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.
In the simplest possible case, assume the relationship between y and pis a linear one, given implicitly. So we have h(y;p)=αy+βp+γ=0.
Solving for h(y,p)=0, one arrives at: y=−1/α(βp+γ).
Note that the solution exists iff α=0.
Also note that the the existence of the solution is equivalent to asking that∂h/∂y=α=0.
Let us say we wish to solve h(y;p)=y3+p2−3yp−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+p2−3sol(p)p−7=0. Differentiating by p,
we get: 3sol(p)2sol′(p)+2p−3sol′(p)p−3sol(p)=0. This gives
us the condition on the derivative:
We have two inputs which are purchaed as x1 units of input 1, x2 units of input 2.
The price of the first input is w1BTC/unit. That of the second input is w2BTC/unit.
We produce an output which is sold at price wBTC/unit.
For a given (x1,x2) units of input, we can produce x1ax2b units of output wherea+b<1.The Coob-douglas function.
The profit is going to be profit(x1x2,w1,w2,w)=w(x1ax2b)−w1x1−w2x2.
We want to select x1,x2 to maximize profits.
Assume we are at break-even: profit(x1,x2,w1,w2,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 (x1) is a functionof the price w1. Moreover, we can show that it's a decreasing functionof the price.
Given a differentiable function f, at a point p, we will have a continuous inversef−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+δ)=f(p)+f′(p)δ. Now since f′(p) islocally invertible, we can solve for δ. y=f(p)+f′(p)δimplies that δ=1/f′(p)[y−f(p+δ)].This gives us the pre-image (p+delta)↦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∗ such that y∗=f(x∗). It follows
the exact same process!
We start with some x[1].
We then find the tangent f′(x[1]).
We draw the tangent at the point x[1] as (y[2]−y[1])=f′(x[1])(x[2]−x[1]).
To find the y∗ we set y[2]=y∗.
This gives us x[2]=x[1]−(y∗−y[1])/f′(x[1]).
Immediately generalizing, we get x[n+1]=x[n]−(y∗−y[n])/f′(x[n]).