How do I use the Taylor series to calculate $\ln(4)$ with an error of $10^{-5}$ with no calculator?

Solution 1:

You could use the fact that $\ln(4)=-2\ln(0.5)$, and approximate $\ln(1-0.5)$ using the Taylor series.

Solution 2:

@Especially Lime gave a good trick $$\log(4)=-2 \log \left(\frac{1}{2}\right)$$ Let us continue writing that $$\frac{1}{2}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}$$ and now, use the fast convergent expansion $$\log \left(\frac{1+x}{1-x}\right)=2\Bigg[\frac {x}1+\frac {x^3}3+\frac {x^5}5+\frac {x^7}7 +\frac {x^9}9 +\cdots\Bigg]$$ So, by hand, the quantity in brackets is $$-\frac {1}3-\frac{1}{81}-\frac{1}{1215}-\frac{1}{15309}-\frac{1}{177147}$$ which is more than enough for an error of $10^{-5}$.

So, we have $$\log(4) \sim -2 \times 2 \Bigg[-\frac {1}3-\frac{1}{81}-\frac{1}{1215}-\frac{1}{15309}-\frac{1}{177147}\Bigg]=\frac{8595212}{6200145}=1.3862921$$ while the exact value is $1.3862944$ corresponding to an error of $2.27\times 10^{-6}$.

Ignoring the last term would not be sufficient.