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mathaddict4471

  • one year ago

challenge calculus question without a calculator use the Maclaurin series to compute e^1 medal and fan rewarded

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  1. mathaddict4471
    • one year ago
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    \[\lim_{n \rightarrow \infty} n \sin (2pien!)\]

  2. mathaddict4471
    • one year ago
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    sry compute equation above using e^1

  3. JamesJ
    • one year ago
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    Is the expression n.sin(2.pi.e.n!) ?

  4. JamesJ
    • one year ago
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    ...which can't be right. What exactly are we taking the limit of?

  5. mathaddict4471
    • one year ago
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    the first thing you said

  6. mathaddict4471
    • one year ago
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    we're taking the limit of n sin i think

  7. mathaddict4471
    • one year ago
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    i'll use wolfram

  8. mathaddict4471
    • one year ago
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    http://www.math.utah.edu/ugrad/calc_challenge/calc-challenge-2006.pdf it's number 6

  9. JamesJ
    • one year ago
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    I see. Ok, let me think.

  10. Hoa
    • one year ago
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    @mathaddict: the equation must be e^x , right? to apply Maclaurin series, how can you apply to e^1

  11. JamesJ
    • one year ago
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    The problem actually says: Using the Maclaurin series for \( e^1 \), calculate \[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

  12. Hoa
    • one year ago
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    got it

  13. JamesJ
    • one year ago
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    Ok. This is what I've got so far. When n = 2, n!.e = 2!(1 + 1 + 1/2 + 1/3! + ...) = 2 + 2 + 1 + 2/3! + ... Now sin(2.pi.e.n!) = sin(2.pi ( 2 + 2 + 1 + 2/3! + ... ) ) = sin(2.pi ( 5 + stuff) ) Recall that sin(a + b) = sin a . cos b + cos a . sin b hence sin(2.pi.e.n!) = sin(2.pi.5).cos(stuff) + cos(2.pi.5).sin(stuff) = 0 .cos(stuff) + 1 .sin(stuff) = sin(stuff) This argument generalizes to \[ \sin(2\pi e n!) = \sin \left[ 2\pi \left( \sum_{i=0}^n n!/i! + \sum_{i=n+1}^{\infty} n!/i! \right) \right] = \sin \left[ 2\pi\sum_{i=n+1}^{\infty} n!/i! \right] \] Thus the limit is equal to this limit \[ \lim_{n \rightarrow \infty} n . \sin\left[ 2\pi\sum_{i=n+1}^{\infty} n!/i! \right] \] Now we want to figure out how to bound the quantity in square brackets and show it goes to zero sufficiently fast to cancel out the n outside altogether or end up with some finite number

  14. Hoa
    • one year ago
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    @JamesJ : are you Math Professor?

  15. JamesJ
    • one year ago
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    In fact that sum is 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + .... < 1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ... and this infinite geometric series is equal to 1/n Hence the sum is bounded above by the limit as n --> infinity of n.sin(2pi/n) = sin(2pi/n)/(1/n) --> 2pi So I imagine the limit is in fact 2pi. Now a little more work is required to show that it is exactly 2pi

  16. JamesJ
    • one year ago
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    Oh, and that's easy because 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + .... > 1/(n+1) and hence the limit is bounded below by the limit of n.sin(2pi/(n+1) which has the same limit. **** Nice question! Yes, I was once.

  17. Hoa
    • one year ago
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    I love your neatness. It means you used to be a Math Professor, right? (sorry for Capitalize the P in professor, because to me, those people deserve that C)

  18. JamesJ
    • one year ago
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    This is very counter-intuitive result. Adding this to the the archives.

  19. Hoa
    • one year ago
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    Can I make question when i have some via message? I don't want to post and wait wait wait..... until someone have free time .

  20. JamesJ
    • one year ago
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    Better to post the question and then mail me with the link.

  21. Hoa
    • one year ago
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    how to post the link. once , other asked me to do that but I didn't know how to send the link

  22. JamesJ
    • one year ago
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    Just copy it from your browser and paste it into a message

  23. Hoa
    • one year ago
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    ok, let me try once

  24. Hoa
    • one year ago
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    do I have to be your fan to send the message? I try reply yours and send message to your profile. both are fail. I cannot submit the mail

  25. JamesJ
    • one year ago
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    Hm, possibly. I'm not sure.

  26. mathaddict4471
    • one year ago
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    thank you soo much for your greatly intuitive answer James, you are the master. best of luck to both of you JamesJ and Hoa

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