Hi,
Considering a fixed-point binary number, and the weight of its bits, is there any formula or method for finding the nth root of such a number? I decided to keep it simple and start by considering only integers and with deducing a pattern for finding theirs squares. I hope that the pattern will be reversible, from squares to square roots. After which, I can go ahead to check that for cubes and cube roots, then ... nth power to nth root. After learning something from integers, I hope that I might be able to pinpoint something for fixed-point numbers.
I have been staring at the following to see whether I can find a pattern for a start:
Decimal number = binary; sq(decimal number) = binary
1 (base 10) = 2^(0 base 10) = 01 (base 2); 1 (base 10) = 2^(0 base 10) = 01 (base 2); -- no shift, no trailing zero
2 (base 10) = 2^(1 base 10) = 10 (base 2); 4 (base 10) = 2^(2 base 10) = 100 (base 2); --shift by one place, two trailing zeros
3 (base 10) = 2^(1 base 10) + 2^(0 base 10) = 11 (base 2); 9 (base 10) = 2^(3 base 10) + 2^(0 base 10) = 1001 (base 2); -- shift by two places, three trailing zeros and no shift, no trailing zero
4 (base 10) = 2^(2 base 10) = 100 (base 2); 16 (base 10) = 2^(4 base 10) = 10000 (base 2); -- shift by two places, four trailing zeros
5 (base 10) = 2^(2 base 10) + 2^(0 base 10) = 0101 (base 2); 25 (base 10) = 2^(4 base 10) + 2^(3 base 10) + 2^(0 base 10) = 11001 (base 2); -- shift by two places, four trailing zeros and shift by one place, three trailing zeros and no shift, no trailing zero.
And so on
Observations:
1) For every odd number, the LSB is 1. For even numbers, the LSB is 0.
2) the powers of 2 seem to play a role in finding a pattern
3) It seems like there are fundamental numbers (like 1, 2, 4, 8, etc which are powers of 2) whose patterns can be followed to build the powers of non-fundamental numbers (like 3, 5, 6, 7, 9, 10, etc which are built from fundamental numbers). Also, it seems like patterns could be deduced from powers of fundamental numbers to build powers of non-fundamental numbers.
4) etc
Please, are there anything you observe? Do you think this might be a good approach to finding a solution to this long-standing problem?
Any feedback is highly appreciated.