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MathGroup Archive 2007

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Re: Simplification with Integers assumption

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74730] Re: [mg74709] Simplification with Integers assumption
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 2 Apr 2007 07:00:04 -0400 (EDT)
  • References: <200704010819.EAA07442@smc.vnet.net>

On 1 Apr 2007, at 09:19, did wrote:

> On Mathematica 5.2 Windows, with the 4 similar commands:
>
> Simplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity} ] ,
> Assumptions -> n > 0]
>
> FullSimplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity} ] ,
> Assumptions -> n > 0]
>
> Simplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity} ] ,
> Assumptions -> n > 0 && n =E2=88=88 Integers]
>
> FullSimplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity}] ,
> Assumptions -> n > 0 && n =E2=88=88 Integers]
>
>
> I get the different answers:
>
> Out[1]= -(-2 + 2^n) n! Gamma[n] Sin[n Pi] / (2 Pi)
>
> Out[2]= -1 + 2^(-1+n)
>
> Out[3]= 0
>
> Out[4]= 0
>
> Outputs 1 & 2 look OK, but 3 & 4 are not. It seems that, with the
> assumption n Integer,
> Mathematica simplifies Sin[n Pi] by 0, omitting that Gamma[-n] is 
> infinite.
> Is it the expected behavior?
>
> In this example, the simplest form can be obtained without imposing n
> Integer (I
> presume it's the correct answer), but in other situations it will be
> required. What
> is the safe way to do it?
>
> Thanks
>
>
The problem is that


Assuming[Element[n,Integers],Simplify[Sin[n Pi]]]

0

But


Simplify[Gamma[n], n =E2=88=88 Integers]

Gamma[n]

(even adding the assumption n<0 will still return Gamma[n]).

Since


Sum[n!/(2*p)!/(n - 2*p)!, {p, 1, Infinity}]

-(((-2 + 2^n)*n!*Gamma[-n]*Sin[n*Pi])/(2*Pi))

simplifying with the assumption ELement[n,Integers] will result in 
Gamma[-n]*0 and since in Mathematica any symbol s (except explicit 
Infitiniy and its variants)  s*0 is always 0, you will get 0.

The best way to avoid the problem is to try to get an expression for 
the sum not involving the Gamma function at all, if possible. In this =

since FullSimplify without the condition Element[n,Integers] returns 
an expression not involving Gamma, you cant then Simplify again using =

the this assumption, although in this particular case it won't make 
any difference. This is one of the cases where one needs to pay 
careful attention to the effect of different assumptions in 
FullSimplify may have on the the expression they are being applied 
to. The danger is that if one of the assumptions implies that one of 
the factors in in a product is zero the whole product will be reduced =

to 0, unless one of the other  factors are explicitly known to 
Mathematica=E2=80=99s evaluator be infinite under the given assumptions. =
An 
example of the latter case is:

FullSimplify[Sum[n!/(2*p)!/(n - 2*p)!, {p, 1, Infinity}], {n == 3}]

While this partiular problem could perhaps be solved by adding the 
knowledge that Gamma[n] is CompexInfinity for any negative integer to =

Simplify and FullSimplify some problems of this kind will always 
remain since Mathematica's evaluator turns x*0 to 0 even if nothing 
is known about x (and in particular when x is an implicit but not 
explicit Infinity).

Andrzej Kozlowski



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