@Kainui I think you'd like this.

- ParthKohli

@Kainui I think you'd like this.

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- ParthKohli

Let \(a_n\) be a sequence of real numbers defined by \(a_1 = t\) and \(a_{n+1} = 4a_n(1 - a_n)\) for \(n > 1\).
Find the last three digits of the number of distinct values of \(t\) such that \(a_{1998} = 0\)?

- ParthKohli

@ganeshie8

- ParthKohli

\[a_1 = t\]\[a_2 = 4t(1-t)\]\[a_3 = 16t(1-t)\left[1 - 4t(1-t) \right] = 16t(1-t)\cdot - (2t-1)^2 = -16t(1-t)(2t-1)^2\]

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## More answers

- ParthKohli

I guess we need to see a pattern and confirm our pattern by induction.

- ParthKohli

If not a pattern in our polynomials, we can at least try to see one in the number of distinct values of \(t\) (and how repeated roots work with increasing \(n\)).

- ParthKohli

n | degree | distinct roots |
----------------------------------------
1 | 1 | 0 |
2 | 2 | 2 |
3 | 4 | 3 |
4 | 8 | 6 |

- ParthKohli

How about trying to see repeated roots?
0, 0, 1, 2, ...

- ganeshie8

just an observation
\[a_{n} =\sin^2{2^{n-1 }\theta} \implies a_{n+1} = 4\sin^2(2^{n-1}\theta)(1-\sin^2(2^{n-1}\theta)) = \sin^2(2^n\theta)\]
It follows that
\[a_{1998}= \sin^2(2^{1997}\theta)\]

- ParthKohli

I noticed that at exactly the same moment as you posted it. Good observation.
However, that would reserve our sequence to [-1,1].

- ParthKohli

It is still a useful thought. Maybe working with \(\tan\) would help us achieve something?

- ParthKohli

Actually it would reserve our sequence to [0,1]...

- ganeshie8

we're interested in [0, 1] only because anything outside spits out numbers that are outside [0, 1] so all the terms of the sequence will be outside the interval [0, 1]

- ParthKohli

Oh, that skipped me! Wow!!!

- ParthKohli

Amazing!

- ParthKohli

So if \(a_1 = t = \sin^2 \theta \) then we're interested in finding the unique values of \(\sin^2 \theta\) such that \(\sin(2^{1997}\theta) = 0\).
I think you just recovered my love for math.

- ParthKohli

\(2^{1997}\theta\) should be a multiple of \(\pi\)...

- ParthKohli

I'm really bad at counting. :|

- ParthKohli

\[\theta = n\pi\]satisfies the equation and returns only one value for \(\sin^2\theta\), i.e., 0. Now we've got to look at fractional values.

- ParthKohli

\[\theta = \pm \pi/2^{1997}, \pm \pi/2^{1996}, \cdots ,\pm \pi/2^1\]Each pair returns one unique value of \(\sin^2\theta\).

- ParthKohli

So far, we've calculated 1998 unique values of \(\sin^2\theta\). Is that it?

- ParthKohli

Oh, nope.

- ganeshie8

\[t\in [0,1] \implies \theta \in [0,\frac{\pi}{2} ]\]
\[\sin(2^{1997}\theta) = 0 \implies \theta = \frac{n\pi}{2^{1997}}\]
\(n=0,1,2,\ldots, 2^{1996}\)

- ParthKohli

Exactly...

- ParthKohli

Oh, I got it. Thanks.

- ParthKohli

I double-counted 0 at \(\theta = 0 \) and \(\pi\). :P

- ParthKohli

1997 distinct values?

- ganeshie8

that would be too less when you compose a quadratic 1998 times

- ganeshie8

also recall the fact that sin is one-to-one in the interval [0, pi/2]

- ParthKohli

\[\theta = \pm \pi/2^{1997}, \pm \pi/2^{1996}, \cdots ,\pm \pi/2^1\]Now what you're saying is that I should multiply \(\pi/2^{1997}\) by \(n = 1, 2, 3,\cdots, 1996\)
then \(\pi/2^{1996}\) by \(n = 1, 2, 3,\cdots, 1995 \) and so on, right?

- ganeshie8

http://gyazo.com/f22b532bddfbe6cd15af74145160b997

- ParthKohli

Oh my, never mind...

- ParthKohli

Goddammit.

- ganeshie8

find the values of \(n\) such that \(\theta \in [0,\pi/2]\)

- ParthKohli

Yes, alright. So \(2^{1997}\) distinct values.

- ParthKohli

(?)

- ParthKohli

Or am I misunderstanding what you're saying?

- ganeshie8

the solution i have says \(2^{1996}+1\)
so double check..

- ParthKohli

Oh my god, what is wrong with me...

- ParthKohli

Obviously, yes.

- ParthKohli

What are the last three digits of that? I don't know how modular arithmetic works.

- ParthKohli

Remind me to get some sleep after this one. Thanks.

- ganeshie8

work \[2^{1996}+1 \pmod{1000}\]

- ParthKohli

Yes, I have no idea how modular arithmetic works.

- ganeshie8

eating dinner.. 10 mnts

- ParthKohli

My ... what a beautiful solution you came up with.

- ganeshie8

you may use chinese remainder thm or go wid binary exponentiation

- ganeshie8

i don't see any other neat way to reduce 2^1996

- ParthKohli

W|A returns 337.

- Kainui

Is it possible to go backwards? Let's say \(b_1=0\) and \(b_{1998} = t\)
\[b_{n+1} = \frac{1 \pm \sqrt{1-b_n}}{2}\]

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