$a_{n+1} = \frac{1}{2}(a_n)^2 - 2a_n +4$

Question

$a_{n+1} = \frac{1}{2}(a_n)^2 - 2a_n +4$

How to find a solution to this nonlinear equation? I've heard that if we have $a_{n+1} = S(a_n)$ then there should be a function $f$ such that $Cf(S(a_n))=f(a_n) \frac{dS}{da_n} $ but is that really the only way? A full solution would be very helpful.

Answer 1

Hint: Let $b_n=a_n-2$ and see what happens when you express $b_{n+1}$ in terms of $b_n$.

Answer 2

Given expression is $$(a_n-2)^2=2(a_{n+1}-2)$$ Let $(a_n-2)=2.e^{t_n}$. Put in above we get $$2t_n-t_{n+1}=0$$ Hence $$\frac{t_{n+1}}{t_n}=2$$ Now find $t_{n+1}$ buy putting different n.