Results 1 to 2 of 2

Math Help - Taylor series

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    2

    Taylor series

    Find the Taylor series expansion for each function for the given value of a.

    f(x) = sqrt[x], a = 4

    f(x) = e^-x, a = 1

    f(x) = x ln x, a = 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    What's the problem? You just directly apply the formula for the taylor series centred around a: f(x)  =  f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + ... + \frac{f^{(n)}(a)}{n!}(x-a)^{n} + ...
    \Rightarrow f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^{n}

    If you have any problems, post exactly what you're having problems with and we'll be happy to help you out.
    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. 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. Taylor Series / Power Series
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 4th 2009, 01:56 PM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum