On 26 Jul, 23:18, Jonathan Bober <[EMAIL PROTECTED]> wrote:
> s(h,k) = h(k - 1)(2k - 1)/(6k) - (k-1)/4 -
> (1/k) sum_{j=1}^{k-1} j*floor(hj/k).
>
> (To compute the floor in the inner sum, you can just use integer
> division.)
Yes, this will be faster. Of course integer division is not faster
than floating point division, but since we are doing a sum from j = 1
to k-1 we only need to know when floor(hj/k) increases by 1. This is
simply a matter of adding h each iteration and seeing if we have gone
over k (subtracting k if we do). If so, floor(hj/k) has increased by 1
and so on.
This gets the inner computation down to about 8 cycles or so on
average. Not enough to beat Mathematica though, unless I made a
mistake in my back of the envelope computation somewhere.
Bill.
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