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Find the limit of a sequence

  1. Jan 21, 2016 #1
    1. The problem statement, all variables and given/known data
    given a geometric sequence sin(x),sin(2x), . . .
    c) find for which values of x∈(0,π) this sequence converges and calculate its limit

    2. Relevant equations
    |q|<1 or -1<q<1


    3. The attempt at a solution
    Ok so in part a) and b) i calculated the quotient and found out that
    q=sin2x/sinx
    q=(2*cos(x)*sin(x))/(sin(x))
    q=2*cos(x)
    so know i tried to figure out for which values x will converge
    |q|<1
    -1<q<1
    -1<2*cos(x)<1 /:2
    -1/2<cos(x)<1/2
    2π/3>x>π/3
    which means that the sequence will converge when the angle is anywhere between 120° and 60°
    However i'm unsure now how to continue to find the limit because I've never seen an example where the value x isin't exactly defined
    could someone show me how this is done or give me some kind of a hit or check if I made a mistake somewhere inbetween
    Thanks
     
    Last edited: Jan 21, 2016
  2. jcsd
  3. Jan 21, 2016 #2

    PeroK

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    What makes you think that is s geometric sequence?

    Also, you use both the terms series and sequence. Which is it?
     
  4. Jan 21, 2016 #3
    I'm really sorry I made a mistake while translating the text into english.
    Also it is written in the problem that this is a geometric sequence
     
  5. Jan 21, 2016 #4

    HallsofIvy

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    And the sequence is sin(x), 2 cos(x)sin(x), 4cos^2(x)sin(x), 8 cos^3(x)sin(x), ... ?
    Yes, the "geometric sequence a, ar, ar^2, ar^3, ... converges for -1< r< 1 and converges to a/(1- r). Here you are saying that a= sin(x) and r= 2cos(x).
     
  6. Jan 21, 2016 #5

    PeroK

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    Perhaps there is a problem in translation, but clearly ##sin(nx)## is not in general a geometric sequence. Take ##x = \pi/2## for example.
     
  7. Jan 21, 2016 #6
    Perhaps it will be better if I write the whole problem out not just the last step I'll try to do that right now
     
  8. Jan 21, 2016 #7
    yes that's the sequence.
    but woudn't that just give me the sum of all elements of the sequence?
    Edit :
    I asked my profesor if he could give me the solutions and he didn't have much time but I did get to take a picture if this helps in any way 20160118_105655.jpg
    so the solutions to 5 c)
    are π/3 ≤ x < 2*π/3 --------- it is the same that I got but I don't know why there is an equal sign?
    x=π/3 ----------- the limit is sqrt(3)/2
    π/3 < x < 2*π/3 the limit is 0 <------------------- i get this because if -1< r< 1 the sequence converges to 0
    Could someone explain the equal sign in the first inequality
     
  9. Jan 21, 2016 #8

    PeroK

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    So, the question is: For which values of x does the geometric sequence converge:

    ##sinx, sinx(2cosx), sinx(2cosx)^2 \dots##
     
  10. Jan 21, 2016 #9
    yes basically
     
  11. Jan 21, 2016 #10

    PeroK

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    Why do you think ##|q| < 1##?
     
  12. Jan 21, 2016 #11
    Well, I though if the series converges the sequence would converge as well.
    Therefore I went ahead and wrote that |q|<1
     
  13. Jan 21, 2016 #12

    PeroK

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    Many sequences converge when the corresponding series does not.

    What about #q =1##?
     
  14. Jan 21, 2016 #13
    Hmmm, now that you pointed it out if q=1 the limit of the sequence would simply be sin(x)
     
  15. Jan 21, 2016 #14

    Ray Vickson

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    Is the sequence
    sin(x), sin(2x), sin(3x), sin(4x), sin(5x), ...
    or is it
    sin(x), sin(x), sin(4x), sin(8x), sin(16x), ....?
     
  16. Jan 21, 2016 #15
    the sequence is
    sin(x), sin(x)*2cos(x), sin(x)*(2*cos(x))2, sin(x)*(2*cos(x))3, sin(x)*(2*cos(x))4, . . .
     
  17. Jan 21, 2016 #16
    i think I get it now
    the sequence
    an=sin(x)*(2*cos(x))n-1 will converge when either |2*cos(x)|<1--------- when that is true the sequence will converge to 0
    or when 2*cos(x)=1
    2*cos(x)>1 this is not possbile because a limit would not exist here
    2*cos(x)=-1 because the limit doesnt exist here either
    am I correct about this?
     
  18. Jan 21, 2016 #17

    PeroK

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    Yes.
     
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