Results 1 to 3 of 3

Math Help - Lagrange Error and Approximating Value Help

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    4

    Lagrange Error and Approximating Value Help

    Hi, I was wondering if anyone could help me understand this concept. I need urgent but specific help. What I mean is that if possible, I would like "baby steps". Step-by-step explanations if you will. I'd really appreciate it. Anyways, here are some of the problems:

    "Find the Taylor polynomial of order four for the function at x=0, and use it to approximate the value of the function at x=0.2."

    1. ) 5 sin ( - x )

    2. ) ( 1 - x )^-2


    As for Lagrange error,

    " 1. ) The approximation Sqrt ( 1 + x ) ≈ 1 + ( x / 2 ) is used when x is small. Estimate the maximum error when |x| < 0.01"


    I hope I am not abusing of your help, though I have a feeling I am. I know it's a lot of questions but I wouldn't feel confident with just one example. Thank you if you read this, or choose to respond.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by BoombaWoomba View Post
    ...Step-by-step explanations if you will. I'd really appreciate it. Anyways, here are some of the problems:

    "Find the Taylor polynomial of order four for the function at x=0, and use it to approximate the value of the function at x=0.2."

    1. ) 5 sin ( - x )

    ...
    You know the general form of the Taylor polynomial:

    f(x) = f(a)+\frac{x-a}{1!}f'(a) + \ldots + \frac{(x-a)^{n-1}}{(n-1)!}f^{(n-1)} + R_n ......... f^{(n-1)} is the (n-1)th derivative of f.

    With your question a = 0.
    Calculate the derivatives:
    Code:
    f(x)     = 5sin(-x) --> f(0)     = 0
    f'(x)    = -5cos(x) --> f'(0)    = -5
    f''(x)   = 5sin(x)  --> f''(0)   = 0
    f'''(x)  = 5cos(x)  --> f'''(0)  = 5
    f''''(x) = -5sin(x) --> f''''(0) = 0
    Now plug in all values:

    f(x) = 0 + (-5)x + 0 + \frac{x^3}{3!} \cdot 5 = -5x + \frac56 x^3

    Calculate f(0.2) = -5 \cdot 0.2 + \frac56 (0.2)^3 = -1 + \frac1{150} = -\frac{149}{150} \approx -0.9933333..

    Check this result by calculating 5 \sin(-0.2) \approx -0.993346...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by BoombaWoomba View Post

    " 1. ) The approximation Sqrt ( 1 + x ) ≈ 1 + ( x / 2 ) is used when x is small. Estimate the maximum error when |x| < 0.01"
    Let f(x) = \sqrt{x+1} then the first older Taylor polynomial is f(0) + f'(0)x = 1 + \frac{x}{2}. Let R_2(1) be the remainder for this at 1. Then, R_2(1) = \frac{f''(y)}{2!}y^2 for some 0<y<1 now f''(y) = -1/4(y+1)^{-3/2}.

    This means, \left| R_2(1)\right| = \frac{1}{8} \cdot (y+1)^{-3/2} y^2 \leq \frac{1}{8} y^2 \leq \frac{1}{8} (.01)^2 = 0.0000125.

    Thus, it means that if we are within .01 of the function we have an accuracy to 4 decimal places.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lagrange error problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 21st 2010, 04:59 PM
  2. approximating error on taylor poly
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 2nd 2009, 10:58 PM
  3. Replies: 1
    Last Post: April 21st 2009, 10:40 PM
  4. Lagrange Error
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 29th 2009, 03:51 PM
  5. Replies: 1
    Last Post: March 19th 2008, 09:52 AM

Search Tags


/mathhelpforum @mathhelpforum