The auxiliary function g, and what it brings

A function built to be tiny

On [0 ; +∞[, the assignment defines:

   g(x) = ln(1 + x) − 2x/(2 + x)

The fraction 2x/(2 + x) did not fall from the sky: it is the best approximation of ln(1 + x) by a quotient of two degree-one polynomials. Saying that g is small is saying that this fraction hugs ln(1 + x).

Computing g'

g is differentiable on [0 ; +∞[: x ↦ ln(1 + x) is, because 1 + x > 0, and x ↦ 2x/(2 + x) is, as a quotient whose denominator never vanishes on [0 ; +∞[.

   g'(x) = 1/(1 + x) − [2(2 + x) − 2x] / (2 + x)²
         = 1/(1 + x) − 4/(2 + x)²
         = [(2 + x)² − 4(1 + x)] / [(1 + x)(2 + x)²]
         = x² / [(1 + x)(2 + x)²]

The numerator x² is non-negative; for x ≥ 0 we have 1 + x ≥ 1 and (2 + x)² ≥ 4, so the denominator is at least 4:

   0  ≤  g'(x)  ≤  x²/4

From g' to g

g(0) = ln(1) − 0 = 0. Integrating the bound on g' between 0 and x:

   0  ≤  g(x)  ≤  x³/12

In other words, for every x ≥ 0:

   2x/(2 + x)  ≤  ln(1 + x)  ≤  2x/(2 + x) + x³/12

The error is of order x³, against x² for the bound of the previous lesson. That extra notch of precision is what will yield limits, and no longer merely bounds.

The identity that makes everything work

One computation worth remembering — it will serve throughout part II:

   1 / (2x/(2 + x)) − 1/x = (2 + x)/(2x) − 1/x = x/(2x) = 1/2

So the reciprocal of the approximation of ln(1 + x) equals exactly 1/x + 1/2. The rest of the problem consists in controlling the gap between 1/ln(1 + x) and that value.