> On Jul 3, 2015, at 8:08 AM, John Nash <john.n...@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-requ...@r-project.org wrote:
>
>> Message: 40
>> Date: Thu, 2 Jul 2015 22:38:45 +0000
>> From: Ravi Varadhan <ravi.varad...@jhu.edu>
>> To: "'Richard M. Heiberger'" <r...@temple.edu>, Aditya Singh
>> <aps...@yahoo.com>
>> Cc: r-help <r-help@r-project.org>
>> Subject: Re: [R] : Ramanujan and the accuracy of floating point
>> computations - using Rmpfr in R
>> Message-ID:
>> <14ad39aaf6a542849bbf3f62a0c2f...@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: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.
David Winsemius, MD
Alameda, CA, USA
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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