[Scilab-users] Problems arising from truncation of %pi

[Stéphane Mottelet]

I do not agree with your answer. Doing *by hand* the shift to a 
principal period  of sin () solves the problem. For example, here is a 
plot of sin(x) vs sin(x-floor(x/2/%pi)*%pi*2) for large values of x:

x = (10^[1:16])*%pi
sx = sin(x-floor(x/2/%pi)*%pi*2)
plot("ln",x,sin(x),'-o',x,sx,'-o')
legend("$\Large\sin x$","$\Large\sin 
\left(x-2\pi\left\lfloor\frac{x}{2\pi}\right\rfloor\right)$",3)

error

I suppose that the hardware implementation used by the compiler when 
using the standard math library does not use this simple trick because 
it would decrease the floating point performance.


S.

Le 06/01/2021 à 09:14, Jean-Yves Baudais a écrit :
> Hello,
>
> Le 05/01/2021 à 09:19, Federico Miyara a écrit :
>> --> sin(%pi)
>>   ans  =
>>     0.0000000000000001224647
>
>
> You face the limited precision of all numerical calculus. See
> --> help %eps
>
> It has no sense to use 23 digits with a precision of 2.22E-16, the 7th 
> last digits are noise.
>
>
>> --> sin(1e10*%pi)
>>   ans  =
>>    -0.0000022393627619559233
>>
>> --> sin(1e15*%pi)
>>   ans  =
>>    -0.2362090532517409080526
>
>
> All is consistent with the definition of the precision. A toy example:
> --> a=2;
> --> b=sqrt(2);
> --> a-b^2
>
> You expect zero because you do symbolic calculus, not Scilab.
>
>
>> The Wolfram Alpha site yields the correct value 0 in all cases (using 
>> their own pi).
>
>
> Because it uses symbolic calculus. So, if you want more precision with 
> Scilab you should change the standard used, but maybe some tricks in 
> your code can solve the problem... Or maybe you need symbolic calculus 
> tool.
>
> --Jean-Yves
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-- 
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688
http://www.utc.fr/~mottelet

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