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Transcript
Notes on Continued Fractions for Math 4400 1. Continued fractions. The continued fraction expansion converts a positive real number α into a sequence of natural numbers. Conversely, a sequence of natural numbers: a0 , a1 , a2 , a3 , . . . is converted into a sequence of rational numbers via: 1 1 1 (∗) a0 , a0 + , a0 + , ··· 1 , a0 + a1 a1 + a2 a1 + a2 +1 1 a3 Thus, for example, the sequence 3, 7, 15, 1 is converted into: 3 1 22 1 333 1 355 3= , 3+ = , 3+ , 3+ = 1 = 1 1 7 7 106 113 7 + 15 7 + 15+ 1 1 which are excellent approximations of π. We will find some useful general properties of continued fractions by replacing numbers a0 , a1 , a2 , a3 , . . . with a sequence of variables: q0 , q1 , q2 , q3 , . . . and considering the sequence of rational functions in many variables: 1 q0 q0 q 1 + 1 1 q 0 q1 q 2 + q0 + q 2 q 0 = , q0 + = , q0 + 1 = 1 q1 q1 q1 q2 + 1 q1 + q2 Definition. Define polynomials fn and gn by: q0 + 1 1 q1 + . . .+ 1 qn = fn (q0 , . . . , qn ) gn (q1 , . . . , qn ) (in lowest terms). Examples. (o) f0 (q0 ) = q0 and g0 = 1. (i) f1 (q0 , q1 ) = q0 q1 + 1 and g1 (q1 ) = q1 . (ii) f2 (q0 , q1 , q2 ) = q0 q1 q2 + q0 + q2 and g2 (q1 , q2 ) = q1 q2 + 1. Proposition 1. (a) (Each g is an f ) gn (q1 , · · · , qn ) = fn−1 (q1 , . . . , qn ). (b) (Recursion) fn (q0 , · · · , qn ) = q0 fn−1 (q1 , · · · , qn )+fn−2 (q2 , · · · , qn ). 1 2 Proof. By definition of the polynomials fn−1 and gn−1 , we have: q1 + 1 1 q2 + . . .+ 1 qn = fn−1 (q1 , . . . , qn ) gn−1 (q2 , . . . , qn ) hence: q0 + 1 1 q1 + . . .+ 1 qn = q0 + 1 fn−1 (q1 ,...,qn ) gn−1 (q2 ,...,qn ) = q0 + gn−1 (q2 , . . . , qn ) fn−1 (q1 , . . . , qn ) q0 fn−1 (q1 , . . . , qn ) + gn−1 (q2 , . . . , qn ) fn−1 (q1 , . . . , qn ) and the denominator gives us (a), and the numerator gives us: = fn (q0 , . . . , qn ) = q0 fn−1 (q1 , . . . , qn ) + gn−1 (q2 , . . . , qn ) which, with gn−1 (q2 , . . . , qn ) = fn−2 (q2 , . . . , qn ) from (a), gives us (b). Using the recursion, we can get a few more of these polynomials: Example. (iii) f3 (q0 , q1 , q2 , q3 ) = q0 (q1 q2 q3 + q1 + q3 ) + (q2 q3 + 1) = q0 q1 q2 q3 + q0 q1 + q0 q3 + q2 q3 + 1 (iv) f4 (q0 , q1 , q2 , q3 , q4 ) = = q0 (q1 q2 q3 q4 + q1 q2 + q1 q4 + q3 q4 + 1) + q2 q3 q4 + q2 + q4 = q 0 q1 q 2 q3 q 4 + q0 q 1 q2 + q 0 q1 q 4 + q0 q 3 q4 + q 2 q3 q 4 + q0 + q 2 + q4 Notice that the first term is always the product of all the q’s. In fact: Euler’s Continued Fraction Criterion: X q0 · · · qn X q0 · · · qn f n = q0 · · · q n + + + ··· q q q q q q i i+1 i i+1 j j+1 i |i−j|>1 i.e. the other terms are obtained by removing consecutive pairs of q’s from the product of all the q’s. Example. f 5 = q0 · · · q5 + q2 q3 q4 q5 + q0 q3 q4 q5 + q0 q1 q4 q5 + q0 q1 q2 q5 + q0 q1 q2 q3 + +q4 q5 + q2 q5 + q2 q3 + q0 q5 + q0 q3 + q0 q1 + 1 Proof. The criterion is true for f0 (q0 ) = q0 and f1 (q0 q1 ) = q0 q1 + 1. By the recursive formula (b) above: fn (q0 , . . . , qn ) = q0 fn−1 (q1 , . . . , qn ) + fn−2 (q2 , . . . , qn ) 3 By induction, fn−1 and fn−2 may be assumed to satisfy the criterion, and it then follows from the formula that fn also satisfies the criterion! Corollary 1. (The palindrome corollary) fn (q0 , . . . , qn ) = fn (qn , . . . , q0 ) Proof. Euler’s criterion defines the same polynomial when the order of the variables is reversed. The most important corollary is now the following. Corollary 2. (An even better recursion) fn (q0 , . . . , qn ) = qn fn−1 (q0 , . . . , qn−1 ) + fn−2 (q0 , . . . , qn−2 ) Proof. Using Corollary 