Results 1 to 7 of 7

Math Help - Taylor Polynomial (Approximation Error)

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    102

    Taylor Polynomial (Approximation Error)

    Hey I'm having trouble with the 2nd part of this question:

    Find the fifth-order Taylor polynomial, then find an interval centered at x = x(o) in which the approximation error  |f(x) - P5(x)| is less than 0.01.

     f(x) = \sqrt {x} : x(o) = 1

    So I found the fifth-order taylor polynomial which is:

     P5(x) = 1 + \frac {1}{2}(x-1) - \frac {1}{8}(x-1)^2 + \frac {1}{16}(x-1)^3 - \frac {5}{128}(x-1)^4 + \frac {7}{256}(x-1)^5

    Now how do I find the approximation error? Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Taylor's inequality

    If |f^{n+1}(x)| \le M for |x-a| \le  d
    Then the remainder R_n(x) of the Taylor series satisfies the inequality

    |R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}


    so \frac{d^6f}{dx^6}=-\frac{945}{64x^{11/2}}

    The 6th derivative has it max at x=1 for all values greater than 1.(it is unbounded as we approch zero from the right) |-\frac{945}{64}|=M

    we want the error to be less than 0.01

    we need to solve

     \frac{945}{64 \cdot 6!}|x-1|^6 \le 0.01

    |x-1|^{6} \le 0.0001190

    |x-1| \le .2218
    Last edited by TheEmptySet; March 18th 2008 at 08:09 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by TheEmptySet View Post
    If |f^{n+1}(x)| \le M for |x-a| \le  d
    Then the remainder R_n(x) of the Taylor series satisfies the inequality

    |R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}


    so \frac{d^6f}{dx^6}=-\frac{945}{64x^{11/2}}

    The 6th derivative has it max at x=1 for all values greater than 1.(it is unbounded as we approch zero from the right) |-\frac{945}{64}|=M

    we want the error to be less than 0.01

    we need to solve

     \frac{945}{6!}|x-1|^6 \le 0.01

    |x-1|^{6} \le 0.007619

    |x-1| \le .4436
    Didn't you forget your 64 in the denominator? Just curious, but other than that I get it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by larson View Post
    Didn't you forget your 64 in the denominator? Just curious, but other than that I get it.
    Yes I did. I have been off today.

    <br />
\frac{945}{64 \cdot 6!}|x-1|^6 \le 0.01<br />


    Thanks

    I will fix the above post
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by TheEmptySet View Post
     \frac{945}{64 \cdot 6!}|x-1|^6 \le 0.01

    |x-1|^{6} \le 0.0001190

    |x-1| \le .2218
    I'm having trouble now with this...

    When you do  \frac{945}{64 \cdot6!} you get...

     \frac{945}{46080} which then equals .0205078125

    Then you divide both sides by that so wouldn't you get...  |x-1|^6 \le \frac{0.01}{.0205078125} which means...  |x-1|^6 \le .487619 right?

    Sorry if I I'm just not understanding it correctly...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    \frac{945}{64 \cdot 6!}|x-1|^6 \le .01

    |x-1|^6 \le \frac{.01 \cdot 64 \cdot 6!}{945}=\frac{460.8}{945}

    |x-1|^6 \le 0.4876  \iff |x-1| \le .8872

    You are correct I just can't read or type today.

    Sorry
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by TheEmptySet View Post
    \frac{945}{64 \cdot 6!}|x-1|^6 \le .01

    |x-1|^6 \le \frac{.01 \cdot 64 \cdot 6!}{945}=\frac{460.8}{945}

    |x-1|^6 \le 0.4876  \iff |x-1| \le .8872

    You are correct I just can't read or type today.

    Sorry
    Thanks for your help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor series approximation and error
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 13th 2011, 10:13 AM
  2. Error in taylor polynomial
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 18th 2010, 11:27 PM
  3. taylor polynomial error
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 17th 2009, 12:00 AM
  4. [SOLVED] Taylor Polynomial approximation
    Posted in the Calculus Forum
    Replies: 8
    Last Post: June 13th 2009, 09:39 PM
  5. Taylor polynomial approximation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 2nd 2008, 01:04 AM

Search Tags


/mathhelpforum @mathhelpforum