# Approximating ln(0.7) using MacLaurin formula

1. Aug 13, 2010

### 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.

Last edited: Aug 13, 2010
2. Aug 13, 2010

### 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.

3. Aug 13, 2010

### 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!

4. Aug 13, 2010

### Dickfore

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

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

The absolute value for this expression is:

$$|R_{5}(x)| = \frac{|x|^{5}}{5(1 + \theta x)^{5}}$$

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

$$\sup{|R_{5}(x)|} = \left\{\begin{array}{ll} \frac{|x|^{5}}{5}} &, 0 < x <1 \\ \frac{|x|^{5}}{5(1 - |x|)^{5}}&, -1 < x < 0 \end{array}\right.$$

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

$$\sup{|R_{5}(-0.3)}| = \frac{0.3^{5}}{5 \times 0.7^{5}} = 0.00289165229$$

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

5. Aug 13, 2010

### Telemachus

Thank you very much Dickfore.