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- Feb 14, 2012

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

Show that $a_n$ has period of 8.

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- #1

- Feb 14, 2012

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

Show that $a_n$ has period of 8.

- Feb 13, 2012

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

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

Show that $a_n$ has period of 8.

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

... 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

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- Feb 14, 2012

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HiIn 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

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.

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- Feb 14, 2012

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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 $n\ge0$.