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Inverse and Implicit Function theorems

  1. Jun 6, 2004 #1
    Hello all,
    Can somebody give a 'simple to understand' proof for the 'Implicit Function Theorem' and the 'Inverse Function Theorem',their significance, applications and some examples.
    I have tried some books on Multivariable calculus, but in vain.....
  2. jcsd
  3. Jun 7, 2004 #2

    Math Is Hard

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    Does this help at all?

    It looks like a pretty deep subject 'cause this guy wrote a whole book on Implicit Function Theorem:

    "The history of the implicit function theorem is a lively and complex story, and is intimately bound up with the development of fundamental ideas in analysis and geometry. This entire development, together with mathematical examples and proofs, is recounted for the first time here. It is an exciting tale, and it continues to evolve."

    um... yeah, Sounds like a real page turner. I think I got stuck next to this guy at a cocktail party once. :zzz:
  4. Jul 22, 2004 #3


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    The implicit function theorem in 2 vbls is pretty easy to understand. how about an explanation instead of a real proof?

    Technically the statement is that if f(x,y) is a function of two variables with continuous partial derivatives at a point (a,b), and if the y partial at (a,b) is non zero, then there is a rectangular neighborhood of (a,b) in which the locus of points where f(x,y) = f(a,b) is a curve in the (x,y) plane that looks like the graph of a differentiable function of x, say y = g(x).

    What this hypothesis means, is that the function f(x,y) divides the (x,y) plane near (a,b) up into differentiable curves where the value of f is constant, and that the curve that passes through (a,b), i.e. the curve of points where f(x,y) has the same value it has at (a,b), is a smooth curve.

    Moreover since the y partial of f at (a,b) is not zero, this means the tangent line to this "level curve" is not vertical. (The gradient vector is the vector perpendicular to this curve, and the y partial not being zero means the gradient does have a y component, so the perpendicular vector, tangent to the curve has an x component, thus it is not vertical.)

    OK, now we see that the hypothesis of the theorem says that the curve of points where f(x,y) has the same value it has at (a,b), is a smooth curve whose tangent vector is not vertical at (a,b). But that just means this curve is not vertical either, i.e. it passes the vertical line test near that point, i.e. is locally the graph of a function.

    Easier?: consider a linear function of two variables f(x,y) = rx+sy+c such that s is not zero. Then since s is not zero, the line defined by setting this expression equal to c, is not vertical, hence is the graph of a function, since we can divide by s and get y as a function of x.

    Now if f is any differentiable function, then near (a,b), f is approximated well by the linear function (df/dx)(a,b) (x-a) + (df/dy)(a,b) (y-b) + f(a,b), i.e. by a linear function rx+sy+c, as above where r is the x derivative and s is the y derivative of f.

    The hypothesis of the implicit function theorem is that the line defined by setting this linear approximation equal to f(a,b) is not vertical. Since that line is tangent to the curve obtained by setting f(x,y) = f(a,b), then that curve is not vertical either. That's it.

    In general the implicit function theorem just says if in the linear approximation to f, at a point p, you can solve for some of the variables in terms of the others, then that is also true for the original f, at least near the given point.

    well maybe I am proving the subject is hard, but a picture would show it isn't in a second.
  5. Jul 23, 2004 #4


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    Lets try an example instead of an explanation. Consider the function
    f(x,y) = x^2 + y^2. For each choice of a constant c >= 0, look at the set of points (x,y) where f(x,y) = c. If c=0 this is one point and if c>0 this is a circle of radius sqrt(c).

    Now look at the hypothesis of the implicit function theorem, i.e. take the y partial of this function. The y partial is 2y, so whenever 2y is not zero, i.e. for points not on the x axis, we can apply the theorem.

    So look at any value of c, at the circle x^2 + y^2 = c, and look at a point with y not zero on this circle. Then the claim is that near this point, a little piece of the circle looks like the graph of a (differentiable) function y(x). I.e. that you can solve for y differentiably in terms of x near that point.

    Well this example is so easy we do not need to be told that, we can actually do it. I.e. we solve for y in x^2 + y^2 = c, as follows:

    y^2 = c-x^2, so y = sqrt(c-x^2). Now, notice that if x took the value sqrt(c), we would be in trouble, since this would not be a differentiable function of x.

    I.e. the derivative is, lets see now..., (-2x)(1/2)/(c-x^2)^(1/2). correct??

    Anyway see this makes no sense if the denominator is zero, i.e. if x = sqrt(c).

    But by our original equation x^2 + y^2 + c, that happens only when y = 0, which we ruled out. So as long as our point has y not zero, then also x is not equal to sqrt(c) and we can solve for y near that point.

    Another way to see it is to look at a circle, and note that away from the two extreme left and right points, where y =0, a small arc of the circle is not vertical, hence indeed looks locally like the graph of a function.

    So we are just saying that at any point on any circle centered at zero, if we pick a point that is not on the x axis, then a little piece of the circle near that point is not vertical, hence is a graph of a fucntion y(x).

    Now if we were willing to let x be a fucntion of y instead of the other way around we could expand the thm to work at any point away from the origin. I.e. if the x partial is not zero, then a little piece of circle is not horizonhtal, hence x is a function of y.

    The only hopeless point from this point of view is the origin, where both partials are zero. From this point of view, the thm just says that at any point where at least one partial is non zero, the "level curve" f(x,y) = c passing through that point is smooth and has a nice tangent line.

    If the tangent line is not vertical, then we can solve for y in terms of x near there, and vice versa if it is not horizontal.

    But at a point where both partials are zero, the set f(x,y) = c, may not look like a curve at all, or the curve may have a kink in it, or cross itself, and in any event the curve has no tangent line, and hence never looks like a graph, either of form y(x) or x(y).

    For example on a topographical map, if you look the level curves of the height function, i.e. the curves where the height is constant, you see that these curves have tangents except where the gradient is zero, i.e. at mountain tops, and valley bottoms, and at crossings between mountains.
  6. Aug 10, 2004 #5


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    This is one of the most important theorems in mathematics and you have asked for a complete elementary treatise on it including proofs, significance and examples. There are plenty of people around that know all this, but you have as yet not responded to what has been offered, so there is little encouragement to continue.
  7. Oct 18, 2010 #6
    This is an amazing explanation mathwonk! Thanks a ton! The implicit function theorem was never so obvious to me.
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