Re: Unexpected result with RSolve?

*To*: mathgroup at smc.vnet.net*Subject*: [mg33636] Re: [mg33629] Unexpected result with RSolve?*From*: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>*Date*: Thu, 4 Apr 2002 19:39:54 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Mathematica is right and you are wrong. If a[n + 1] == n a[n] and a[0]==1 then a[1]==0 a[0] which is 0, and hence a[n] is 0 for all n>0. Of course the correct equation is: In[11]:= RSolve[{a[n + 1] == (n+1) a[n], a[0] == 1}, a[n], n] Out[11]= {{a[n]->n!}} Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Thursday, April 4, 2002, at 08:09 AM, 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 > > > > >