[VA] SRC #009 - Pi Day 2021 Special

[robve]

Quote:c. Another most famous transcendental constant also appearing everywhere is e = 2.718... is there any other simpler way to get π from e which does not involve complex numbers? Yes, there is, simply evaluate:

$$ \pi = 4(\arctan \mathrm{e} - \arctan \frac{\mathrm{e}-1}{\mathrm{e}+1}) $$

My HP-71B gives:

>RADIANS
>4*ATAN(EXP(1))-ATAN((EXP(1)-1)/(EXP(1)+1))
3.1415926536

Looks good for sure!

Why is this equal to π? Use the famous identity and the corresponding series

$$ \frac{\pi}{4} = \arctan 1 = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} $$

We also have

$$ \arctan u - \arctan v = \arctan\frac{u-v}{1+uv} $$

Therefore

$$ \arctan \mathrm{e} - \arctan \frac{\mathrm{e}-1}{\mathrm{e}+1} = \arctan \frac{\mathrm{e}-(\mathrm{e}-1)/(\mathrm{e}+1)}{1+\mathrm{e}(\mathrm{e}-1)/(\mathrm{e}+1)} $$

Multiplying both sides by e+1 then expanding and cancelling terms:

$$ \arctan \frac{\mathrm{e}(\mathrm{e}+1)-(\mathrm{e}-1)}{\mathrm{e}+1+\mathrm{e}(\mathrm{e}-1)} = \arctan \frac{\mathrm{e}^2+1}{\mathrm{e}^2+1} = \arctan 1 $$

Bingo!

- Rob