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Convergence prove question

  1. Mar 11, 2009 #1
    [tex]a_1=\sin x [/tex]
    [tex]-\infty<x<\infty[/tex]
    [tex]a_{n+1}=\sin a_n [/tex]

    prove that [tex]a_n[/tex] convergent and find

    [tex]\lim _{n->\infty}a_n=?[/tex]
    the solution that i saw is that because
    [tex]a_1=\sin x [/tex] then its bounded from 1 to -1

    so

    |[tex]a_{n+1}[/tex]|<|[tex]\sin a_n[/tex]|=<|[tex]a_n[/tex]|

    so its non increasing and it goes to 0.

    but the teacher says that its not a proof
    why its not a proof
    how to solve it correctly??
     
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

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    Hi transgalactic! :smile:
    Because you haven't proved it's non increasing …

    and even when you do, you'll need to prove it doesn't decrease to a limit > 0 :wink:
     
  4. Mar 11, 2009 #3
    yes i did
    i showed the inequality and i written about the property of sinus

    what else do i need to write??
     
  5. Mar 11, 2009 #4

    tiny-tim

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    no … you only wrote it :rolleyes:

    how would you prove it? :smile:
     
  6. Mar 11, 2009 #5
    i dont need to prove the inequality its obvious

    no matter what number i will put into x its sinx will be from 1 to -1
    and there is a theorem for which sin x and x have approximately the same value near 0

    and i put absolute value because we cant look at it from - infinity too

    how to prove this inequality
    ??
     
  7. Mar 11, 2009 #6

    tiny-tim

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    he he he :rofl:

    it isn't obvious!
    irrelevant!
    how does that help?

    you need to prove that sinx is smaller than x
     
  8. Mar 11, 2009 #7
    ok
    i prove that sinx <x
    f(x)=sinx-x
    f'(x)=cosx-1
    so for all x that differs 0 f'(x)<0 so f(x) is decreasing
    f(0)=0
    so on x=0 its 0 and decreasing
    so f(x) <0 everywhere except x=0
    so sin<x
    what next
    ??
     
  9. Mar 11, 2009 #8

    tiny-tim

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    messy, but correct :approve:

    ok, now it easily follows that an is decreasing …

    but you still need to prove that it decreases to 0, and not to some number > 0 :smile:
     
  10. Mar 11, 2009 #9
    [tex]
    a_{n+1}=\sin a_n
    [/tex]
    a1=sin x where -infinity<x<+infinity

    i dont know how to prove that its decreasing
    i cant use the function method that i used before with derivatives and stuff

    i can only say an observation
    ??
     
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