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Homework Help: Maclaurin Series f(x) = (2x)/(1+x2)

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Write the Machlaurin series for :

    f(x) = (2x)/(1+x2)

    2. Relevant equations


    3. The attempt at a solution

    I tried finding all the derivatives (aka f(x), f'(x), f''(x), etc..) but the equations started getting longer and longer and would always result in 0 when x=0. This was on a final exam last year, and I don't think they would have made students do long computations like that

    Also, I tried just finding the derivatives for (1-x)-1, but again, I ended up with a bunch of zeros

    Help?
     
  2. jcsd
  3. Dec 10, 2009 #2
    Hmmm... I'm not sure, but would it be cheating to use that

    [tex] 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r} [/tex]

    for [tex]|r| < 1[/tex]?

    If you do that, you wouldn't need any calculus at all (except for the uniqueness of Maclaurin series).
     
  4. Dec 10, 2009 #3
    I was thinking of doing that in the first place, but on another exam there was a similar question and part c was Find f^2007(0) :(
     
  5. Dec 10, 2009 #4
    In that case, you can work backwards, because the coefficient of xn in the series gives you [tex]\frac{f^{n}(0)}{n!}[/tex]
     
  6. Dec 10, 2009 #5
    so my series would just be: Summation of [(-1)n2x2n+1]?

    I'm not sure how I would work backwards though :|
     
  7. Dec 10, 2009 #6
    Usually, you would compute [tex] \frac{f^{2007}(0)}{2007!} [/tex] to find the coefficient of x2007 in a Maclaurin series. But in this case, you know the coefficient. What does that tell you about [tex] \frac{f^{2007}(0)}{2007!} [/tex] ?
     
  8. Dec 10, 2009 #7
    I'm really not getting it..

    do i isolate for f(2007) by equating it to the summation? (without the summation term in front)
     
  9. Dec 10, 2009 #8
    You're on the right track. What is the coefficient of x2007 in this problem?
     
  10. Dec 10, 2009 #9
    the summation divided by x^-2007?
     
  11. Dec 10, 2009 #10
    * without the negative
     
  12. Dec 10, 2009 #11
    Actually, you found that

    [tex]\frac{2x}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n2x^{2n+1}[/tex]

    When you write out the summation, you get

    [tex]\frac{2x}{1+x^2} = 2 x-2 x^3+2 x^5-2 x^7+2 x^9-2 x^{11} + \cdots[/tex]

    From this, can you find the coefficient of x2007?
     
    Last edited: Dec 10, 2009
  13. Dec 10, 2009 #12
    is it just -2?


    (i'm so sorry I logged off btw, my computer overheat :(
     
  14. Dec 10, 2009 #13
    Yep!

    Now remember that the Maclaurin series of f(x) can be found by:

    [tex]f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 +\cdots[/tex]

    So the coefficient of x2007 must be

    [tex]\frac{f^{2007}(0)}{2007!}[/tex]

    But you know that the coefficient is -2. So what does this tell you about f2007(0)?
     
  15. Dec 10, 2009 #14
    it's equalled to 2?..and so f(2007) = 2*2007!?
     
  16. Dec 10, 2009 #15
    Close! It's actually f(2007)(0) = -2 * 2007!, but I think you got the idea. ;)
     
  17. Dec 10, 2009 #16
    OMG..it actually makes sense!!

    Thank you so much for your help!! :D
     
  18. Dec 11, 2009 #17

    Mark44

    Staff: Mentor

    If all you need to do is get the Maclaurin series for 2x/(1 + x^2), there's something you can do that's much simpler than what I've seen in this thread - just polynomial long division to divide 2x by 1 + x^2. Doing this, I get 2x - 2x^3 + 2x^5 -...
     
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