Let ${a_n}$ be the sequence of real numbers defined by $a_1=t$, $a_{n+1}=4a_n(1-a_n)$ for $n \ge 1$. For how many distinct values of $t$ do we have $a_{1998}=0$?
There are one attractive fixed point in $\displaystyle x_{1}= \frac{3}{4}$ and one repulsive fixed point in $\displaystyle x_{0}=0$, so that 0 is a repulsive fixed point. That means that 0 can be 'captured' for n=2 only if is...
$\displaystyle a_{1} = t = - \Delta_{1} \implies t - t^{2}= 0\ (2)$
... and that is for t=0 or t=1...
But it is required to find the set of t for which 0 is capured for $\displaystyle n \le 1998$ and that requires further study...
In my previous post I wrote that the two possible values of t for which 0 is captured at n=1 are t=0 and t=1...
all right!... now we have to find the possible values of t for which the 1 can be captured at n=1, that means that the 0 will be captured at n=2. These values of t are the solutions of the equation...
$\displaystyle 4\ t^{2} - 4\ t + 1 = 0\ (1)$
... that is the only $\displaystyle t = \frac{1}{2}$. Now we have to search the values of t for which $\frac{1}{2}$ is captured at n=1 so that 0 will be captured at n=3. These values of t are the solutions of the equation...
$\displaystyle 4\ t^{2} -4\ t + \frac{1}{2}= 0\ (2)$
... that are $\displaystyle t = \frac{1}{2} \pm \frac{1}{\sqrt{8}}$. Each of these solutions will generate two solution for n=4 and so one till to n=1998 so that the required number of values of t is...
$\displaystyle N = 2 + \sum_{k=0}^{1996} 2^{k} = 2^{1996}+1\ (3)$
I do hope that nobody ask me to compute them all! ...
we see that sin ^ 2 y/2 and cos^2 y /2 are roots of equation
2)
this has got 2 solutions for all y except sin y = 1
now counting
a1 = 0 is a solution as all of the values a0 on wards 0
a1 = 1 is a solution as all of the values a2 onwards 0
a1 to a1996 = 1/2 then next value = 1 and so a1998 = 0
a1 =1/2 = one solution
a2 = 1/2 2 solution
a1996 = 1/2 so 2^1995 solution
adding we get 2^1996+ 1 solution
now I predict the the solution is sin^2 ((npi)/(k)/2)
In order to have \(\displaystyle a_{1998}=0\), we need \(\displaystyle \sin^2 2^{1997}\alpha=0\), i.e. \(\displaystyle \alpha=\frac{k \pi}{2^{1997}}\) where \(\displaystyle k \in Z\).
Therefore, being bounded by the range of $0 \le \theta \le \frac{\pi}{2}$, we get \(\displaystyle 2^{1996}+1\) such values of $t$ such that \(\displaystyle a_{1998}=0\).