Hi Rich, The Wolfram answer is correct. http://mathworld.wolfram.com/RamanujanConstant.html
There is no code for Wolfram alpha. You just go to their web engine and plug in the expression and it will give you the answer. http://www.wolframalpha.com/ I am not sure that the precedence matters in Rmpfr. Even if it does, the answer you get is still wrong as you showed. Thanks, Ravi -----Original Message----- From: Richard M. Heiberger [mailto:r...@temple.edu] Sent: Thursday, July 02, 2015 6:30 PM To: Aditya Singh Cc: Ravi Varadhan; r-help Subject: Re: [R] : Ramanujan and the accuracy of floating point computations - using Rmpfr in R There is a precedence error in your R attempt. You need to convert 163 to 120 bits first, before taking its square root. > exp(sqrt(mpfr(163, 120)) * mpfr(pi, 120)) 1 'mpfr' number of precision 120 bits [1] 262537412640768333.51635812597335712954 ## just the last four characters to the left of the decimal point. > tmp <- c(baseR=8256, Wolfram=8744, Rmpfr=8333, wrongRmpfr=7837) > tmp-tmp[2] baseR Wolfram Rmpfr wrongRmpfr -488 0 -411 -907 > You didn't give the Wolfram alpha code you used. There is no way of verifying the correct value from your email. Please check that you didn't have a similar precedence error in that code. Rich On Thu, Jul 2, 2015 at 2:02 PM, Aditya Singh via R-help <r-help@r-project.org> wrote: > > Ravi > > I am a chemical engineer by training. Is there not something like law of > corresponding states in numerical analysis? > > Aditya > > > > ------------------------------ > On Thu 2 Jul, 2015 7:28 AM PDT Ravi Varadhan wrote: > >>Hi, >> >>Ramanujan supposedly discovered that the number, 163, has this interesting >>property that exp(sqrt(163)*pi), which is obviously a transcendental number, >>is real close to an integer (close to 10^(-12)). >> >>If I compute this using the Wolfram alpha engine, I get: >>262537412640768743.99999999999925007259719818568887935385... >> >>When I do this in R 3.1.1 (64-bit windows), I get: >>262537412640768256.0000 >> >>The absolute error between the exact and R's value is 488, with a relative >>error of about 1.9x10^(-15). >> >>In order to replicate Wolfram Alpha, I tried doing this in "Rmfpr" but I am >>unable to get accurate results: >> >>library(Rmpfr) >> >> >>> exp(sqrt(163) * mpfr(pi, 120)) >> >>1 'mpfr' number of precision 120 bits >> >>[1] 262537412640767837.08771354274620169031 >> >>The above answer is not only inaccurate, but it is actually worse than the >>answer using the usual double precision. Any thoughts as to what I am doing >>wrong? >> >>Thank you, >>Ravi >> >> >> >> [[alternative HTML version deleted]] >> >>______________________________________________ >>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >>https://stat.ethz.ch/mailman/listinfo/r-help >>PLEASE do read the posting guide >>http://www.R-project.org/posting-guide.html >>and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
