Tuesday, September 1, 2009

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 = 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 = 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 ex (although to be honest I am still confused about how that works):

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