RE: Accuracy and Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg36899] RE: Accuracy and Precision*From*: "DrBob" <drbob at bigfoot.com>*Date*: Tue, 1 Oct 2002 04:45:37 -0400 (EDT)*Reply-to*: <drbob at bigfoot.com>*Sender*: owner-wri-mathgroup at wolfram.com

Actually, we don't know whether SetAccuracy "succeeds", because we don't know how inexact those numbers really are. If they are known to more digits than shown in the original post, they should be entered with as much precision as they deserve. If not, there's no trick or algorithm that will give the result precision, because the value of f really is "in the noise". That is, we have no idea what the value of f should be. Mathematica's failing is in returning a value without pointing out that it has no precision. Bobby -----Original Message----- From: Allan Hayes [mailto:hay at haystack.demon.co.uk] To: mathgroup at smc.vnet.net Subject: [mg36899] Re: Accuracy and Precision Andrzej, Bobby, Peter It looks as if using SetAccuracy succeeds here because the inexact numbers that occur have finite binary representations. If we change them slightly to avoid this then we have to use Rationalize: 1) Using SetAccuracy Clear[a,b,f] f=SetAccuracy[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b^8+a/(2*b), Infinity]; a=77617.1; b=33096.1; a=SetAccuracy[a,Infinity];b=SetAccuracy[b,Infinity]; f -1564032114908486835197494923989618867972540153873951942813115514949 3891236234\ 525007719168693704591197760187988046304361497869199129319625743010292363 1246 75\ /10867106143970760551000357827554793888198143135975649579607989867743572 8240 16\ 0653953612982932181371232436367739737604096 2) Rewriting as fractions a=776171/10; b=330961/10; f=33374/100*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+54/10*b^8+a/(2*b) -(5954133808997234115690303589909929091649391296257/ 41370125000000) 3) Using Rationalize Clear[a,b,f] f=Rationalize[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b^8+a/(2*b), 0]; a=77617.1; b=33096.1; a=Rationalize[a,0];b=Rationalize[b,0]; f -(5954133808997234115690303589909929091649391296257/ 41370125000000) I use Rationalize[. , 0] besause of results like Rationalize[3.1415959] 3.1416 Rationalize[3.1415959,0] 31415959/10000000 -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Andrzej Kozlowski" <andrzej at tuins.ac.jp> wrote in message news:an8s8i$6pk$1 at smc.vnet.net... > Well, first of of all, your using SetAccuracy and SetPrecision does > nothing at all here, since they do not change the value of a or b. You > should use a = SetAccuracy[a, Infinity] etc. But even then you won't > get the same answer as when you use exact numbers because of the way > you evaluate f. Here is the order of evaluation that will give you the > same answer, and should explain what is going on: > > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121* > b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity]; > > a = 77617.; > > > b = 33096.; > > a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity]; > > f > > 54767 > -(-----) > 66192 > > Andrzej Kozlowski > Toyama International University > JAPAN > > > > On Sunday, September 29, 2002, at 03:55 PM, Peter Kosta wrote: > > > Could someone explain what is going on here, please? > > > > In[1]:= > > a = 77617.; b = 33096.; > > > > In[2]:= > > SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; > > SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; > > > > In[4]:= > > f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + > > a/(2*b) > > > > In[5]:= > > SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; > > > > In[6]:= > > f > > > > Out[6]= > > -1.1805916207174113*^21 > > > > In[7]:= > > a = 77617; b = 33096; > > > > In[8]:= > > g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + > > (55/10)*b^8 + a/(2*b) > > > > In[9]:= > > g > > > > Out[9]= > > -(54767/66192) > > > > In[10]:= > > N[%] > > > > Out[10]= > > -0.8273960599468214 > > > > Thanks, > > > > PK > > > > > > > >