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Math Help - Implicit Function Theorem

  1. #1
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    Implicit Function Theorem

    I have an idea on this one, tell me what you guys think.

    (1) Find conditions on the function f which will allow you to solve the equation f(f(x,y),y) = 0 for y as a function of x near (0,0).

    I think all that is required of f is that:

    (1) f is continuous and differentiable
    (2) the partial of f with respect to y is non-zero, which will make the required determinant non-zero for use in the implicit function theorem. This just seems too easy. Any thoughts?
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  2. #2
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    Quote Originally Posted by joeyjoejoe View Post
    I have an idea on this one, tell me what you guys think.

    (1) Find conditions on the function f which will allow you to solve the equation f(f(x,y),y) = 0 for y as a function of x near (0,0).

    I think all that is required of f is that:

    (1) f is continuous and differentiable
    (2) the partial of f with respect to y is non-zero, which will make the required determinant non-zero for use in the implicit function theorem. This just seems too easy. Any thoughts?
    Say that f:\mathbb{R}^2 \to \mathbb{R} is a \mathcal{C}^1 function. Define g:\mathbb{R}^2 \to \mathbb{R} by g(x,y) = f(f(x,y),y). You are being to ask to find the conditions so that you can solve g(x,y)=0 not f(x,y) = 0. By the implict function theorem it is sufficient to have \tfrac{\partial g}{\partial y} \not = 0 at (0,0). Now notice that g(x,y) = f(u(x,y),v(x,y)) where u(x,y) = f(x,y) and v(x,y) = y. By the chain rule this means \tfrac{\partial g}{\partial y} = \tfrac{\partial f}{\partial u}\cdot \tfrac{\partial u}{\partial y} + \tfrac{\partial f}{\partial v}\cdot \tfrac{\partial v}{\partial y}. Therefore, \tfrac{\partial g}{\partial y} = f_1 (f(x,y),y)\cdot f_1(x,y) + f_2(f(x,y),y). At (0,0) this tells us g_y(x,y) = f_1(f(0,0),0)\cdot f_1(0,0) + f_2(f(0,0),0)\cdot f_2(f(0,0),0)\not = 0. Where f_1 is the partial derivative to the first component and f_2 is the partial derivative to the second component.
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