# Implicit Function Theorem example in Baby Rudin

I am looking at example 2.29 of Baby Rudin (page 227) of my edition to illustrate the implicit function theorem. This is what the example is:

Take $n= 2$ and $m=3$ and consider $\mathbf{f} = (f_1,f_2)$ of $\Bbb{R}^5$ to $\Bbb{R}^2$ given by $$\begin{eqnarray*} f_1(x_1,x_2,y_1,y_2,y_3) &=& 2e^{x_1} + x_2y_1 -4y_2 + 3 \\ f_2(x_1,x_2,y_1,y_2,y_3) &=& x_2\cos x_1 - 6x_1 + 2y_1 - y_3 \end{eqnarray*}.$$ If $\mathbf{a} = (0,1)$ and $\mathbf{b} = (3,2,7)$, then $\mathbf{f(a,b)} = 0$. With respect to the standard bases, the derivative of $f$ at the point $(0,1,3,2,7)$ is the matrix $$[A] = \left[\begin{array}{ccccc} 2 & 3 & 1 & -4 & 0 \\ -6 & 1 & 2 & 0 & -1 \end{array}\right].$$ Hence if we observe the $2 \times 2$ block $$\left[\begin{array}{cc} 2 & 3 \\ -6 & 1 \end{array}\right]$$ it is invertible, and so by the implicit function theorem there exists a $C^1$ mapping $\mathbf{g}$ defined on a neighbourhood of $(3,2,7)$ such that $\mathbf{g}(3,2,7 ) = (0,1)$ and $\mathbf{f}(\mathbf{g}(\mathbf{y}),\mathbf{y}) = 0$.

Now what I don't understand is from such a $\mathbf{g}$, how does this mean that I can solve the variables $x_1$ and $x_2$ for $y_1,y_2,y_3$ locally about $(3,2,7)$?

Also if I wanted to carry out this computation explicitly, how can I do it? We do not have a nice and shiny linear system to solve unlike problem 19 of the same chapter.

Thanks.

• The inverse and implicit function theorems are not really constructive, as far as I can tell. Being able to say that zero sets of certain smooth functions are automatically manifolds is a nice tool, though. There's some good discussion here. Commented Jun 16, 2012 at 8:08
• @DylanMoreland Right Thanks. However even for the existence bit, how do I know from the existence of $g$ that I can solve $x_1$ and $x_2$ in terms of $y_1,y_2$ and $y_3$?
– user38268
Commented Jun 16, 2012 at 8:09
• Maybe I didn't understand your question. I guess I don't have a better answer than, "You put $\mathbf y$-values into $\mathbf g$ and it spits $\mathbf x$-values out." It seems to me, sadly, that Rudin's example is more about checking that the hypotheses are satisfied. Commented Jun 16, 2012 at 8:16
• @DylanMoreland It's ok then, because the exercise at the end of the chapter does allow us to do an explicit computation.
– user38268
Commented Jun 16, 2012 at 8:19
• Interesting. Which exercise, specifically? Commented Jun 16, 2012 at 8:23

With regards to your second question I do not think the implicit function theorem gives you an explicit way to solve the system. Consider the function $y = xe^x$. Even trying to solve this in terms of $y$ requires the Lambert W function.
What is extremely important in the two theorems (apart from the existence claim) are the assertions of the uniqueness/invertibility of the solution in a whole neighbourhood (having topological consequences, e.g., the inverse functions theorem shows that $C^1$ maps with invertible derivative in one point are locally open) and regularity of the functions the existence of which is guaranteed (i.e. $f(x,y)\in C^1 (C^k)$ and the assumptions of the theorem are fulfilled $\Rightarrow g\in C^1 (C^k)$ if $f(x, g(x))=0$).