ParthKohli
  • ParthKohli
@Kainui I think you'd like this.
Mathematics
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • 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
  • ParthKohli
@ganeshie8
ParthKohli
  • 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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ParthKohli
  • ParthKohli
I guess we need to see a pattern and confirm our pattern by induction.
ParthKohli
  • 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
  • ParthKohli
n | degree | distinct roots | ---------------------------------------- 1 | 1 | 0 | 2 | 2 | 2 | 3 | 4 | 3 | 4 | 8 | 6 |
ParthKohli
  • ParthKohli
How about trying to see repeated roots? 0, 0, 1, 2, ...
ganeshie8
  • 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
  • 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
  • ParthKohli
It is still a useful thought. Maybe working with \(\tan\) would help us achieve something?
ParthKohli
  • ParthKohli
Actually it would reserve our sequence to [0,1]...
ganeshie8
  • 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
  • ParthKohli
Oh, that skipped me! Wow!!!
ParthKohli
  • ParthKohli
Amazing!
ParthKohli
  • 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
  • ParthKohli
\(2^{1997}\theta\) should be a multiple of \(\pi\)...
ParthKohli
  • ParthKohli
I'm really bad at counting. :|
ParthKohli
  • 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
  • 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
  • ParthKohli
So far, we've calculated 1998 unique values of \(\sin^2\theta\). Is that it?
ParthKohli
  • ParthKohli
Oh, nope.
ganeshie8
  • 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
  • ParthKohli
Exactly...
ParthKohli
  • ParthKohli
Oh, I got it. Thanks.
ParthKohli
  • ParthKohli
I double-counted 0 at \(\theta = 0 \) and \(\pi\). :P
ParthKohli
  • ParthKohli
1997 distinct values?
ganeshie8
  • ganeshie8
that would be too less when you compose a quadratic 1998 times
ganeshie8
  • ganeshie8
also recall the fact that sin is one-to-one in the interval [0, pi/2]
ParthKohli
  • 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
  • ganeshie8
http://gyazo.com/f22b532bddfbe6cd15af74145160b997
ParthKohli
  • ParthKohli
Oh my, never mind...
ParthKohli
  • ParthKohli
Goddammit.
ganeshie8
  • ganeshie8
find the values of \(n\) such that \(\theta \in [0,\pi/2]\)
ParthKohli
  • ParthKohli
Yes, alright. So \(2^{1997}\) distinct values.
ParthKohli
  • ParthKohli
(?)
ParthKohli
  • ParthKohli
Or am I misunderstanding what you're saying?
ganeshie8
  • ganeshie8
the solution i have says \(2^{1996}+1\) so double check..
ParthKohli
  • ParthKohli
Oh my god, what is wrong with me...
ParthKohli
  • ParthKohli
Obviously, yes.
ParthKohli
  • ParthKohli
What are the last three digits of that? I don't know how modular arithmetic works.
ParthKohli
  • ParthKohli
Remind me to get some sleep after this one. Thanks.
ganeshie8
  • ganeshie8
work \[2^{1996}+1 \pmod{1000}\]
ParthKohli
  • ParthKohli
Yes, I have no idea how modular arithmetic works.
ganeshie8
  • ganeshie8
eating dinner.. 10 mnts
ParthKohli
  • ParthKohli
My ... what a beautiful solution you came up with.
ganeshie8
  • ganeshie8
you may use chinese remainder thm or go wid binary exponentiation
ganeshie8
  • ganeshie8
i don't see any other neat way to reduce 2^1996
ParthKohli
  • ParthKohli
W|A returns 337.
Kainui
  • 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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