3 ms·
I’m not sure that’s efficient. The number of digits in that tenth power grows very rapidly (59049^¹⁰ already has 47 digits; (59049¹⁰)¹⁰ around 477) and there’s
by Someone 3y ago
I’m not sure that’s efficient. The number of digits in that tenth power grows very rapidly (59049^¹⁰ already has 47 digits; (59049¹⁰)¹⁰ around 477) and there’s the risk that rounding introduces errors (did you really do that with “a 4 function calculator”?)
On the other hand, (hand-waving) it takes many generated digits before the “far away” get shifted left of the decimal point, so numerical analysis probably can show you don’t need them all to reach a target number of digits.
- nickcw 3y agoThe intermediate numbers never get bigger than 10^10 - if you look at the python program I posted you might find that easier to see.
- Someone 3y agoThat’s true, but the number of digits to the right of the decimal point grows rapidly.
- mikewarot 3y ago>did you really do that with “a 4 function calculator”? I have in the past, this time I used the 4 function calculator mode of Windows 10's calculator. x^2, MS, x^2, x^2 * MR = on the loops, then divide by 1+n zeros, repeat. I used NotePad++ to record the data as I went. Doing it for a binary logarithm would be a lot easer, because then it's square and optionally divide by 2.
- Someone 3y ago> this time I used the 4 function calculator mode of Windows 10's calculator AFAIK, that calculator has “Infinite precision for basic arithmetic operations (addition, subtraction, multiplication, division) so that calculations never lose precision” (https://github.com/microsoft/calculator https://github.com/microsoft/calculator)