MathGroup Archive 2012

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

Search the Archive

Re: Recurrence Equation RSolve, no error shown, no solution?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg129135] Re: Recurrence Equation RSolve, no error shown, no solution?
  • From: Dana DeLouis <dana01 at icloud.com>
  • Date: Mon, 17 Dec 2012 02:59:16 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Delivered-to: l-mathgroup@wolfram.com
  • Delivered-to: mathgroup-newout@smc.vnet.net
  • Delivered-to: mathgroup-newsend@smc.vnet.net

> RSolve[{w[n + 1] == w[n] - 0.02/w[n], w[0] == 2}, w[n], n]
> Could someone tell me if there is possible to solve this and what's my error?

Hi.  Just to add 2 cents on "what's my error"
Here, the next value of x is based on current value of x:

	x - 1/(50 x)

We see that the values get smaller as we loop.
We also note that if the next value of x were zero, then we get a divide by zero error.

So, our values can only decrease until:

Solve[x-1/(50 x)==0]
{{x->-(1/(5 Sqrt[2]))},{x->1/(5 Sqrt[2])}}

%//N
{{x->-0.141421}, {x -> 0.141421}}

So, we can only decrease until our value reaches 1/(5 Sqrt[2]),
then we calculate zero for the next value, and encounter infinity, or div by 0.

Hense, I don't believe this is a True Recurrence Equation as one may have hoped.

Looks like the loop is only valid for about 102 loops.  (At machine precision)

NestWhileList[#-1/(50 #)&,  2.,  # > 1/(5 Sqrt[2]) &] ;

% // Length
102

= = = = = = = = = =
HTH  :>)
Dana DeLouis
Mac & Mathematica 8
= = = = = = = = = =




On Tuesday, December 11, 2012 7:54:32 PM UTC-5, Fran Alvarez wrote:
> Hi all!
> 
> I'm looking for the solution of:
> 
> 
> 
> RSolve[{w[n + 1] == w[n] - 0.02/w[n], w[0] == 2}, w[n], n]
> 
> 
> 
> but the output is the same on the input and no error is shown. I've read the help that comes with Mathematica (I'm using 7.01 version for Mac) but no way. It seems to me that the main difference with the majority of examples shown in the help is the w[n] that appears as a denominator at the equation's RHS. I've seen complex rational examples at the help though, and RSolve manages to give a solution.
> 
> Could someone tell me if there is possible to solve this and what's my error?
> 
> Thank you!
> 
> Fran =C1lvarez




  • Prev by Date: Re: Mathematica strange behaviour finding a cubic root
  • Next by Date: Re: Working with indefinite number of variables
  • Previous by thread: Re: Recurrence Equation RSolve, no error shown, no solution?
  • Next by thread: Clean output of mathematica