Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: creating functions from the result of Solve[]

• To: mathgroup at smc.vnet.net
• Subject: [mg41862] Re: [mg41826] creating functions from the result of Solve[]
• From: Tomas Garza <tgarza01 at prodigy.net.mx>
• Date: Sat, 7 Jun 2003 00:08:53 -0400 (EDT)
• References: <200306061350.JAA13089@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

If you want to solve a difference equation - or recurrence relation - you
may use the AddOn package DiscreteMath`RSolve`.

In[1]:=
Clear[y]
In[2]:=
<< "DiscreteMath`RSolve`"

In[3]:=
RSolve[3*y[n - 2] + 6*y[n - 1] - 2*y[n] == 0, y[n], n]

Out[3]=
{{y[n] -> -2*If[n >= -2, (1/Sqrt[15])*(2^(-4 - n)*
        (2*((3 - Sqrt[15])^(2 + n) - (3 + Sqrt[15])^
            (2 + n))*(C[1] - 3*C[2]) +
         ((3 - Sqrt[15])^(3 + n) - (3 + Sqrt[15])^
            (3 + n))*C[2])), 0]}}

You must supply initial values in order to determine the two constants c[1]
and c[2]. From there on, you shouldn't have any difficulty to proceed.

Tomas Garza
Mexico City
----- Original Message ----- 
From: "Okke" <kroosu at tref.nl>
To: mathgroup at smc.vnet.net
Subject: [mg41862] [mg41826] creating functions from the result of Solve[]


> hello,
>
> i have an equation and i want to make a function from the result of
> Solve[].
>
> for example:
>
> IN: Solve[3y[n - 2] + 6y[n - 1] - 2y[n] == 0, y[n]]
>
> OUT: {{y[n] -> (3*y[-2 + n] + 6*y[-1 + n])/2}}
>
> and now i'd like to have the function
> y[n_/;n>=0] := (3*y[-2 + n] + 6*y[-1 + n])/2
>
> y[n]/.Flatten[Solve[eqn==0,y[n]]]       is possible, but
> y[n_]:=y[n]/.Flatten[Solve[eqn==0,y[n]]] isn't
>
>
> is there anybody who could help me get this to work?
>
> -- 
> Okke
> Experience is that marvelous thing that enables you to recognize a
> mistake when you make it again.
> -- F. P. Jones
>
>

• References:
  • creating functions from the result of Solve[]
    • From: kroosu@tref.nl (Okke)