## Monday, August 31, 2009

### Connecting ln(x) and 1/x

I have more from Strang. We're going to look at the the limit of Δy/Δx in the context of the exponential. Following the book, we use the symbol h for Δx, the small change in x. So we have:

 `dy/dx = lim(h->0) [y(x+h) - y(x)] / h]`

In this case, y(x) = bx so we want

 `dy/dx = lim(h->0) [ bx+h - bx ] / h ]`

As he says, the key idea is that we can split the first part

 `bx+h = bx bh`

and then factor out bx and move it outside the limit:

 `dy/dx = bx lim(h->0) [bh - 1] / h ]`

The term in the limit is something. It depends on what the limit converges too. However, the important thing is that it is not a variable but a constant. Thus,

 `dy/dx = c bx`

This is the equation that I used in the last post. Starting from here there is a quick demonstration of something that we proved in a more roundabout fashion before. Start with the equation above and invert it:

 `dx/dy = 1 / c bx`

but

 `y = bxx = logb ydx/dy = 1/cyd/dy logb(y) = 1/cy`

Beautiful! Strang says "that proof was...powerful too quick." So he does the following.

 `y = bxf(y) = xf(bx) = xf'(bx) (c bx) = 1 # this is the chain rulef'(bx) = 1 / (c bx) # identify bx as yf'(y) = 1/cy`

f(y) converts y to x, it takes the logarithm to base b of x.