Approximating ln(0.7) using MacLaurin formula

[Telemachus]

Hi. Well, I have a problem with this one. It asks me to approximate \ln(0.7) using MacLaurin polynomial of fourth degree. And estimate the error.

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

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

So then

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

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

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

Where -0.3<\alpha<0

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

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.

 

[Mark44]

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.

 

[Telemachus]

Thanks Mark. As in the error I got \alpha 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!

 

[Dickfore]

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

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

Where -0.3<\alpha<0

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

The correct expression for the remainder R_{5}(x) in Lagrange's form is:

<br /> R_{5}(x) = -\frac{x^{5}}{5 (1 + \theta x)^{5}}, \; 0 &lt; \theta &lt; 1, \; |x| &lt; 1<br />

The absolute value for this expression is:

<br /> |R_{5}(x)| = \frac{|x|^{5}}{5(1 + \theta x)^{5}}<br />

The upper bounds for this expression (with respect to \theta) are:

<br /> \sup{|R_{5}(x)|} = \left\{\begin{array}{ll}<br /> \frac{|x|^{5}}{5}} &amp;, 0 &lt; x &lt;1 \\<br /> <br /> \frac{|x|^{5}}{5(1 - |x|)^{5}}&amp;, -1 &lt; x &lt; 0<br /> \end{array}\right.<br />

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

<br /> \sup{|R_{5}(-0.3)}| = \frac{0.3^{5}}{5 \times 0.7^{5}} = 0.00289165229<br />

Since this result is between 0.0005 and 0.005, your approximation has no more than 2 significant decimals with certainty.

 

[Telemachus]

Thank you very much Dickfore.