Some of the easiest to find series for computing \(\pi\) can be found from inverse trigonometric functions. We can evaluate inverse trigonometric functions at special angles and obtain series, some of which converge exponentially. Exponentially converging series provide a way to compute \(\pi\) efficiently with binary splitting and require \(O(n)\) terms to compute \(n\) digits.
We can find it by starting from the derivative. If \(y=\arcsin(x)\), then \(\sin(y)=x\) and differentiating implicitly: \[\cos(y){dy\over dx}=1 \Rightarrow {dy\over dx}={1\over\cos(y)} ={1\over\sqrt{1-\sin^2(y)}}={1\over\sqrt{1-x^2}}\] Then we can find the integral (which is defined for \(x\in[-{\pi\over2},{\pi\over2}]\)). \[\arcsin(x)=\int_0^x{dt\over\sqrt{1-t^2}}=\int_0^x(1-t^2)^{-1/2}dt\] To turn this into a summation, use the generalized binomial theorem. \[=\int_0^x\left[\sum_{k=0}^\infty{-1/2\choose k}(-t^2)^k\right]dt =\int_0^x\left[\sum_{k=0}^\infty{-1/2\choose k}(-1)^k t^{2k}\right]dt\] The generalized binomial coefficient can be simplified. \[{-1/2\choose k}={{-1\over2}\cdot{-3\over2}\cdot{-5\over2} \ldots{-(2k-1)\over2}\over k!}={(-1)^k(2k-1)!!\over 2^k\cdot k!}\] Then we can multiply \((2k)!=2^k\cdot k!\) to find another way to write it. \[={(-1)^k(2k)!!(2k-1)!!\over4^k\cdot(k!)^2}={(-1)^k(2k)!\over4^k\cdot(k!)^2} ={2k\choose k}{(-1)^k\over4^k}\] Then substitute this result into the integral and we continue. \[=\int_0^x\left[\sum_{k=0}^\infty{2k\choose k}{t^{2k}\over4^k}\right]dt =\sum_{k=0}^\infty\left[{2k\choose k}{1\over4^k}\int_0^x{t^{2k}}dt\right] =\sum_{k=0}^\infty{2k\choose k}{x^{2k+1}\over4^k(2k+1)}\] This power series converges when \(|x|\leq1\). To see the convergence rate, use Stirling's approximation on the part with factorials. \[{(2k)!\over(k!)^2}{x^{2k+1}\over4^k(2k+1)}\sim {\sqrt{4\pi k}\left(2k\over e\right)^{2k}\over2\pi k\left(k\over e\right)^{2k}} ={1\over\sqrt{\pi k}}\left({2k\over e}{e\over k}\right)^{2k} ={4^k\over\sqrt{\pi k}}\] So from this, we can see that the \(k\)th term is asymptotically \[{x^{2k+1}\over\sqrt{\pi k}(2k+1)}\propto{x^{2k+1}\over k^{3/2}}\] Now we can use the special angles \(x={1},{\sqrt{3}\over2},{\sqrt{2}\over2} ,{1\over2}\) to obtain some \(\pi\) formulas.
| Angle | Formula | Convergence |
|---|---|---|
| \[\pi\over2\] | \[\pi=2\sum_{k=0}^\infty{2k\choose k}{1\over4^k(2k+1)}\] | Very slow, error is \(\propto k^{-1/2}\) |
| \[\pi\over3\] | \[\pi={3\sqrt{3}\over2}\sum_{k=0}^\infty{2k\choose k} \left({3\over16}\right)^k{1\over2k+1}\] | ~0.125 decimal digits per term |
| \[\pi\over4\] | \[\pi=2\sqrt{2}\sum_{k=0}^\infty{2k\choose k} {1\over8^k(2k+1)}\] | ~0.301 decimal digits per term |
| \[\pi\over6\] | \[\pi=3\sum_{k=0}^\infty{2k\choose k}{1\over16^k(2k+1)}\] | ~0.602 decimal digits per term |
Similarly, we can find the derivative of \(y=\arctan(x)\) implicitly. \[\tan(y)=x\Rightarrow\sec^2(x){dy\over dx}=1 \Rightarrow{dy\over dx}={1\over\sec^2(x)}={1\over1+\tan^2(x)}={1\over1+x^2}\] Then the integral is defined for \(x\in(-\infty,\infty)\) \[\arctan(x)=\int_0^x{dt\over1+t^2}\] We can turn this into a summation with the geometric summation formula (or the generalized binomial theorem, but than gives the same result) \[=\int_0^x\left[\sum_{k=0}^\infty(-1)^kt^{2k}\right]dt =\sum_{k=0}^\infty\left[(-1)^k\int_0^x t^{2k}dt\right] =\sum_{k=0}^\infty{(-1)^kx^{2k+1}\over2k+1}\] There are no factorials in this one, so the convergence is determined just by the power \(x^{2k+1}\). It's an alternating series and converges for \(x\in[-1,1]\). Because of the more restricted convergence interval, we only can use 2 of the special angles to obtain \(\pi\) formulas: \(x={\sqrt{2}\over2},{\sqrt{3}\over3}\).
| Angle | Formula | Convergence |
|---|---|---|
| \[{\pi\over4}\] | \[\pi=4\sum_{k=0}^\infty{(-1)^k\over2k+1}\] | Very slow, error is \(\propto k^{-1}\) |
| \[{\pi\over6}\] | \[\pi=2\sqrt{3}\sum_{k=0}^\infty{(-1)^k\over3^k(2k+1)}\] | ~0.477 decimal digits per term |
It is possible to create formulas that use other angles \(\theta\) such that \(\sin(\theta)\) or \(\tan(\theta)\) have a square root. Such series can be summed in \(\mathbb{Z}[\sqrt{n}]\) or \(\mathbb{Q}[\sqrt{n}]\), which adds complexity that would outweight the fewer terms needed. We could even go further to smaller angles with deeper nested square roots, but this adds complexity of summing in a ring/field with more square root terms to track.
