Given the following (discrete time) system
$x(k+1)=r-rx(k)$ where $ r>=0 $ is a parameter
- Find for which $r>=0$ all solutions of this system converge to a fixed point
- Verify if there exist $r>=0$ such that this system admits solutions of period 2
For 1.:
Is there a way to this without Matlab? I couldn't find out how...
For 2. :
I found that I should write $x(k+2)=r−rx(k+1)$, substitute in my expression for $x(k+1)$ and see if there are solutions with $x(k+2)=x(k)$ but I can't get it right. If someone can show me how to do this part it would be great.
Thanks !
Notice that by your definition of the system $$x(k+1)=r-rx(k)=r-r(r-rx(k-1))=r-r^2+r^3+..+(-1)^{k}r^{k+1}+(-1)^{k+1}r^{k+1}x(0)$$ But on the other hand \begin{align}r-r(r-rx(k-1))&=r-r^2+r^3+..+(-1)^{k}r^{k+1}+(-r)^{k+1}x(0)\\&=r(1-r+r^2+..+(-1)^kr^k)+(-1)^{k+1}r^{k+1}x(0)\\& =r\frac{1-(-r)^{k+1}}{1-(-r)}+(-r)^{k+1}x(0)\end{align} Now if the system has a fixed point than $$\lim_{k\to\infty}x(k+1)=\lim_{k\to\infty}x(k)$$ Clearly the series converges for $|r|<1$ so in the limit one has $$\lim_{k\to\infty}x(k+1)=\lim_{k\to\infty}\Big\{r\frac{1-(-r)^{k+1}}{1-(-r)}+(-r)^{k+1}x(0)\Big\}=\frac{r}{1+r}$$ In fact you could get the last result by only assuming that if there is a fixed point then $x(k+1)=x(k)=x^*$ to get $x^*=\frac{r}{1+r}$. However you would not know for what values this would be valid so the series approach gives you the insight. Regarding the second part notice that if there is a solution of period two i.e. $$x(k+2)=x(k)\Rightarrow x(k)=r-rx(k+1)\Rightarrow x(k)=r-r(r-rx(k))\Rightarrow x(k)=\frac{r-r^2}{1-r^2}$$ The last result is equivalent to $$x(k)=\frac{r}{1+r}$$ for all $k$. Any $|r|>0$ would be a solution.