I have the following equation which I am trying to find the closed form for: $$ x_{n+1} = \frac{1}{2}-\frac{x_{n}}{2}$$
So far rearranging and substituting has yielded the following equations:
$$2x_{n+1}+ x_{n-1} = 1$$
$$2x_{n+1} = x_{n}+ x_{n-1}$$ $$2x_{n+1} = 1- x_{n}$$ $$ 6x_{n+1} - x_{n}-2x_{n-1} =1$$ $$4x_{n+1} - x_{n-1}=1$$ I've just been trying to solve a jumble of these equations which hasn't yielded much so any help would be appreciated. Thanks!
On
Let $ n $ be a positive integer, wa have that : \begin{aligned} \left(\forall k<n\right),\ x_{n-k}&=\frac{1}{2}-\frac{x_{n-k-1}}{2}\\ \iff \left(\forall k<n\right),\ \left(-\frac{1}{2}\right)^{k}x_{n-k}&=\frac{1}{2}\left(-\frac{1}{2}\right)^{k}+\left(-\frac{1}{2}\right)^{k+1}x_{n-k-1}\\ \iff \sum_{k=0}^{n-1}{\left(\left(-\frac{1}{2}\right)^{k}x_{n-k}-\left(-\frac{1}{2}\right)^{k+1}x_{n-k-1}\right)}&=\frac{1}{2}\sum_{k=0}^{n-1}{\left(-\frac{1}{2}\right)^{k}}\\ \iff x_{n}-\left(-\frac{1}{2}\right)^{n}x_{0}&=\frac{1}{2}\frac{1-\left(-\frac{1}{2}\right)^{n}}{1-\frac{1}{2}} \end{aligned}
And thus : $$ \left(\forall n\in\mathbb{N}\right),\ x_{n}=\left(-\frac{1}{2}\right)^{n}x_{0}+1-\left(-\frac{1}{2}\right)^{n} $$
If $y_n=x_n-\frac 1 3$ then $y_{n+1}=-\frac {y_n} 2$ and you can iterate this easily.
[If $x_n$ converges the limit has to be $\frac 1 3$ from the given recurrence relation. This suggests that you look at $x_n-\frac 1 3$].