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Infinite limit of cos sum

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data
    ScreenShot038.jpg

    I have no idea how to start. Any hints?

    Thanks in advance.
     
  2. jcsd
  3. Jul 9, 2011 #2

    micromass

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    Hi flyerpower! :smile:

    The trick is to write this sum in a Riemann sum. So, write the sum as something of the form

    [tex]\sum_{k=1}^n{f(x_k)\Delta x}[/tex]

    The limit of such a sum is an integral. So if you can write your sum as a Riemann sum, then you can find integrals to calculate the limit.
     
  4. Jul 9, 2011 #3

    I like Serena

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    Hmm, nice problem! :smile:

    @MM: I tried it your way, but couldn't find such a function. Do you have a suggestion how to get there?

    What I did, was to write the cosine function as a Taylor expansion with only 1 term and an upper estimate for the remainder term. From there I could find the limit....
     
    Last edited: Jul 9, 2011
  5. Jul 9, 2011 #4
    I've already tried finding a function suitable for Riemann but, unfortunately, i couldn't come up with a result. I'll keep trying if you say it can be solved using riemann sums:).

    @serena what value did you come up with by using taylor series?
     
  6. Jul 9, 2011 #5

    micromass

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    I'd try ILSerena's method, it is much easier. :smile:
    It should work with riemann sums, but I think it will get a bit complicated...
     
  7. Jul 9, 2011 #6

    I like Serena

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    1/3.

    (You're not doing some online homework test for extra credit, for which you only need the answer I hope?)
     
  8. Jul 9, 2011 #7
    No :), actually i'm preparing for an important math exam and i'm just practicing. I wanted to know your result and try with taylor series.
     
  9. Jul 9, 2011 #8
    So, i kinda guessed it using my intuition.

    So that sum expands like this : cos PI/(2n+1) + cos PI/(2n+2) + ... cos PI/(2n+n)

    then if we apply the limit n -> infinity :

    PI/(2n+1) -> 0, so does PI/(2n+2) .... PI/(2n+n), so every term converges to cos(0) = 1

    We will have 1+1+1+1+1.....+1 = n

    And then if we plug it into the initial limit we have n/(3n+1) which converges to 1/3.
    Is my intuition correct ?:)


    Also if you may write here the Taylor series method i would really appreciate. I don't know how to write that cosine function with only one term if Taylor series are infinite polynomial series :).
     
  10. Jul 9, 2011 #9

    I like Serena

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    Yes, your intuition is correct! :smile:

    The main thing the Taylor series remainder term provides, is to show that the remainder vanishes.

    For reference, the Taylor series is explained on wiki:
    http://en.wikipedia.org/wiki/Taylor_series

    In particular on this page you can find that the expansion for the cosine is:
    [tex]\cos x = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - ...[/tex]

    Sadly the wiki article does not explain about the remainder term, but it is explained in this wiki article:
    http://en.wikipedia.org/wiki/Taylor's_theorem

    Long story short, we have:
    [tex]1 - \frac {x^2} {2!} \le \cos x \le 1[/tex]

    If you fill that in, you should find your limit (although it's still not trivial :wink:).
     
    Last edited: Jul 9, 2011
  11. Jul 9, 2011 #10
    Thank you for the explanation, i got it.
    Taylor series seem to be very useful in many situations, i'll take a deeper look on them even though i didn't study that at school.
     
  12. Jul 9, 2011 #11

    I like Serena

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    Well, you wouldn't.
    That is, not in high school (assuming that is what you mean).
    It's college/university material, which is not taught in high school (afaik).
    This problem has university in mathematics or physics written over it.
    How did you get by it?
     
  13. Jul 9, 2011 #12
    I just finished high school and now i'm preparing for college admission.
    As for this problem, i found it in the math book that i use for practice.
     
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