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You are not logged in. #1 20120830 19:49:27
Spigot algorithms and log(2)Hi; First we multiply both sides of 1) by 2^49. The RHS is split into two parts. This term on the right is the important one. We can replace 4) with This form can be rapidly computed using mods and only needs floating point arithmetic of the single precision variety. So 3) becomes: Okay, let's do the number crunching and let's try to get it right. We are only interested in the fractional part so this becomes, ( You may have noticed the smearing error in the above calculation and how to repair it.) Now we need that number in binary. So according to our calculations the 50th  56th binary digit of log(2) is 1111001 Checking this we find that we have agreement between our prediction in 7) and the answer from the 50th to the 56th digit. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #2 20120830 22:37:08
Re: Spigot algorithms and log(2)Hi bobbym The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20120831 02:48:05
Re: Spigot algorithms and log(2)Hi; and then see what you get. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 