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MathGroup Archive 2006

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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.


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