Re: Floating-Point Computing

*To*: mathgroup at smc.vnet.net*Subject*: [mg93958] Re: Floating-Point Computing*From*: "Nasser Abbasi" <nma at 12000.org>*Date*: Sat, 29 Nov 2008 04:33:42 -0500 (EST)*References*: <ggofnk$rsh$1@smc.vnet.net>*Reply-to*: "Nasser Abbasi" <nma at 12000.org>

"Antonio" <aneves at gmail.com> wrote in message news:ggofnk$rsh$1 at smc.vnet.net... > 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]]]] > > Any ideas? > > ANtonio > The above are 2 different implementations. In C, there you did not do pre allocation of all the numbers being square rooted. It is done in a loop, one at a time. In Mathematica, you are asking to pre-allocate a list whose length is 1000000000, then sqrt each entry in the list, then add the result. I think you are comparing apples and oranges in terms of implementation. This is what I get on my PC using gcc 3.4.4 (using the above C code) $ ./a.exe Finished 21081851083598.38281250000000 And this is what I get using Mathematica version 7, code rewritten to use a for loop as in C In[22]:= f = 0.; Timing[For[i = 0, i <= 1000000000, i++, f = f + Sqrt[N[i]]]; f] Out[23]= {3249.547, 2.1081851083598383*^13} ps. I noticed if I do Sqrt[ N[i] ] vs Sqrt[ i ], it runs faster. Nasser