Results 1 to 4 of 4

Math Help - power series

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    1

    Exclamation power series

    I have a problem that i can't figure out.
    a.) Consider the function f(x) = 1/(1+x^2). Write this functio as power series centered at 0.


    b.) Now, use the first part find a power series for g(x) = arctan(x) also centered at x.

    E is a summation.
    I think i have found the first part but the second part is giving me problems i came out with this answer E(-x^(2n+2))/ (2n+2) but the answer is apparently E(-x^2n)/ (2n+2)

    if someone could explain to me why that would be great!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,725
    Thanks
    1478
    Quote Originally Posted by snorris103 View Post
    I have a problem that i can't figure out.
    a.) Consider the function f(x) = 1/(1+x^2). Write this functio as power series centered at 0.


    b.) Now, use the first part find a power series for g(x) = arctan(x) also centered at x.

    E is a summation.
    I think i have found the first part but the second part is giving me problems i came out with this answer E(-x^(2n+2))/ (2n+2) but the answer is apparently E(-x^2n)/ (2n+2)

    if someone could explain to me why that would be great!!
    I don't see how anyone can tell you why your answer is wrong if you don't tell us how you got that.
    In particular what did you get for (a)?

    Here'es a guess: What is the anti-derivative of x^{2n}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    (a) Use the fact that: \frac{1}{1-y} = \sum_{n=0}^{\infty} y^n \qquad |y| < 1

    Here, imagine: y = x^2

    So: f(x) = \frac{1}{1+x^2} = \frac{1}{1 - \left(- x^2\right)} = \sum_{n=0}^{\infty} \left(-x^2\right)^n = \sum_{n=0}^{\infty}(-1)x^{2n}

    ____________________

    (b) Recall that: \arctan (x) = \int \frac{1}{1+x^2} dx

    So: \arctan (x) = \int \left(\sum_{n=0}^{\infty} (-1)x^{2n}\right) dx = \sum_{n=0}^{\infty} \left( (-1)^n \int x^{2n} dx\right)

    The power series of \arctan x should be: \arctan (x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} \qquad |x| < 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by snorris103 View Post
    I have a problem that i can't figure out.
    a.) Consider the function f(x) = 1/(1+x^2). Write this functio as power series centered at 0.


    b.) Now, use the first part find a power series for g(x) = arctan(x) also centered at x.

    E is a summation.
    I think i have found the first part but the second part is giving me problems i came out with this answer E(-x^(2n+2))/ (2n+2) but the answer is apparently E(-x^2n)/ (2n+2)

    if someone could explain to me why that would be great!!
    Part a is the sum of the geometric series with r=-x^2

    so its series is \sum_{n=0}^\infty (-1)^nx^{2n}

    This series is valid on (-1,1)

    Note that the derivative of g(x)=\tan^{-1}(x)

    g'(x)=\frac{1}{1+x^2}

    Now from part a we know that

    g'(x)=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}

    Now we can integrate the equation(There are some technical points that you may need to justify to exchange a sum and an integral) to get

    g(x)=\tan^{-1}(x)=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}

    Since g(0)=0 we get

    g(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  2. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  3. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 09:01 AM
  4. Multiplying power series - Maclaurin series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 7th 2009, 11:24 PM
  5. Replies: 10
    Last Post: April 18th 2008, 10:35 PM

Search Tags


/mathhelpforum @mathhelpforum