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MathGroup Archive 2002

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Re: Accuracy and Precision

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36923] Re: Accuracy and Precision
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Wed, 2 Oct 2002 03:32:03 -0400 (EDT)
  • References: <anbofj$e7u$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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: [mg36923] 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\
> >
> > 52500771916869370459119776018798804630436149786919912931962574301029236
> > 3
> > 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
> >>>
> >>>
> >>>
> >>
> >>
> >
> >
> >
> >
> >
> >
> >
> >
>
>
>










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