Re: Trying to get closed form solution to simple recurrence.

*To*: mathgroup at smc.vnet.net*Subject*: [mg100083] Re: [mg100017] Trying to get closed form solution to simple recurrence.*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 22 May 2009 23:40:09 -0400 (EDT)*References*: <200905220542.BAA18819@smc.vnet.net>

negatron wrote: > Product[(100 - n)/100, {n, 1, x}] > > ((100!/(100 - (x + 1))!)*(1/100)^(x + 1)) > > So the two above are equivalent (perhaps the second can do with less > brackets). > > What I'm trying to figure out is how to get Mathematica to solve such > a problem. That is, input the first expression and get the second. > > I'm aware of the RSolve function and I've used it in the past, but I > can't seem to figure out a way how to express this properly, or more > likely it doesn't look like such a problem can be solved by this > particular function. > > Is such a thing solvable by any function in Mathematica? It seems RSolve does not handle this directly. A workaround is to take logs, use RSolve on that, and exponentiate. rsol = RSolve[{ fl[n]==Log[100-n]-Log[100]+fl[n-1], fl[1]==Log[99/100]}, fl[n], n]; In[30]:= InputForm[f[n] = Exp[fl[n]/.First[rsol]]] Out[30]//InputForm= (-1/100)^(1 + n)*Pochhammer[-100, 1 + n] In[31]:= InputForm[fnew = FunctionExpand[f[n], Assumptions->Element[n,Integers]]] Out[31]//InputForm= (9881297415446727147594496649775206852319571477668037853762810667968023095834\ 839075329261976769165978884198811117* (-1)^(2*n)*2^(95 - 2*n)*25^(11 - n))/Gamma[100 - n] Notice that we have an factor of (-1)^(2*n) which "obviously" is one. I do not use simplification at this point because it is easier to see this is equivalent to the desired form, at least once you see that form evaluated. In[40]:= InputForm[gg = ((100!/(100 - (n + 1))!)*(1/100)^(n + 1))] Out[40]//InputForm= (9881297415446727147594496649775206852319571477668037853762810667968023095834\ 839075329261976769165978884198811117*2^(95 - 2*n)*25^(11 - n))/(99 - n)! In[41]:= FullSimplify[fnew-gg, Assumptions->Element[n,Integers]] Out[41]= 0 Notice that Product already will handle the original input. In[4]:= InputForm[Product[(100 - j)/100, {j,1,n}]] Out[4]//InputForm= (-1/100)^n*Pochhammer[-99, n] If you use FunctionExpand it will give the form to which gg evaluates, along with the (unfortunate, needless) factor of (-1)^(2*n). You can get rid of it by using Simplify[...,Assumptions->Element[n,Integers]]. Daniel Lichtblau Wolfram Research

**References**:**Trying to get closed form solution to simple recurrence.***From:*negatron <lokieffect@gmail.com>