## Tuesday, September 1, 2009

I have one last post about e. For the previous ones, see here, here, here, here, here and here.

I'm almost done. In fact the best advice I have for you is to get the online version of Strang's Calculus, (or better yet, buy your own copy), and read Chapter 6.

In the last post we came to:

 `y = bxdy/dx = c bx`

and flipping things around to look at x as a function of y:

 `x = logbydx/dy = 1/cy`

It turns out that the constant c = logb e, and then of course, c = 1 when b = e. We can get an expression for that. Since I get a little freaked out looking at dx/dy, let's express the log function in the normal way, where y = logb x. So now we want dy/dx. As before, call Δx, the small change in x, h, then:

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

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

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

At the particular value x = 1:

 `dy/dx = lim(h->0) [ logb(1+h) - logb1 ] / h ]logb1 = 0dy/dx = lim(h->0) [ logb(1+h) ] / h ] = lim(h->0) logb [(1+h)1/h ]`

If we substitute n = 1/h

 ` = lim(h->0) logb [(1+1/n)n ]`

The part in the brackets is e:

 `e = (1+1/n)n`

There are other definitions in the book, but this one about e as the limit of an infinite series is certainly fundamental. In exactly the same way we can develop:

 `ex = (1+x/n)n`

and this will give us the famous series for ex (although to be honest I am still confused about how that works):