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anonymous

  • one year ago

A limit (challenge) problem for smart people. In the question n is an integer Condition:- Try to solve it in not more than 3-4 steps.

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  1. ganeshie8
    • one year ago
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    Is \(n\) an integer ?

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    Yes n is an integer. :)

  4. dan815
    • one year ago
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    0

  5. anonymous
    • one year ago
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    no

  6. dan815
    • one year ago
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    infinity

  7. dan815
    • one year ago
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    1

  8. anonymous
    • one year ago
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    no

  9. dan815
    • one year ago
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    pi

  10. dan815
    • one year ago
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    e,

  11. anonymous
    • one year ago
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    u need to show your method , don't just give out answers like a idiot

  12. dan815
    • one year ago
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    lulz :)

  13. dan815
    • one year ago
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    maybe complex numbers i dont like the root in there

  14. anonymous
    • one year ago
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    no real sollution :)

  15. dan815
    • one year ago
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    you can get real solution from that

  16. dan815
    • one year ago
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    is it indeterminate?

  17. dan815
    • one year ago
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    is the answer even converging

  18. anonymous
    • one year ago
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    no there is a answer

  19. dan815
    • one year ago
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    okay then ill work on it : )

  20. anonymous
    • one year ago
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    It dosen't have stupid answers like it does not exist etc

  21. TrojanPoem
    • one year ago
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    No.name is right. It got answer.

  22. anonymous
    • one year ago
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    You got the answer @TrojanPoem

  23. anonymous
    • one year ago
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    ?

  24. TrojanPoem
    • one year ago
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    Nearly , but more than 3 steps.

  25. anonymous
    • one year ago
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    its ok tell your answer

  26. ganeshie8
    • one year ago
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    im getting pi but im still messing wid it

  27. anonymous
    • one year ago
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    @ganeshie8 you are correct !

  28. dan815
    • one year ago
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    psh i already guessed pi lol

  29. anonymous
    • one year ago
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    Ok , before i post my sollution , how did u get it @ganeshie8

  30. ganeshie8
    • one year ago
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    bit embarrassing i kno |dw:1434985625298:dw|

  31. TrojanPoem
    • one year ago
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    haha @ganeshie8 Like ++

  32. anonymous
    • one year ago
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    You made my day @ganeshie8 :)

  33. ganeshie8
    • one year ago
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    I'm glad you find it was entertaining :) please provide a hint as im kinda stuck..

  34. anonymous
    • one year ago
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    no problem , never mind Well the hint is itself the whole sollution , so let me post the sollution itself ok?

  35. dan815
    • one year ago
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    no odont

  36. ganeshie8
    • one year ago
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    ^ give us some time

  37. anonymous
    • one year ago
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    Yeah don't give up so easily , u will get it!

  38. anonymous
    • one year ago
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    After giving hint it is easy , so i will not give any hint :)

  39. anonymous
    • one year ago
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    Ok a small hint

  40. ganeshie8
    • one year ago
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    nope wait almost there

  41. anonymous
    • one year ago
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    ok ok

  42. anonymous
    • one year ago
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    Don't even think of binomial expansions , you will get the answer no doubt , but that's a waste of time and space

  43. ganeshie8
    • one year ago
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    \[\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ \end{align}}\]

  44. ganeshie8
    • one year ago
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    next i want to use sinx/x limit but im gonna need more paper, one sec..

  45. dan815
    • one year ago
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    its so weird i cant believe the 1 is skewing it form 2pi to pi

  46. dan815
    • one year ago
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    for a big number sqrt(1+n^2) is pretty much sqrt(n^2)

  47. dan815
    • one year ago
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    so i said it was just n*sin(2pi*n) and thats inf * 0 so from lohopitals ud get 2pi

  48. dan815
    • one year ago
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    but that +1 in there changed it to pi O_O still working

  49. ganeshie8
    • one year ago
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    I just figured out a really neat way to solve this w/o using lhopital xD

  50. dan815
    • one year ago
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    are you squaring and sqrting the whole limit

  51. dan815
    • one year ago
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    or k^2=1+n^2!

