Powers of \(x\) show up when we integrate \({1\over1-x}\) which turns into an integral of powers \(x^k\). If instead we had \({1\over1-x(1-x)}\), we would turn it into integrals of \(x^k(1-x)^k\), the beta integral. We can generalize this a bit with a constant and for \(c>0\), the following integral can be turned into formulas for \(\pi\). Note that the integral is proper when \(0<c<4\). \[\int_0^1{dx\over1-cx(1-x)}=\sum_{k=0}^\infty c^k\int_0^1 x^k(1-x)^kdx =\sum_{k=0}^\infty{c^k(k!)^2\over(2k+1)!}\] Now we need to actually evaluate the integral so we can find what constants the summation can be used to approximate. Start by completing the square \[=\int_0^1{dx\over1-cx+cx^2}={1\over c}\int_0^1{dx \over\left(x-{1\over2}\right)^2+\left({1\over c}-{1\over4}\right)}\] Let \(d=\sqrt{4-c\over4c}\) so now we can integrate \[{1\over c}\int_0^1{d\left(x-{1\over2}\right) \over\left(x-{1\over2}\right)^2+d^2} ={1\over cd}\arctan\left({x-{1\over2}\over d}\right)\Big|_0^1\] \[={1\over cd}\left(\arctan\left({1\over2d}\right) -\arctan\left({-1\over2d}\right)\right) ={2\over cd}\arctan\left({1\over2d}\right)\] So if we wanted to find a series for \(\arctan(r)\), we would need \[r={1\over2d}=\sqrt{c\over4-c}\Rightarrow c={4r^2\over1+r^2}\] Then our constant part is \[{2\over cd}={2\over{4r^2\over1+r^2}{1\over2r}}={1+r^2\over r}\] Using this to rewrite the constants, we find \[\arctan(r)={r\over1+r^2}\sum_{k=0}^\infty{(k!)^2\over(2k+1)!} \left({4r^2\over1+r^2}\right)^k\] For convergence, we need \(\left|4r^2\over1+r^2\right|<4\) which is true for all real numbers \(r\), so this series converges anywhere, but converges fastest near \(0\).
We can do this with the usual special angles. Since we also only have \(r^2\) in the summation ratio part, we might also like to see if other special angles, which require more complicated square roots, can give nice formulas. Below are formulas found for the usual special angles.
| Angle | Formula | Digits Per Term |
|---|---|---|
| \[\pi\over3\] | \[\pi={3\sqrt{3}\over4}\sum_{k=0}^\infty {3^k(k!)^2\over(2k+1)!}\] | ~0.125 |
| \[\pi\over4\] | \[\pi=2\sum_{k=0}^\infty{2^k(k!)^2\over(2k+1)!}\] | ~0.301 |
| \[\pi\over6\] | \[\pi={3\sqrt{3}\over2}\sum_{k=0}^\infty{(k!)^2\over(2k+1)!}\] | ~0.602 |
And formulas with other special angles that require more complicated square root expressions.
| Angle | Formula | Digits Per Term |
|---|---|---|
| \[\pi\over5\] | \[\pi={5\over8}\left(3+\sqrt{5}\right)\sqrt{5-2\sqrt{5}} \sum_{k=0}^\infty{(k!)^2\over(2k+1)!} \left({5-\sqrt{5}\over2}\right)^k\] | ~0.462 |
| \[\pi\over8\] | \[\pi=2\sqrt{2}\sum_{k=0}^\infty{(k!)^2\over(2k+1)!} \left(2-\sqrt{2}\right)^k\] | ~0.834 |
| \[\pi\over10\] | \[\pi={5\over4}\left(5+\sqrt{5}\right)\sqrt{1-{2\sqrt{5}\over5}} \sum_{k=0}^\infty{(k!)^2\over(2k+1)!} \left({3-\sqrt{5}\over2}\right)^k\] | ~1.020 |
| \[\pi\over12\] | \[\pi=3\sum_{k=0}^\infty{(k!)^2\over(2k+1)!} \left(2-\sqrt{3}\right)^k\] | ~1.174 |
The usual special angles give us formulas which look like similar complexity to the ones from \(\arcsin\). In a way, they have the \({2k\choose k}\) part flipped. The extra special angles all give us formulas which require powers of something with a square root, which is disappointing. The \(\pi/12\) formula does not require computing a weird square root and produces almost 4 times as many digits per term as the \(\pi/4\) formula, so that might lead to the curiosity of whether its faster convergence is enough to overcome the added complexity of summing a series in \(\mathbb{Z}[\sqrt{3}]\) or \(\mathbb{Q}[\sqrt{3}]\).