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Sin(npi/2) non-convergence to 0

  1. Nov 2, 2012 #1
    Hello,

    I'd like to show that sin(npi/2) does not converge. I have managed to show that it does not converge to +-1, but how may I also prove that it does not converge to 0?
     
  2. jcsd
  3. Nov 2, 2012 #2

    Dick

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    sin(0*pi/2)=0, sin(1*pi/2)=1, sin(2*pi/2)=0, sin(3*pi/2)=(-1), sin(4*pi/2)=0. Now just kind of extrapolate.
     
  4. Nov 2, 2012 #3
    I have to demonstrate it using the non-existence of limit defintion. Hence, using epsilon etc.
     
  5. Nov 2, 2012 #4

    Dick

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    How did you show it doesn't converge to +/-1? Why is it harder to show it doesn't converge to 0?
     
  6. Nov 2, 2012 #5

    micromass

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    Sure. But what Dick is suggesting is that you actually calculate [itex]\sin(n\pi /2)[/itex] and to describe the sequence without sines. This makes it way easier to prove that a limit does not exist.
     
  7. Nov 2, 2012 #6
    In order to prove that it does not converge to +-1, I simply defined it as (-1)^n and then showed that for any even n (or odd n, in the case of limit being +1) |(-1)^n - 1| >= epsilon (taking epsilon=1).
    But what would be the general term, the equivalent of (-1)^n, for 0? I mean, I'd still have to prove that the sequence does not converge to 0 in order for my proof to be complete.
     
  8. Nov 3, 2012 #7

    Dick

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    Now why would you 'define' sin(pi*n/2) to be (-1)^n when it isn't true? That would seem like a losing strategy for proving something?
     
    Last edited: Nov 3, 2012
  9. Nov 3, 2012 #8
    It is true and is equal to (-1)^n for all odd positions, i.e. n=2k+1.
     
  10. Nov 3, 2012 #9

    Dick

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    I'd write that as (-1)^k but alright. Your sequence is 0 for even n. It's also true |(-1)^k-0|>=1. Can't you use the same argument? There's no rule against comparing even terms with odd terms.
     
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