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Power Series Problem!

  1. Jul 20, 2011 #1
    1. The problem statement, all variables and given/known data
    a) Determine the series of the given function. In the first box after the summation symbol, type in -1 or 1 indicating whether the series is alternating or not.
    b) Write out the sum of the first four nonzero terms of the series representing this function.
    c) Determine the interval of convergence. The outside boxes require the endpoints and the inside boxes require the symbol < or <=.

    For:
    g(x)=arctan(x/sqrt(6))

    2. Relevant equations



    3. The attempt at a solution
    I already got a.) which is sum from n=0 to infinity [ (-1)^n *(x/sqrt(6))^(2n+1) ] / (2n+1)
    I think I got b.) not too sure if this one is right but i got (x/sqrt(6))-(x^3/(3*6^(3/2)))+(x^5/(5*6^(5/2)))-(x^7/(7*6^(7/2))).
    And so I just need someone to check b for me and I dont even know what to do for the interval of convergence.
     
  2. jcsd
  3. Jul 20, 2011 #2

    lanedance

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    ok, difficult to check without the function ;)

    as for determining the radius of convergence, I would check your notes for the method or read the following
    http://en.wikipedia.org/wiki/Radius_of_convergence

    getting the radius is reasonably straight forward but remember to check the boundary points
     
  4. Jul 20, 2011 #3
    I have the function there? The thing is, I wasn't in class and this is work from the spring semester and the book just doesnt seem to help at all
     
  5. Jul 20, 2011 #4

    lanedance

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    ok, i can see it up there now

    so you have
    [tex] f(x) = arctan(\frac{x}{\sqrt{6}}) [/tex]
    [tex] f '(x) = \frac{1}{sqrt(1 - (\frac{x}{\sqrt{6}})^2)}\frac{1}{\sqrt{6}}[/tex]

    [tex] f(0) = arctan(0) = 0[/tex]
    [tex] f '(x) = \frac{1}{\sqrt{6}} [/tex]

    which agrees with the first 2, is this the method you used?
     
  6. Jul 20, 2011 #5
    I don't remember doing an derivative stuff .. I'm not even sure why/when you decided to do all that. I'm pretty lost with series stuff. I tried this problem a few different times but one of the times to get a.) I took the integral of the function. So what exactly does that all mean, what you did?

    oh, wait. no i think i basically did that. cause i said that the function equals the integral of 1/(1+(x/sqrt(6))^2) so thats the same right?

    and i thought the derivative of arctan(x) = 1/(1+x^2) ??
     
    Last edited: Jul 20, 2011
  7. Jul 21, 2011 #6

    lanedance

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    ok, so how did you get your series?
     
  8. Jul 21, 2011 #7
    I said that arctan(x/sqrt(6))= [itex]\int[/itex] [itex]\frac{1}{(1+x^2/6)}[/itex]dx = [itex]\int[/itex] [itex]\frac{1}{(1-(-x^2/6))}[/itex]dx = [itex]\sum[/itex](-1)n*[itex]\int[/itex][itex]\frac{x^2}{6}[/itex]dx = [itex]\sum[/itex]from n=0 to infinity (-1)n*(x(2n+1)) / (6n * (2n+1))
     
  9. Jul 21, 2011 #8

    lanedance

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    do you know you can write a whole equation in tex tags?

    ok so you wrote it as integral, then used a geometric series to expand (i'm guessing your middle step is missing a power of n). I haven't checked all the steps but that's a valid approach and quicker than finding the derivatives.
     
  10. Jul 22, 2011 #9
    okay ... so .. that's what i did. and i know the answer is right. but i still need help on c. b is right. i got the answer for c i just dont know how to do it
     
  11. Jul 23, 2011 #10

    lanedance

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    did you read the wiki post? where are you stuck?
     
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