Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2002
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2002

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

Search the Archive

Re: recognizing integer numbers

  • To: mathgroup at smc.vnet.net
  • Subject: [mg34040] Re: [mg33924] recognizing integer numbers
  • From: BobHanlon at aol.com
  • Date: Sat, 27 Apr 2002 00:57:12 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 4/23/02 8:27:20 AM, manuel_ballester at yahoo.com writes:

>I need some help with the following. I am trying mathematica to solve
>some sums for me. The thing is that I don't know how to tell
>mathematica that certain numbers are non-negative integers and some
>answers that I would expect as (for example) n! are given as
>Gamma[n-1] and so on, sometimes Hypergeometric functions are also
>involved. Example:
>
>Sum[Binomial[2*m+1,k]*( a^k*b^(2*m+1-k)+ b^k*a^(2*m+1-k) ),{k,0,m}]
>
>if m is a natural number then the answer to this is simply
>(a+b)^(2*m+1)
>
>but since I don't know how to say this to mathematica it gives me an
>answer with gamma and hypergeometric functions. I already tried
>Simplify and FullSimplify.
>

s1 = FullSimplify[Sum[Binomial[2*m+1,k]*
          (a^k*b^(2*m+1-k)+b^k*a^(2*m+1-k)),
        {k,0,m}], Element[m, Integers]];

The same sum in reverse order is

s2 = FullSimplify[Sum[(Binomial[2*m+1,k]*
              (a^k*b^(2*m+1-k)+b^k*a^(2*m+1-k))) /.
 
          k -> (m-k),{k,0,m}], Element[m, Integers]];

The result is then

Simplify[(s1+s2)/2]

(a + b)^(2*m + 1)


Bob Hanlon
Chantilly, VA  USA


  • Prev by Date: Re: invisible graphics
  • Next by Date: Re: Dynamic referencing AND hyperlinking for numbered equations
  • Previous by thread: Re: recognizing integer numbers
  • Next by thread: Re: Re: recognizing integer numbers