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Question about the implicit function theorem

  1. Feb 16, 2016 #1
    I won't post the whole rigorous statement of the theorem, but basically the theorem states that
    If ##F(x,y) = 0## on a neighborhood of the form ##[x-\delta ,x+\delta ]\times [y- \epsilon ,y+\epsilon ]## and if ##\frac{\partial F(x,y)}{\partial y} \neq 0##, then there exists a function ##y=\phi (x)\: s.t. F(x,\phi (x)) = 0 \: \forall x \in [x-\delta ,x+\delta ]##. My question is about ##\phi '(x)##, which can be obtained by deriving ##F(x,\phi (x))## w.r.t. x as follows:
    [tex]
    \frac{d}{dx}F(x,\phi (x)) = F_x(x,\phi (x)) + F_y(x,\phi (x))\phi '(x) = 0\\\rightarrow \phi '(x) = -\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}
    [/tex]
    Why can we set ##\frac{d}{dx}F(x,\phi (x)) = 0##? By definition, by composing F with ##\phi##, we are on a curve where F is 0 for all x(and so it makes sense that all its derivatives w.r.t. x are 0). But ##\frac{\partial F(x,y)}{\partial x}## isn't necessarily equal to 0 in that same neighborhood. What is the difference between the two? Isn't ##F(x,\phi (x)) \in F(x,y)##? I'm quite confused
     
    Last edited: Feb 16, 2016
  2. jcsd
  3. Feb 16, 2016 #2

    Samy_A

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    You can set ##\frac{d}{dx}F(x,\phi (x)) = 0## because, as you said, ##F(x,\phi (x))=0## (in an interval).

    Maybe use a trivial example to see what is going on here:
    ##F(x,y)=x²+2y##
    Compute ##\phi(x)##, and verify that ##\phi '(x) = -\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}## by computing all the terms in this equation.
     
  4. Feb 16, 2016 #3
    I see now. Thanks Samy! :)
     
  5. Feb 16, 2016 #4
  6. Feb 16, 2016 #5
    Hello zinq. For a function on the xy-plane to be implicit, doesn't it mean that it has to be in the form ##F(x,y) = 0##? Thank you for the link btw
     
  7. Feb 16, 2016 #6
    Gianmarco, yes — in order to have an implicit function, you need to have an equation of the form F(x, y) = 0.

    What I am pointing out is that you do not want to assume that F(x, y) is equal to 0 for all (x, y) in an entire product neighborhood, like

    [x−δ, x+δ] × [y−ϵ, y+ϵ],​

    as you did.

    You have a function F defined on some (generally assumed to be) open set of the plane R2.

    Let's call that open set — the domain of F — by the name U. Then F is a function

    F: U → R

    Suppose we make a graph of F(x, y) in 3D:

    z = F(x, y)​

    which means, of course, the set

    G = {(x, y, z) | (x, y) ∈ U and z = f(x, y)}​

    which we have called G. Now you are interested in the points (x, y) of U where

    F(x, y) = 0.​

    One way to picture this is to think of the plane P defined by

    z = 0​

    and where that plane intersects the graph

    z = F(x, y).​

    Typically, the intersection

    G ∩ P = {(x, y, z) | (x, y) ∈ U and z = F(x, y) = 0}​

    of these two surfaces in 3-space (P being just a plane) will be a curve. Not necessarily, but typically.

    It would not be a curve if the value of F(x, y) is 0 on an entire product neighborhood such as

    [x−δ, x+δ] × [y−ϵ, y+ϵ].​

    In that unlikely case, that intersection

    G ∩ P = {(x, y, z) | (x, y) ∈ U and z = F(x, y) = 0}​

    will contain a set that contains the entire product neighborhood

    [x−δ, x+δ] × [y−ϵ, y+ϵ]​

    at the level z = 0.

    In order to make sense of this, it would be best to draw pictures of what is going on.
     
    Last edited: Feb 16, 2016
  8. Feb 17, 2016 #7
    Actually, your explanation was very clear. My mistake was not thinking of ##F(x,y)## in terms of a 3D function intersected with the plane z = 0. I have another question though. When could this theorem be useful? I see that you could study the level curves of any 3D function of the form ##F(x,y) - z = 0##, with z fixed, provided that either of its partial derivatives is not zero. But since the theorem only guarantees you that, given ##\frac{\partial F(x_0,y_0)}{\partial y} \neq 0## and ##F(x_0,y_0) = 0## for some point ##(x_0, y_0)##, the graph of the level curve IS a function, but it doesn't say that you can algebraically express y in terms of x (for instance in ##F(x,y)=y^2+x-e^y## which, at the point ##(1,0)## satisfies ##F(x,y) = 0## and has ##\frac{\partial F}{\partial y} \neq 0##), then how can this be useful? Is it because it makes it possible to study the critical points w.r.t. a certain axis?
     
  9. Feb 17, 2016 #8
    When you have an equation like

    F(x, y) = 0​

    and the derivative condition is satisfied so you know that (at least locally) there exists a function

    y = φ(x)​

    that the implicit function satisfies:

    F(x, φ(x)) = 0,​

    then to actually find an algebraic expression for φ(x) is not as important in practical applications as being able to solve for φ(x) numerically, to a sufficient degree of accuracy for whatever the purpose is.

    Of course, it's always nice if you can get an explicit expression for φ(x), but it is usually not necessary.
     
  10. Feb 17, 2016 #9
    I see what you mean! You could numerically approximate the ##\phi (x)## with a taylor expansion. That's why we care about finding the value of ##\phi '(x)##. Thanks a lot zinq, that was enlightening :D
     
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