Re: simplifying a summation / integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg64828] Re: simplifying a summation / integral*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Sun, 5 Mar 2006 03:18:46 -0500 (EST)*Organization*: The Open University, Milton Keynes, UK*References*: <dubjgg$gpp$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Patrik wrote: > Jean, Paul: > > Thanks for your comments. > > Yes I tried those and got the hypergeometricPFQ functions. But, I need to take the derivative of the summation and hence do not mind making approximations if that helps get a simple looking derivative (the derivative is part of a first order condition and hence I've to substitute the derivative in an equation & solve for x). The hypergeometricPFQ is not suited for that. > > Thanks, > Kartik Indeed, if you take the derivative the following function f in the variable x and depending on the parameter q, you get a simple derivative which is the same as the derivative of the sum after simplification. Hope this is what you are looking for. In[1]:= f[q_][x_] := Sum[Binomial[q, r]*(x^r/r), {r, 1, q}] In[2]:= f[q][x] Out[2]= q*x*HypergeometricPFQ[{1, 1, 1 - q}, {2, 2}, -x] In[3]:= f[q]'[x] Out[3]= q -1 + (1 + x) ------------- x In[4]:= D[Sum[Binomial[q, r]*(x^r/r), {r, 1, q}], x] Out[4]= q -1 + (1 + x) q (------------- - HypergeometricPFQ[{1, 1, 1 - q}, q x {2, 2}, -x]) + q HypergeometricPFQ[{1, 1, 1 - q}, {2, 2}, -x] In[5]:= Simplify[%] Out[5]= q -1 + (1 + x) ------------- x Best regards, /J.M.