Vieta's Formula for Pi

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This proof is about Vieta's Formula for Pi in the context of Real Analysis. 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

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


Thus:

\(\ds \frac 1 {2^{n - 1} \sin \frac \pi {2^n} }\) \(=\) \(\ds \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}\)
\(\ds \leadsto \ \ \) \(\ds \frac 2 \pi \times \frac {\pi / 2^n} {\map \sin {\pi / 2^n} }\) \(=\) \(\ds \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}\) multiplying top and bottom by $\pi$
\(\ds \leadsto \ \ \) \(\ds \pi \times \frac {\map \sin {\pi / 2^n} } {\pi / 2^n}\) \(=\) \(\ds 2 \times \frac 1 {\cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n} }\)


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

\(\ds \cos \frac \pi {2^{k} }\) \(=\) \(\ds \sqrt {\frac {1 + \map \cos {\pi / 2^{k - 1} } } 2 }\)
\(\ds \) \(=\) \(\ds \sqrt {\frac {2 + 2 \map \cos {\pi / 2^{k - 1} } } {2 \times 2} }\) multiplying top and bottom by $2$
\(\ds \) \(=\) \(\ds \frac {\sqrt {2 + 2 \map \cos {\pi / 2^{k - 1} } } } 2\)
\(\ds \) \(=\) \(\ds \frac {\sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 2} } } } } 2\)
\(\ds \) \(=\) \(\ds \frac {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 3} } } } } } 2\)

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 $\dfrac {\sin x} x$ at Zero we have:

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

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

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


The result follows after some algebra.

$\blacksquare$


Also known as

Vieta's formula for pi is also known as Vieta's product, or Viète's product.

Some sources refer to it merely as Vieta's formula, or Viète's formula, but this can be confused with Viète's formulas in the context of polynomial theory.

The form Viète's formula for pi can of course also be seen.


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 $1592$.

Some sources suggest that it was $1593$.


Sources

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