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MathGroup Archive 2003

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Re: Positive Integer Assumptions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg43774] Re: Positive Integer Assumptions
  • From: bobhanlon at aol.com (Bob Hanlon)
  • Date: Sat, 4 Oct 2003 02:04:53 -0400 (EDT)
  • References: <blj5ra$1gi$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

f[x_,n_,m_]:=n!/((n-m)!(m-1)!) x^(m-1) (1-x)^(n-m)

Use Assumptions

Integrate[x^k * f[t,n,m],{t,0,x}, 
  Assumptions->{Element[{k,n,m}, Integers], 
      k >0, n >0, m >0, 0<x<1}]

(120*x^(k + m)*Hypergeometric2F1[m - 5, m, m + 1, x])/
  (m*(5 - m)!*(m - 1)!)

Although fewer assumptions can be used

Integrate[x^k * f[t,n,m],{t,0,x}, 
  Assumptions->{m >0, 0<x<1}]

(120*x^(k + m)*Hypergeometric2F1[m - 5, m, m + 1, x])/
  (m*(5 - m)!*(m - 1)!)

Or turn off GenerateConditions

Integrate[x^k * f[t,n,m],{t,0,x}, 
  GenerateConditions->False]

(120*x^(k + m)*Hypergeometric2F1[m - 5, m, m + 1, x])/
  (m*(5 - m)!*(m - 1)!)

Table[%, {m,8}] // Simplify

{x^(k + 1)*(x^4 - 5*x^3 + 10*x^2 - 10*x + 5), 
  x^(k + 2)*(-4*x^3 + 15*x^2 - 20*x + 10), 
  x^(k + 3)*(6*x^2 - 15*x + 10), (5 - 4*x)*x^(k + 4), 
  x^(k + 5), 0, 0, 0}


Bob Hanlon

In article <blj5ra$1gi$1 at smc.vnet.net>, Stewart Mandell <stewart at rentec.com>
wrote:

<< How do I specify assumptions that k,n,m are all positive integers in the
Integral

f[x_, n_,m_] := n!/((n-m)!(m-1)!) x^(m-1) (1-x)^(n-m)

 Integrate[ x^k  f[t, n, m], {t, 0, x}]


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