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Find a power series for the function

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a power series representation for the function and determine the radius of convergence.

    f(x) = arctan(x/3)


    2. Relevant equations
    not really any equations ... just intervals


    3. The attempt at a solution

    here's what i did:

    f'(x) = [itex]\frac{1}{1+(x/3)^2}[/itex] = [itex]\frac{1}{1+(x^2/9)}[/itex]

    arctan(x/3) = [itex]\int\frac{1}{1+(x^2/9)}dx[/itex] = [itex]\int\frac{1}{1-(-x^2/9)}dx[/itex] = ∫ (Ʃ (-1)n* [itex]\frac{x^(2n)}{9}[/itex])dx = ∫ (1/9 - x2/9 + x4/9 - x6/9 + ...)dx

    = C + x/9 - x3/27 + x5/45 - x763 + ...

    To find C:

    make x = 0 so that C = arctan(0) = 0.
    so, arctan(x/3) = x/9 - x3/27 + x5/45 - x763 + ...

    = Ʃ (-1)n*(x^(2n+1) / (18n+9))
    from n=0 to infinity


    is this all right ??
     
  2. jcsd
  3. Feb 23, 2012 #2

    Dick

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    In your power series it isn't just (x^2)^n. It's (x^2/9)^n.
     
  4. Feb 23, 2012 #3
    I'm not sure if I know where/what you mean. Can you be a little more specific?
     
  5. Feb 23, 2012 #4

    Dick

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    1/(1+(x^2/9)) is equal to the sum of (-x^2/9)^n. Not the sum of (-x^2)^n/9.
     
  6. Feb 23, 2012 #5
    Oh ok. So you mean the 9 has an exponent of n too?

    So you get arctan(x/3) = x - x^3/27 + x^5/405 - x^7/5103 + ...

    so it would be summation (-1)^n * x^(2n+1) / ?
    I don't know about the bottom part I can't figure it out
     
  7. Feb 23, 2012 #6

    Dick

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    Just keep track of how you are getting the numbers on the bottom. There is another error as well. arctan(x/3) isn't equal to [itex]\int \frac{1}{1+(\frac{x}{3})^2} dx[/itex]. To do the integral you do a substitution of u=x/3. There's another factor coming from the du.
     
  8. Feb 23, 2012 #7
    oh =[

    ok so when you have arctan(x/3) = ∫ [itex]\frac{1}{1+(x/3)^2}[/itex]dx and then you do the u=x/3 and then 3*du=dx so then you have 3*∫ [itex]\frac{1}{1+u^2}[/itex]dx
    what do i do next?
     
  9. Feb 23, 2012 #8

    Dick

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    You don't have arctan(x/3)=[itex]\int \frac{1}{1+(\frac{x}{3})^2} dx[/itex]. That's what I'm telling you. You need to fix that before you proceed. Use the u substitution to figure out what the integral actually is. Then use that to write arctan(x/3) correctly as an integral.
     
    Last edited: Feb 23, 2012
  10. Feb 23, 2012 #9
    is the correct integral just the same thing just with just a 3 in front of the integral?
     
  11. Feb 23, 2012 #10

    Dick

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    If [itex]\int \frac{1}{1+(\frac{x}{3})^2} dx[/itex]=3*arctan(x/3), what's arctan(x/3)?
     
  12. Feb 23, 2012 #11
    there'd be a 1/3 times the integral
     
  13. Feb 23, 2012 #12

    Dick

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    Yes, the 1/3 is the factor you are missing in your power series.
     
  14. Feb 23, 2012 #13
    so its this summation (-1)^n * x^(2n+1) / ?
    just with a 3 on the bottom? but there has to more to the bottom, i cant figure that out, there doesnt seem to be a pattern that i can catch for the bottom
     
  15. Feb 23, 2012 #14

    Dick

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    There are going to be three things on the bottom, the 3, a power of 3^2 and something that comes from integrating x^(2n). This is really not that hard.
     
  16. Feb 23, 2012 #15
    So there should be the 3, 2n+1, and I don't get the other part
     
  17. Feb 24, 2012 #16

    Dick

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    I was thinking of the 9^n. What do you get? Compare with the series you gave in post 5 after you add the extra 1/3.
     
    Last edited: Feb 24, 2012
  18. Feb 24, 2012 #17
    Wait, are the numbers on the bottom still right? Or does the 1/3 affect that?
     
  19. Feb 24, 2012 #18

    Dick

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    No, they are not right yet. You have to multiply by 1/3.
     
  20. Feb 26, 2012 #19
    ok so just

    arctan(x/3) = x/3 - x^3/81 + x^5/1215 - x^7/15309 + ...

    so like we said, there is a 3, 2n+1, and you said something about something that is related to the 9^n is on the bottom too?
     
  21. Feb 26, 2012 #20

    Dick

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    I really think you can work this out by yourself if you starting thinking about it.
     
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