Results 1 to 3 of 3

Math Help - Talyor Poly. of Degree 2n

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    5

    Talyor Poly. of Degree 2n

    f(x) = cosx, n \in \mathbb{N}

    Find the Taylor Poly. of P_{2n}:= {P_{2n}}^{f,0}
    I know x_0 = 0 and f^{(k)}(x) = cosx and f^{(k)}(0) = cos(0) = 1 and P_{2n} = \displaystyle\sum_{k=0}^{2n} \frac{f^{(k)}(x_0)}{k!} \cdot (x-x_0)^k = f(0) + \sum_{k=1}^{2n} \frac{f^{(k)}(0)}{k!} \cdot x^k = 1+ \sum_{k=1}^{2n} \frac{x^k}{k!}

    So I am confused on what I should do from here to continue (if what I did previously was correct) because I know this is not complete but I do not know where to go from here, give me a boost please. Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by ABigSmile View Post
    f(x) = cosx, n \in \mathbb{N}

    Find the Taylor Poly. of P_{2n}:= {P_{2n}}^{f,0}
    I know x_0 = 0 and f^{(k)}(x) = cosx and f^{(k)}(0) = cos(0) = 1 and P_{2n} = \displaystyle\sum_{k=0}^{2n} \frac{f^{(k)}(x_0)}{k!} \cdot (x-x_0)^k = f(0) + \sum_{k=1}^{2n} \frac{f^{(k)}(0)}{k!} \cdot x^k = 1+ \sum_{k=1}^{2n} \frac{x^k}{k!}

    So I am confused on what I should do from here to continue (if what I did previously was correct) because I know this is not complete but I do not know where to go from here, give me a boost please. Thank you
    f^{k}(0)=1?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    147
    Quote Originally Posted by ABigSmile View Post
    f(x) = cosx, n \in \mathbb{N}

    Find the Taylor Poly. of P_{2n}:= {P_{2n}}^{f,0}
    I know x_0 = 0 and f^{(k)}(x) = cosx and f^{(k)}(0) = cos(0) = 1 and P_{2n} = \displaystyle\sum_{k=0}^{2n} \frac{f^{(k)}(x_0)}{k!} \cdot (x-x_0)^k = f(0) + \sum_{k=1}^{2n} \frac{f^{(k)}(0)}{k!} \cdot x^k = 1+ \sum_{k=1}^{2n} \frac{x^k}{k!}

    So I am confused on what I should do from here to continue (if what I did previously was correct) because I know this is not complete but I do not know where to go from here, give me a boost please. Thank you
    First, your statement is not correct:

    f(x) = cosx --> f(0) = 1
    f'(x) = -sinx --> f'(0) = 0
    f''(x) = -cosx --> f''(0) = -1
    f'''(x) = sinx --> f'''(0) = 0

    and repeat
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 5th 2012, 12:41 AM
  2. Replies: 4
    Last Post: April 7th 2011, 11:08 AM
  3. min poly = char poly
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: May 26th 2010, 09:30 AM
  4. poly
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: May 9th 2010, 07:32 AM
  5. Taylor Poly's and l'Hop
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 6th 2008, 04:01 PM

Search Tags


/mathhelpforum @mathhelpforum