Thank you Prof. Maechler! Indeed 0.1 is not the same as 1/10 and there are
many ways to get more precise versions of it. I will look into how to take
inputs as strings. Thank you!
Best,
Khue Tran
On Sat, Jul 20, 2024 at 4:40 AM Duncan Murdoch <murdoch.dun...@gmail.com>
wrote:
> Just for your info:
>
> > exp(mpfr("0.1", 152))
> 1 'mpfr' number of precision 152 bits
> [1] 1.1051709180756476248117078264902466682245471948
>
> So it looks like your approach of parsing the numbers yourself makes sense.
>
> You can find the R parser here:
>
> https://github.com/wch/r-source/blob/b64422334a8269535718efd9a1f969c94b103056/src/main/gram.y#L2577-L2705
>
> I don't know how much of that you want to replicate, but I suppose
> handling the weird cases (e.g. 0x1p1) the way R does will make it easier
> for your users.
>
> Duncan Murdoch
>
> On 2024-07-18 4:29 p.m., Khue Tran wrote:
> > Hi,
> >
> > I am trying to create an Rcpp package that involves arbitrary precise
> > calculations. The function to calculate e^x below with 100 digits
> precision
> > works well with integers, but for decimals, since the input is a double,
> > the result differs a lot from the arbitrary precise result I got on
> > Wolfram.
> >
> > I understand the results are different since 0.1 cannot be represented
> > precisely in binary with limited bits. It is possible to enter 1 then 10
> > and get the multiprecision division of these two integers to attain a
> more
> > precise 0.1 in C++, but this method won't work on a large scale. Thus, I
> am
> > looking for a general solution to get more precise inputs?
> >
> > The dummy example is as follows:
> >
> > library(Rcpp)
> >
> > sourceCpp(code = '
> > #include <boost/multiprecision/cpp_dec_float.hpp>
> > #include <boost/math/special_functions/expm1.hpp>
> >
> > // [[Rcpp::depends(BH)]]
> > // [[Rcpp::export]]
> > std::string calculateExp100(double x) {
> > boost::multiprecision::cpp_dec_float_100 bx = x;
> > boost::multiprecision::cpp_dec_float_100 expx = exp(bx);
> > return expx.str(50);
> > }
> > ')
> >
> > [1] represents the R output and [W] for Wolfram results
> >
> > calculateExp100(1) # Agrees with Wolfram answer all the way to the last
> > digit
> > [1] "2.7182818284590452353602874713526624977572470937"
> > [W] 2.7182818284590452353602874713526624977572470936999595749669676277...
> >
> > calculateExp100(0.1) # Differs pretty significantly from Wolfram's answer
> > [1] "1.1051709180756476309466388234587796577416634163742"
> > [W] 1.1051709180756476248117078264902466682245471947375187187928632894...
> >
> > I am currently trying to get precise inputs by taking strings instead of
> > numbers then writing a function to decompose the string into a rational
> > with the denominator in the form of 10^(-n) where n is the number of
> > decimal places. I am not sure if this is the only way or if there is a
> > better method out there that I do not know of, so if you can think of a
> > general way to get precise inputs from users, it will be greatly
> > appreciated!
> >
> > Thank you!
> > Best,
> > Khue Tran
> >
>
>
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