Near 1 fractions and computing log(2)

 Near 1 fractions are:

2/1   3/2   4/3  5/4  ...

i.e. fractions  (n+1)/n  where  n  is an arbitrary natural number.

When  n  gets larger, there is a tendency for a computation of  log(n) to get faster while a direct computation of  log(2)  is relatively slow. Thus, in order to speed up computing  log(2)  we may try to represent  2  as a product of near 1 fractions that have large denominators (hence numerators too). Let's start modestly:

2  =  (3/2) * (4/3)

hence

log(2)  =  log(3/2) + log(4/3)

One should be able to compute  log(3/2)  faster than  log(2);  and a computation of  log(4/3)  should be still faster. However, it's not clear that computing both logarithms  log(3/2)  and  log(4/3) takes less time than the single direct calculation of  log(2). So, we would like to use fractions with larger denominators.

There is only representation of  2  as a product of two near 1 fractions (namely the one above). Are we stuck?

Not really because we can decompose  2  into a product of two powers of near 1 fractions. This works as follows:

3/2  =  (4/3) * (9/8)

and now:

2  =  (4/3)^2 * (9/8)

hence

log(2)  =  2*log(4/3) + log(9/8)

This time there is a good chance that by applying the above formula we can compute  log(2). faster than by a direct computation. I'll leave the detailed discussion of the computing efficiency for later.



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