For any 0 < ε <1 and a ∈R , consider the Kepler equation x − ε sin x =

1.4.11. For any 0 < ε <1 and a ∈R , consider the Kepler equation x − ε sin x =

a . Define x 0 =a , x 1 =a+ ε sin x 0 ,...,x n =a+ ε sin x n −1 . Prove that there exists ξ = lim n → ∞ x n and that ξ is the unique solution of Kepler’s equation.

Solution. We first observe that the sequence (x n ) n ≥0 is bounded. This follows from

|x n | = |a + ε sin x n −1 | ≤ |a| + ε < |a| + 1.

We prove that the sequence (x n ) n ≥0 converges using Cauchy’s criterion. More precisely, we show that for any δ > 0, there exists N( δ ) ∈ N such that for all n ≥ N( δ ) and for any p ∈ N we obtain |x n −x n +p |< δ .

38 1 Sequences We have |x n −x n +p |= ε |sinx n −1 − sinx n +p−1 |

n =2 +p−1

ε sin n −1 −x n +p−1

x n −1 +x

cos

2 ≤2 ε 2 |x n −1 −x n ≤2 +p−1 · 2 = ε |x n −1 −x n +p−1 |.

ε sin n −1 −x n +p−1

But 0 |x 2 −x p |= ε |sinx p −1 |≤ ε , |x 1 −x p +1 |≤ ε · ε = ε . This implies that

|x n −x n

|≤ ε −1 n +p−1 < δ ,

provided that n > log δ / log ε [we have used here our assumption ε ∈ (0,1)]. Set ξ = lim n → ∞ x n . It follows that ξ − ε sin ξ = a, which shows that ξ is a solution of Kepler’s equation.

In order to prove the uniqueness, let ξ 1 , ξ 2 be distinct solutions of Kepler’s equa- tion. We have

ε sin 1 − ξ 2 ξ 1 + ξ | 2 1 − 2 |= |sin 1 − sin 2 |=2 · cos

2 2 =| ξ 1 − ξ 2 |.⊓ ⊔ This contradiction shows that the Kepler equation has a unique solution.

≤2 ε sin

The following quadratic Diophantine equation (that is, the unknowns are positive integers) is a special case of the Pell equation x 2 −Dy 2 = m (where D is a nonsquare positive integer) and m is an integer. It seems that the English mathematician John Pell (1610–1685) has nothing to do with this equation. According to Lenstra [69], Euler mistakenly attributed to Pell a solution method that had been found by another English mathematician, William Brouncker (1620–1684). An exposition for solving the Pell equation may be found even in Euler’s Algebra [25, Abschn. 2, Cap. 7].

Usually, Pell-type equations of this form are solved by finding the continued √ fraction of D . This method is also due to Euler (see [84, Chap. 7]). Our proof is based on an iterative process. The idea of computing by recursion is as old as counting itself. It occurred in primitive form in the efforts of the Babylonians as early as 2000 B.C. to extract roots and in more explicit form around 450 B.C. in the Pythagoreans’ study of figurative numbers, since in modern notation the tri- angular numbers satisfy the difference equation t n =t n −1 + n, the square numbers the equation s n =s n −1 + 2n − 1, and so forth. The Pythagoreans also used a sys- tem of difference equations x n =x n −1 + 2y n −1 ,y n =x n −1 +y n −1 to generate large √

2. In his attempts to compute the circumference of a circle, Archimedes (about 250 B.C.) √ employed equations of the form P 2n = 2p n P n /(p n +P n ), p 2n = p n P 2n to compute

solutions of Pell’s equation x 2 − 2y 2 = 1, and thereby approximations of

1.4 Qualitative Results 39 the perimeters P n and p n of the circumscribed polygon of n sides and the inscribed

polygon of n sides, respectively. Other familiar ancient discoveries about recurrence include the Euclidean algorithm and Zeno’s paradox. Euclid also studied geometric series, although the general form of the sum was not obtained until around 1593 by Vieta.