1 and Proposition 1 (b), we have: fn (q0 , . . . , qn ) = fn (qn , . . . , q0 ) = qn fn−1 (qn−1 , . . . , q0 )+fn−2 (qn−2 , . . . , q0 ) = = qn fn (q0 , . . . , qn−1 ) + fn−2 (q0 , . . . , qn−2 ) This corollary, together with the initial conditions: f0 (q0 ) = q0 , f1 (q0 , q1 ) = q0 q1 + 1 is going to turn out to be extremely useful. Back to Numbers. We now apply the polynomial results to continued fractions associated to natural numbers. Definition. Given a sequence of natural numbers a0 , a1 , a2 , . . . , let: (a) An := fn (a0 , a1 , . . . , an ) and (b) Bn := gn (a1 , . . . , an ) = fn−1 (a1 , . . . , an ) Gathering together what we have done with polynomials, we have: Proposition 2. (a) The sequence of rational numbers (*) coming from the sequence of natural numbers a0 , a1 , a2 , . . . is: A0 a0 A1 a0 a1 + 1 A2 A3 = , = , , , ... B0 1 B1 a1 B2 B3 (b) The numbers An and Bn satisfy the Fibonacci-like rule: An = an An−1 + An−2 and Bn = an Bn−1 + Bn−2 for n ≥ 2 Proof. (a) is from the definition, and (b) follows from Corollary 2. Example. Taking the sequence 3, 7, 15, 1 again, we have: A0 = 3, A1 = 22, A2 = 15 · 22 + 3 = 333, A3 = 1 · 333 + 22 = 355 B0 = 1, B1 = 7, B2 = 15 · 7 + 1 = 106, B3 = 1 · 106 + 6 = 113 giving the sequence of rational approximations to π that we saw earlier. 4 Another Example. Take the sequence 1, 1, 1, 1, . . . . We have: A0 = 1, A1 = 2, A2 = 2+1 = 3, A3 = 3+2 = 5, A4 = 5+3 = 8, . . . B0 = 1, B1 = 1, B2 = 1+1 = 2, B3 = 2+1 = 3, B4 = 3+2 = 5, . . . which are two copies of the Fibonacci sequence offset by one. The associated sequence of rational numbers: 3 5 8 13 21 1, 2, , , , , ,... 2 3 5 8 13 √ converges pretty rapidly to the golden mean φ = (1 + 5)/2. 2. Convergence. We now have some powerful tools for analyzing the sequence of rational numbers that arise from a continued fraction. Proposition 3. Let a0 , a1 , a2 , . . . be a sequence of natural numbers, and let An and Bn be the natural numbers from Proposition 2. Then: (a) Each quadruple of numbers An , Bn , An+1 , Bn+1 satisfies: An+1 Bn − An Bn+1 = (−1)n (b) Each pair (An , Bn ) is relatively prime so An /Bn is in lowest terms. (c) If the sequence a0 , a1 , a2 , . . . is infinite, then there is a limit: An 1 1 An = α and each α − < < 2 lim n→∞ Bn Bn Bn Bn+1 B n Proof. We prove (a) by induction. First of all: A1 B0 − A0 B1 = (a0 a1 + 1) − a0 a1 = (−1)0 = 1 Using Proposition 2 (b), we have: An+2 Bn+1 −An+1 Bn+2 = (an+2 An+1 +An )Bn+1 −An+1 )(an+2 Bn+1 +Bn ) = An Bn+1 − An+1 Bn = (−1)(An+1 Bn − An Bn+1 ) which proves it. Then (b) follows directly from (a), since any common factor of An and Bn would be a factor of (−1)n . Finally, for (c), we notice that (a) also gives us: An+1 An (−1)n − = Bn+1 Bn Bn Bn+1 which means that the differences of consecutive terms in the sequence: A0 A1 A2 A3 , , , , ... B0 B1 B2 B3 are alternating in sign and decreasing to zero. This implies that the sequence has a limit, and that the limit is between any two consecutive terms, which gives (c). 5 Example. The sequence 1, 2, 2, 2, . . . gives: A0 = 1, A1 = 3, A2 = 7, A3 = 17, A4 = 41 B0 = 1, B1 = 2, B2 = 5, B3 = 12, B4 = 29 which give rational numbers: A0 A1 A2 A3 A4 = 1, = 1.5, = 1.4, ≈ 1.417, ≈ 1.414 B0 B1 B2 B3 B4 √ that alternate above and below 2, and seem to converge to it. Getting to the point (finally) which is to convert α into ai ’s. Case 1. r0 is a rational number > 1 in lowest terms r1 Apply the Euclidean algorithm to the (relatively prime) pair (r0 , r1 ): r2 r0 = a0 + r0 = a0 r1 + r2 ; r1 r1 r1 r3 r1 = a1 r2 + r3 ; = a1 + r2 r2 .. . rn−1 1 rn−1 = an−1 · rn + 1; = an−1 + rn rn rn = an · 1 + 0; rn = an But the right column gives us: 1 1 1 α = a0 + r1 = a0 + 1 = · · · = a0 + 1 a1 + r2 a1 + . r2 r3 .. 