Re: prove of ln

From: JoeS (jhs_at_math.brown.edu)
Date: 12/09/04

Date: 9 Dec 2004 11:32:20 -0800

That's really clever. You can do it less explicitly by using the first
two terms of Taylor series with the remainder term expressed as an
integral.

But why not simply use the Mean Value Theorem for f(z) = ln(z). This
gives

(f(x) - f(1)) / (x - 1) = f'(y)
for some y (depending on x) satisfying 1 < y < x.

Hence

(ln(x) - ln(1))/(x - 1) = 1/y for some y between 1 and x.

Then use ln(1)=1 and multiply through by x - 1 to get

ln(x) = (x - 1) / y for some y satisfying 1 < y < x.

This immediately gives both inequalities.
ln(x) < x - 1 since y > 1,
and
ln(x) > (x - 1)/x since y < x.