Re: hi,friends(8)

*To*: mathgroup at smc.vnet.net*Subject*: [mg92269] Re: hi,friends(8)*From*: "Szabolcs HorvÃt" <szhorvat at gmail.com>*Date*: Thu, 25 Sep 2008 05:32:13 -0400 (EDT)*References*: <gb7o7i$nas$1@smc.vnet.net> <48D77722.5070300@gmail.com>

2008/9/24 ddkkjj <noble_cbc at hotmail.com>: > > >> Date: Mon, 22 Sep 2008 12:44:50 +0200 >> From: szhorvat at gmail.com >> To: noble_cbc at hotmail.com >> Subject: Re: hi,friends(8) >> >> China -->hk wrote: >> > hi,friends: >> > Can we use Mathematica to solve the following recurrence function >> > problem? >> > f[1,1]=1,f[m,n+1]=f[m,n]+2,f[m+1,n]=2*f[m,n],m,n are positive >> > integers.f[m,n]=? >> > thanks, >> > >> >> Recurrence equations can be solved with RSolve, but there is a little >> problem with your set of equations: >> >> f[1,1] = 1 >> f[1,2] = f[1,1] + 2 = 3 >> f[2,2] = 2*f[1,2] = 6 >> >> f[1,1] = 1 >> f[2,1] = 2*f[1,1] = 2 >> f[2,2] = f[2,1] + 2 = 4 > > Thank you. > yes,sure,the Problem is wrong--i.e.: not consistency! So i want Mathematica > to prove it's wrong!how can we? Well, it does tell you that the system is overdetermined, doesn't it? In[1]:= RSolve[{f[m, n + 1] == f[m, n] + 2, f[m + 1, n] == 2*f[m, n], f[1, 1] == 1}, f[m, n], {m, n}] During evaluation of In[1]:= RSolve::overdet: There are fewer dependent variables than equations, so the system is overdetermined. Out[1]= RSolve[{f[m, 1 + n] == 2 + f[m, n], f[1 + m, n] == 2 f[m, n], f[1, 1] == 1}, f[m, n], {m, n}]