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Re: Bug in SymbolicSum, or am I just stupid?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg5637] Re: [mg5593] Bug in SymbolicSum, or am I just stupid?
  • From: Dave Snead <dsnead at pacbell.net>
  • Date: Wed, 1 Jan 1997 21:05:01 -0500
  • Organization: Saguaro Software
  • Sender: owner-wri-mathgroup at wolfram.com

Sean Luke wrote:
> 
> I'm wondering if I'm just doing something really dumb, or if there
> is some profound reason why Mathematica 2.2 seems to be finding different
> symbolic infinite sumations for the same expression:
> 
> Mathematica 2.2 for SPARC
> Copyright 1988-94 Wolfram Research, Inc.
>  -- Terminal graphics initialized --
> 
> In[1]:= <<Algebra`SymbolicSum`
> 
> In[2]:= ?SymbolicSum
> SymbolicSum[f, {i, imin, imax}] attempts to find the value of Sum[f, {i, imin,
>    imax} ] for symbolic imin,imax. SymbolicSum[f, {i, imax}] evaluates the sum
>    of f with i running from 1 to imax.
> 
> In[2]:= Simplify[(2)( 1/(2^n )  - ((n+1)/2)(1/(2^n))]  (* expression 1 *)
> 
>         3 - n
> Out[2]= -----
>            n
>         2 2
> 
> In[3]:= Simplify[(2 - (n+1)/2) (1/(2^n))]  (* expression 2--should be same *)
> 
>         3 - n
> Out[3]= -----
>            n
>         2 2
> 
> In[4]:= SymbolicSum[(2)( 1/(2^n) ) - ((n+1)/2)(1/(2^n)),{n,1,Infinity}]  (*1*)
> 
>         1
> Out[4]= -
>         2
> 
> In[5]:= SymbolicSum[(2 - (n+1)/2) (1/(2^n)),{n,1,Infinity}]  (*2*)
> 
>           3
> Out[5]= -(-)
>           4
> 
> In[6]:= SymbolicSum[Out[3],{n,1,Infinity}]   (* simplified *)
> 
>           3
> Out[6]= -(-)
>           4
> 
> Thanks very much anyone.  Sorry if this is a FAQ or obnoxious newbie question.
> 
> Sean Luke
> U Maryland at College Park, very late a night
> seanl at cs.umd.edu


This looks like a bug in 2.2:

Mathematica 2.23 under Windows 95 gives

In[1]:= <<Algebra`SymbolicSum`

In[2]:= SymbolicSum[(3-n)/2^(n+1),{n,1,Infinity}]
           
          3
Out[2]= -(-)
          4

while Mathematica 3.0 (which has apparently incorporated 
the SymbolicSum functionality into Sum) gives

In[1]:= Sum[(3-n)/2^(n+1),{n,1,Infinity}]

        1
Out[1]= -
        2

which is clearly the correct answer

Dave Snead
dsnead at pacbell.net


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