[R] : Ramanujan and the accuracy of floating point computations - using Rmpfr in R

Ravi Varadhan
2015-07-03T19:01:52Z

Thank you all.  I did think about declaring `pi' as a special constant, but
for some reason didn't actually try it.
Would it be easy to have the mpfr() written such that its argument is
automatically of extended precision? In other words, if I just called:
mpfr(exp(sqrt(163)*pi, 120), then all the constants, 163, pi, are automatically
of 120 bits precision.

Is this easy to do?

Best,
Ravi
 ________________________________________
From: David Winsemius <dwinsemius at comcast.net>
Sent: Friday, July 3, 2015 2:06 PM
To: John Nash
Cc: r-help; Ravi Varadhan
Subject: Re: [R] : Ramanujan and the accuracy of floating point computations -
using Rmpfr in R

> On Jul 3, 2015, at 8:08 AM, John Nash <John.Nash at uottawa.ca>
wrote:
>
>
>
>
> Third try -- I unsubscribed and re-subscribed. Sorry to Ravi for extra
> traffic.
>
> In case anyone gets a duplicate, R-help bounced my msg "from
non-member" (I changed server, but not email yesterday,
> so possibly something changed enough). Trying again.
>
> JN
>
> I got the Wolfram answer as follows:
>
> library(Rmpfr)
> n163 <- mpfr(163, 500)
> n163
> pi500 <- mpfr(pi, 500)
> pi500
> pitan <- mpfr(4, 500)*atan(mpfr(1,500))
> pitan
> pitan-pi500
> r500 <- exp(sqrt(n163)*pitan)
> r500
> check <-
"262537412640768743.99999999999925007259719818568887935385..."
> savehistory("jnramanujan.R")
>
> Note that I used 4*atan(1) to get pi.

RK got it right by following the example in the help page for mpfr:

Const("pi", 120)

The R `pi` constant is not recognized by mpfr as being anything other than
another double .


There are four special values that mpfr recognizes.

?
Best;
David


> It seems that may be important,
> rather than converting.
>
> JN
>
> On 15-07-03 06:00 AM, r-help-request at r-project.org wrote:
>
>> Message: 40
>> Date: Thu, 2 Jul 2015 22:38:45 +0000
>> From: Ravi Varadhan <ravi.varadhan at jhu.edu>
>> To: "'Richard M. Heiberger'" <rmh at
temple.edu>, Aditya Singh
>>      <aps6dl at yahoo.com>
>> Cc: r-help <r-help at r-project.org>
>> Subject: Re: [R] : Ramanujan and the accuracy of floating point
>>      computations - using Rmpfr in R
>> Message-ID:
>>      <14ad39aaf6a542849bbf3f62a0c2f38f at
DOM-EB1-2013.win.ad.jhu.edu>
>> Content-Type: text/plain; charset="utf-8"
>>
>> 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:rmh at 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 at
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]]
>>>>>>
>>>>>> ______________________________________________
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>>>>>> PLEASE do read the posting guide
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>>>>>> and provide commented, minimal, self-contained,
reproducible code.
>>>> ______________________________________________
>>>> R-help at 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.
>> ------------------------------