Welcome to our community

Be a part of something great, join today!

Show that a_{n+8} = a_n

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Suppose \(\displaystyle \left({a_n}\right)_{n=1}^\infty\) be recursively defined by $a_0>1$, $a_1>0$ and $a_2>0$,

\(\displaystyle a_{n+3}=\frac{1+a_{n+1}+a_{n+2}}{a_n}\) for $n=0,1,2,\cdots$,

Show that $a_n$ has period of 8.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: Show that a_{n+8}=a_n

Suppose \(\displaystyle \left({a_n}\right)_{n=1}^\infty\) be recursively defined by $a_0>1$, $a_1>0$ and $a_2>0$,

\(\displaystyle a_{n+3}=\frac{1+a_{n+1}+a_{n+2}}{a_n}\) for $n=0,1,2,\cdots$,

Show that $a_n$ has period of 8.
In the paper of Lothar Berg 'Nonlinear difference equation with periodic solutions' [2006] it is explained that given a difference equation like...


$\displaystyle x_{n+1} = f(x_{n},x_{n-1},...,x_{n-k})\ (1)$


... it admits a periodic solution of periodo p if it exists an equilibrium point $x^{*}$ such that...


$\displaystyle x^{*} = f(x^{*},x^{*},...,x^{*})\ (2)$


... and, defining...


$\displaystyle f_{i}= \frac{\partial f}{\partial u_{i}} (x^{*},x^{*},...,x^{*})\ (3)$

... all the rouths of the polynomial...


$\displaystyle \lambda^{k+1} - f_{0} \lambda^{k} - ... - f_{k-1} \lambda - f_{k}\ (4)$

... are simple p-th routh of unity. In Your case is...

$\displaystyle x_{n+1}= \frac{1 + x_{n} + x_{n-1}}{x_{n-2}}\ (5)$

... the characteristic polynomial is...

$\displaystyle \lambda^{3} - \frac{1}{x^{*}}\ (\lambda^{2} + \lambda) + 1\ (6)$


... with $\displaystyle x^{*} = 1 \pm \sqrt{2}$ and the roths of (6) are...


$\displaystyle \lambda = -1, \lambda = \frac {1-i}{\sqrt{2}}, \lambda = \frac{1+i}{\sqrt{2}}, \lambda = - \frac{1+i}{\sqrt{2}}, \lambda =- \frac{1-i}{\sqrt{2}}\ (7)$


... so that the periodicity p=8 is demonstrated...

Kind regards

$\chi$ $\sigma$

P.S. the Lotahr's article is ...

http://ftp.math.uni-rostock.de/pub/romako/heft61/lothar.pdf

 
Last edited:
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Show that a_{n+8}=a_n

In the paper of Lothar Berg 'Nonlinear difference equation with periodic solutions' [2006] it is explained that given a difference equation like...


$\displaystyle x_{n+1} = f(x_{n},x_{n-1},...,x_{n-k})\ (1)$


... it admits a periodic solution of periodo p if it exists an equilibrium point $x^{*}$ such that...


$\displaystyle x^{*} = f(x^{*},x^{*},...,x^{*})\ (2)$


... and, defining...


$\displaystyle f_{i}= \frac{\partial f}{\partial u_{i}} (x^{*},x^{*},...,x^{*})\ (3)$

... all the rouths of the polynomial...


$\displaystyle \lambda^{k+1} - f_{0} \lambda^{k} - ... - f_{k-1} \lambda - f_{k}\ (4)$

... are simple p-th routh of unity. In Your case is...

$\displaystyle x_{n+1}= \frac{1 + x_{n} + x_{n-1}}{x_{n-2}}\ (5)$

... the characteristic polynomial is...

$\displaystyle \lambda^{3} - \frac{1}{x^{*}}\ (\lambda^{2} + \lambda) + 1\ (6)$


... with $\displaystyle x^{*} = 1 \pm \sqrt{2}$ and the roths of (6) are...


$\displaystyle \lambda = -1, \lambda = \frac {1-i}{\sqrt{2}}, \lambda = \frac{1+i}{\sqrt{2}}, \lambda = - \frac{1+i}{\sqrt{2}}, \lambda =- \frac{1-i}{\sqrt{2}}\ (7)$


... so that the periodicity p=8 is demonstrated...

Kind regards

$\chi$ $\sigma$

P.S. the Lotahr's article is ...

http://ftp.math.uni-rostock.de/pub/romako/heft61/lothar.pdf

Hi chisigma,

Thanks for participating and thanks for the pdf link too, that's a wonderful reading material to say the least...

I'll only post the solution to this problem later, I just feel there are others who still want to attempt to it. :D
 
  • Thread starter
  • Admin
  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Show that a_{n+8}=a_n

Here is the solution provided by others which I think is worth sharing at MHB:

We're given \(\displaystyle a_{n+3}=\frac{1+a_{n+1}+a_{n+2}}{a_n}\) for $n=0,1,2,\cdots$

First we multiply the equation by $a_n$ to eliminate the fraction and get

$a_{n+3}a_{n}=1+a_{n+1}+a_{n+2}$---(1)

If we replace $n$ by $n-1$, the above equation becomes

$a_{n+2}a_{n-1}=1+a_{n}+a_{n+1}$---(2)

And subtracting the equations (1) and (2) yields

$a_{n+3}a_{n}-a_{n+2}a_{n-1}=a_{n+2}-a_{n}$

Collecting the like terms and factoring out the common factor we now have

$a_{n}(1+a_{n+3})=a_{n+2}(1+a_{n-1})$

Adding the term $a_{n}a_{n+2}$ to both sides we get

$a_{n}(1+a_{n+2}+a_{n+3})=a_{n+2}(1+a_{n-1}+a_{n})$---(*)

And by applying the given recursive equation to (*) we obtain

$a_{n}a_{n+1}a_{n+4}=a_{n+2}a_{n+1}a_{n-2}$

$a_{n}a_{n+4}=a_{n+2}a_{n-2}$---(3)

Replace $n$ by $n-2$ to get

$a_{n-2}a_{n+2}=a_{n}a_{n-4}$---(4)

By comparing the equations (3) and (4) we notice that

$a_{n}a_{n+4}=a_{n}a_{n-4}$

$\therefore a_{n+4}=a_{n-4}$, $n\ge4$

This implies $ a_{n}=a_{n+8}$ for $n\ge0$.