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

[Federico Miyara]

Antoine,

Thanks for the link, very interesting.

Regards,

Federico Miyara

On 06/01/2021 05:22, Antoine Monmayrant wrote:
>
> Hello Frederico,
>
> Like Christophe, I am not sure this has anything to do with the 
> implementation of sin().
> It seems to be a known limitation of numerical calculations using 
> floating numbers.
> In particular, even with en hypothetical ideal value of %pi, because 
> of the conversion to a double, %pi*1e15 is converted to the nearest 
> available double and this "shift" or error as compared to the ideal 
> infinitely-precise value is increasing with the size of your number.
> Thus you make a x1e15 bigger error when using "%pi*1e15" than when 
> using "%pi".
> As Christophe as said, there is not much you can do, apart from 
> resorting to symbolic calculation (what Alpha does).
>
> You can see that by using nearfloat to get the distance between two 
> doubles:
>
>     nearfloat("succ", %pi)-%pi
>      ans  =   4.441D-16
>     nearfloat("succ", %pi*1e15)-%pi*1e15
>      ans  =   0.5
>
> As I said, the error on the sin() argument is getting x1e15 bigger!
>
> I'm sorry if my explanation of floating point numbers is not really good.
> This one was a revelation for me: 
> https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html .
> It helped me understand why it's usually a good idea to use "reduced 
> units" for your calculations (ie keep everything close to 1 when 
> possible).
>
> I hope it helps,
>
> Cheers,
>
> Antoine
> On 05/01/2021 09:19, Federico Miyara wrote:
>>
>> Dear all,
>>
>> In order to test the FFT on a periodic signal whose period is an 
>> exact submultiple of the FFT length I found a frequency fo satisfying 
>> this for the chosen sample rate and window length, and generated the 
>> following signal:
>>
>> x = sin(2*%pi*fo*t);
>>
>> where t is a time vector. This should be a perfectly periodic 
>> discrete signal but it isn't because the sin() function has a 
>> (virtually) exact period of pi, while %pi is exact to 16 digits only.
>>
>> For instance, we have (23 digits shown)
>>
>> --> sin(%pi)
>>  ans  =
>>    0.0000000000000001224647
>>
>> --> sin(1e10*%pi)
>>  ans  =
>>   -0.0000022393627619559233
>>
>> --> sin(1e15*%pi)
>>  ans  =
>>   -0.2362090532517409080526
>>
>> The Wolfram Alpha site yields the correct value 0 in all cases (using 
>> their own pi).
>>
>> Questions:
>>
>> 1) How is the sin() function extended to very large values of the 
>> argument? My first guess would be to compute a quarter cycle using 
>> Taylor and then extend it by symmetry and periodicity, but with which 
>> period? The best approximation available is 2*%pi. Or it is possible 
>> to use extended precision internally?
>>
>> 2) Is there a way to get a periodic discrete sine other than 
>> extending it periodically with the desired period?
>> Wouldn't this create a slight glitch at the frontier between cycles?
>>
>> Regards,
>>
>> Federico Miyara
>>
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