Results 1 to 4 of 4

Math Help - Taylor series, is this correct?

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    151

    Taylor series, is this correct?

    Find Taylor series of f around x=0 where

    f(x)=\int_0^{x^2}\arctan(t)dt.

    Well first of I derived the series expansion of arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}.

    Using that:

    f(x)=\int_0^{x^2}\arctan(t)dt=\int_0^{x^2}\left(\s  um_{n=0}^{\infty}\frac{(-1)^n}{2n+1}t^{2n+1}\right)dt=\sum_{n=0}^{\infty}\f  rac{(-1)^n}{2n+1}\int_0^{x^2}t^{2n+1}dt

    =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{(x^2)^{2n+2}}{2n+2}=\sum_{n=0}^{\  infty}\frac{(-1)^n}{(2n+1)(2n+2)}x^{4n+4}

    Is this correct? I could not make wolfram alpha show me the series expansion of f(x).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    69
    I can't see anything wrong here. If you want to be really certain, here's some options.

    Integrating argtan and plugging in the appropriate limits we get:
     f(x) = x^2 arctan(x^2) - \frac{1}{2} ln(x^4 + 1) .
    You could at this point just pick some representative values of x, plug them into f(x) above, and plug them into the series you obtained (evaluated over, say, the first few thousand terms) to see if you get roughly the same values. Even better, you can plot f(x) and plot your series (again, really a truncated version of your series) to see if the plots line up.

    Alternatively, we know that  arctan(x^2) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1} x^{4n+2} and  ln(x^4+1) = \sum_{n = 1}^{\infty} \frac{(-1)^{n+1}}{n}x^{4n} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{n+1}x^{4n+4}. Plug these into the f(x) above and I think you'll arrive at the same answer you got.
    Last edited by nehme007; December 3rd 2009 at 10:34 AM. Reason: Power series for ln(x^4+1) was messed up, should be correct now
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    186
    Question does this not work for other methods, such as sin x, cos x, ln x? where you have the basic series expansion of x and you can substitute different types of x, such as x^2 assuming the series expansion you use was of a positive power not ^-x or such, ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2009
    Posts
    69
    Yes, and you can replace x with anything you want, the only caveat being that you must update the region of convergence as well. For example:
     arctan(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1}x^{2n+1},|x| \le 1
    To get the expansion for arctan(5x), for example, replace x with 5x everywhere, including the ROC:
     arctan(5x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1}x^{10n+5}, |5x| \le 1 .
    So the new series converges only for  |x| \le 1/5 .

    Quote Originally Posted by RockHard View Post
    Question does this not work for other methods, such as sin x, cos x, ln x? where you have the basic series expansion of x and you can substitute different types of x, such as x^2 assuming the series expansion you use was of a positive power not ^-x or such, ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Getting stuck on series - can't develop Taylor series.
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 5th 2010, 08:32 AM
  2. Series sum question, is this correct?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: January 29th 2010, 02:33 AM
  3. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  4. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 09:01 AM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum