Solving Implicit Functions: F1(x), F2(x) near 0

[asif zaidi]

Hello:
I thought I had this but in doing the problem I realized I didn't (or maybe I didn't).
One of the problems was that in class and notes all examples were done in terms of f(x,y). Obviously in h/w, the problem is given as f(x1,x2,x3) - just to confuse me !

Problem statement

a- For the following equation, decide whether they implicitly determine functions near 0
b- If the equation implicitly determine a function, compute the partial derivatives at 0

Equation:

F1(x) = ( x1$$^{2}$$ + x2$$^{2}$$ + x3$$^{2}$$ )$$^{3}$$ - x1 + x3 = 0;
F2(x) = cos (x1$$^{2}$$ + x2$$^{4}$$) + exp(x3) - 2 = 0

Solution

For Part a-

I know I have to compute the determinant of a matrix. So for this example I played out with x1,x2,x3 and saw which determinant would not be 0. I came up with the following matrix.

B(x) = [partial_der_F1 (x1) partial_der_F1 (x3) ; partial_der_F2(x1) partial_der_F2(x3) ]. Evaluating this at 0, I get the following matrix

[-1 1; 0 1] and the determinant of this matrix is -1 != 0. So I can take the inverse of this matrix

Thus I can say that for each x2, the following function F(phi1(x2), x2, phi2(x2) ) = 0 where phi1 and phi2 are unique functions.

I think I have part a right. It is part b I am having problems with

For part b:

The formula for the derivative given in class notes is -B(x,g(x))$$^{-1}$$ A(x,g(x))

B = matrix calculated above
A = partial derivative of F1, F2 (2x1 matrix) with respect to x2.

Now my question is do I have to evaluate this at x2 = 0 or can I just leave it in terms of phi1(x2), x2, phi2(x2).

Whats confusing me is that if I compute A wrt x2 and evaluate at x2=0, I will get a 0 matrix.


Thanks in advance



Asif

 

[benorin]

I looked over the baby Rudin treatment of this theorem (pg. 224-228) and my conclusion is that the formula given for a derivative in your class notes holds when evaluated at the point. Thus $$\left(\phi_1^\prime (0),\phi_2^\prime (0)\right)=(0,0)$$ is my conclusion if your calculations are correct.