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RE: Accuracy and Precision
*To*: mathgroup at smc.vnet.net
*Subject*: [mg36936] RE: Accuracy and Precision
*From*: "DrBob" <drbob at bigfoot.com>
*Date*: Wed, 2 Oct 2002 03:33:03 -0400 (EDT)
*Reply-to*: <drbob at bigfoot.com>
*Sender*: owner-wri-mathgroup at wolfram.com
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
Subject: [mg36936] 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: [mg36936] 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: [mg36936] 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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