Given the sequence $\{z_n\}_0 ^ \infty$ with $z_{n+1}-z_n=a(z_n-z_{n-1})$, $0<|a|<1$. Express lim $z_n$ in terms of $z_0$ and $z_1$.

Question

Any help on how to approach this problem would be greatly appreciated. I have tried to rewrite $a$ and $z_n$ in different ways but none lead to good results

Given the sequence $\{z_n\}_0 ^ \infty$ with $z_{n+1}-z_n=a(z_n-z_{n-1})$ where $0<|a|<1$. Express lim $z_n$ in terms of $z_0$ and $z_1$.

Please put the question into the body itself as well as the title. — The Count, Mar 29 '17 at 00:45

Simply Beautiful Art · Answer 1 · 2017-03-29T12:32:00.520

0

Thanks to this, it is relatively straight-forward to solve

$$t^2-t=at-a$$

$$t_0=a,t_1=1$$

And thus, we have

$$z_n=c_0(t_0)^n+c_1(t_1)^n=c_0a^n+c_1$$

By letting $n=0,1$, we may then solve for $c_0$ and $c_1$ and we are done.

$$z_0=c_0+c_1$$

$$z_1=c_0a+c_1$$

$$c_0=\frac{z_1-z_0}{1-a}$$

$$c_1=\frac{z_0(2-a)-z_1}{1-a}$$

edited Mar 29 '17 at 12:32

answered Mar 29 '17 at 00:47

Simply Beautiful Art

74,685

Thanks! Would this also be solveable using a similar method as used here? http://math.stackexchange.com/questions/332993/find-lim-n-to-infty-z-n-if-z-n1-frac12-left-z-n-frac1 – Arthur W Mar 29 '17 at 00:54
I doubt it. What you have is a linear equation, so... yeah. – Simply Beautiful Art Mar 29 '17 at 00:59
@MyGlasses ah yes, thanks for the catch. – Simply Beautiful Art Mar 29 '17 at 12:32

Given the sequence $\{z_n\}_0 ^ \infty$ with $z_{n+1}-z_n=a(z_n-z_{n-1})$, $0<|a|<1$. Express lim $z_n$ in terms of $z_0$ and $z_1$.

1 Answers1