nLab
trigonometric identities and the irrationality of pi

Contents

Goal

This page deals with the question whether it might be possible to prove the irrationality of π\pi by using trigonometric identities.

Details

We begin with an infinite sequence of functions of some variables θ 1,θ 2,θ 3,\theta_1,\theta_2,\theta_3,\dots:

f 0(θ 1,θ 2,θ 3,) = evenn0(1) n/2 |A|=n iAsinθ i iAcosθ i f 1(θ 1,θ 2,θ 3,) = oddn1(1) (n1)/2 |A|=n iAsinθ i iAcosθ i f 2(θ 1,θ 2,θ 3,) = evenn2(1) (n2)/2n |A|=n iAsinθ i iAcosθ i f 3(θ 1,θ 2,θ 3,) = oddn3(1) (n3)/2(n1) |A|=n iAsinθ i iAcosθ i f 4(θ 1,θ 2,θ 3,) = evenn4(1) (n4)/2n(n2) |A|=n iAsinθ i iAcosθ i f 5(θ 1,θ 2,θ 3,) = oddn5(1) (n5)/2(n1)(n3) |A|=n iAsinθ i iAcosθ i f 6(θ 1,θ 2,θ 3,) = evenn6(1) (n6)/2n(n2)(n4) |A|=n iAsinθ i iAcosθ i f 7(θ 1,θ 2,θ 3,) = oddn7(1) (n7)/2(n1)(n3)(n5) |A|=n iAsinθ i iAcosθ i \begin{aligned} f_0(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{even}\;n \ge 0} (-1)^{n/2} \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_1(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{odd}\;n \ge 1} (-1)^{(n-1)/2} \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_2(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{even}\;n \ge 2} (-1)^{(n-2)/2} n \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_3(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{odd}\;n \ge 3} (-1)^{(n-3)/2} (n-1) \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_4(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{even}\;n \ge 4} (-1)^{(n-4)/2} n(n-2) \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_5(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{odd}\;n \ge 5} (-1)^{(n-5)/2} (n-1)(n-3) \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_6(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{even}\;n \ge 6} (-1)^{(n-6)/2} n(n-2)(n-4) \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ f_7(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{\text{odd}\;n \ge 7} (-1)^{(n-7)/2} (n-1)(n-3)(n-5) \sum_{|A| = n} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i \\ & \vdots \end{aligned}

etc. In each case we could have said evenn0\text{even}\;n \ge 0 or oddn0\text{odd}\;n \ge 0, since the coefficients kill off the terms in which nn is smaller than the index—the coefficients are the unique simplest expressions that do that.

  • Each of these is a symmetric function of θ 1,θ 2,θ 3,\theta_1,\theta_2,\theta_3,\dots.

  • 0 is an identity element for each of these, i.e. we have f k(0,θ 2,θ 3,θ 4,)=f k(θ 2,θ 3,θ 4,)f_k(0,\theta_2,\theta_3,\theta_4,\dots) = f_k(\theta_2,\theta_3,\theta_4,\dots).

  • Lemma:

    f k(θ 1,θ 2,θ 3,θ 4,θ 5,)f k(θ 1+θ 2,θ 3,θ 4,θ 5,)={k ifk2is even k1 ifk2is odd}sinθ 1sinθ 2f k2(θ 3,θ 4,θ 5,) f_k(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\dots) - f_k(\theta_1+\theta_2,\theta_3,\theta_4,\theta_5,\dots) = \left.\begin{cases} k & \text{if}\;k \ge 2\;\text{is even} \\ k - 1 & \text{if}\;k \ge 2\;\text{is odd} \end{cases} \right\} \cdot \sin\theta_1\sin\theta_2 f_{k-2}(\theta_3,\theta_4,\theta_5,\dots)

    and if k=0k = 0 or 11 then the difference is 0.

