Given a binary number that repeats, for example 0.(0011) or 0.0(101), how would one go about converting it to decimal?
What I've been able to dig up so far is the simple method for converting a terminating binary number to decimal, as below:
res(N+2) = res(N+1) / 2 + res(N)
where res is the result after step N, and N is the current iteration (N=0; n->(num binary digits)). Applying that repeatedly to a nonterminating binary number gives a good approximation, for example
dec:0.4 || bin: 0.(0110):
0 / 2 + 0 = 0
0 / 2 + 0 = 0
0 / 2 + 1 = 1
1/2 / 2 + 1 = 3/2
3/2 / 2 + 0 = 3/4
3/4 / 2 + 0 = 3/8
3/8 / 2 + 1 = 19/16
19/16 / 2 + 1 = 51/32
51/32 / 2 + 0 = 51/64
51/64 / 2 + 0 = 51/128 = 0.3984
which is approximately 0.4.
So, I've got a means of calculating approximates, but I'm struggling with finding a way to express this. I've started trying to write it as a series I can compute at the limit as n->inf without too much success so far.
One way to get an exact answer is using infinite geometric series. The infinite sum of powers of a fraction r, for exponents 1 to infinity, 0 <= r < 1, is r/(1-r).
In your example, 0.(0011), 0.0011 represents the fraction 3/16. Factor out the 3 and you get r=1/16. r/(1-r) = (1/16)/(15/16) = 1/15. Multiply that by the 3 you factored out and you get your answer: 3/15 = 1/5 = 0.2.