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MathGroup Archive 1997

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Re: linear equations with indexed variables?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6963] Re: [mg6896] linear equations with indexed variables?
  • From: jpk at max.mpae.gwdg.de
  • Date: Wed, 30 Apr 1997 22:25:23 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

> From hxie at gradin.cis.upenn.edu Fri Apr 25 23:44:18 1997
> Date: Fri, 25 Apr 1997 14:00:30 -0400 (EDT)
> From: hxie at gradin.cis.upenn.edu (Hong-liang Xie)
> To: mathgroup at smc.vnet.net
> Subject: [mg6896] linear equations with indexed variables?
> 
> I am new to Mathematica and I need to solve a system of 
> linear equations whose variables are indexed (i.e., x[1], 
> x[2], x[3]), for example,
> 
>        3x[i] = 2x[i+1] + x[i-1] ( 1 <= i <= 3)
>        x[4 ]  = 1
>        x[0]   = 0
> 
> The above is a system of 5 linear equations with
> 5 variables x[0], x[1], x[2], x[3], x[4].  I can certainly
> rewrite it into 5 equations using 5 variables x0, x1, x2, 
> x3, x4.  However, if the range of i gets bigger, or, worse, 
> if each x has two indexes as in x[i,j], this rewriting could 
> get of hand quickly.  I tried different equation solving 
> functions in Mathematica but with no luck.  I would therefore 
> appreciate help from experts here on how to solve this kind of 
> equations directly.  Thanks a lot!
> 
> Hong
> CIS Dept
> Univ of Pennsylvania
> 
> 
> 
> 
> 
> 
Hi Hong,

You had a system of recurence equations with constant coeficients and
a boundary problem for this system. The recurence equations with constant
coefficients can be solved usualy by the ansatz x(i)=zeta^i. By insertion
of this ansatz in Your recurence equation  You get the two solutions
zeta=1/2 and zeta=1. This step looks in Mathematica 3.0 so

In[1]:= fdeqn=3 x[i]==2 x[i+1]+x[i-1];

In[2]:= fdsol=Solve[
               Cancel[\[Zeta]^(-i)*#] & /@ 
                 (fdeqn /. x[i_]:>\[Zeta]^i),
                \[Zeta]
               ]
Out[2]=
{{\[Zeta] -> 1/2}, 
  {\[Zeta] -> 1}}
  
With this solutions for zeta You can construct the general solution
of Your recurence equation

as x[i]=C[1]*zeta[1]^i+C[2]*zeta[2]^i

The two arbitary constants C[1] and C[2] must be determined from the
boundary conditons. We make our solution a bit more Mathematica like
with

In[3]:=
sol=x[i_]:>
    Evaluate[Plus
         @@(Map[Apply[Times,#] &, 
            Transpose[{{C[1],C[2]},\[Zeta] ^i/.fdsol}]])]

Out[3]=
 x[i_] :> C[1]/2^i + C[2]
 
Now lets determine C[1] and C[2]

In[4]:=
boundsol=Solve[bound={x[4]==1,x[0]==0} /. sol,{C[1],C[2]}]

Out[4]=
{{C[1] -> -(16/15), 
   C[2] -> 16/15}}
   
 Now one gets the final solution
 for Your equations with
 
 In[5]:= finalsol= First[x[i] /. sol /.boundsol]
 
 Out[5]=16/15 - 2^(4 - i)/15
 
 That's all.
 
 Hope that helps
 Jens


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