  • Identities:

    f 0(θ 1,θ 2,θ 3,) =cos(θ 1+θ 2+θ 3+). f 1(θ 1,θ 2,θ 3,) =sin(θ 1+θ 2+θ 3+). If i1θ i=π,thenf 2(θ 1,θ 2,θ 3,) = i1sin 2θ i. If i1θ i=π,thenf 3(θ 1,θ 2,θ 3,) = i1sin(2θ i)2. \begin{aligned} f_0(\theta_1,\theta_2,\theta_3,\dots) &= \cos(\theta_1 + \theta_2 + \theta_3 + \cdots). \\ f_1(\theta_1,\theta_2,\theta_3,\dots) &= \sin(\theta_1 + \theta_2 + \theta_3 + \cdots). \\ \text{If}\;\sum_{i \ge 1} \theta_i = \pi,\;\text{then}\;f_2(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{i \ge 1} \sin^2 \theta_i. \\ \text{If}\;\sum_{i \ge 1} \theta_i = \pi,\;\text{then}\;f_3(\theta_1,\theta_2,\theta_3,\dots) &= \sum_{i \ge 1} \frac{\sin(2\theta_i)}{2}. \end{aligned}

There is a simple geometric interpretation of the expression on the right side of the identity involving f 3f_3. Suppose θ i,i1\theta_i, i \ge 1 are the angles between adjacent diagonals of a polygon inscribed in a circle of unit diameter. For present purposes we consider the sides of the polygon to be “diagonals”. (All but two of the θ i\theta_i appear at each vertex. For example, in an octagon, at one vertex the angles are θ 1,,θ 6\theta_1,\dots,\theta_6 at the next vertex they are θ 2,,θ 7\theta_2,\dots,\theta_7; at the next they are θ 3,,θ 8\theta_3,\dots,\theta_8; and then at the next they are θ 4,,θ 8,θ 1\theta_4,\dots,\theta_8,\theta_1, and so on. All eight appear at each vertex if one draws a tangent line to the circle and includes the angles between sides of the polygon and the tangent line.) Then the value of each side of the identity involving f 3f_3 is half the area of the polygon. Each term on the right side of the identity is half the signed area of a triangle having one vertex at the center of the circle. As one goes counterclockwise around the polygon, the signed area is positive or negative as the ray from the center is then turning counterclockwise or clockwise. The terms on the left side may perhaps not admit so simple an interpretation when the number of nonzero θ i\theta_i is more than 4. The special case involving only three nonzero θ i\theta_i is the well-known (?) identity that says that if α+β+γ=π\alpha+\beta+\gamma=\pi then 4sinαsinβsinγ=sin(2α)+sin(2β)+sin(2γ)4\sin\alpha\sin\beta\sin\gamma = sin(2\alpha) + \sin(2\beta) + \sin(2\gamma).

(Is there any simple geometric interpretation of the identity involving f 2f_2?)

A question

Are these four identities the first four terms in a sequence that continues?

Michael Hardy: A student in India whom I encountered via the internet told me that the problem of proving the “conditional identity” that 4sinαsinβsinγ=sin(2α)+sin(2β)+sin(2γ)4\sin\alpha\sin\beta\sin\gamma = sin(2\alpha) + \sin(2\beta) + \sin(2\gamma) if α+β+γ=π\alpha+\beta+\gamma=\pi recurs perennially on the joint entrance exam of the Indian Institutes of Technology. Hence “well known”.

