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The following function is given: $$f:\Bbb R^3 \rightarrow \Bbb R^2, \left(\begin{matrix}x \\ y \\ z\\ \end{matrix}\right) \mapsto \left(\begin{matrix}-2x^2 + y^2 + z^2 \\ x^2+e^{y-1} - 2y \end{matrix}\right)$$

The first task is to determine whether the function can be solved in terms of $y$ and $z$ at the point $(1, 1, 1)^T$, when $f = (0, 0)^T$. Should this be possible, the next task would be to calculate the derivative at the aforementioned point.

Attempt at a solution:

$f\left(\begin{matrix}1 \\ 1 \\ 1\\ \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0\\ \end{matrix}\right)$ is clear.

Calculate the determinant of the following Jacobian matrix $$\left| \begin{matrix} \frac{\partial f_1(1,1,1)}{\partial y} && \frac{\partial f_1(1,1,1)}{\partial z} \\ \frac{\partial f_2(1,1,1)}{\partial y} && \frac{\partial f_2(1,1,1)}{\partial z}\end{matrix}\right|,$$where $f_1(x,y,z)=-2x^2 + y^2 + z^2$, $f_2(x,y,z)=x^2+e^{y-1} - 2y$.

Thus, we have$$\left| \begin{matrix} 2y^2 && 2z^2 \\ e^{y-1}-2 && 0 \end{matrix}\right|_{(1,1,1)}=\left| \begin{matrix} 2 && 2 \\ -1 && 0 \end{matrix}\right|=2 \neq 0$$

Thus, by the implicit function theorem, there exist open neighborhoods $U\subseteq \Bbb R$ and $V\subseteq \Bbb R^2$ with $1\in U$ and $(1,1)^T \in V$ and a continuously differentiable function $g:U \rightarrow V$ such that for all $(x,y) \in U \times V$ the following holds:$$f(x,y) = 0 \iff y=g(x)$$

It's the next task that I'm not 100% sure on. Do I need to calculate the partial of $f$ w.r.t. $y$ and then again w.r.t. $z$? How does one do this?

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Suppose you know that $x = 9/10$ (a representative number near $x = 1$, i.e., a candidate for an element of $U$). Then if $f(x, y, z) = 0$, what do you know about $y$ and $z$? From the second term, you know that $$ (9/10)^2+e^{y-1} - 2y = 0\\ 0.81 + e^{y-1} - 2y = 0 $$ If you're willing to guess that the solution for $y$ is near $1$, you can approximate $e^{y-1}$ with $1 + (y-1)$ (the first two terms of the Taylor series) to convert this to $$ 0.81 + (1 + (y-1)) - 2y \approx 0 \\ 0.81 \approx y $$ thus determining $y$ from a known value of $x$.

More generally, you can see that there's a unique solution for $y$: the one-dimensional implicit value theorem applied to $f_2(x, y)$ near the point $(x, y) = (1, 1)$ says so, since $\frac{\partial f_2}{\partial y} (1, 1) = -1$, as you already computed. So there's a function $h$, defined on a neighborhood $U$ of $x = 1$, with the property that $$ f_2(x, h(x)) = 0 $$ for $x \in U$.

Now continuing with the example, knowing $x = 9/10$ and $y \approx 0.81 $, look at the first term: from that, you can solve for $z$. It'll be a square root of some kind, and one of the two roots will be near $+1$ and the other near $-1$ so you pick the $+1$ root.

Continuing with the general analysis instead of the single instance, we have that $h(x)$ is a number such that $x^2 + e^{h(x) - 1} - 2h(x) = 0$ (for $x$ near $0$); we can then build the required function $g$ via

$$ g(x) = \begin{bmatrix} h(x) \\ \sqrt{2x^2 - h(x)^2} \end{bmatrix} $$

Does that help? The fact that you can't explicitly write $h$ isn't a problem -- you know from the 1D implicit function theorem that it exists.

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  • $\begingroup$ There are several things I do not quite understand. Why was $x= \frac {1}{10}$ chosen, and how does this allow me to solve for $y$? $\endgroup$ – Swiftprotector Jan 18 '14 at 8:24
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    $\begingroup$ Good questions. I should have chosen $x = 9/10$; I meant it to be a representative number near $1$ (which is the 'center' point of $U$), but instead chose it near $0$. Sorry. For the second, look at $f_2(x, y) = x^2 + e^{y-1} - 2y$ near $x = y = 1$. Because $\frac{\partial f_2}{ \partial_y} (1, 1) = -1$, you can apply the 1D implicit function theorem to find the function $y = h(x)$ near $x = 1$ such that $f_2(x, h(x)) = 1$; I was trying to illustrate this with a particular $x_0$ near $1$ to get things started. I'll edit a little to clean this up. $\endgroup$ – John Hughes Jan 18 '14 at 12:21
  • $\begingroup$ Ok, so you proved that $y=h(x)$ in that neighborhood, and thus you have a function $g(x)$, which is dependent on $y$ and $z$. Shouldn't, however, $z=\sqrt{2x^2 - h(x)^2}$ and the two entries in the vector for $g$ be switched? $\endgroup$ – Swiftprotector Jan 18 '14 at 13:42
  • $\begingroup$ @Swiftprotector: you're right again. That'll teach me to write answer fast, looking only at the screen and not doing it on paper. I'll edit appropriately. $\endgroup$ – John Hughes Jan 18 '14 at 14:38
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I think what you want is a function $(y,z)=g(x)$, that is a curve in a parametric representation instead of the intersection of two surfaces $f_1(x,y,z)=0, f_2(x,y,z)=0$. In your statement:

" ...$g:U→V$ such that for all $\underline{(x,y)\in U\times V}$ the following holds: $\underline{f(x,y)=0}$ iff $\underline{y=g(x)}$..."

you should probably say: $(x,(y,z))\in U\times V$ and $f(x,y,z)=(0,0)$ iff $(y,z)=g(x)$.

Your calculation of the condition of existence is correct. This being cleared out you just do brute force calculation, pretending $f(x,g(x))=0$ wherever $g(x)$ is defined. Then $0=\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial (y,z)}\cdot \frac{dg}{dx}$ and you can solve it for the vector $\frac{dg}{dx}$ plugging in the values of $x,y,z$ in your point.

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