Results 1 to 6 of 6

Math Help - Help with lagrange problem

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    228

    Help with lagrange problem

    The third degree Taylor polynomial of f around x=2 is...

    T(x) = 7 - 9(x-2)^2 - 3(x-2)^3


    The fourth derivative satisfies the inequality [abs(f''''(x))] <_ 6 for all x in the closed interval [0,2]. Use the Lagrange error bound on the approximation of f(0) via T(0) to explain why f(0) is negative.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by Kaitosan View Post
    The third degree Taylor polynomial of f around x=2 is...

    T(x) = 7 - 9(x-2)^2 - 3(x-2)^3


    The fourth derivative satisfies the inequality [abs(f''''(x))] <_ 6 for all x in the closed interval [0,2]. Use the Lagrange error bound on the approximation of f(0) via T(0) to explain why f(0) is negative.
    T(0) = 7 - 36 + 24 = -5<br />

    error = |f(0) - T(0)| \leq \left|\frac{6(-2)^4}{4!}\right| = 4

    therefore,

    -4 \leq f(0) - T(0) \leq 4

    -4 \leq f(0) - (-5) \leq 4

    -4 \leq f(0) + 5 \leq 4

    -9 \leq f(0) \leq -1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    228
    **applauds**



    Thank you!

    As a sidenote, I think you were supposed to input x=2 as your "c" but it wouldn't have made a difference anyways. Right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by Kaitosan View Post
    **applauds**



    Thank you!

    As a sidenote, I think you were supposed to input x=2 as your "c" but it wouldn't have made a difference anyways. Right?
    you've got the wrong idea ...

    "c" is where the Taylor polynomial is centered, not the function input ... the function input is x = 0.

    T(x) = 7 - 9(x-2)^2 - 3(x-2)^3 ... c = 2

    T(0) = 7 - 9(0-2)^2 - 3(0-2)^3

    the Lagrange error bound is \frac{M(x-c)^{n+1}}{(n+1)!}

    ... \frac{6(0-2)^4}{4!}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    228
    Yeah, that's right. I mis-worded, sorry.

    I've a question. "C" is supposed to be a number between "a" (the number that the series is centered around) and "x" (the actual output of the function) that maximizes the Lagrange error.... right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by Kaitosan View Post
    Yeah, that's right. I mis-worded, sorry.

    I've a question. "C" is supposed to be a number between "a" (the number that the series is centered around) and "x" (the actual output of the function) that maximizes the Lagrange error.... right?
    notation differs ... check your textbook.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lagrange Multiplier Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 27th 2011, 04:25 AM
  2. Lagrange Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2010, 04:37 AM
  3. Lagrange Proof Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2010, 04:38 PM
  4. Lagrange Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 22nd 2009, 09:21 PM
  5. lagrange problem set
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: May 8th 2008, 07:43 PM

Search Tags


/mathhelpforum @mathhelpforum