| It is pretty much necessary to have some understanding of calculus (preferably multivariable calculus or calculus on manifolds) for the purposes of understanding this node; however, I've tried to hardlink enough definitions for it to be as intelligible as possible.
Consider a circle of radius 1 centred around the origin in R2. Then the points in the circle are those points (x,y) that satisfy the equation x2 + y2 = 1. This is the same as saying that, if we define a function f(x,y) = x2 + y2 - 1, then the circle is made up of all the points (x,y) where f(x,y) = 0. We would like to be able to find a function g such that setting y = g(x) gives us this unit circle (in other words, f(x,g(x)) = 0 for every real number x between 0 and 1). By rearranging the formula, we can obtain y2 = 1 - x2; however, this equation gives two different roots for y (namely, the positive and the negative root). Therefore, we instead define two different functions, g1 and g2, by g1(x) = sqrt(1 - x2) and g2(x) = - sqrt(1 - x2). These differentiable functions are said to be defined implicitly by the equation f(x,y) = 0.
The existence of these functions demonstrates that for every x in (-1,1), there exists a unique number g1(x) in (0,1] and a unique number g2(x) in [-1,0) such that f(x,g1(x)) = f(x,g2(x)) = 0. In fact, an even stronger property applies; (1) for every x' in (-1,1), there exists an open interval A containing x' and a function g: A -> R such that f(x,g(x)) = 0 for all x in A. However, this property does not hold for x = 1 and x = -1, even though (1,0) and (-1,0) are points on the unit circle.
In the example above, f is a function that takes points in R2 (or R X R) to points in R. We wish to find a general condition that tells us when property (1) (as described above) applies, not only for functions from R2 to R, but for any function from Rn X Rm to Rm. In other words, we wish to find a condition where if f is a function from Rn X Rm to Rm, and we are given points (a1, ... , an) in Rn and (b1, ... , bm) in Rm for which each component function fi: Rn X Rm -> R of f satistifies fi(a1, ... , an, b1, ... , bm) = 0, we can find a function g for which there exist an open set A containing (a1, ... , an) and an open set B containing (b1, ... , bm) such that the function g: A -> B satisfies f(x,g(x)) = 0 for all x in A. This condition is given by
The Implicit Function Theorem
(Proof taken from Michael Spivak's Calculus on Manifolds (1965), Cambridge, Mass.: Perseus Books)
Suppose f: Rn X Rm -> Rm is continuously differentiable in an open set containing (a,b) and f(a,b) = 0. Let M be the m X m matrix (Dn + jfi(a,b)), where i and j take values between 1 and m inclusive. Then if det(M) does not equal 0, there is an open set A in Rn containing a and an open set B in Rm containing b with the following property: for each x in A there is a unique g(x) in B such that f(x,g(x)) = 0. The function g is differentiable.
Proof: Define F: Rn X Rm -> Rn X Rm by F(x,y) = (x,f(x,y)). Then det(F'(a,b)) = det(M), which is not equal to 0. By the inverse function theorem there is an open set W in Rn X Rm containing (a,b), which we may take to be of the form A X B (that is, all the points (x,y) for which x is in some open set A in Rn and y is in some open set B in Rm), such that F: A X B -> W has a differentiable inverse h: W -> A X B. Clearly h is of the form h(x,y) = (x,k(x,y)) for some differentiable function k (since F is of this form). Let p: Rn X Rm -> Rm be defined by p(x,y) = y; then p composed with F = f. Therefore
f(x,k(x,y))
= f(h(x,y))
= p(F(h(x,y)))
But h is the inverse of F; therefore F(h(x,y)) = (x,y)
Therefore f(x,k(x,y)) = p(x,y) = y. Thus f(x,k(x,0)) = 0; in other words we can define g(x) = k(x,0). This gives us the desired function g. |
