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MathGroup Archive 2002

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RE: Unexpected result with RSolve?

  • To: mathgroup at
  • Subject: [mg33631] RE: [mg33629] Unexpected result with RSolve?
  • From: "Curt Fischer" <cfisher at>
  • Date: Thu, 4 Apr 2002 19:39:48 -0500 (EST)
  • Sender: owner-wri-mathgroup at

-----Original Message-----
From: wouter.van.den.broeck at
To: mathgroup at
[mailto:wouter.van.den.broeck at] 
Subject: [mg33631] [mg33629] Unexpected result with RSolve?


There's probably a sound solution, but i'm struggling to grab it:

RSolve[{a[n + 1] == n a[n], a[0] == 1}, a[n], n]

{{a[n] -> 0}}

where, i believe it 'should' return
{{a[n] -> (n-1)!}}

Anyone care to give me some directions as to why this 'unexpected'
Dear Wouter,

The result Mathematica gives for your example is correct.  If a[0]==1,
then you can solve for a[1] using the equation a[0+1]==0 a[0], which
obviously gives a[1]==0.  And so on for the other terms.  One recursion
relation which has the solution you were expecting, {a[n]->(n-1)!}, is 

In2:  RSolve[{a[n+1]==n a[n],a[1]==1},a[n],n]

Out2: {{a[n]->If[n\[GreaterEqual]1,(-1+n)!,0]}}

In this case a[0] is zero because of the definition of the factorial
Curt Fischer
Tokyo Institute of Technology
Dept. of Bioengineering

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