Results 1 to 2 of 2

Thread: Taylor's theorem

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    297

    Taylor's theorem

    Suppose that $\displaystyle f(x)=\ln (1+x)$

    1) Express $\displaystyle f(x)$ in the form $\displaystyle p_1(x)+R_2(x)$, where $\displaystyle p_1$ is the first Taylor polynomial for $\displaystyle f$ about 0 and $\displaystyle R_2$ is the Lagrange formula for the remainder.

    $\displaystyle f(x)=p_1(x)+R_2(x)$

    $\displaystyle p_1(x)=f(0)+f'(0)x$

    $\displaystyle \Longrightarrow p_1(x)=x$

    $\displaystyle R_2(x)=\frac{f''(c)}{2!}x^2$

    $\displaystyle f''(x)=-\frac{1}{(1+x)^2}\Rightarrow f''(c)=-\frac{1}{(1+c)^2}$

    $\displaystyle f(x)=x-\frac{x^2}{2(1+c)^2}$

    2) Suppose that $\displaystyle x\in [-0.1, 0.1]$ and consider the approximation $\displaystyle \ln (1+x) \approx x$. Use the answer in the first part to show that an upper bound for the absolute error in this approximation is $\displaystyle \frac{1}{162}$.

    $\displaystyle |Error|=|f(x)-p_1(x)|$

    $\displaystyle =|R_2(x)|$

    How would I do this question?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,780
    Thanks
    3028
    Quote Originally Posted by acevipa View Post
    Suppose that $\displaystyle f(x)=\ln (1+x)$

    1) Express $\displaystyle f(x)$ in the form $\displaystyle p_1(x)+R_2(x)$, where $\displaystyle p_1$ is the first Taylor polynomial for $\displaystyle f$ about 0 and $\displaystyle R_2$ is the Lagrange formula for the remainder.

    $\displaystyle f(x)=p_1(x)+R_2(x)$

    $\displaystyle p_1(x)=f(0)+f'(0)x$

    $\displaystyle \Longrightarrow p_1(x)=x$

    $\displaystyle R_2(x)=\frac{f''(c)}{2!}x^2$

    $\displaystyle f''(x)=-\frac{1}{(1+x)^2}\Rightarrow f''(c)=-\frac{1}{(1+c)^2}$

    $\displaystyle f(x)=x-\frac{x^2}{2(1+c)^2}$

    2) Suppose that $\displaystyle x\in [-0.1, 0.1]$ and consider the approximation $\displaystyle \ln (1+x) \approx x$. Use the answer in the first part to show that an upper bound for the absolute error in this approximation is $\displaystyle \frac{1}{162}$.

    $\displaystyle |Error|=|f(x)-p_1(x)|$

    $\displaystyle =|R_2(x)|$

    How would I do this question?
    Well, you have already determined that $\displaystyle R_2(x)= -\frac{x^2}{(1+ c)^2}$ for some c between -0.1 and 0.1 so that $\displaystyle |R_2(x)|= \frac{x^2}{(1+ c)^2}$. Now what is the largest possible value for that? Remember that a fraction achieves it maximum value when the numerator is maximum and the denominator is minimum.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor's Theorem
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Nov 10th 2011, 06:45 PM
  2. Taylor theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 2nd 2011, 06:39 AM
  3. Taylor's theorem and e
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 3rd 2009, 02:05 AM
  4. Taylor's Theorem
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Apr 23rd 2009, 07:52 AM
  5. Taylor's Theorem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Nov 23rd 2007, 01:15 PM

Search Tags


/mathhelpforum @mathhelpforum