So I have this equation
$4y_{k+1} = 2y_k$.
If i divide both sides by 4, I get
$$y_{k+1}=\frac12y_k$$
However, the equation can be re-written as
so that
$$y_{1}=\frac12y_0$$
and
$$y_{2}=\frac12 y_1=\frac1{2^2}y_0$$
and
$$y_{3}=\frac12 y_2=\frac1{2^3}y_0$$ $$\cdots$$
This is a difference equation, where the initial condition first is y=0 and then y=2.
Can somebody please explain how , from $$y_{k+1}=\frac12y_k$$ i can get to the following equations?
Notice that $$y_n={1\over 2}y_{n-1}={1\over 2^2}y_{n-2}={1\over 2^3}y_{n-3}=\cdots ={1\over 2^k}y_{n-k}$$let $k=n$ therefore $$y_n={1\over 2^n}y_0$$