Re: Accuracy problem in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg49112] Re: [mg49061] Accuracy problem in Mathematica*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 1 Jul 2004 05:26:22 -0400 (EDT)*References*: <200406300934.FAA05317@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 30 Jun 2004, at 18:34, aaaa wrote: > 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/(1-a^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 best thing to do is to use exact input for a and N to specify how many digits of precision you want in your answer. Thus: In[1]:= n=1000; In[2]:= a=1/2; In[3]:= N[1/(1-a^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}],20] Out[3]= 0.035624797050814481592 This looks right. Andrzej Kozlowski Chiba, Japan http://www.mimuw.edu.pl/~akoz/