Inversion formulas

i=1 cosθ i =f 0+12f 2+124f 4+1246f 6+12468f 8+ j=1 sinθ j i:ijcosθ i =f 1+12f 3+124f 5+1246f 7+12468f 9+ |A|=2 iAsinθ i iAcosθ i =12(f 2+12f 4+124f 6+1246f 8+12468f 10+) |A|=3 iAsinθ i iAcosθ i =12(f 3+12f 5+124f 7+1246f 9+12468f 11+) |A|=4 iAsinθ i iAcosθ i =124(f 4+12f 6+124f 8+1246f 10+12468f 12+) |A|=5 iAsinθ i iAcosθ i =124(f 5+12f 7+124f 9+1246f 11+12468f 13+) \begin{aligned} \prod_{i=1}^\infty \cos\theta_i & = f_0 + \frac{1}{2} f_2 + \frac{1}{2 \cdot 4} f_4 + \frac{1}{2 \cdot 4 \cdot 6} f_6 + \frac{1}{2 \cdot 4 \cdot 6 \cdot 8} f_8 + \cdots \\ \sum_{j=1}^\infty \sin\theta_j \prod_{i : i \neq j} \cos\theta_i & = f_1 + \frac{1}{2} f_3 + \frac{1}{2\cdot 4} f_5 + \frac{1}{2\cdot 4\cdot 6} f_7 + \frac{1}{2 \cdot 4 \cdot 6 \cdot 8} f_9 + \cdots \\ \sum_{|A| = 2} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i & = \frac{1}{2} \left(f_2 + \frac{1}{2} f_4 + \frac{1}{2\cdot 4} f_6 + \frac{1}{2\cdot 4 \cdot 6} f_8 + \frac{1}{2\cdot 4\cdot 6\cdot 8} f_{10} + \cdots\right) \\ \sum_{|A| = 3} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i & = \frac{1}{2} \left(f_3 + \frac{1}{2} f_5 + \frac{1}{2\cdot 4} f_7 + \frac{1}{2\cdot 4 \cdot 6} f_9 + \frac{1}{2\cdot 4\cdot 6\cdot 8} f_{11} + \cdots\right) \\ \sum_{|A| = 4} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i & = \frac{1}{2\cdot 4} \left(f_4 + \frac{1}{2} f_6 + \frac{1}{2\cdot 4} f_8 + \frac{1}{2\cdot 4 \cdot 6} f_{10} + \frac{1}{2\cdot 4\cdot 6\cdot 8} f_{12} + \cdots\right) \\ \sum_{|A| = 5} \prod_{i \in A} \sin\theta_i \prod_{i \notin A} \cos\theta_i & = \frac{1}{2\cdot 4} \left(f_5 + \frac{1}{2} f_7 + \frac{1}{2\cdot 4} f_9 + \frac{1}{2\cdot 4 \cdot 6} f_{11} + \frac{1}{2\cdot 4\cdot 6\cdot 8} f_{13} + \cdots\right) \\ & \vdots \end{aligned}

What Leonhard Euler and Mary Cartwright did

Leonhard Euler used something similar to the identities for f 0f_0 and f 1f_1 in order to derive the power-series expansion?s of the sine? and cosine?. The idea is that to find sinx\sin x, one lets x=θ++θx = \theta+\cdots+\theta, where θ\theta is infinitely small and the number of terms is an infinitely large integer, and then apply the identities for f 0f_0 and f 1f_1. More precisely, Euler wrote cos(nθ)\cos(n\theta) and sin(nθ)\sin(n\theta) as functions of sinθ\sin\theta and cosθ\cos\theta and then supposed nn is infinitely large and nθn\theta is finite. Then sinθ\sin\theta becomes θ\theta and cosθ\cos\theta becomes 1.

If one does with f kf_k, for k0k \ge 0, what Euler did with f 0f_0 and f 1f_1, one gets an infinite sequence of power series in xx. The odd-indexed functions of xx in this sequence are the functions J nJ_n that appear in Cartwright’s proof of the irrationality of π\pi. But in defining those functions, Cartwright used neither trigonometric expressions nor power series. (The proof was published in an appendix to the third edition of Harold Jeffreys’ book ‘’Scientific Inference’’. It is not in later editions. It appears in Wikipedia’s article on the proof of the irrationality of π\pi.)

The main question

Could a proof of the irrationality of π\pi using only finitary trigonometric identities be lurking somewhere in all this? ({f k(θ 1,,θ )} k,\left\{f_k(\theta_1,\dots,\theta_\ell)\right\}_{k,\ell} is a doubly indexed sequence each term of which is the a sum of finitely many products of finitely many factors.)

Revised on June 2, 2010 17:11:11 by Toby Bartels (64.89.48.241)