In view of what we can compute by hand, on a piece of paper, without having to use a computer or a calculator, how far can we go with the evaluation of $\log$-functions and fractional powers?
More concretely, are there practical methods, that work in general well enough, for computing the followings?
- $\log(x)$ (natural log used here) for $x\in ]0,\infty[$ (e.g take $x=3,$ which has a natural log of $\approx 1.09$)
- $x^{\alpha},$ ($x$ being any real number) for $\alpha$ not being a natural number. So e.g. $\alpha=1/2,1/3,\cdots$
It would be neat to learn about possible ways of maybe first simplifying these calculations, translating them into more feasible calculations (by hand), and then take it from there. Or if direct methods exist that work for certain values or powers. This surely begs the question, how people used to do these calculations before computers were around.
Best Answer
The well known series
$$\log(1+x) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k}$$
converges for $-1< x\leq 1$, so it cannot be used to calculate $\log(3)$ with directly. However, it is possible to extend the radius of convergence using a conformal mapping. If we put $y = \frac{1+x}{1-x}$, then we have:
$$\log(y) = \log(1+x) - \log(1-x) = 2\sum_{k=0}^{\infty}\frac{x^{2k+1}}{2k+1}$$
Since $x = \frac{y-1}{y+1}$, this means that
$$\log(y) = 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{y-1}{y+1}\right)^{2k+1}$$
and this converges for all positive y. So, you can directly insert $y = 3$ in this series and compute $\log(3)$ quite accurately using only a few terms:
$$\log(3) =1 + \frac{1}{3\times 4} + \frac{1}{5\times 16} + \frac{1}{7\times 64} + \frac{1}{9\times 256} + \frac{1}{11\times 1024} +\cdots$$
So, with 6 terms we get 5 significant figures. But for larger $y$ the series will converge more slowly, it's then more convenient to use the above series to construct a series for $\log(y+1) -\log(y) = \log\left(1+\frac{1}{y}\right)$:
$$\log(1+y) = \log(y)+ 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{1+2y}\right)^{2k+1}$$
The series now converges faster when computing $\log(n)$ for $n$ larger than 2, but you then need to know $\log(n-1)$. However, it is then possible to compute several logarithms simultaneously in terms of only fast converging series. E.g. to compute $\log(2)$, $\log(3)$ and $\log(5)$ simultaneously, we can use $2^4 = 16 = 15+1 = 3\times 5 +1$, $3^4 = 81 = 80+1 = 2^4\times 5 +1$, and $5^2 = 24+1 = 3\times 2^3+1$, this yields:
$$ \begin{split} &4\log(2) - \log(3) - \log(5) &= 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{31}\right)^{2k+1}\\ &4\log(3) - 4\log(2) - \log(5) &= 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{161}\right)^{2k+1}\\ &2\log(5) - 3\log(2) - \log(3) & = 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{49}\right)^{2k+1} \end{split} $$
So, you can then solve for the 3 logarithms using these 3 equations involving fast converging series.