A Universe of Sorts

§ 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: $x^2 + y^2 = 1$ to $y = \sqrt{1 - x^2}$ .

If we have a function

y = g(p, q)

, we can write this as

y - g(p, q) = 0

. This can be taken as an implicit function

h(y; p, q) = y - g(p, q)

. We then want to recover the explicit version of

y = g'(p, q)

such that

h(g'(p, q); p, q) = 0

. That is, we recover the original explicit formulation of

y = g'(p, q)

in a way that satisfies

h

§ The 1D linear equation case

In the simplest possible case, assume the relationship between

y

and

p

is a linear one, given implicitly. So we have

h(y; p) = \alpha y + \beta p + \gamma = 0

. Solving for

h(y,p) = 0

, one arrives at:

y = -1/\alpha (\beta p + \gamma)

Note that the solution exists iff $\alpha \neq 0$ .
Also note that the the existence of the solution is equivalent to asking that $\partial h / \partial y = \alpha \neq 0$ .

§ The circle case

In the circle case, we have

h(y; p) = p^2 + y^2 - 1

. We can write

y = \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. $y+ = +\sqrt{p^2 - 1}$ ; $y- = -\sqrt{p^2 - 1}$ .

In this case, we have $\partial h / \partial y = 2y$ , which changes signfor the two solutions. If $y^\star > 0$ , then $(\partial h / \partial y)(y^\star = 0)$ .Similarly for the negative case.

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

Let us say we wish to solve

h(y; p) = y^3 + p^2 - 3 yp - 7 = 0

. Let's assume that we have a solution

y = sol(p)

around the point

(y=3, p=4)

. Then we must have:

sol(p)^3 + p^2 - 3 sol(p) p - 7 = 0

. Differentiating by

p

, we get:

3 sol(p)^2 sol'(p) + 2p - 3 sol'(p) p - 3 sol(p) = 0

. This gives us the condition on the derivative:

\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)^2 - 3p \neq 0)

. This quantity is again

\partial h / \partial y

§ Application to economics

We have two inputs which are purchaed as $x_1$ units of input 1, $x_2$ units of input $2$ .
The price of the first input is $w_1 BTC/unit$ . That of the second input is $w_2 BTC/unit$ .
We produce an output which is sold at price $w BTC/unit$ .
For a given $(x_1, x_2)$ units of input, we can produce $x_1^a x_2^b$ units of output where $a + b < 1$ .The Coob-douglas function.
The profit is going to be $profit(x_1 x_2, w_1, w_2, w) = w(x_1^a x_2^b) - w_1 x_1 - w_2 x_2$ .
We want to select $x_1, x_2$ to maximize profits.
Assume we are at break-even: $profit(x_1, x_2, w_1, w_2, w) = 0$ .
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 ( $x_1$ ) is a functionof the price $w_1$ . Moreover, we can show that it's a decreasing functionof the price.

§ Inverse function: Function to Bijection

Given a differentiable function $f$ , at a point $p$ , we will have a continuous inverse $f^{-1}(p)$ if the derivative $f'(p)$ is locally invertible.

The intuition is that we can approximate the original function with a linearfunction. $y = f(p + \delta) = f(p) + f'(p) \delta$ . Now since $f'(p)$ islocally invertible, we can solve for $\delta$ . $y = f(p) + f'(p) \delta$ implies that $\delta = 1/f'(p) [y - f(p + \delta) ]$ .This gives us the pre-image $(p + delta) \mapsto y$ .

The fact that $1/f'(p)$ is non-zero is the key property. This generalizesin multiple dimensions to saying that $f'(p)$ is invertible.

One perspective we can adopt is that of Newton's method. Recall that Newton's method allows us to find

x^*

for a fixed

y^*