Estimate ln(3) using Taylor Expansion up to 3rd order (without the use of a calculator).

$$f(x)=ln(x)$$ $$f'(x)=1/x$$ $$f''(x)=-1/x^2$$ $$f'''(x)=2/x^3$$


If I let a=2, I can solve for the first to third order, but I will then have to estimate f(2)=ln(2). Using the same method as above to estimate ln(2), i.e. a=1, I will have a rather inaccurate estimate of ln(2).

Is there any other ways to estimate ln(3) more accurately?


A chance is given by solving $e^x=3$ with the Newton's method with starting point $x_0=1$.

We have: $$ x_{k+1} = x_k-\frac{\exp(x_k)-3}{\exp(x_k)}=x_k-1+3\exp(-x_k)$$ and the difference between $\log 3$ and: $$ x_2 = -1+ 3 e^{-1} + 3 e^{-3/e} $$ is yet less than $2\cdot 10^{-5}$. Another chance is given by considering that:

$$ \log\frac{1+x}{1-x}=2\operatorname{arctanh} x = \sum_{n\geq 0}\frac{2\,x^{2n+1}}{2n+1} $$ for any $x$ such that $|x|<1$, and $\log 3$ is the LHS of the previous identity in $x=\frac{1}{2}$, hence:

$$ \log 3 = \sum_{n\geq 0}\frac{1}{(2n+1)4^n} \tag{1}$$

that converges pretty fast. Also notice that $(1)$ is just: $$ \log 3 = \int_{0}^{2}\frac{dx}{1+x}=2\int_{0}^{1}\frac{dx}{1+2x}=\int_{0}^{1}\frac{dx}{1-\frac{x^2}{4}}$$ in disguise. Since: $$ 3 = \frac{6}{4}\cdot\frac{4}{2}=\frac{1+1/5}{1-1/5}\cdot\frac{1+1/3}{1-1/3}$$ we also have:

$$ \log 3 = \frac{2}{15}\sum_{n\geq 0}\frac{1}{2n+1}\left(\frac{5}{9^n}+\frac{3}{25^n}\right)\tag{2}$$

that converges even faster than $(1)$.

  • 1
    $\begingroup$ This is a nice one ! Cheers $\endgroup$ – Claude Leibovici Mar 7 '15 at 12:39

Here I'm going to pick $a=1$ so that $\log(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^{n}$. Applying the ratio test shows that this series converges for $0<x<2$. If you do not know about convergence of series, it means that the above series I wrote only works for values of $x$ between $0$ and $2$ (inclusive); generically, any $x$ values outside a series' radius of converges produces $\pm \infty$ i.e. divergence.

Here is the trick: let $x = 3^{-1}$. Clearly $0 < \frac{1}{3}<2$ so the above series converges for this particular $x$ value, then apply rules of logs: $\log 3^{-1} = - \log{3}$. In other words, $\log(3) = - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{3}-1\right)^{n}$.

Of course, you can truncate this series at the third order if that is the approximation you are looking for.


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.