Vieta's Formula for Pi

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This proof is about an approximation formula for pi. For other uses, see Viète's Formulas.

Theorem

$\pi = 2 \times \dfrac 2 {\sqrt 2} \times \dfrac 2 {\sqrt {2 + \sqrt 2} } \times \dfrac 2 {\sqrt {2 + \sqrt {2 + \sqrt 2} } } \times \dfrac 2 {\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } } \times \cdots$


Proof

\(\displaystyle 1\) \(=\) \(\displaystyle \sin \frac \pi 2\) Sine of Half-Integer Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle 2 \sin \frac \pi 4 \cos \frac \pi 4\) Double Angle Formula for Sine
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {2 \sin \frac \pi 8 \cos \frac \pi 8} \cos \frac \pi 4\) Double Angle Formula for Sine
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {2 \paren {2 \sin \frac \pi {16} \cos \frac \pi {16} } \cos \frac \pi 8} \cos \frac \pi 4\) Double Angle Formula for Sine
\(\displaystyle \) \(=\) \(\displaystyle \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle 2^{n - 1} \sin \frac \pi {2^n} \cos \frac \pi {2^n} \cos \frac \pi {2^{n - 1} } \cdots \cos \frac \pi 8 \cos \frac \pi 4\)


Thus:

\(\displaystyle \frac 1 {2^{n - 1} \sin \frac \pi {2^n} }\) \(=\) \(\displaystyle \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 2 \pi \times \frac {\pi / 2^n} {\map \sin {\pi / 2^n} }\) \(=\) \(\displaystyle \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}\)


Then we have from the Half Angle Formula for Cosine that:

\(\displaystyle \cos \frac \pi {2^{k} }\) \(=\) \(\displaystyle \frac {\sqrt {2 + 2 \map \cos {\pi / 2^{k - 1} } } } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 2} } } } } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 3} } } } } } 2\)

So we can replace all the instances of $\cos \dfrac \pi 4$, $\cos \dfrac \pi 8$, etc. with their expansions in square roots of $2$.


Finally, we note that from Limit of Sine of X over X we have:

$\displaystyle \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$

As $n \to \infty$, then, we have that $\dfrac \pi {2^n} \to 0$, and so:

$\displaystyle \lim_{n \mathop \to \infty} \frac {\map \sin {\pi / 2^n} } {\pi / 2^n} = 1$


The result follows after some algebra.

$\blacksquare$


Also known as

This result is also known as Vieta's (or Viète's) product.


Source of Name

This entry was named for Franciscus Vieta.


Historical Note

Franciscus Vieta, or François Viète as he is otherwise known, published his formula for $\pi$ in $1592$.


Sources