Re: limited accuracy for tanh-sinh integrator

From: kila suelika (semeegozy_at_[hidden])
Date: 2020-08-21 12:30:58

*First problem - more precision*
Check the stopping condition of boost.math, it says the tolerance is
\sqrt{\epsilon}. So if you use 100-digits precision, then tolerance is
roughly 10^-50. For more precision, you have to use more digits such as
1000 digits by using BF= number<cpp_dec_float<1000> >; . Full code:

#include<iostream>
#include<boost/math/quadrature/tanh_sinh.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>

using namespace std;
using namespace boost::math::quadrature;
using namespace boost::multiprecision;
using BF= number<cpp_dec_float<1000> >;

int main(){
auto f2 = [](BF x) { return x/BF(sqrt(x*x-BF("0.25"))); };
tanh_sinh<BF> integrator;

std::cout << std::setprecision(std::numeric_limits<BF>::max_digits10)
     << "Comp.: " << integrator.integrate(f2, BF("0.5"), sqrt(BF("1.25")))
<< std::endl;

    return 0;
};

Be patient, this run about 10 mins in my single core server.

*Second problem - error*
I didn't reproduce it. Just modify my code and everything is OK.

On Fri, Aug 21, 2020 at 9:18 AM Anirban Pal via Boost-users <
boost-users_at_[hidden]> wrote:

> Thank you for your response. I’ve run into a few more issues with
> accuracy. Here is my function, which has a singularity at 0.5 and I'm
> integrating it from 0.5 to sqrt(1.25):
>
> using BF = boost::multiprecision::cpp_bin_float_100;
>
> auto f1 = [](BF x) { return x/BF(sqrt(x*x-BF("0.25"))); };
> BF Q1 = integrator.integrate(f1, BF("0.5"), BF(sqrt(BF("1.25"))));
>
> This gives me a value of Q1 which is close to the true value of 1 with an
> error of O(10^-51). I am curious if this should be lower, like O(10^-100).
> (I hope I’m not being greedy!)
>
> Nevertheless, if I try a related function.
>
> auto f2 = [](BF x) { return x/BF(sqrt(x*x-BF("0.25"))) - x; };
> BF Q2 = integrator.integrate(f2, BF("0.5"), BF(sqrt(BF("1.25"))));
>
> or an equivalent one:
>
> auto f3 = [](BF x) { return (x - x*BF(sqrt(x*x-BF("0.25"))) )/BF(sqrt(x*x-BF("0.25"))); };
> BF Q3 = integrator.integrate(f3, BF("0.5"), BF(sqrt(BF("1.25"))));
>
> I get an error message:
>
> boost_test5: /usr/local/include/boost/math/quadrature/tanh_sinh.hpp:196: boost::math::quadrature::tanh_sinh<Real, Policy>::integrate(F, Real, Real, Real, Real*, Real*, std::size_t*)::<lambda(Real, Real)> [with F = main(int, char**)::<lambda(BF)>; Real = boost::multiprecision::number<boost::multiprecision::backends::cpp_bin_float<100> >; Policy = boost::math::policies::policy<boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy>; result_type = boost::multiprecision::number<boost::multiprecision::backends::cpp_bin_float<100> >]: Assertion `position != a' failed.
> Aborted (core dumped)
>
> I'm not sure how I could improve my code. I tried using
> boost::multiprecision::sqrt but i didn't notice any difference.
>
> Thank you for your attention.
>
>
>
> On Mon, 17 Aug 2020 at 04:47, Paul A. Bristow via Boost-users
> boost-users_at_[hidden] <http://mailto:[e-mail]>
> wrote:
>
>
>>
>> > -----Original Message-----
>> > From: Boost-users <boost-users-bounces_at_[hidden]> On Behalf Of
>> John Maddock via Boost-users
>> > Sent: 16 August 2020 21:24
>> > To: Anirban Pal via Boost-users <boost-users_at_[hidden]>
>> > Cc: John Maddock <jz.maddock_at_[hidden]>
>> > Subject: Re: [Boost-users] limited accuracy for tanh-sinh integrator
>> >
>> >
>> > On 16/08/2020 21:15, Anirban Pal via Boost-users wrote:
>> > >
>> > > Hello,
>> > >
>> > > I’m using the tanh-sinh integrator to integrate a simple function f(x)
>> > > = 0.26*x from 3.0 to 4.0.
>> > > The exact result is 0.91. With the integrator I’m getting a result
>> > > accurate to only 10^-18 with cpp_bin_float_100 multiprecision.
>> > >
>> > You have the double precision constant 0.26 in your code, and since
>> this is an inexact binary value,
>> > everything that depends on that constant is inherently limited to
>> double precision.  Try constructing it
>> > as Real(26) / 100.
>>
>> Everyone falls into this pit, and in my case, repeatedly ☹
>>
>> You need to be ever vigilant not to use double precision items by mistake.
>>
>> Using numeric_limits<>::max_digits10 is useful because it shows the
>> digits beyond double at about 17 decimal digits.  Being random should raise
>> suspicions.
>>
>> If you are only using multiprecision types, then always using decimal
>> digit strings like "1.23456" is simple.
>>
>> If you can use a fraction like Real(1)/1000 that also works fine.
>>
>> You may also find the macro BOOST_MATH_TEST_VALUE  in
>> I:\boost\libs\math\test\test_value.hpp helpful if you want to make code
>> portable over different floating-point types from fundamental to
>> multiprecision?
>>
>> Paul
>>
>>
>> _______________________________________________
>> Boost-users mailing list
>> Boost-users_at_[hidden]
>> https://lists.boost.org/mailman/listinfo.cgi/boost-users
>>
> --
> Anirban Pal
>
>
>
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