Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/260

This series converges for $x=1$, and we can find $\log 2$ by placing $x=1$ in (A), giving

$\log 2=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .$

But this series is not well adapted to numerical computation, because it converges so slowly that it would be necessary to take 1000 terms in order to get the value of $\log 2$ correct to three decimal places. A rapidly converging series for computing logarithms will now be deduced.

By the theory of logarithms,

(B)

$\log {\frac {1+x}{1-x}}=\log \left(1+x\right)-\log \left(1-x\right)$

By 8, §1

Substituting in (B) the equivalent series for $\log(1+x)$ and $log(1-x)$ found in Exs. 6 and 7 §145, we get^[1]

(C)	$\log {\frac {1+x}{1-x}}=2\left[x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots \right],$

which is convergent when $x$ is numerically less than unity. Let

(D)	${\frac {1+x}{1-x}}={\frac {M}{N}},$ whence $x={\frac {M-N}{M+N}},$

and we see that $x$ will always be numerically less than unity for all positive values of $M$ and $N$. Substituting from (D) into (C), we get

(E)	${\begin{aligned}\log {\frac {M}{N}}&=\log M-\log N\\&=2\left[{\frac {M-N}{M+N}}+{\frac {1}{3}}\left({\frac {M-N}{M+N}}\right)^{3}+{\frac {1}{5}}\left({\frac {M-N}{M+N}}\right)^{5}+\cdots \right],\\\end{aligned}}$

a series which is convergent for all positive values of $M$ and $N$; and it is always possible to choose $M$ and $N$ so as to make it converge rapidly.

Placing $M=2$ and $N=1$ in (E), we get

$\log 2=2\left[{\frac {1}{3}}+{\frac {1}{3}}\cdot {\frac {1}{3^{3}}}+{\frac {1}{5}}\cdot {\frac {1}{3^{5}}}+{\frac {1}{7}}\cdot {\frac {1}{3^{7}}}+\cdots \right]=0.69314718\cdots .$

[ Since $\log N=\log 1=0$, and ${\tfrac {M-N}{M+N}}={\tfrac {1}{3}}$.]

1. The student should notice that we have treated the series as if they were ordinary sums, but they are not; they are limits of sums. To justify this step is beyond the scope of this book.