how to solve recurrence relations with exponents
solve $\large a_n = 1-2a_{n-1}^2~;~a_0 = -\frac{1}{3}$

Question

anonymous · Answer

it's always useful to try substitution until you see a patern

rational · Answer

like this ?
$$\large  \begin{align} \  a_n &=  1-2a_{n-1}^2 \~\&= 1 - 2(1-2a_{n-2}^2)^2\~\\end{align}$$

anonymous · Answer

exactly. it's going to be very tedious because of the square exponent

ikram002p · Answer

why is difficult ?

ikram002p · Answer

its*

rational · Answer

can this be a solution :

$$\large  
a_n =  -\cos(2^n \arccos(-1/3)) 

$$

even if this is a solution, this doesn't look that exciting... im feeling there should be a way to find solution without trig -.-

ikram002p · Answer

how did u got that ?

rational · Answer

from  double angle identity $$\cos (2\theta) = 2\cos^2\theta - 1$$

ikram002p · Answer

u know we could always check revercly , then find limit when n goes to $\infty$
that would be extra solution