  • $\begingroup$ The parameters to $h$ are $f(z),z$ which have size $2,1$ respectively, hence the $3$. $\phi$ just formally represents the parameters so I can use the chain rule. Does this explain it? $\endgroup$ – copper.hat Jun 10 '15 at 17:18
  • $\begingroup$ No, I am sorry. Our professor used the determinant in order to find the result for a similar exercise, but the Jacobi matrix of h was $n x n$ for that exercise. Is there any link you can refer me to that might help me understand your answer? $\endgroup$ – Husky653 Jun 10 '15 at 17:32
  • $\begingroup$ Write out the derivative of the function $z \mapsto h(f(z),z)$ (which is $\mathbb{R} \to \mathbb{R}^2)$. It is a little cumbersome notationally, but the result must be zero. The expression involves a $2 \times 2$ matrix multiplying $Df(z)$ and some other quantity. When you equate to zero and solve for $Df(z)$, you will get the above. $\endgroup$ – copper.hat Jun 10 '15 at 17:57

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