4 ms·
The distance you cover with step b is about b/(log(10) 10^k). The exact distance is log10((10^k + b)/10^k). That's a funny way of saying fracPart(log10(10^k + b
by SamReidHughes 7y ago
The distance you cover with step b is about b/(log(10) 10^k). The exact distance is log10((10^k + b)/10^k). That's a funny way of saying fracPart(log10(10^k + b)).
The number of steps tells you how big the last step is.
- markisus 7y agoOkay I think I buy it. Great proof, thanks!