Vieta's Formula for Pi
(Redirected from Viète'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:
\(\ds \frac 1 {2^{n - 1} \sin \frac \pi {2^n} }\) | \(=\) | \(\ds \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 2 \pi \times \frac {\pi / 2^n} {\map \sin {\pi / 2^n} }\) | \(=\) | \(\ds \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}\) |
Then we have from the Half Angle Formula for Cosine that:
\(\ds \cos \frac \pi {2^{k} }\) | \(=\) | \(\ds \frac {\sqrt {2 + 2 \map \cos {\pi / 2^{k - 1} } } } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 2} } } } } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 3} } } } } } 2\) |
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
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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.9$: A proof of Vieta's Formula
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Viète's product