Re: summing a series in mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg64061] Re: summing a series in mathematica
- From: Patrik <hosanagar at gmail.com>
- Date: Sun, 29 Jan 2006 23:10:16 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hi Peter,
Thanks a lot. Your values are quite representative. P, Q & d are positive integers with P<=d.
When I set P=12, you'll see that you get a complex number. Interestingly, Beta [-0.5, 1+1/d, Q] is also a complex number & I thought that was causing it. But the complex number probably got multiplied with another complex number to produce a real number when P=3.
I've been able to make some progress by doing this by hand. I have a reasonable functional form for the derivative now.
Thanks again for your help.
> Hi Patrik,
>
> what are the domains of P, Q and d?
>
> I've got no problem, evaluating your sum at e.g.
> g=1/2:
>
> In[1]:= $Assumptions={Element[{P,Q,R},
> Integers],P>0<R,Q>0<g<1};
> In[2]:= f=Q!/(R! (Q-R)!) (g^R) ((1-g)^(Q-R)) (P(1+2 R
> d-P)/(1+R d));
> In[3]:=
> fsum=Sum[f//Evaluate,{R,1,Q}]//FunctionExpand//FullSim
> plify
> Out[3]= (-1 + (1 - g)^Q)*(-1 + P)*P - ((1 - g)^(1/d +
> Q)*P*(1 + P)*Q*Beta[g/(-1 + g), 1 + 1/d,
> Q])/(-g)^d^(-1)
> In[4]:= fsum/.{g->1/2,d->3,Q->17,P->12}//FullSimplify
> Out[4]= 12265917559597029/692132625121280
> In[5]:= N[%]
> Out[5]= 17.72191790185835
>
> confused greetings,
> Peter
>