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MathGroup Archive 2001

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Re: Possible Incorrect Summation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31070] Re: Possible Incorrect Summation
  • From: "Alan Mason" <swt at austin.rr.com>
  • Date: Sat, 6 Oct 2001 03:33:11 -0400 (EDT)
  • References: <9pjgai$3fe$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

I think the answer is correct, although Mathematica refuses to simplify it
to the desired result.  Version 4.1 gives the following:
In[1]:=
ans = Sum[ (-1)^k Binomial[2 n, k] Binomial[2k, k] Binomial[4n - 2k, 2n -
k],{k,0,2n}]
Out[1]=
\!\(\(16\^n\ \[Pi]\)\/\(Gamma[1\/2 - n]\^2\ Gamma[1 + n]\^2\)\)
In[2]:=
FullSimplify[ans - Binomial[2n, n]^2]
Out[2]=
\!\(\(-\(\(16\^n\ \[Pi]\ Tan[n\ \[Pi]]\^2\)\/\(Gamma[1\/2 - n]\^2\ Gamma[1 +
\
n]\^2\)\)\)\)

The second output is clearly zero for n an integer, showing that Dixon's
identity is obeyed.

Unfortunately, one already has to know about Dixon's identity in order to
formulate the second input.  Perhaps you can coax Mathematica to simplify
the result as desired by Unprotecting Gamma and writing a suitable rule.

Your example shows a limitation of Mathematica's simplification rules:
In[27]:=
ans = Sum[(-1)^k Binomial[2 n,k] Binomial[2k,k] Binomial[4n-2k,2n-
          k],{k,0,2n}]
Out[27]=
\!\(\(16\^n\ \[Pi]\)\/\(Gamma[1\/2 - n]\^2\ Gamma[1 + n]\^2\)\)
In[28]:=
FullSimplify[ans, n \[Element] Integers && n > 0,
  ComplexityFunction\[Rule] ByteCount]
Out[28]=
\!\(\(16\^n\ \[Pi]\)\/\(\(n!\)\^2\ Gamma[1\/2 - n]\^2\)\)
In[29]:=
bn = Binomial[2n, n]^2
In[31]:=
FullSimplify[ans - bn,  n \[Element] Integers && n > 0,
  ComplexityFunction\[Rule] ByteCount]
Out[31]=
0
In[32]:=
FullSimplify[bn,  n \[Element] Integers && n > 0, ComplexityFunction\[Rule]
ByteCount]
Out[32]=
\!\(Binomial[2\ n, n]\^2\)
In[33]:=
ByteCount[%]
Out[33]=
104
In[34]:=
FullSimplify[ans,  n \[Element] Integers && n > 0, ComplexityFunction\[Rule]
ByteCount]
Out[34]=
\!\(\(16\^n\ \[Pi]\)\/\(\(n!\)\^2\ Gamma[1\/2 - n]\^2\)\)
In[35]:=
ByteCount[%]
Out[35]=
320
Thus bn = Binomial[2n, n]^2 is simpler than ans, but Mathematica isn't
finding Binomial.  A clue as to why is in the online help -- Binomial[n, m]
is defined in terms of Gamma for arbitrary (complex) n, m.  Seems a rule is
needed to revert back to Binomial when n, m are positive integers.

Alan


"Richard Palmer" <mapsinc at bellatlantic.net> wrote in message
news:9pjgai$3fe$1 at smc.vnet.net...
>
> I believe
>
>     Sum[ (-1)^k Binomial[2 n, k] Binomial[2k, k] Binomial[4n - 2k, 2n -
> k],{k,0,2n}]
>
>  is a special case of Dixon's Identity and should equal Binomial[2n,n]^2.
> Previously the SymbolicSum package returned a value which could be
> simplified to the result.  The latest version of Mathematica returns a
value
> which is is not this result as can be seen by graphing the difference.
What
> has changed?
>
>
>



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