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- Feb 14, 2012

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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 $a_{1998}=0$?

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- Thread starter anemone
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- Feb 13, 2012

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$\displaystyle \Delta_{n}= a_{n+1} - a_{n} = 3\ a_{n} - 4\ a^{2}_{n}= f(a_{n})\ (1)$

The f(x) is represented in the pitcure...

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...

Kind regards

$\chi$ $\sigma$

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- Mar 31, 2013

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gives a2 = 1 and then a3 = a4 ... a1998 = 0

it appears that for a1 upto a1997 = 1

for each ofthem there is a different a1 and a1= 0 gives 1998

for example a1 = 1/2 shall give an a0 = (cos pi/4 + 1/2) so it appears much more solution.

- Feb 13, 2012

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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! ...

Kind regards

$\chi$ $\sigma$

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- Mar 31, 2013

- 1,350

1st we should solve

4x(1-x) = sin^2 y

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)

where n is from 0 to 2^1996 and k = 2^1996

this should be the solution

the proof I shall post later

- Mar 31, 2013

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sin^2 ((npi)/(k)/2)

where n is from 0 to 2^1996 and k = 2^1996

where n is from 0 to 2^1996 and k = 2^1996

f(sin^2 y) = sin ^2 2y

apply f 1997 times and we get

sin^2 (npi) = 0

done

apply f 1997 times and we get

sin^2 (npi) = 0

done

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- Feb 14, 2012

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Thanks for participating and the answer that both of you have found is correct! Well done!

The solution that I wanted to share here (which isn't my own solution) is as follows:

Observe that if $0 \le f(x) \le 1$, then $0 \le x \le 1$.

Hence if $a_{1998}=0$, then we must have $0 \le t \le 1$.

Now choose $0 \le \theta \le \frac{\pi}{2}$ such that $\sin \theta= \sqrt{t}$.

Observe also that for any $\alpha \in R$,

\(\displaystyle a_1=\sin^2 \alpha\)

\(\displaystyle a_2=f(\sin^2 \alpha)=4\sin^2 \alpha(1-\sin^2 \alpha)=4\sin^2 \alpha\cos^2 \alpha=\sin^2 (2\alpha)\)

\(\displaystyle a_3=f(\sin^2 2\alpha)=4\sin^2 2\alpha(1-\sin^2 2\alpha)=4\sin^2 2\alpha\cos^2 2\alpha=\sin^2 4\alpha=\sin^2 (2^2\alpha)\)

\(\displaystyle a_4f(\sin^2 4\alpha)=4\sin^2 4\alpha(1-\sin^2 4\alpha)=4\sin^2 4\alpha\cos^2 4\alpha=\sin^2 8\alpha=\sin^2 (2^3\alpha)\)

i.e. \(\displaystyle a_{n}=\sin^2 (2^{n-1}\alpha)\)

\(\displaystyle \therefore a_{1998}=\sin^2 (2^{1997}\alpha)\)

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 \(\displaystyle a_{1998}=0\).