Re: [math-fun] How can I accurately calculate (p/q)^n mod 1, for large n
23 Apr
2011
23 Apr
'11
11:02 p.m.
Yes, Julian was right, in fact what I was calculating was Floor[100*Mod[(3/2)^k,1]] for long sequences of consecutive k. the error was just a typo in the math-fun query. --Dan Sometimes the brain has a mind of its own.
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Dan Asimov