Results 1 to 2 of 2

Math Help - taylor series

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    44

    taylor series

    Can anybody help me with this problem please?
    How do I find the taylor series of f(x) = (27 - x)^(1/3) centered at 0? Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jun 2009
    Posts
    32
    Quote Originally Posted by sidi View Post
    Can anybody help me with this problem please?
    How do I find the taylor series of f(x) = (27 - x)^(1/3) centered at 0? Thanks
    F(x) = f(0) + f(0)' * x / 1! +f(0)'' * x^2 / 2! + f(0)''' * x^3 /3! + .... (1)
    now
    f(0) = 3
    f(x)' = -1/3 *( 27 - x ) ^ ( -2/3) ,,, --> f(0)' = - 1 / 27
    f(x)'' = 2/9 *( 27 - x ) ^ ( -5 /3) ,,, --> f(0)'' = 2 / 2187
    f(x)''' = -10/27 *( 27 - x ) ^ ( -8/3) ,,, --> f(0)''' = 10 / 177147
    .....................

    now by (1) we get
    f(x)= 3 - (1/27) x + 2/2187 x^2 / 2! - 10/ 177147 x^3 / 3! + ....
    f(x)= 3 - (1/27) x + 1/2187 x^2 - 5 / 531441 x^3 + ....



    rem:- at taylor series , when x centered at 0 , it`s becam`s Maclaurin series
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor Series using Geometric Series.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 1st 2010, 04:17 PM
  2. 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
  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