0
$\begingroup$

I am having problems with the following exercise:

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

$\endgroup$
1
$\begingroup$

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 $Df(z) = - \begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} \end{bmatrix}^{-1} {\partial h(\phi(z)) \over \partial z}$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.