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Re: Re: Floating-Point Computing

  • To: mathgroup at smc.vnet.net
  • Subject: [mg93952] Re: [mg93927] Re: Floating-Point Computing
  • From: "Hobbs, Sylvia (DPH)" <Sylvia.Hobbs at state.ma.us>
  • Date: Sat, 29 Nov 2008 04:32:37 -0500 (EST)
  • References: <ggofnk$rsh$1@smc.vnet.net> <200811281211.HAA01366@smc.vnet.net>

List is my formal name. You can just call me Sylvia.

> -----Original Message-----
> From: mathgroup-adm at smc.vnet.net [mailto:mathgroup-adm at smc.vnet.net] =
On Behalf
> Of Szabolcs Horv=E1t
> Sent: Friday, November 28, 2008 7:11 AM
> To: mathgroup at smc.vnet.net
> Subject: [mg93927] Re: Floating-Point Computing
>
> Antonio wrote:
> > Dear List,
> >
> > I came across this Sun article on floating point:
> > http://developers.sun.com/solaris/articles/fp_errors.html
> >
> > which calculates the following code:
> >
> > main()
> > {
> >  register i;
> >  double f=0.0;
> >  double sqrt();
> >
> >  for(i=0;i<=1000000000;i++)
> >  {
> >     f+=sqrt( (double) i);
> >  }
> >  printf("Finished %20.14f\n",f);
> > }
> >
> > whose correct answer using Euler-Maclaurin formula is:
> > f=21081851083600.37596259382529338
> >
> > Tried to implement in mathematica, just for fun, so that later I =
could
> > play with precision, but the kernel has no more memory and shuts =
down.
> >
> > N[Total[Sqrt[Range[1000000000]]]]
> >
>
> If all those numbers are stored as 32-bit integers in a packed array,
> they would take up about 4*10^9/1024^3 ~= 3.7 GB space.  That is =
already
> too much for a 32-bit machine, and more than most present-day personal
> computers can handle.  Mapping Sqrt to the list unpacks it, eating up
> even more memory.  Creating a symbolic sum with Total and only
> numericising at the end doesn't help either.
>
> The only reasonable approach here is to write code that is equivalent =
to
> the C version, i.e. doesn't pre-compute the list of numbers to be
> summed.  But that will run too slowly in Mathematica.  So Mathematica =
is
> not really the right tool for summing 10^9 numbers.
>
> Two possible versions:
>
> fun1[n_Integer] := Module[{f = 0.},
>    Do[f += N@Sqrt[i], {i, n}];
>    f
>    ]
>
> fun2 = Compile[{{n, _Integer}}, Module[{f = 0.},
>     Do[f += N@Sqrt[i], {i, n}];
>     f
>     ]
>    ]
>
> fun2[10^9] can finish in a reasonable time, but it only works with
> machine precision numbers (because of Compile).
>
> If you sum fewer numbers, use Total@Sqrt@N@Range[n] instead of
> N@Total@Sqrt@Range[n].



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