Any good approximation methods of $\ln(2)$?

Question

If you do a Taylor polynomial for $\ln(x)$ at 1 you can approximate: $$\ln(2) \approx \sum^n_{k=1} \frac{(-1)^k}{k}$$ The problem is that this converges really slowly, for an error of at most $\frac{1}{n}$ you need to sum $n$ terms.
Are there better approximations?

Answer 1

Similar to @Lutz Lehmann's answer $$\log(2)=-\log \left(\frac{1}{2}\right)=-\log \left(\frac{1-\frac{1}{3}}{1+\frac{1}{3}}\right)=2\sum_{n=0}^\infty\frac {(-1)^{n+1} }{(2n+1)\,3^{2n+1}}$$ which is alternating.

So, writing $$\log(2)=2\sum_{n=0}^p\frac {(-1)^{n+1} }{(2n+1)\,3^{2n+1}}+2\sum_{n=p+1}^\infty\frac {(-1)^{n+1} }{(2n+1)\,3^{2n+1}}$$ looking for $p$ in order to have $k$ exact significant figures, you need to solve for $p$ the equation $$\frac {2}{(2p+3)\,3^{2p+3}}=10^{-k}\implies p=\frac{W\left(2 \log (3) 10^k\right)}{2 \log (3)}-\frac{3}{2}$$ where $W(.)$ is Lambert function.

A very good approximation is just $p \sim k-\log(10)$ (which is not very fast).

Much faster would be to use one of the Machin like formulae for $\log(2)$ $$144 \tanh ^{-1}\left(\frac{1}{251}\right)+54 \tanh ^{-1}\left(\frac{1}{449}\right)-38 \tanh ^{-1}\left(\frac{1}{4801}\right)+62 \tanh ^{-1}\left(\frac{1}{8749}\right)$$ and use the series expansion $$\tanh ^{-1}(x)=\sinh ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=\sum_{n=0}^\infty \frac{(-1)^n (2 n)! }{4^n(2 n+1) (n!)^2} \left(\frac{x}{\sqrt{1-x^2}}\right)^{2 n+1}$$ Computing the partial sums from $0$ to $p$, this gives $$\left( \begin{array}{cc} p & \text{partial sum} \\ 0 & \color{red} { 0.69314}879768524097705624889400775230453401054328895 \\ 1 & \color{red} { 0.6931471805}4888240122015467945517617637052089006247 \\ 2 & \color{red} { 0.693147180559945}41250031627789944252986567946384249 \\ 3 & \color{red} { 0.69314718055994530941}612343717574660943915747172139 \\ 4 & \color{red} { 0.6931471805599453094172321}3439904033679757948070744 \\ 5 & \color{red} { 0.693147180559945309417232121458}01731170065313419101 \\ 6 & \color{red} { 0.6931471805599453094172321214581765}7010956709552748 \\ 7 & \color{red} { 0.693147180559945309417232121458176568075}47342767602 \\ 8 & \color{red} { 0.693147180559945309417232121458176568075500134}71847 \\ 9 & \color{red} { 0.6931471805599453094172321214581765680755001343602}5 \\ 10 & \color{red}{ 0.69314718055994530941723212145817656807550013436026} \end{array} \right)$$

Edit

The advantage of using the alternating series $$\tanh ^{-1}(x)=\sum_{n=0}^\infty \frac{(-1)^n (2 n)! }{4^n(2 n+1) (n!)^2} \left(\frac{x}{\sqrt{1-x^2}}\right)^{2 n+1}$$ is that $$ \left|\frac{b_{n+1}}{b_n}\right|=\frac{4 n^2+4 n+1}{2 (n+1) (2 n+3)}\frac{x}{\sqrt{1-x^2}} \to \frac{x}{\sqrt{1-x^2}}$$ which is very fast with so small values of $x$.

Moreover, we can easily know how many terms are required in order to have $$R_n=\frac{(2 n)! }{4^n(2 n+1) (n!)^2} \left(\frac{x}{\sqrt{1-x^2}}\right)^{2 n+1} \leq 10^{-k}$$ Using Stirling approximation and truncating to $O\left(\frac{1}{n^2}\right)$, we finally have $$n \geq -\frac{3 }{2 \log (a)}W(-A)\quad \text{with} \quad a=\frac{x}{\sqrt{1-x^2}}\quad \text{and} \quad A=\frac{\log(a)} 3\sqrt[3]{\frac{2 }{ \pi }10^{2 k}}$$

For example, using $x=\frac 1 {100}$ and $k=50$, this gives $n=\lceil 23.6945\rceil=24$.

Checking $$R_{23}=2.49\times 10^{-49}\qquad \text{and} \qquad R_{24}=2.34\times 10^{-51}$$

Answer 2

Maybe apply Newton's method to the equation $e^x -2=0$? The numerical scheme could be something like $x_0=1, \quad x_{n+1} = x_n - \dfrac{e^{x_n}-2}{e^{x_n}}=x_n-1+2e^{-x_n}$. The convergence will be fast, but it implies that you can accurately compute the exponential. The error committed when approximating $\ln 2$ by $x_3$ is close to $0.4 \times 10^{-6}$.

Answer 3

Since $\log(2)=-\log\left(\frac12\right)$ and since the Taylor series of the $\log$ function converges fast at $\frac12$, you can use this fact to compute $\log(2)$ quite fast.

Answer 4

Another frequently used expansion is $$ \ln(2)=\ln(\frac43)-\ln(\frac23)=\sum_{k=0}^\infty\frac2{3(2k+1)\cdot9^k} $$ There are other decompositions with arguments closer to $1$ (similar to the Euler-Machin like formulas for $\pi=4\arctan(1)$), but it is an open question if there is one that gives faster than this kind of linear convergence.

Answer 5

Use Simpson's rule at $n=10$ and the definite integral definition of natural logarithm.

Answer 6

This answer adds onto PierreCarre's answer and is too long for a comment.

The cost of multiple evaluations of the exponential function when applying root-finding methods to $e^x-2$ are rapidly decreasing. Note that for Newton's method, one has

$$x_{n+1}=x_n-\Delta x_n,\quad\Delta x_n=1-2\exp(-x_n)$$

so that the next exponential function evaluated is

$$\exp(-x_{n+1})=\exp(-x_n+\Delta x_n)=\exp(-x_n)\exp(\Delta x_n)$$

where $\exp(-x_n)$ is already known and $\exp(\Delta x_n)$ can be computed extremely rapidly. As $x_n$ becomes more accurate, $\Delta x_n$ decreases, so less and less terms are needed to compute $\exp(\Delta x_n)$.

This essentially leaves only the initial evaluation of $\exp(x_0)$, which takes roughly $d/\log_{10}(d)$ iterations to compute to $d$ digits.

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Additional notes:

An improved initial estimate should be taken from the other answers, using only their first few terms.

Since higher order derivatives are all just $e^x$, Householder methods such as Halley's method may be easily used to give faster iterative refinement.