  52. ganeshie8
    • one year ago
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    nvm its not working, was trying to interpret the given expression as a variation of area of regular polygon formula... but no success

  53. ganeshie8
    • one year ago
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    as \(n\to\infty\), the regular polygon becomes an unit circle evaluating the limit to \(\pi\) thought it would be that simple but looks tricky to conclude like that

  54. dan815
    • one year ago
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    ya not sure what im doing wrong, i keep getting 2pi again

  55. dan815
    • one year ago
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    look at my work, can u spot something wrong?

  56. ganeshie8
    • one year ago
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    okie

  57. dan815
    • one year ago
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    |dw:1434988291816:dw|

  58. dan815
    • one year ago
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    first is that step legal

  59. dan815
    • one year ago
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    can i just square and sqrt the lmit like that, now this lets me use lhopitals rule a couple times with no problems

  60. dan815
    • one year ago
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    because k has remained an integer

  61. ganeshie8
    • one year ago
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    that is okay because sqrt(x) is a continuous function \[\lim \sqrt{f(x)} = \sqrt{\lim f(x)}\] But aren't you assuming \(\sqrt{1+n^2}\) is an integer which is incorrect ? let me go read it agian..

  62. dan815
    • one year ago
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    oh true okay throw that out then

  63. anonymous
    • one year ago
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    Does this count? Step 1: Make a conjecture about the limit value (\(\pi\)). Step 2: Prove it.

  64. ganeshie8
    • one year ago
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    i smell epsilon delta but it smells good to me :)

  65. anonymous
    • one year ago
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    I was actually thinking monotone convergence, but showing either condition might be a bit tricky. As for an \(\epsilon\) proof, not sure how much analysis would be required relative to monotone convergence. We have to find some \(N\) for which \(n>N\) yields \[\left|n\sin\left(2\pi\sqrt{1+n^2}\right)-\pi\right|<\epsilon\]

  66. anonymous
    • one year ago
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    (It should go without saying that \(\epsilon>0\))

  67. anonymous
    • one year ago
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    My point of view is: When a function has the limit where n tends to infinity, then the result from sine will eventually tend to undefinition, meaning the function becomes undefined.

  68. ganeshie8
    • one year ago
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    \[\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ &=\lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} +\mathcal{O}(1/n^2)\right) ~~~\color{gray}{(\text{taylor})}\\~\\ &= \lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} \right) + 0 \\~\\ &=\pi \end{align}}\]

  69. xapproachesinfinity
    • one year ago
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    can you do it without taylor?

  70. ganeshie8
    • one year ago
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    yeah just use \(\sin x\approx x\) ;p

  71. ganeshie8
    • one year ago
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    (jk)

  72. ParthKohli
    • one year ago
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    Amazing!

  73. anonymous
    • one year ago
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    Well shall i give my sollution?

  74. ganeshie8
    • one year ago
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    Yes waiting

  75. anonymous
    • one year ago
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  76. ganeshie8
    • one year ago
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    wow thats exact same as my solution

  77. anonymous
    • one year ago
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    I cannot see your sollution , it is in latex form , so i am not able to understand it. nice if u solved it

  78. ganeshie8
    • one year ago
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    what do you mean it is in latex form

  79. anonymous
    • one year ago
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    Like the only the code is visible not the output

  80. ganeshie8
    • one year ago
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    Ahh ok some problem with rendering here it is |dw:1435050449486:dw|

  81. anonymous
    • one year ago
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    nice but what is taylor

  82. anonymous
    • one year ago
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    thnx

  83. ganeshie8
    • one year ago
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    Here taylor series of most common functions |dw:1435050691842:dw|

  84. anonymous
    • one year ago
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    Oh i know these expansions

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