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Homework Help: Approximating ln(0.7) using MacLaurin formula

  1. Aug 13, 2010 #1
    Hi. Well, I have a problem with this one. It asks me to approximate [tex]\ln(0.7)[/tex] using MacLaurin polynomial of fourth degree. And estimate the error.

    So I used:
    [tex]f(x)=\ln(x+1)[/tex] [tex]f'(x)=\displaystyle\frac{1}{1+x}[/tex] [tex]f''(x)=\displaystyle\frac{-1}{(1+x)^2}[/tex] [tex]f'''(x)=\displaystyle\frac{2}{(1+x)^3}[/tex] [tex]f^4(x)=\displaystyle\frac{-6}{(1+x)^4}[/tex] [tex]f^5(x)=\displaystyle\frac{24}{(1+x)^5}[/tex]

    And the MacLaurin polynomial I got:
    [tex]P_4(x)=x-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^3}{3}-\displaystyle\frac{x^4}{4}[/tex]

    So then

    [tex]P_4(-0.3)=-0.3-\displaystyle\frac{0.3^2}{2}-\displaystyle\frac{0.3^3}{3}-\displaystyle\frac{0.3^4}{4}=-0.356025[/tex]

    Here is the problem, when calculating the error I got:

    [tex]R_5(-0.3)=\displaystyle\frac{-0.3^5}{5(1+\alpha)^5}[/tex]

    Where [tex]-0.3<\alpha<0[/tex]

    So if I use [tex]\alpha=0[/tex] I should get an upper boundary for the error, but [tex]\displaystyle\frac{-03^5}{5}{\leq{\epsilon}\Rightarrow{-0.000486{\leq{\epsilon}}[/tex]

    But with wolframalpha: http://www.wolframalpha.com/input/?i=ln(0.7)+0.356025
    So the error is wrong, but I can't find where is my mistake. It should give something like 0.0006 where I got 0.0004.

    Bye there.
     
    Last edited: Aug 13, 2010
  2. jcsd
  3. Aug 13, 2010 #2

    Mark44

    Staff: Mentor

    The largest value for f(5)(x) on the interval [-.3, 0] occurs at -.3, not at 0. For x > -1, this function is decreasing. f(5)(-.3) is about 143, and f(5)(0) = 24.
     
  4. Aug 13, 2010 #3
    Thanks Mark. As in the error I got [tex]\alpha[/tex] in the denominator I thought I should use zero to get the bigger value for the error. Now I see you're right, cause it gives less than 1 for the denominator.

    Greetings!
     
  5. Aug 13, 2010 #4
    The correct expression for the remainder [itex]R_{5}(x)[/itex] in Lagrange's form is:

    [tex]
    R_{5}(x) = -\frac{x^{5}}{5 (1 + \theta x)^{5}}, \; 0 < \theta < 1, \; |x| < 1
    [/tex]

    The absolute value for this expression is:

    [tex]
    |R_{5}(x)| = \frac{|x|^{5}}{5(1 + \theta x)^{5}}
    [/tex]

    The upper bounds for this expression (with respect to [itex]\theta[/itex]) are:

    [tex]
    \sup{|R_{5}(x)|} = \left\{\begin{array}{ll}
    \frac{|x|^{5}}{5}} &, 0 < x <1 \\

    \frac{|x|^{5}}{5(1 - |x|)^{5}}&, -1 < x < 0
    \end{array}\right.
    [/tex]

    Since your [itex]x = -0.3[/itex], you need to use the second expression and the result is:

    [tex]
    \sup{|R_{5}(-0.3)}| = \frac{0.3^{5}}{5 \times 0.7^{5}} = 0.00289165229
    [/tex]

    Since this result is between 0.0005 and 0.005, your approximation has no more than 2 significant decimals with certainty.
     
  6. Aug 13, 2010 #5
    Thank you very much Dickfore.
     
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