[Date Index]
[Thread Index]
[Author Index]
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.
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**
| |