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Re: Accuracy problem in Mathematica
 To: mathgroup at smc.vnet.net
 Subject: [mg49118] Re: Accuracy problem in Mathematica
 From: ab_def at prontomail.com (Maxim)
 Date: Fri, 2 Jul 2004 02:01:18 0400 (EDT)
 References: <cbu21o$5ea$1@smc.vnet.net>
 Sender: ownerwrimathgroup at wolfram.com
"aaaa" <aaa at huji.ac.il> wrote in message news:<cbu21o$5ea$1 at smc.vnet.net>...
> Hello,
>
> I'm having a problem with a calculation in Mathematica which I can't
> solve. I have an expression which I know to be (from analytical reasons)
> always between 0 and 1. It's a function of a and n ( n being natural and
> a rational) and it looks like this:
>
> 1/(1a^2)^n +
>
> Sum[((2*n  k  1)!/((n  1)!*(n  k)!*2^(2*n  k)))*(1/(1 + a)^k  1/(1
>  a)^k), {k, 1, n}]
>
>
>
> Let's say a=0.5.
>
> Now, when I try to calculate for small n, it's ok. When calculating for
> large n's (around 400 and above) I'm starting to get wrong results (the
> number not being between 0 and 1). The problem is that the first term
> (the first line before the sum) is VERY VERY close to the negative of
> the second term (the sum), and it's getting closer as n grows. When
> using large n's, Mathematica says they are the same number or even that
> the last term is bigger (which means the whole expression becomes
> negative)  which is wrong. It's a matter of accuracy, and I'm not sure
> how I can fix it.
>
> Can anybody help me?
>
>
>
> Itamar
The straightforward way is to perform all calculations using
arbitraryprecision numbers:
In[1]:=
f[a_, n_]:=
1/(1a^2)^n +
NSum[((2*n  k  1)!/
((n  1)!*(n  k)!*2^(2*n  k)))*(1/(1 + a)^k  1/(1  a)^k),
{k, 1, n},
Method>Fit, NSumTerms>n, WorkingPrecision>150]
f[1/2, 1000] // Timing
Out[2]=
{7.172 Second, 0.0356247970508144815921869}
Actually in this case it wouldn't make much difference if we simply
used Sum with highprecision values of a and n. As this would also
take significant time, there is a more contrived way which relies on
evaluating the sum symbolically:
In[3]:=
exact =
1/(1a^2)^n +
Sum[ ((2*n  k  1)!/
((n  1)!*(n  k)!*2^(2*n  k)))*(1/(1 + a)^k  1/(1  a)^k),
{k, 1, n}]
Out[3]=
(1  a^2)^(n)  (2^(1  2*n)*Pi*Csc[2*n*Pi]*
(HypergeometricPFQRegularized[{1, 1  n}, {2  2*n}, (2/(1 + a))] +
a*HypergeometricPFQRegularized[{1, 1  n}, {2  2*n}, (2/(1 + a))]

HypergeometricPFQRegularized[{1, 1  n}, {2  2*n}, 2/(1 + a)] +
a*HypergeometricPFQRegularized[{1, 1  n}, {2  2*n}, 2/(1 +
a)]))/
((1 + a)*(1 + a)*(1 + n)!*Gamma[n])
The problem is that Mathematica returns an expression which is
undefined for integer values of n, it is correct only as the limit
when n>n0 and n0 is integer. What we can do is to evaluate this
expression slightly off the integer point:
In[4]:=
f2 = Function @@ {{a,n}, exact /. n > n + 1`150*^150};
Timing[f2[1/2, 1000]]
Out[5]=
{0.594 Second, 0.0356247970508144815922 + 0.``172*I}
Taking the 1`150*^150 magic number is of course just heuristics, but
it seems to work quite well.
As a side note, here's why I used the Function@@L method above.
Compare the next two outputs:
In[6]:=
Clear[y, func]
y = x;
func[x_] = y;
DownValues[func]
Out[9]=
{HoldPattern[func[x_]] :> x}
In[10]:=
Clear[y, func]
Module[{y = x},
func[x_] = y];
DownValues[func]
Out[12]=
{HoldPattern[func[x$_]] :> x}
The resulting functions are different; in my opinion, this is an
inconsistency, so I just try to avoid such constructs whenever
possible, and Function@@{x,expr} or Set@@{f[x_],expr} bypasses the
variable renaming.
Maxim Rytin
m.r at inbox.ru
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