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Problem 1, page 78 of Munkres (Analysis on Manifolds):

Let $f: \mathbb{R}^3 \to \mathbb{R}^2$ be of class $C^1$; write $f$ in the form $f(x,y_1,y_2)$. Assume that $f(3,-1,2) = \mathbf{0}$ and $$ Df(3,-1,2) = \begin{pmatrix} 1 & 2 & 1 \\ 1 & -1 & 1 \end{pmatrix}.$$

(a) Show that there is a function $g: B \to \mathbb{R}^2$ of class $C^1$ defined on an open set $B$ in $\mathbb{R}$ such that $f(x,g_1(x),g_2(x)) = \mathbf{0}$ for $x \in B$ and $g(3) = (-1,2)$.

(b) Find $Dg(3)$.

(c) Discuss the problem of solving the equation $f(x,y_1,y_2) = \mathbf{0}$ for an arbitrary pair of unknowns in terms of the third, near the point $(3,-1,2)$.

Here's what I have so far:

Let $b=(-1,2)$ so that $a = (3,-1,2)$. Write $f(x,y_1,y_2)$ with $y = (y_1,y_2)$ then,

a =3, and b = (-1,2)

and determinant partial of $f$ w.r.t partial of $y (3,-1,2) =$ ?

$$ \det \begin{pmatrix} \frac{\partial f_1}{\partial y_1}(a,b) & \frac{\partial f_1}{\partial y_2}(a,b) \\ \frac{\partial f_2}{\partial y_1}(a,b) & \frac{\partial f_2}{\partial y_2}(a,b) \end{pmatrix} $$

Derivative of partial of f = [partial of f w.r.t x partial of f w.r.t. y] implies Df(3,b) = (Stuck on evaluating this), but I know it is the expression above which I have wrote

and what is partial of f1 w.r.t. y1 (a,b), partial of f2 w.r.t. y2 (a,b), partial of f1 w.r.t. y2 (a,b), and partial of f2 w.r.t. y2(a,b)?

For part b:

Dg(3) = -{partial of f w.r.t y(3,b)]^-1 [partial of f w.r.t. x1] at (3,b) =?

for part c, I thought of taking 2 variables u and v s.t. partial of f w.r.t partial of (u,v) is not equal to zero.

Since $f(a)$ is zero then by the implicit function theorem, there is a neighborhood $B$ of $(-1,2)$ in $\mathbb{R}^2$ and a unique function $g: B \to \mathbb{R}^3$ so that $g(a) = b$.

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Your question is extremely difficult to read; please edit it to use proper TeX markup, see meta.math.stackexchange.com/questions/107/… (you can right-click on a formula in any question or answer and choose "Show math as TeX-commands" in the menu to show the TeX-markup that generated it). –  Martin Wanvik Feb 8 '12 at 13:38
    
The question is correctly formatted in LATEX except what I tried isnt. I need someone to check over this Thanks –  James R. Feb 8 '12 at 21:46
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Yes, I can see that part of it is correct TeX-markup (heck, I wrote most of it). The problem is precisely the parts that isn't, which is why I asked you to fix it. It is very difficult to help you, if you can't be bothered to comply with a simple request to express yourself clearly. –  Martin Wanvik Feb 8 '12 at 22:48
    
I see you've started to edit your post; good! Some hints to help you along: to see how to use the various matrix environments properly, see en.wikibooks.org/wiki/LaTeX/Mathematics#Matrices_and_arrays. Note that the wikibook uses \[ and \] to delimit a displayed formula; on this site, you either use double dollar signs, such as $$ \int e^x dx $$ or an extra backslash \\[ \int e^x dx \\] (compare this with $ \int e^x dx $, to understand how a displayed formula differs from an inline one; see also en.wikibooks.org/wiki/LaTeX/…) –  Martin Wanvik Feb 9 '12 at 12:48
    
Can you please help me only this one time on this problem? I need to know this for a test on Friday. And then, I will format it correctly. Thanks –  James R. Feb 9 '12 at 18:55

1 Answer 1

up vote 2 down vote accepted

It is quite difficult to determine exactly what you're asking here, but the following will hopefully be helpful.

For question (a), it is just a matter of applying the implicit function theorem. Let us first get all of these vectors straight. As the theorem is stated in Munkres' book, we view $f: \mathbb{R}^{1+2} \to \mathbb{R}^2$, so that $k = 1$ and $n = 2$. We are also given the point $(3,-1,2)$, which is supposed to be written as $(a,b)$ with $a \in \mathbb{R}^1$ and $b \in \mathbb{R}^2$, so $a = 3$ and $b = (-1,2)$.

To apply the theorem, there is a single condition to check, namely that $$ \det \frac{\partial f}{\partial y}(a,b) = \det \begin{pmatrix} \frac{\partial f_1}{\partial y_1}(a,b) & \frac{\partial f_1}{\partial y_2}(a,b) \\ \frac{\partial f_2}{\partial y_1}(a,b) & \frac{\partial f_2}{\partial y_2}(a,b) \end{pmatrix} \neq 0,$$ with $y_1$ and $y_2$ as in the problem statement. So, you need to determine this matrix, and compute its determinant (the total derivative matrix is given in the problem statement, and that includes all partial derivatives at the point $(a,b)$, so it is just a matter of figuring out which elements belong in your matrix).

Edit: I believe that the problem you're having is still in interpreting the matrix that is given to you in the problem statement; in order to do this, all you need to know is how to compute the total derivative $Df$ for an arbitrary $C^1$-function $f: \mathbb{R}^{n} \to \mathbb{R}^m$. In this case, $f$ can be written as a vector of $m$-functions, i.e. $f = (f_1, f_2, \ldots, f_m)$, where each $f_i, 1 \leq i \leq m$ is a real-valued function in $n$ variables, i.e. $f_i(x_1, x_2, \ldots, x_n)$. So each function has $n$ partial derivatives, and the $m$'th row in the matrix is going to be the partial derivatives of $f_m$, so $$ Df = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \ldots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \ddots & & \frac{\partial f_2}{\partial x_n} \\ \vdots & & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \ldots & \frac{\partial f_m}{\partial x_{n-1}} & \frac{\partial f_m}{\partial x_n} \\ \end{bmatrix}$$ The way to remember this is to note that the derivative is supposed to be a linear map $Df(x): \mathbb{R}^n \to \mathbb{R}^m$ also, so that the number of columns has to be $n$ (in order for the matrix-vector multiplication to be defined), so the variables vary with columns and the function vary with rows.

The matrix given in the problem statement is $$ Df(3,-1,2) = \begin{bmatrix} 1 & 2 & 1 \\ 1 & -1 & 1 \end{bmatrix},$$ and what we're after here is the submatrix consisting of the partial derivatives with respect to the last two variables $y_1$ and $y_2$, and those are $$ \frac{\partial f}{\partial y}(a,b) = \begin{bmatrix} 2 & 1 \\ -1 & 1 \end{bmatrix}$$ This should not be difficult to understand. Hence the determinant in question is $3$, so the implicit function theorem can be applied.

For problem (b), you can see from the formula that you are also going to need the inverse of the matrix you computed the determinant for; for $2\times 2$-matrices, the easiest way to do this is to use Cramer's rule, see http://en.wikipedia.org/wiki/Invertible_matrix#Methods_of_matrix_inversion

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