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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}]


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