Additional series found with \(\arcsin(x)\) or \(\arccos(x)={\pi\over2}-\arcsin(x)\). Because of the angle complement, \(\theta\) and \({\pi\over2}-\theta\) give us the same series results.
| Angle (\(\theta\)) | \(\sin(\theta)\) and \(\cos(\theta)\) |
Formula from \(\arcsin\) and Formula from \(\arccos\) |
Decimal Digits Per Term |
|---|---|---|---|
| \[{\pi\over5}\] | \[{1\over2}\sqrt{{1\over2}\left(5-\sqrt{5}\right)}\] | \[\pi={5\over2}\sqrt{{1\over2}\left(5-\sqrt{5}\right)} \sum_{k=0}^\infty{2k\choose k}{\left(5-\sqrt{5}\right)^k \over32^k(2k+1)}\] | ~0.462 |
| \[{1\over4}\left(1+\sqrt{5}\right)\] | \[\pi={5\over6}\left(\sqrt{5}+1\right)\sum_{k=0}^\infty{2k\choose k} {\left(3+\sqrt{5}\right)^k\over32^k(2k+1)}\] | ~0.184 | |
| \[{\pi\over8}\] | \[{\sqrt{2-\sqrt{2}}\over2}\] | \[\pi=4\sqrt{2-\sqrt{2}}\sum_{k=0}^\infty{2k\choose k} {\left(2-\sqrt{2}\right)^k\over16^k(2k+1)}\] | ~0.834 |
| \[{\sqrt{2+\sqrt{2}}\over2}\] | \[\pi={4\over3}\sqrt{2+\sqrt{2}}\sum_{k=0}^\infty{2k\choose k} {\left(2+\sqrt{2}\right)^k\over16^k(2k+1)}\] | ~0.069 | |
| \[{\pi\over10}\] | \[{1\over4}\left(\sqrt{5}-1\right)\] | \[\pi={5\over2}\left(\sqrt{5}-1\right)\sum_{k=0}^\infty{2k\choose k} {\left(3-\sqrt{5}\right)^k\over32^k(2k+1)}\] | ~1.020 |
| \[{1\over2}\sqrt{{1\over2}\left(5+\sqrt{5}\right)}\] | \[\pi={5\over4}\sqrt{{1\over2}\left(5+\sqrt{5}\right)} \sum_{k=0}^\infty{2k\choose k} {\left(5+\sqrt{5}\right)^k\over32^k(2k+1)}\] | ~0.044 | |
| \[{\pi\over12}\] | \[{\sqrt{3}-1\over2\sqrt{2}}\] | \[\pi=3\sqrt{2}\left(\sqrt{3}-1\right)\sum_{k=0}^\infty{2k\choose k} {\left(2-\sqrt{3}\right)^k\over16^k(2k+1)}\] | ~1.174 |
| \[{1+\sqrt{3}\over2\sqrt{2}}\] | \[\pi={3\over5}\sqrt{2}\left(1+\sqrt{3}\right)\sum_{k=0}^\infty {2k\choose k}{\left(2+\sqrt{3}\right)^k\over16^k(2k+1)}\] | ~0.030 |
Additional series found with \(\arctan(x)\).
| Angle (\(\theta\)) | \(\tan(\theta)\) | Formula | Decimal Digits Per Term |
|---|---|---|---|
| \[{\pi\over5}\] | \[\sqrt{5-2\sqrt{5}}\] | \[\pi=5\sqrt{5-2\sqrt{5}}\sum_{k=0}^\infty {(-1)^k\left(5-2\sqrt{5}\right)^k\over2k+1}\] | ~0.277 |
| \[{\pi\over8}\] | \[\sqrt{2}-1\] | \[\pi=8\left(\sqrt{2}-1\right)\sum_{k=0}^\infty {(-1)^k\left(3-2\sqrt{2}\right)^k\over2k+1}\] | ~0.766 |
| \[{\pi\over10}\] | \[\sqrt{1-{2\sqrt{5}\over5}}\] | \[\pi=10\sqrt{5-2\sqrt{5}\over5}\sum_{k=0}^\infty {(-1)^k\left(5-2\sqrt{5}\right)^k\over5^k(2k+1)}\] | ~0.976 |
| \[{\pi\over12}\] | \[2-\sqrt{3}\] | \[\pi=12\left(2-\sqrt{3}\right)\sum_{k=0}^\infty {(-1)^k\left(7-4\sqrt{3}\right)^k\over2k+1}\] | ~1.144 |
These methods would converge faster if we could find a smaller \(x\) to use. We need \(x\) to be in a nice form (at worst, a square root of a rational) and for \(\arcsin(x)\) or \(\arctan(x)\) to be \(\pi\) times some number which is easy to compute, so at worst we may want \(\pi\) times a square root of a rational. It turns out we cannot find any other such nice values to use. There are smaller angles like \(\pi/12\) which can be described by radicals, but not in a single term. It is possible to sum those with binary splitting over rings like \(\mathbb{Z}[\sqrt{3}]\) or fields like \(\mathbb{Q}[\sqrt{3}]\) but that adds complexity which would outweigh the smaller number of terms required. We can extend further to smaller angles whose values nest square roots more deeply, but then we increase complexity further by having to track more radical terms.