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Re: hi,friends(8)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg92196] Re: hi,friends(8)
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Mon, 22 Sep 2008 07:26:19 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <gb7o7i$nas$1@smc.vnet.net>

China -->hk wrote:

> 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]=?

Mathematica can be used to solve partial difference equations; however, 
  one may have to do some work to get an answer.

*RSolve* is the function you are looking for. *Eliminate* is used to 
rewrite two equations as only one. We find a general solution (w/o using 
  the condition f[1,1] == 1). Finally, using some intuition, we find a 
suitable function for the coefficient C[1]. So we have,


     f[m,n] == -4 (-1 + 2^m) + 5 2^(-1 + m) Cos[(m + n) Pi]


Below is the step by step process describe above.

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}]

In[2]:= Eliminate[{f[m, n + 1] == f[m, n] + 2,
   f[m + 1, n] == 2*f[m, n]}, f[m, n]]

Out[2]= 4 + f[1 + m, n] == 2 f[m, 1 + n]

In[3]:= RSolve[{4 + f[1 + m, n] == 2 f[m, 1 + n], f[1, 1] == 1},
  f[m, n], {m, n}]

Out[3]= RSolve[{4 + f[1 + m, n] == 2 f[m, 1 + n], f[1, 1] == 1},
  f[m, n], {m, n}]

In[4]:= RSolve[4 + f[1 + m, n] == 2 f[m, 1 + n], f[m, n], {m, n}]

Out[4]= {{f[m, n] -> -4 (-1 + 2^m) + 2^(-1 + m) C[1][m + n]}}

In[5]:= f[m, n] /. % /. {m -> 1, n -> 1}

Out[5]= {-4 + C[1][2]}

In[6]:= f[m, n] /. %%[[1]] /. C[1] -> Function[k, 5 Cos[Pi k]]

Out[6]= -4 (-1 + 2^m) + 5 2^(-1 + m) Cos[(m + n) \[Pi]]

In[7]:= % /. {m -> 1, n -> 1}

Out[7]= 1

In[8]:= Plot3D[%%, {m, 1, 5}, {n, 1, 5}, PlotRange -> All]

Out[8]= [...  Graphic deleted  ...]


Regards,
-- Jean-Marc


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