Paradox du Jour
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A brief working note by Phil, labeled Paradox 143.56 and dated 10.7.16, from his Lagrange multipliers folder. It sets two claims against each other: the gradient of f vanishes at a stationary point, yet a surface normal cannot be null. He resolves it by distinguishing the gradient in N dimensions from the gradient of g = u - f(r) in N+1 dimensions, which is never null.
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Paradox 143.56 10.7.16
1. On the one hand:
Given a function f(r) on EN-1 and no constraints, a stationary point (sometimes called a critical point) is a value of r for which (N-1)f(r) = 0.
2. On the other hand:
Let r be a stationary point for f(r) where f(r) = K. The surface f(r) = K has a normal f(r) which is a vector in EN. No normal vector can be a null vector because it is the normal vector!
3. Paradox. Item one say f(r) = 0 at a stationary point, item two says this is impossible.
Idea A. Look at an example of 1. Think z = f(x,y) which is g(x,y,z) = z - f(x,y) = 0. There is a 2D normal f which can vanish at the north pole of a surface z = f(x,y). But the 3D gradient of g is this: g = and this is not 0. So maybe you need to be clearer about which gradient you are talking about. The surface f(r) = K is a surface of dimension N-1 in EN. The object f is a vector in EN. I think it is true that this vector is never null.
I think my item 1 above is wrong.
Let f(r) be a function on EN. Let (r, u) be a point in EN+1 where u = f(r) is regarded as the (N+1)th coordinate. Define g(r,u) ≡ u- f(r). The equation g(r,u) = 0 describes a surface in EN+1. This surface has a normal (N+1)g(r,u) which is a vector in EN+1. This gradient is never null for a smooth surface. A critical point of the function f(r) is one where (N)f(r) = 0. This gradient is null at a stationary point of the function f(r).
Given a function f(r) and no constraints, a stationary point (sometimes called a critical point) is a value of r for which f(r) = 0.