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 $\frac{1+x}{1-x}=\frac 32 \implies x=\frac 15$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.