Re: Accuracy and Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg36932] Re: Accuracy and Precision*From*: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>*Date*: Wed, 2 Oct 2002 03:32:42 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

You are of course right, I forgot that Mathematica does not try to keep precision or accuracy of machine arithmetic computations. But I think the actual precision you set need not be higher than machine precision, it just has to be set explicitely (is that right?). For example: In[1]:= Clear[f,a,b,k] In[2]:= k = $MachinePrecision; In[3]:= 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),k]; In[4]:= a=77617.;b=33096.; In[5]:= a=SetAccuracy[a,k]; In[6]:= b=SetAccuracy[b,k]; In[7]:= f Out[7]= \!\(\(-5.51716400890319`-2.8311*^19\)\) In[8]:= Accuracy[f] From In[8]:= Accuracy::mnprec: Value -23 would be inconsistent with $MinPrecision; \ bounding by $MinPrecision instead. Out[8]= -20 Andrzej On Wednesday, October 2, 2002, at 01:45 AM, Allan Hayes wrote: > Andrzej > > Yes, like you I took the original question to be about how to get the > result > of using the naive rational versions in place of the inexact numbers. > Bobby raises the question of how we might know accuracy of the result. > > You answer this with >> a = 77617.; b = 33096.; >> >> In[2]:= >> 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[3]:= >> f >> >> Out[3]= >> -1.1805916207174113*^21 >> >> In[4]:= >> Accuracy[%] >> >> Out[4]= >> -5 > > However this computation is done in machine arithmetic, which means > that > Mathematica keeps no check on the accuracy and precision of the > computation, > and Mathematica gives the default accuracy value without any real > signifcance: > > $MachinePrecision - Log[10,Abs[f]]//Round > > -5 > > To get meaningful accuracy and precision values we need to force the > computation to be in bignums (bigfloat, arbitrary precision) > arithmetic. > Mathematica then keeps track of the accuracy and precision that it can > guarantee. > > Clear[f,a,b,k] > k=50; > > f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b^8+a/ > (2*b),k]; > a=77617.;b=33096.; > a=SetAccuracy[a,k]; > b=SetAccuracy[b,k]; > f > > -0.82739605995 > > Accuracy[f] > > 11 > > Precision[f] > > 11 > > Which tells us that the error in the computed value of f is not more > than > > 10^-11 > > > The above results are rounded. More accurately we get > > Accuracy[f, Round->False] > > 11.4956 > > Precision[f, Round->False] > > 11.4133 > > There are several related issues here: > - is the precision (accuracy) of rhe input known? > - how does Mathematica interpret what one gives it? > - how does Mathematica go about the calculation? > > -- > Allan > > --------------------- > Allan Hayes > Mathematica Training and Consulting > Leicester UK > www.haystack.demon.co.uk > hay at haystack.demon.co.uk > Voice: +44 (0)116 271 4198 > Fax: +44 (0)870 164 0565 > > > "Andrzej Kozlowski" <andrzej at platon.c.u-tokyo.ac.jp> wrote in message > news:anbofj$e7u$1 at smc.vnet.net... >> It seems clear to me that what Allan and what you mean by "succeeds" >> here refer to quite different things and your objection is therefore >> beside the point. There are obviously two ways in which one can >> interpret the original posting. The first interpretation, which Allan >> and myself adopted, was that the question concerned purely the >> computational mechanism of Mathematica. Or, to put it in other words, >> it was "why are the results of these two computations not the same?". >> In this sense "success" means no more than making Mathematica return >> the same answer using the two different routes the original poster >> used. >> You on the other hand choose to assume that the posting shows that its >> author does not understand not just the mechanism of significance >> arithmetic used by Mathematica but also computations with inexact >> numbers in general. I do not think this is necessarily the correct >> assumption. I also don't see that Mathematica is doing anything wrong. >> After all, one can always check the accuracy of your answer: >> >> In[1]:= >> a = 77617.; b = 33096.; >> >> In[2]:= >> 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[3]:= >> f >> >> Out[3]= >> -1.1805916207174113*^21 >> >> In[4]:= >> Accuracy[%] >> >> Out[4]= >> -5 >> >> which tells you that it can't be very reliable. What more do you >> demand? >> >> Andrzej >> >> >> Andrzej Kozlowski >> Yokohama, Japan >> http://www.mimuw.edu.pl/~akoz/ >> http://platon.c.u-tokyo.ac.jp/andrzej/ >> >> On Tuesday, October 1, 2002, at 02:59 AM, DrBob wrote: >> >>> 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 >>> Sent: Monday, September 30, 2002 11:59 AM >>> Subject: [mg36932] 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\ >>> >>> 525007719168693704591197760187988046304361497869199129319625743010292 >>> 36 >>> 3 >>> 1246 >>> 75\ >>> >>> / >>> 108671061439707605510003578275547938881981431359756495796079898677435 >>> 72 >>> 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 >>>>> >>>>> >>>>> >>>> >>>> >>> >>> >>> >>> >>> >>> >>> >>> >> >> >> > > > > > > > > > > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/

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