which the exponents of 2 are greater than or equal to zero, and a "tail", in which the exponents of 2 are negative: We are only interested in the fractional part of this value, so we can replace each of the summands in the "head" by Calculating each of these terms and adding them to a running total where we again only keep the fractional part, we have: {| class="wikitable" |- ! k ! A = 27−k ! B = A mod k ! C = ! Sum of C mod 1 |- | 1 | 64 | 0 | 0 | 0 |- | 2 | 32 | 0 | 0 | 0 |- | 3 | 16 | 1 | | |- | 4 | 8 | 0 | 0 | |- | 5 | 4 | 4 | | |- | 6 | 2 | 2 | | |- | 7 | 1 | 1 | | |} We add a few terms in the "tail", noting that the error introduced by truncating the sum is less than the final term: {| class="wikitable" |- ! k ! D = ! Sum of D ! Maximum error |- | 8 | | | |- | 9 | | | |- | 10 | | | |} Adding the "head" and the first few terms of the "tail" together we get: so the 8th to 11th binary digits in the binary expansion of ln(2) are 1, 0, 1, 1. Note that we have not calculated the values of the first seven binary digits – indeed, all information about them has been intentionally discarded by using modular arithmetic in the "head" sum. The same approach can be used to calculate digits of the binary expansion of ln(2) starting from an arbitrary nth position. The number of terms in the "head" sum increases linearly with n, but the complexity of each term only increases with the logarithm of n if an efficient method of modular exponentiation is used. The precision of calculations and intermediate results and the number of terms taken from the "tail" sum are all independent of n, and only depend on the number of binary digits that are being calculated – single precision arithmetic can be used to calculate around 12 binary digits, regardless of the starting position. References Further reading Arndt, Jorg; Haenel, Christoph, unleashed, Springer Verlag,