5.5 $^*$Brouncker's Formula For $\ln (2)$

The following calculation of $\ln (2)$ is due to William Brouncker (1620-1684)[22, page 54].

Let $${ P_{2^n} = \{ x_0, x_1, \cdots , x_{2^n}\} }$$ denote the regular partition of the interval $[1,2]$ into $2^n$ equal subintervals. Let

$$K(2^n) = I_1^2(\Big[{1\over t}\Big], P_{2^n}) = \bigcup_{i=1}^{2^n} B(x_{i-1},x_i;0,{1\over x_i}).$$

We can construct $K(2^{n+1})$ from $K(2^n)$ by adjoining a box of width $${1\over 2^{n+1}}$$ to the top of each box $B(x_{i-1},x_i;0,{1\over x_i})$ that occurs in the definition of $K(2^n)$ (see figures a) and b)).

We have

$$\alpha(K(1)) = \alpha(B(1,2;0,{1\over 2})) = 1\cdot {1\over 2} = {1\over 2}.$$

From figure a) we see that

$$\begin{eqnarray*} \alpha(K(2)) &=& \alpha(K(1)) + \alpha(B({2\over 2},{3\over 2}... ...ver 3} - {1\over 4}\Big) \\ &=& {1\over 2} + {1\over 3\cdot 4}. \end{eqnarray*}$$

From figure b) we see that

$$\begin{eqnarray*} \alpha(K(4)) &=& \alpha(K(2)) + \alpha(B({4\over 4},{5\over 4}... ...2} + {1\over 3\cdot 4} + {1\over 5\cdot 6} + {1\over 7\cdot 8}. \end{eqnarray*}$$

In general we will find that

$$\alpha(K(2^n)) = \sum_{j=1}^{2^n} {1\over (2j-1)(2j)}.$$

Now

$$0 \leq \alpha(S_1^2(\Big[{1\over t}\Big])) - \alpha(K(2^n)) \leq (1-{1\over 2})\mu(P_{2^n}),$$

i.e.

$$0 \leq \ln(2) - \sum_{j=1}^{2^n} {1\over (2j-1)(2j)} \leq {1\over 2^{n+1}}.$$

Thus

$$\ln(2) = \sum_{j=1}^{2^n}{1\over (2j-1)(2j)} \mbox{ with an error smaller than } {1\over 2^{n+1}}.$$

We can think of $\ln (2)$ as being given by the ``infinite sum"

$$\ln (2)={1\over {1\cdot 2}}+{1\over {3\cdot 4}}+{1\over {5\cdot 6}}+{1\over {7\cdot 8}}+\cdots .$$ (5.82)

Equation (5.82) is sometimes called Mercator's expansion for $\ln (2)$, after Nicolaus Mercator, who found the result sometime near 1667 by an entirely different method.

Brouncker's calculation was published in 1668, but was done about ten years earlier [22, pages 56-56].

Brouncker's formula above is an elegant result, but it is not very useful for calculating: it takes too many terms in the sum to get much accuracy. Today, when a logarithm can be found by pressing a button on a calculator, we tend to think of ``$\ln (2)$" as being a known number, and of Brouncker's formula as giving a ``closed form" for the sum of the infinite series $${ {1\over {1\cdot 2}}+{1\over {3\cdot 4}}+{1\over {5\cdot 6}}+\cdots}$$.