Implicit Function Theorem Question!

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Implicit Function Theorem
I've been having a lot of trouble understanding the statement of the theorem and its proof, so I would like to see if I did the following question below correctly.

The problem

Let f : R² → R be given by f(x,y,z) = sin(xyz) + e^[2x + y(z - 1)]. Show that the level set {f = 1} can be solved as x = x(y,z) near (0,0,0) and compute ∂x/∂y (0,0) and ∂x/∂z (0,0).

SOLUTION!

Let G(x,y,z) = f(x,y,z) - 1 and note that G(0,0,0) = 0.

DG = [yzcos(xyz) + e^[2x + y(z - 1)]; xzcos(xyz) + (z - 1)e^[2x + y(z - 1)]; xycos(xyz) + e^[2x + y(z - 1)]; 1 x 3 matrix

I have ∂G/∂x = yzcos(xyz) + 2e^[2x + y(z - 1)]. Now at (0,0,0) I have it equal to 1 and hence the determinant is nonzero, so I can apply the Implicit Function Theorem.

So Dx(0,0) = -[∂G/∂x]^(-1) ⋅ [∂G/∂(y,z)] = -[1]^(-1) ⋅ [-1 1] = [1 -1]

Therefore, ∂x/∂y (0,0) = 1 and ∂x/∂z (0,0) = -1.

Is this all right?
 

Answers and Replies

  • #2
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Munkres has f: R^k x R^n → R^n, and then writes f as f(x,y) where x ∈ R^k and y ∈R^n and then goes on to solve for y in terms of the x. However, this isn't as general as one could be, so this is what's been confusing me, but I think I have finally understood it (hopefully). So if anyone could verify whether what I did above is right or wrong would be great. Thanks!
 
  • #3
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An explanation of the entire theorem and maybe with a more complicated (i.e. not the cliche unit circle lol) example would be nice as well!
 
  • #4
mathwonk
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try these notes from my riemann surfaces/algebraic curves class. the point is the implicit function theorem tells you when a curve in the plane defined by an equation like f(x,y)=0, is "smooth", i.e. looks like a graph near each point, either of y(x) or of x(y).

I.e. it tells you when a small nbhd of a point on the curve can be projected isomorphically onto either the y or axis. E.g. the IFT fails at the mid point of a figure eight, but holds near every other point.

start reading the day 1-3 notes on page 3.
 

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Thanks for the links. I'll read them...and I hadn't thought of it like that "it tells you when a small nbhd of a point on the curve can be projected isomorphically onto either the y or x-axis" :)
 
  • #6
mathwonk
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yeah, if ∂f/ƒy is not zero it projects locally isomorphically ointo the x axis, i.e. locally it is a function of x, i.e. locally each point is determined by its x coordinate, and vice versa.

my notes help explain how to remember this, which sounds backwards. The point is that if z = f(x,y) and ∂z/∂y is not zero, then z and y are invertible functions of each other, i.e. they are interchangeable, so instead of x and y, we could use x and z as coordinates.

then y=0, the x axis, gets mapped isomorphically into the set z=0, i.e. f(x,y) = 0, the curve. i.e. locally the curve is isomorphic to the x axis when ∂f/ƒy is non zero.

oh well i guess it is still confusing.
 

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