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Why does sin converge not the cos?

  1. Oct 11, 2006 #1
    thanks...................
     
  2. jcsd
  3. Oct 11, 2006 #2
    alternating series
    for example
    Sigma (-1)^n sin(Pi/n) compare with Signma(-1)^n cos (Pi/n)
     
  4. Oct 11, 2006 #3

    StatusX

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    What do the terms look like as n->infinity?
     
  5. Oct 11, 2006 #4
    With calculator, I can see the difference.

    But how do you analyze this without using calculator?
     
  6. Oct 11, 2006 #5

    StatusX

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    As n gets large, pi/n goes to zero.
     
  7. Oct 11, 2006 #6
    For sin -0+1-sqrt(3)/2+sqrt(2)/2 ------>althernating series ---->conver.
    con -1+0-1/2+sqrt(2)/2-sqrt(3)/2 ---------increasing---diver.
     
  8. Oct 11, 2006 #7
    oh..i see now.

    So the lim for Sin(pi/n) = 0-->can't be determined...if con or div.
    lim for Cos (pi/n) = 1 by div. test, it's div.
     
    Last edited: Oct 11, 2006
  9. Oct 11, 2006 #8

    matt grime

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    Uh? Sorry, what the heck is this all about? I've not understood a single word of what's been typed.
     
  10. Oct 11, 2006 #9
    To check Sigma (-1)^n sin(Pi/n) and Signma(-1)^n cos (Pi/n) are divergent or convergent.
     
  11. Oct 11, 2006 #10
    Analyzing it without calculator.
     
  12. Oct 11, 2006 #11

    StatusX

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    Actually you can say whether the sin series converges because it is an alternating series.
     
  13. Oct 11, 2006 #12
    Thanks.........................
     
  14. Oct 11, 2006 #13

    matt grime

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    And that bears what relation to your first post? We are not psychics. Post full questions.

    For the question, as has been noted, one has terms that do not tend to zero, so the sum can't converge, and one passes the alternating series test.

    Note that it is impossible to work out the answer *with* a calculator.
     
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