# Show that the signal $x_n = A \cos(\omega n)$ can be fully predicted by a system with two weights $w_1,w_2$

I am trying to solve the following exercise:

Show that the signal $x_n = A\cos(\omega n)$ can be fully predicted by a system with two weights $w_1,w_2$ (i.e. $x_n = w_1 x_{n-1} + w_2 x_{n-2}$). Find $w_1,w_2$.

Some ideas came up but, even if they are right, I can't seem to order them in the right direction:

1. I can see that $x'' = -\omega^2x$. Does it connected somehow to what is asked?

2. I guess it has something to do with stationary signals. If yes, how can I prove that this signal is stationary? (Also, somehow I can't find in the internet a clear definition of "stationary signal". Just vage ideas like "not depended on time". What does this mean mathematically for a signal to be "not depended on time"?)

3. Maybe it is concerned with some trigonometry of this form: $$x_{n+2} = A\cos((n+2)\omega) = A\cos(n\omega + 2\omega) = A\cos(n\omega)\cos(2\omega) - A\sin(n\omega)\sin(2\omega) = B\cos(n\omega) + C\sin(n\omega)$$ I could go on with this development but I am realy not sure what am I looking for.

• I'd be useful for you to contextualize your use of the work "system", as it's very general. From your tags, I'd guess a neural network, but it'd be worth editing in a description. Commented Mar 23, 2016 at 14:58
• I think you can just write $x_n = w_1 x_{n-1} + w_2 x_{n-2}$ explicitly, i.e. replace the $x_n$ with the cosine expression and then see if the equation holds (for example the $A$ cancels out immediately). I quickly tried and got that it holds if $w_1 + w_2 e^{2 i \omega} = 1$ and $w_1 e^{- i \omega} + w_2 e^{-2 i \omega} = 1$, which can be solved for $w_1$ and $w_2$. Maybe I made a mistake somewhere but like this it should work. Since if you can write an expression down in closed form you proved not only that it can be predicted but also how. Commented Mar 23, 2016 at 15:03
• @Matthew Drury, the context is linear perceptron (supervised learning). Since I don't fully understand the question myself, I am not sure beyond what I wrote. But thanks. Let me know if you will have some more insights. Commented Mar 23, 2016 at 15:28

This is basically equivalent to, given $\cos(a)$, $\cos(a-\omega)$, predict $\cos(a+\omega)$. $$\cos(a-\omega)=\cos(a)\cos(\omega)+\sin(a)\sin(\omega)$$ $$\cos(a+\omega)=\cos(a)\cos(\omega)-\sin(a)\sin(\omega)$$ So let $w_1=2\cos(\omega)$, $w_2=-1$, we have $$w_1\cos(a)+w_2\cos(a-\omega)=2\cos(\omega)\cos(a)-\cos(a)\cos(\omega)-\sin(a)\sin(\omega)\\ =\cos(a)\cos(\omega)-\sin(a)\sin(\omega)=\cos(a+\omega).$$ The rest should be easy.