RE: Accuracy and Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg36940] RE: Accuracy and Precision*From*: "DrBob" <drbob at bigfoot.com>*Date*: Wed, 2 Oct 2002 03:33:38 -0400 (EDT)*Reply-to*: <drbob at bigfoot.com>*Sender*: owner-wri-mathgroup at wolfram.com

>>In any case the answer you get is meaningless. Precisely. Bobby -----Original Message----- From: Andrzej Kozlowski [mailto:andrzej at tuins.ac.jp] To: mathgroup at smc.vnet.net Subject: [mg36940] Re: Accuracy and Precision If you don't wont error messages all you need to do is to set MinPrecision low enough: In[1]:= $MinPrecision = -3; 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]= -5.517164009`-2.8311*^19 In[8]:= Accuracy[f] Out[8]= -23 In[9]:= Precision[f] Out[9]= -3 In any case the answer you get is meaningless. On Wednesday, October 2, 2002, at 10:33 AM, DrBob wrote: > I think I'd prefer that Mathematica gave me a clue -- without being > asked explicitly in JUST the right way that only you know and had > forgotten -- that there's a precision problem. > > Oddly enough, when you DO ask it nicely, you get error messages, but if > you're not aware there's a problem, it lets you go on your merry way, > working with noise. > > Bobby > > -----Original Message----- > From: Andrzej Kozlowski [mailto:andrzej at tuins.ac.jp] To: mathgroup at smc.vnet.net > Sent: Tuesday, October 01, 2002 8:05 PM > Cc: 'Allan Hayes'; mathgroup at smc.vnet.net > Subject: [mg36940] Re: Accuracy and Precision > > So? It's just as it should be, isn't it? > > Andrzej > > Andrzej Kozlowski > Toyama International University > JAPAN > http://sigma.tuins.ac.jp/~andrzej/ > > > > On Wednesday, October 2, 2002, at 09:02 AM, DrBob wrote: > >> Go one step further: >> >> Clear[f, a, b, k] >> k = $MachinePrecision; >> 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 >> Accuracy[f] >> Precision[f] >> >> -5.517164009`0*^19 >> >> Accuracy::mnprec:Value -23 would be inconsistent with $MinPrecision; >> bounding by $MinPrecision instead. >> >> -20 >> >> 0 >> >> Accuracy::"mnprec":"Value \!\(-23\) would be inconsistent with >> $MinPrecision; \ >> bounding by $MinPrecision instead." >> >> See that? NO precision. >> >> Bobby >> >> -----Original Message----- >> From: Andrzej Kozlowski [mailto:andrzej at platon.c.u-tokyo.ac.jp] To: mathgroup at smc.vnet.net >> Sent: Tuesday, October 01, 2002 5:53 PM >> Cc: drbob at bigfoot.com >> Subject: [mg36940] Re: Accuracy and Precision >> >> 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: [mg36940] 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/ >> >> >> >> > > > > > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/

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