And finally about e

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 = bx dy/dx = c bx

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

x = logby dx/dy = 1/cy

It turns out that the constant c = log_b 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 = log_b 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) = log_bx 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 = 0 dy/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 e^x (although to be honest I am still confused about how that works):