Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1997
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1997

[Date Index] [Thread Index] [Author Index]

Search the Archive

D[f,x] vs f'[x]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg7682] D[f,x] vs f'[x]
  • From: Gianluca Gorni <gorni at dimi.uniud.it>
  • Date: Sun, 29 Jun 1997 22:17:17 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

I have stumbled into a marked difference between
the derivation by D[] and the derivation as f',
that I don't quite understand.

Mathematica 3.0 Kernel for Power Macintosh

Let f be an infinite sum, which has a closed form:

In[1]:= f=Sum[x^n/(4n^2+4n+2),{n,0,Infinity}]//Simplify

         1   I                         1   I       3   I
Out[1]= (- + -) (-I HypergeometricPFQ[{- - -, 1}, {- - -}, x] +
         4   4                         2   2       2   2

                          1   I       3   I
>      HypergeometricPFQ[{- + -, 1}, {- + -}, x])
                          2   2       2   2

The derivative is no problem:

In[2]:= D[f,x]//Simplify

         I                     1   I       3   I
Out[2]= (- (HypergeometricPFQ[{- - -, 1}, {- - -}, x] -
         4                     2   2       2   2

                            1   I       3   I
>        HypergeometricPFQ[{- + -, 1}, {- + -}, x])) / x
                            2   2       2   2


However, if I define a function g[x] by that sum, using immediate
assignment (= instead of :=),


In[3]:= g[x_]=Sum[x^n/(4n^2+4n+2),{n,0,Infinity}]//Simplify

         1   I                         1   I       3   I
Out[3]= (- + -) (-I HypergeometricPFQ[{- - -, 1}, {- - -}, x] +
         4   4                         2   2       2   2

                          1   I       3   I
>      HypergeometricPFQ[{- + -, 1}, {- + -}, x])
                          2   2       2   2

then the derivative g'[x] is Indeterminate!


In[4]:= g'[x]

Infinity::indet:
                                1    2 I                       5   I
                             (-(-) - ---) Sqrt[5] ? Sign[Gamma[- - -]]
                                5     5                        2   2
   Indeterminate expression (-----------------------------------------) +
                                                   3   I
                                        Sign[Gamma[- - -]]
                                                   2   2
     <<1>> encountered.

Out[4]= Indeterminate


I am stumped. Is there a simple explanation?

               Gianluca Gorni


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Gianluca Gorni
Universita` di Udine
Dipartimento di Matematica e Informatica
via delle Scienze 208
I-33100 Udine UD
Italy

Ph.:(39) (432) 558422    Fax:(39) (432) 558499
mailto:gorni at dimi.uniud.it
http://www.dimi.uniud.it




  • Prev by Date: Making command called HoldTemporary
  • Next by Date: Re: Error in basic integrals
  • Previous by thread: Making command called HoldTemporary
  • Next by thread: Re: Error in basic integrals