Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - Taylor Polynomial e^x * cos(x)

  1. #1
    Junior Member
    Joined
    Jul 2011
    Posts
    28

    Taylor Polynomial e^x * cos(x)

    Question: Find the Taylor Polynomial P_2 (x) for e^x cos(x) about x_0 = 0.

    Here's what I did:

    P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - \frac{2e^x sin(x) (x)^2 }{2!}
    P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - e^x sin(x) (x)^2
    P_2 (x) = e^x (cos(x) + x cos(x) - x sin(x) - x^2 sin(x))

    The answer the book has is:

    P_2 (x) = 1 + x

    I'm not seeing how it simplifies to this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,444
    Thanks
    1863

    Re: Taylor Polynomial e^x * cos(x)

    You are not distinguishing between x_0= 0 and the variable x. The Taylor's series about x_0 is f(x_0)+ f'(x_0)(x- x_0)+ \frac{f''(x_0)}{2!} (x- x_0)^2 or, with x_0= 0,
    f(0)+ f'(0)x+ \frac{f''(0)}{2}x^2
    Thanks from deezy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602

    Re: Taylor Polynomial e^x * cos(x)

    Quote Originally Posted by deezy View Post
    Question: Find the Taylor Polynomial P_2 (x) for e^x cos(x) about x_0 = 0.

    Here's what I did:

    P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - \frac{2e^x sin(x) (x)^2 }{2!}
    P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - e^x sin(x) (x)^2
    P_2 (x) = e^x (cos(x) + x cos(x) - x sin(x) - x^2 sin(x))

    The answer the book has is:

    P_2 (x) = 1 + x

    I'm not seeing how it simplifies to this?
    Since you only want a Taylor Polynomial up to order 2, the easiest thing to do is to write the first few terms of the MacLaurin series for \displaystyle \begin{align*} e^x \end{align*} and \displaystyle \begin{align*} \cos{x} \end{align*} and multiply them together.

    \displaystyle \begin{align*} e^x &= 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!} + \dots \\ \\ \cos{x} &= 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots + \frac{(-1)^n x^{2n}}{(2n)!} + \dots \\ \\ e^x\cos{x} &= \left( 1 + x + \dots \right) \left( 1 - \frac{x^2}{2} + \dots \right) \\ &= 1 + x - \frac{x^2}{2} + \dots \end{align*}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. taylor polynomial help!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 21st 2011, 02:46 PM
  2. taylor polynomial help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 13th 2009, 04:11 AM
  3. taylor polynomial
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 4th 2009, 07:23 PM
  4. Taylor Polynomial
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 21st 2008, 11:41 AM
  5. Replies: 9
    Last Post: April 3rd 2008, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum