§ The implicit and inverse function theorem

I keep forgetting the precise conditions of these two theorems. So here I'm writing it down as a reference for myself.

§ Implicit function: Relation to function

  • Example 1: x2+y2=1x^2 + y^2 = 1 to y=1x2y = \sqrt{1 - x^2}.
If we have a function y=g(p,q)y = g(p, q), we can write this as yg(p,q)=0y - g(p, q) = 0. This can be taken as an implicit function h(y;p,q)=yg(p,q)h(y; p, q) = y - g(p, q). We then want to recover the explicit version of y=g(p,q)y = g'(p, q) such that h(g(p,q);p,q)=0h(g'(p, q); p, q) = 0. That is, we recover the original explicit formulation of y=g(p,q)y = g'(p, q) in a way that satisfies hh.

§ The 1D linear equation case

In the simplest possible case, assume the relationship between yy and pp is a linear one, given implicitly. So we have h(y;p)=αy+βp+γ=0h(y; p) = \alpha y + \beta p + \gamma = 0. Solving for h(y,p)=0h(y,p) = 0, one arrives at: y=1/α(βp+γ)y = -1/\alpha (\beta p + \gamma).
  • Note that the solution exists iff α0\alpha \neq 0.
  • Also note that the the existence of the solution is equivalent to asking thath/y=α0\partial h / \partial y = \alpha \neq 0.

§ The circle case

In the circle case, we have h(y;p)=p2+y21h(y; p) = p^2 + y^2 - 1. We can write y=±p21y = \pm \sqrt{p^2 - 1}. These are two solutions, not one, and hence a relation, not a function.
  • We can however build two functions by taking two parts. y+=+p21y+ = +\sqrt{p^2 - 1};y=p21y- = -\sqrt{p^2 - 1}.
  • In this case, we have h/y=2y\partial h / \partial y = 2y, which changes signfor the two solutions. If y>0y^\star > 0, then (h/y)(y=0)(\partial h / \partial y)(y^\star = 0).Similarly for the negative case.

§ Assuming that a solution for h(y,p)h(y, p) exists

Let us say we wish to solve h(y;p)=y3+p23yp7=0h(y; p) = y^3 + p^2 - 3 yp - 7 = 0. Let's assume that we have a solution y=sol(p)y = sol(p) around the point (y=3,p=4)(y=3, p=4). Then we must have: sol(p)3+p23sol(p)p7=0sol(p)^3 + p^2 - 3 sol(p) p - 7 = 0. Differentiating by pp, we get: 3sol(p)2sol(p)+2p3sol(p)p3sol(p)=03 sol(p)^2 sol'(p) + 2p - 3 sol'(p) p - 3 sol(p) = 0. This gives us the condition on the derivative:
3sol(p)2sol(p)+2p3sol(p)p3sol(p)=0sol(p)[3sol(p)23p]=3sol(p)2psol(p)=[3sol(p)2p]/[3(sol(p)23p)] \begin{aligned} &3 sol(p)^2 sol'(p) + 2p - 3 sol'(p) p - 3 sol(p) = 0 \\ &sol'(p)\left[ 3 sol(p)^2 - 3p \right] = 3 sol(p) - 2p \\ &sol'(p) = [3 sol(p) - 2p] / \left[ 3(sol(p)^2 - 3p) \right] \\ \end{aligned}
The above solution exists if 3(sol(p)23p0)3(sol(p)^2 - 3p \neq 0). This quantity is again h/y\partial h / \partial y.

§ Application to economics

  • We have two inputs which are purchaed as x1x_1 units of input 1, x2x_2 units of input 22.
  • The price of the first input is w1BTC/unitw_1 BTC/unit. That of the second input is w2BTC/unitw_2 BTC/unit.
  • We produce an output which is sold at price wBTC/unitw BTC/unit.
  • For a given (x1,x2)(x_1, x_2) units of input, we can produce x1ax2bx_1^a x_2^b units of output wherea+b<1a + b < 1.The Coob-douglas function.
  • The profit is going to be profit(x1x2,w1,w2,w)=w(x1ax2b)w1x1w2x2profit(x_1 x_2, w_1, w_2, w) = w(x_1^a x_2^b) - w_1 x_1 - w_2 x_2.
  • We want to select x1,x2x_1, x_2 to maximize profits.
  • Assume we are at break-even: profit(x1,x2,w1,w2,w)=0profit(x_1, x_2, w_1, w_2, w) = 0.
  • The implicit function theorem allows us to understand how any variable changeswith respect to any other variable. It tells us that locally, for example,that the number of units of the first input we buy (x1x_1) is a functionof the price w1w_1. Moreover, we can show that it's a decreasing functionof the price.

§ Inverse function: Function to Bijection

  • Given a differentiable function ff, at a point pp, we will have a continuous inversef1(p)f^{-1}(p) if the derivative f(p)f'(p) is locally invertible.
  • The intuition is that we can approximate the original function with a linearfunction. y=f(p+δ)=f(p)+f(p)δy = f(p + \delta) = f(p) + f'(p) \delta. Now since f(p)f'(p) islocally invertible, we can solve for δ\delta. y=f(p)+f(p)δy = f(p) + f'(p) \deltaimplies that δ=1/f(p)[yf(p+δ)]\delta = 1/f'(p) [y - f(p + \delta) ].This gives us the pre-image (p+delta)y(p + delta) \mapsto y.
  • The fact that 1/f(p)1/f'(p) is non-zero is the key property. This generalizesin multiple dimensions to saying that f(p)f'(p) is invertible.
One perspective we can adopt is that of Newton's method. Recall that Newton's method allows us to find xx^* for a fixed yy^* such that y=f(x)y^* = f(x^*). It follows the exact same process!
  • We start with some x[1]x[1].
  • We then find the tangent f(x[1])f'(x[1]).
  • We draw the tangent at the point x[1]x[1] as (y[2]y[1])=f(x[1])(x[2]x[1])(y[2] - y[1]) = f'(x[1])(x[2] - x[1]).
  • To find the yy^* we set y[2]=yy[2] = y^*.
  • This gives us x[2]=x[1](yy[1])/f(x[1])x[2] = x[1] - (y^* - y[1])/f'(x[1]).
  • Immediately generalizing, we get x[n+1]=x[n](yy[n])/f(x[n])x[n+1] = x[n] - (y^* - y[n]) / f'(x[n]).

§ References