# Estimating value and error with Taylor polynomial

I am trying to estimate $$ln(\frac{3}{2})$$ to three decimal places. I was trying to use Taylor series for $$ln(x+1)$$ with Lagrange's form of reminder. As

$$(ln(x+1))^{(n)} = \frac{(-1)^{n+1}(n-1)!}{(x+1)^{n}}$$

the n-th reminder has the following form:

$$r_n(x) = \frac{(-1)^{n}}{(c+1)^{n}}x^{n}$$

In our case $$c\in (0,\frac{1}{2})$$, so we substitute c = 0 as we want to limit from the top. Putting $$\frac{1}{2}$$ as x we get

$$|r_n(\frac{1}{2})| < \frac{1}{n2^n}$$ To make sure $$|r_n(1/2)| < 0,001$$ we solve $$\frac{1}{n2^n} < 0,001$$ which gives n = 8, $$r_8(\frac{1}{2}) \approx 0.000488281$$. That would mean that I would have to compute seven elements of Taylor polynomial at x = $$\frac{1}{2}$$, which seems like a rather tedious task. This brings me to my question: are my computations correct?

You are correct exploiting tha alternate series. However, you can make it much faster using $$\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n$$ $$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$ $$\log \left(\frac{1+x}{1-x}\right)=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}$$ Now $$\frac{1+x}{1-x}=\frac 32 \implies x=\frac 15$$