Finding Recursive Definition for the following:

Question

How would i start off to find a recursive definition for

$X_{0}$=.19

$X_{1}$=.1919

$X_{2}$=.191919

...

$X_{n+1}$= what goes here?

Answer 1

Firstly, we see how $x(n+1)$ can be obtained from $x(n)$.

By intuition, it is "like $x(n)$ but with one more $19$".

To make $x(n+1)$, we can use $x(n)$ to "match up" with all the $19$.

$x(n+1)=0.19\color{red}{19\dots19}=0.19+0.00\color{red}{19\dots19}=0.01x(n)+0.19$

Answer 2

Or, we can do this way

$$x(n+1)=x(n)+0.19\cdot0.01^n\quad(x(0)=0)$$

And I noticed that if we combine this with the Element118's result, we can also easily get the general expression for $x(n)$ as

$$x(n+1)=x(n)+0.19\cdot0.01^n = 0.01x(n)+0.19$$ $$x(n)=\frac{0.19\cdot(1-0.01^n)}{0.99}$$