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Calculate approximations to log 2

by RoRi

Theorem 6.5 (page 240 of Apostol) states that if 0 < x < 1 and if m \geq 1 then

    \[ \log \frac{1+x}{1-x} = 2 \left( x + \frac{x^3}{3} + \cdots + \frac{x^{2m-1}}{2m-1} \right) + R_m (x), \]

with

    \[ \frac{x^{2m+1}}{2m+1} < R_m (x) \leq \frac{2-x}{1-x} \frac{x^{2m+1}}{2m+1}. \]

Let m = 5 and x = \frac{1}{3} and apply this theorem to approximate \log 2 with sufficient precision to obtain

    \[ 0.6931460 < \log 2 < 0.6931476. \]

By the theorem we have

    \[ \log \frac{1+x}{1-x} = 2 \left( x + \frac{x^3}{3} + \cdots + \frac{x^{2m-1}}{2m-1} \right) + R_m (x), \]

where

    \[ \frac{x^{2m-1}}{2m-1} < R_m (x) \leq \left( \frac{2-x}{1-x} \right) \left( \frac{x^{2m+1}}{2m+1} \right). \]

Therefore, letting x = \frac{1}{3} and m = 5 we have

    \[ \log \frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2 \left( \frac{1}{3} + \frac{(1/3)^3}{3} + \frac{(1/3)^5}{5} + \frac{(1/3)^7}{7} + \frac{(1/3)^9}{9} \right) + R_5 \left( \frac{1}{3} \right). \]

Therefore,

    \[ 0.693146 < \log 2 < 0.6931476. \]

One comment

  1. Anonymous says:

    The inequality for the lower bound of R_m(x) is incorrect, it should be 2m+1 not 2m-1 both in the denominator and the exponent. Typo probably.

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