Re: [Numpy-discussion] Re: numpy.sin

[Alexander B. <ale...@gm...>]

I agree with Robert.  In fact on the FPUs such as x87, where floating
point registers have extended precision, sin(x % (2%pi)) will give you
a less precise answer than sin(x).  The improved presision that you
see is illusory because gives that 64 bit pi is not a precise
mathematical pi, sin(2*pi) is not 0.

On 5/25/06, Robert Kern <rob...@gm...> wrote:
> Alan G Isaac wrote:
> > On Thu, 25 May 2006, Robert Kern apparently wrote:
>
> >>That your demonstration results in the desired exact 0.0
> >>for multiples of 2*pi is an accident. The results for
> >>values other than integer multiples of pi will be as wrong
> >>or more wrong.
> >
> > It seems that a continuity argument should undermine that as
> > a general claim.  Right?
>
> Let me clarify. Since you created your values by multiplying the floating=
-point
> approximation pi by an integer value. When you perform the operation % (2=
*pi) on
> those values, the result happens to be exact or nearly so but only becaus=
e you
> used the same approximation of pi. Doing that operation on an arbitrary v=
alue
> (like 1000000) only introduces more error to the calculation. Floating-po=
int
> sin(1000000.0) should return a value within eps (~2**-52) of the true,
> real-valued function sin(1000000). Calculating (1000000 % (2*pi)) introdu=
ces
> error in two places: the approximation pi and the operation %. A floating=
-point
> implementation of sin(.) will return a value within eps of the real sin(.=
) of
> the value that is the result of the floating-point operation (1000000 % (=
2*pi)),
> which already has some error accumulated.
>
> --
> Robert Kern
>
> "I have come to believe that the whole world is an enigma, a harmless eni=
gma
>  that is made terrible by our own mad attempt to interpret it as though i=
t had
>  an underlying truth."
>   -- Umberto Eco
>
>
>
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