Re: Unexpected result with RSolve?

• To: mathgroup at smc.vnet.net
• Subject: [mg33637] Re: [mg33629] Unexpected result with RSolve?
• From: Rob Pratt <rpratt at email.unc.edu>
• Date: Thu, 4 Apr 2002 19:39:56 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```On Wed, 3 Apr 2002 wouter.van.den.broeck at vub.ac.be wrote:

> Hey,
>
> 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]
>
> returns:
> {{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'
> result?
>
> (note: I started of with the standard textbook example:
> RSolve[{a[n] == n a[n-1], a[0] == 1}, a[n], n]
> which returns
> {{a[n] -> n!}}
> so there's no need to correct me in this direction, I just explored a
> bit, and wonder why i'm getting that result with the variation above)
>
> TIA
>
> wouter vdb

If you use your recursive formula to compute a[n] by hand, you will find
that a[1] = 0*a[0] = 0, a[2] = 1*a[1] = 0, a[3] = 2*a[2] = 0, ....  But
RSolve's answer of a[n] = 0 is a little misleading since it is true only
when n >= 1.

If you use a[1] == 1 instead of a[0] == 1, RSolve will return your desired

Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill

rpratt at email.unc.edu

http://www.unc.edu/~rpratt/

```

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