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Series cos(x)^n/3^n boundary question

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum^{infinity}_{n=0}\frac{cos(x)^n}{3^n}[/tex]

    Find the values of x for which the series converges.
    (z,p)

    Find the sum of the series for those values of x.


    2. Relevant equations
    Geometric Series Sum (a/(1-r))
    Geometric Series a*r^n

    3. The attempt at a solution
    I believe I have b correct, so I'm going to lead with b solution:
    [tex]\sum^{infinity}_{n=0}\frac{cos(x)^{n}}{3^{n}}[/tex]

    [tex]\sum^{infinity}_{n=0}(\frac{cos(x)}{3})^{n}[/tex]
    This series is a geometric series a = 1 and r = [tex]\frac{cos(x)}{3}[/tex].
    Since, [tex]\left|r\right|=\left|\frac{cos(x)}{3}\right|<1[/tex], it converges and gives
    [tex]\frac{1}{1-\frac{cos(x)}{3}}[/tex]

    Attempt at a.
    Since the series starts at n = 0, the first term, is always going to be 1. Find the values of x for which the series converges. Thus,
    [tex]\left|\frac{cos(x)}{3}\right|<1[/tex]

    [tex]-1<\frac{cos(x)}{3}<1[/tex]

    [tex]-3<cos(x)<3[/tex]

    [tex]arccos(-3)<x<arccos(3)[/tex]

    But, the issue is the the domain of arccos(x) is -1<x<1... I'm not necessarily sure what I'm doing wrong for this particular problem.

    I've tried -1/3 (since the range of cosine is -1 to 1) and -3, both were incorrect.

    As always any help or further explanation is welcome and will be greatly appreciated!

    Sincerely,

    NastyAccident
     
  2. jcsd
  3. Oct 6, 2009 #2

    lanedance

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    Homework Helper

    hi NastyAccident, may computer is being a bit silly with tex to read, but hope this helps

    first does the series below converge?
    [tex]\sum^{infinity}_{n=0}\frac{1}{3^{n}}[/tex]

    then maybe have a think about absolute convergence & the absolute value of cos
     
  4. Oct 6, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What you did at first is correct and completely answers the question: that is a geometric series in cos(x) and converges as long as -1< cos(x)/3< 1 which is the same as -3< cos(x)< 3. Since the range of cosine is only -1 to 1, that converges for all x and the sum is
    [tex]\frac{1}{1- \frac{cos(x)}{3}}= \frac{3}{3- cos(x)}[/tex]
     
  5. Oct 6, 2009 #4
    Ahh! So, I should of stopped on the step before I took the arccos of each side.

    Now, a new question that I have is why did you just leave it as cos(x)? The normal thought is to create an inequality like -1<x<1 that has x solely in the middle. Since there was a trigonometry function, is it just custom to have an inequality like -1<cos(x)<1?

    From after that step, I can easily understand everything else: such as, cos(x) range is from -1 to 1 and range for this particular series would be [-1/3,1/3]. This fits inside the -3<cos(x)<3 inequality.

    Many thanks for your help Ivy!

    Sincerely,

    NastyAccident
     
    Last edited: Oct 6, 2009
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