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ParthKohli

  • one year ago

@Kainui I think you'd like this.

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  1. ParthKohli
    • one year ago
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    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 \(a_{1998} = 0\)?

  2. ParthKohli
    • one year ago
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    @ganeshie8

  3. ParthKohli
    • one year ago
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    \[a_1 = t\]\[a_2 = 4t(1-t)\]\[a_3 = 16t(1-t)\left[1 - 4t(1-t) \right] = 16t(1-t)\cdot - (2t-1)^2 = -16t(1-t)(2t-1)^2\]

  4. ParthKohli
    • one year ago
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    I guess we need to see a pattern and confirm our pattern by induction.

  5. ParthKohli
    • one year ago
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    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 \(n\)).

  6. ParthKohli
    • one year ago
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    n | degree | distinct roots | ---------------------------------------- 1 | 1 | 0 | 2 | 2 | 2 | 3 | 4 | 3 | 4 | 8 | 6 |

  7. ParthKohli
    • one year ago
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    How about trying to see repeated roots? 0, 0, 1, 2, ...

  8. ganeshie8
    • one year ago
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    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 \[a_{1998}= \sin^2(2^{1997}\theta)\]

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

  10. ParthKohli
    • one year ago
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    It is still a useful thought. Maybe working with \(\tan\) would help us achieve something?

  11. ParthKohli
    • one year ago
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    Actually it would reserve our sequence to [0,1]...

  12. ganeshie8
    • one year ago
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    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]

  13. ParthKohli
    • one year ago
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    Oh, that skipped me! Wow!!!

  14. ParthKohli
    • one year ago
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    Amazing!

  15. ParthKohli
    • one year ago
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    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.

  16. ParthKohli
    • one year ago
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    \(2^{1997}\theta\) should be a multiple of \(\pi\)...

  17. ParthKohli
    • one year ago
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    I'm really bad at counting. :|

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

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

  20. ParthKohli
    • one year ago
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    So far, we've calculated 1998 unique values of \(\sin^2\theta\). Is that it?

  21. ParthKohli
    • one year ago
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    Oh, nope.

  22. ganeshie8
    • one year ago
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    \[t\in [0,1] \implies \theta \in [0,\frac{\pi}{2} ]\] \[\sin(2^{1997}\theta) = 0 \implies \theta = \frac{n\pi}{2^{1997}}\] \(n=0,1,2,\ldots, 2^{1996}\)

  23. ParthKohli
    • one year ago
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    Exactly...

  24. ParthKohli
    • one year ago
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    Oh, I got it. Thanks.

  25. ParthKohli
    • one year ago
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    I double-counted 0 at \(\theta = 0 \) and \(\pi\). :P

  26. ParthKohli
    • one year ago
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    1997 distinct values?

  27. ganeshie8
    • one year ago
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    that would be too less when you compose a quadratic 1998 times

  28. ganeshie8
    • one year ago
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    also recall the fact that sin is one-to-one in the interval [0, pi/2]

  29. ParthKohli
    • one year ago
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    \[\theta = \pm \pi/2^{1997}, \pm \pi/2^{1996}, \cdots ,\pm \pi/2^1\]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?

  30. ganeshie8
    • one year ago
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    http://gyazo.com/f22b532bddfbe6cd15af74145160b997

  31. ParthKohli
    • one year ago
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    Oh my, never mind...

  32. ParthKohli
    • one year ago
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    Goddammit.

  33. ganeshie8
    • one year ago
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    find the values of \(n\) such that \(\theta \in [0,\pi/2]\)

  34. ParthKohli
    • one year ago
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    Yes, alright. So \(2^{1997}\) distinct values.

  35. ParthKohli
    • one year ago
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    (?)

  36. ParthKohli
    • one year ago
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    Or am I misunderstanding what you're saying?

  37. ganeshie8
    • one year ago
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    the solution i have says \(2^{1996}+1\) so double check..

  38. ParthKohli
    • one year ago
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    Oh my god, what is wrong with me...

  39. ParthKohli
    • one year ago
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    Obviously, yes.

  40. ParthKohli
    • one year ago
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    What are the last three digits of that? I don't know how modular arithmetic works.

  41. ParthKohli
    • one year ago
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    Remind me to get some sleep after this one. Thanks.

  42. ganeshie8
    • one year ago
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    work \[2^{1996}+1 \pmod{1000}\]

  43. ParthKohli
    • one year ago
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    Yes, I have no idea how modular arithmetic works.

  44. ganeshie8
    • one year ago
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    eating dinner.. 10 mnts

  45. ParthKohli
    • one year ago
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    My ... what a beautiful solution you came up with.

  46. ganeshie8
    • one year ago
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    you may use chinese remainder thm or go wid binary exponentiation

  47. ganeshie8
    • one year ago
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    i don't see any other neat way to reduce 2^1996

  48. ParthKohli
    • one year ago
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    W|A returns 337.

  49. Kainui
    • one year ago
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    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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