So I have the difference equation $4y_{k+1} = 2y_k$.
I need to solve this equation, first with the initial condition $y_0 = 0$ and then with the initial condition $y_0 = 2$.
The professor wrote on the answer sheet that applying iteratively the recurrence relation it is easy to check that:
$y_k= (1/2)^k y_0 $.
Since I missed the lecture where he explained this, can somebody explain how to come about with this question?
I imagine that once I get the equation, then the initial condition $y_0 = 0$ gives us the particular solution $y_k = 0$. Thus the other initial condition $y_0 = 2$ is verified by the solution $y_k = 2(1/2)^k$.
Any help on how to get to the equation $y_k= (1/2)^k y_0$ will be appreciated.
$$y_{k+1}=\frac12y_k$$ so that
$$y_{1}=\frac12y_0$$
and
$$y_{2}=\frac12 y_1=\frac1{2^2}y_0$$
and
$$y_{3}=\frac12 y_2=\frac1{2^3}y_0$$ $$\cdots$$