A community for students.
Here's the question you clicked on:
 0 viewing
ParthKohli
 one year ago
@Kainui I think you'd like this.
ParthKohli
 one year ago
@Kainui I think you'd like this.

This Question is Closed

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Let \(a_n\) 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
 one year ago
Best ResponseYou've already chosen the best response.5\[a_1 = t\]\[a_2 = 4t(1t)\]\[a_3 = 16t(1t)\left[1  4t(1t) \right] = 16t(1t)\cdot  (2t1)^2 = 16t(1t)(2t1)^2\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5I guess we need to see a pattern and confirm our pattern by induction.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5If 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
 one year ago
Best ResponseYou've already chosen the best response.5n  degree  distinct roots   1  1  0  2  2  2  3  4  3  4  8  6 

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5How about trying to see repeated roots? 0, 0, 1, 2, ...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2just an observation \[a_{n} =\sin^2{2^{n1 }\theta} \implies a_{n+1} = 4\sin^2(2^{n1}\theta)(1\sin^2(2^{n1}\theta)) = \sin^2(2^n\theta)\] It follows that \[a_{1998}= \sin^2(2^{1997}\theta)\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5I noticed that at exactly the same moment as you posted it. Good observation. However, that would reserve our sequence to [1,1].

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5It is still a useful thought. Maybe working with \(\tan\) would help us achieve something?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Actually it would reserve our sequence to [0,1]...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we'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
 one year ago
Best ResponseYou've already chosen the best response.5Oh, that skipped me! Wow!!!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5So 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
 one year ago
Best ResponseYou've already chosen the best response.5\(2^{1997}\theta\) should be a multiple of \(\pi\)...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5I'm really bad at counting. :

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[\theta = n\pi\]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
 one year ago
Best ResponseYou've already chosen the best response.5\[\theta = \pm \pi/2^{1997}, \pm \pi/2^{1996}, \cdots ,\pm \pi/2^1\]Each pair returns one unique value of \(\sin^2\theta\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5So far, we've calculated 1998 unique values of \(\sin^2\theta\). Is that it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[t\in [0,1] \implies \theta \in [0,\frac{\pi}{2} ]\] \[\sin(2^{1997}\theta) = 0 \implies \theta = \frac{n\pi}{2^{1997}}\] \(n=0,1,2,\ldots, 2^{1996}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Oh, I got it. Thanks.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5I doublecounted 0 at \(\theta = 0 \) and \(\pi\). :P

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.51997 distinct values?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that would be too less when you compose a quadratic 1998 times

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2also recall the fact that sin is onetoone in the interval [0, pi/2]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[\theta = \pm \pi/2^{1997}, \pm \pi/2^{1996}, \cdots ,\pm \pi/2^1\]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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Oh my, never mind...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2find the values of \(n\) such that \(\theta \in [0,\pi/2]\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Yes, alright. So \(2^{1997}\) distinct values.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Or am I misunderstanding what you're saying?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the solution i have says \(2^{1996}+1\) so double check..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Oh my god, what is wrong with me...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5What are the last three digits of that? I don't know how modular arithmetic works.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Remind me to get some sleep after this one. Thanks.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2work \[2^{1996}+1 \pmod{1000}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Yes, I have no idea how modular arithmetic works.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2eating dinner.. 10 mnts

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5My ... what a beautiful solution you came up with.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you may use chinese remainder thm or go wid binary exponentiation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i don't see any other neat way to reduce 2^1996

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0Is it possible to go backwards? Let's say \(b_1=0\) and \(b_{1998} = t\) \[b_{n+1} = \frac{1 \pm \sqrt{1b_n}}{2}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.