Proving π Using an Infinite Product of Cosines: Euler’s Formula Revisited

Euler's remarkable infinite product formula for the sinc function has been a cornerstone in mathematical analysis. This article aims to delve into a beautiful derivation that uses this formula to prove a fundamental value of π. The proof, which involves the manipulation of an infinite product of cosines, is both elegant and insightful.

Introduction to Euler's Infinite Product

Let's start with Euler's formula for the sinc function:

[ frac{sin(x)}{x} prod_{k1}^infty cosleft(frac{x}{2^k}right) ]

This formula is attributed to both Euler and Viète. It expresses the sinc function, a fundamental trigonometric entity, as an infinite product. By isolating (x), we can explore deeper into the nature of π through this product.

Isolating x to Connect π

By rearranging the formula, we can express (x) in terms of an infinite product:

[ x frac{sin(x)}{prod_{k1}^infty cosleft(frac{x}{2^k}right)} ]

Substituting (x frac{pi}{2}) into this equation yields:

[ frac{pi}{2} frac{sinleft(frac{pi}{2}right)}{prod_{k1}^infty cosleft(frac{frac{pi}{2}}{2^k}right)} ]

Simplifying the sine term, we get:

[ pi 2 times frac{1}{prod_{k1}^infty cosleft(frac{pi}{2^{k-1}}right)} ]

This expression can be further simplified to:

[ pi 2 times frac{1}{cosleft(frac{pi}{2^2}right) times cosleft(frac{pi}{2^3}right) times cosleft(frac{pi}{2^4}right) cdots} ]

Applying the Half-Angle Formula

The half-angle formula for the cosine function, [ frac{1 - cos(theta)}{2} cos^2left(frac{theta}{2}right) ], plays a crucial role in the derivation. By applying this formula iteratively, we can express each cosine term in the product:

[ cosleft(frac{pi}{2^2}right) frac{sqrt{2}}{2} ] [ cosleft(frac{pi}{2^3}right) frac{sqrt{2 sqrt{2}}}{2} frac{sqrt{2 cosleft(frac{pi}{2^2}right)}}{2} ] [ cosleft(frac{pi}{2^4}right) frac{sqrt{2 sqrt{2 sqrt{2}}}}{2} frac{sqrt{2 cosleft(frac{pi}{2^3}right)}}{2} ]

Each iteration halfs the angle, leading to the product:

[ pi 2 times frac{1}{frac{sqrt{2}}{2} times frac{sqrt{2 sqrt{2}}}{2} times frac{sqrt{2 sqrt{2 sqrt{2}}}}{2} cdots} ]

Simplifying, we get:

[ pi 2 times frac{2}{sqrt{2}} times frac{2}{sqrt{2 sqrt{2}}} times frac{2}{sqrt{2 sqrt{2 sqrt{2}}}} times cdots ]

Introducing Pk and Telescoping Product

Define [ P_k sqrt{2 sqrt{2 cdots sqrt{2}}} quad text{(k radicals)} ]. We observe that:[ P_{k-1} sqrt{2 P_k} ]. If we let [ P_k 2 cos(theta_k) ], the relation becomes:

[ P_{k-1} sqrt{2 2 cos(theta_k)} sqrt{4 cos^2left(frac{theta_k}{2}right)} 2 cosleft(frac{theta_k}{2}right) ]

This implies that each iteration halves the angle [ theta_k ], so [ P_k 2 cosleft(frac{pi}{2^{k-1}}right) ]. For [ P_1 sqrt{2} ], we have [ theta_1 frac{pi}{4} ].

The Product and Limit Analysis

Consider the product [ S_n prod_{k1}^{n} frac{P_k}{2} prod_{k1}^{n} cosleft(frac{pi}{2^{k-1}}right) ]. Define [ S_k(alpha) prod_{r1}^{k} cosleft(frac{alpha}{2^{r-1}}right) ]. We then have:

[ 2 S_{k-1}(alpha) sinleft(frac{alpha}{2^{k-2}}right) S_k(alpha) sinleft(frac{alpha}{2^{k-1}}right) ]

Which further implies:

[ prod_{k1}^{n} frac{S_{k-1}(alpha)}{S_k(alpha)} prod_{k1}^{n} left(frac{1}{2} cdot frac{sinleft(frac{alpha}{2^{k-1}}right)}{sinleft(frac{alpha}{2^{k-2}}right)}right) ]

This product is telescoping, leading to:

[ S_{n-1}(alpha) frac{S_1(alpha)}{2^n} cdot frac{sinleft(frac{alpha}{2}right)}{sinleft(frac{alpha}{2^{n-1}}right)} ]

To find [ lim_{n to infty} S_{n-1}(pi) ], we note:

[ lim_{n to infty} S_{n-1}(pi) lim_{n to infty} frac{sinleft(frac{pi}{2}right)}{2^{n-1} sinleft(frac{pi}{2^{n-1}}right)} frac{2}{pi} lim_{n to infty} frac{frac{pi}{2^{n-1}}}{sinleft(frac{pi}{2^{n-1}}right)} frac{2}{pi} ]

Using the standard limit property: [ lim_{x to 0} frac{sin x}{x} 1 ], we conclude:

[ frac{2}{pi} frac{sqrt{2}}{2} cdot frac{sqrt{2 sqrt{2}}}{2} cdot frac{sqrt{2 sqrt{2 sqrt{2}}}}{2} cdots ]

Conclusion

This proof, involving an infinite product of cosines, not only demonstrates a unique way to derive π but also showcases the power of trigonometric identities and series in mathematical exploration. The elegant application of Euler's formula and the half-angle formula provides a profound insight into the nature of π.