       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:
>
>
> 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:= f[q_][x_] := Sum[Binomial[q, r]*(x^r/r), {r, 1, q}]

In:= f[q][x]

Out= q*x*HypergeometricPFQ[{1, 1, 1 - q}, {2, 2}, -x]

In:= f[q]'[x]

Out=
q
-1 + (1 + x)
-------------
x

In:= D[Sum[Binomial[q, r]*(x^r/r), {r, 1, q}], x]

Out=
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:= Simplify[%]

Out=
q
-1 + (1 + x)
-------------
x

Best regards,
/J.M.

• Prev by Date: Re: Multiple application of LinearFilter
• Next by Date: Re: Trouble with tables
• Previous by thread: Re: simplifying a summation / integral
• Next by thread: No mesh with Graphics3D