1 a + 1 α= n−1 an which exactly tells us that r0 An = r1 Bn for the (finite!) sequence a0 , a1 , . . . , an of natural numbers. α= Bonus! From Proposition 3(a), we get: r0 Bn−1 − r1 An−1 = (−1)n−1 which tells us precisely how to solve the equation: r0 x + r1 y = 1 with integers (x, y). 6 Example. α = 61/48. 61 = 1 · 48 + 13 48 = 3 · 13 + 9 13 = 1 · 9 + 4 9=2·4+1 4=4·1+0 The algorithm terminates with an output of: 1, 3, 1, 2, 4. Rebuilding: A0 1 A1 4 A2 5 A3 14 A4 61 = , = , = , = , = B0 1 B1 3 B2 4 B3 11 B4 48 gives the bonus equation 61 · 11 − 48 · 14 = −1. Case 2. α > 1 is irrational. In this case there is no Euclidean algorithm, but we may define: • [α] is the round down of α to the nearest integer, and • {α} = α − [α] is the fractional part of α. Then we get an infinite sequence of natural numbers a0 , a1 , a2 , . . . : 1 1 , letting a0 = [α] and α1 = α1 {α} 1 1 α1 = a1 + , letting a1 = [α1 ] and α2 = α2 {α1 } 1 1 etc. α2 = a2 + , letting a2 = [α2 ] and α3 = α3 {α2 } Claim. The sequence of rational numbers {An /Bn } coming from the sequence {an } converges to the number α that we started with. α = a0 + Proof of Claim. From the definitions above, we have: 1 1 1 α = a0 + = a0 + = ··· 1 = a0 + α1 a1 + α2 a1 + a +1 1 2 α3 and our polynomial results tell us that: fn+1 (a0 , . . . , an , αn+1 ) α= fn (a1 , . . . , an , αn+1 ) which, by Corollary 2, tells us that: αn+1 fn (a0 , . . . , an ) + fn−1 (a0 , . . . , an−1 ) αn+1 An + An−1 = α= αn+1 fn−1 (a1 , . . . , an ) + fn−2 (a1 , . . . , an−1 ) αn+1 Bn + Bn−1 An But it is easy to check that this number is between B and n since this is true for all n, it follows that α is the limit! An−1 , Bn−1 and 7 3. Periodic Continued Fractions. A purely periodic continued fraction is associated to a sequence of natural numbers of the form: a0 , a1 , . . . , an , a0 , a1 , . . . , an , a0 , a1 , . . . , an , . . . If we let: An n→∞ Bn α = lim then we get: α = a0 + 1 a1 + . 1 . .+ 1 1 an + α from which we may conclude, as in the proof of the claim above, that: αAn + An−1 α= αBn + Bn−1 so α is a root of the quadratic equation: Bn x2 + (Bn−1 − An )x − An−1 = 0 and one can solve for α with the quadratic formula. Examples. (i) Expansions of the form: a, a, a, a, a, . . . give α = a + 1/α so α is a root of the equation x2 − ax − 1 = 0 and since α > 1, we conclude that: √ a + a2 + 4 α= 2 √ (a) When a = 1 we get the golden mean, α = (1 + 5)/2. √ (b) When a = 2k is even, we get α = k + k 2 + 1 so: √ k, 2k, 2k, 2k, . . . is the expansion for k 2 + 1 (ii) Expansions of the form: 2k, k, 2k, k, . . . give α which is a root of: 0 = B1 x2 + (B0 − A1 )x − A0 = k(x2 − 2kx − 2) and therefore: √ √ 2k + 4k 2 + 8 = k + k2 + 2 α= 2 Thus it follows that: √ k, k, 2k, k, 2k, . . . is the expansion of k 2 + 2 8 (iii) Expansions of the form: 2k, 1, 2k, 1, 2k, 1 . . . give α which is a root of x2 − 2kx − 2k = 0, so: √ √ 2k + 4k 2 + 8k α= = k + k 2 + 2k 2 p √ and square roots k 2 + 2k = (k + 1)2 − 1 have expansions: k, 1, 2k, 1, 2k, . . . (iv) Expansions of the form: 2k, 2, 2k, 2, 2k, . . . give α which is a root of 2(x2 − 2kx − k) = 0 so: √ √ 2k + 4k 2 + 4k α= = k + k2 + k 2 √ and square roots of the form k 2 + k have expansions: k, 2, 2k, 2, 2k, . . . Question. Which α > 1 have purely periodic continued fractions? We know each such α is an irrational root of a quadratic equation. Definition. Suppose ax2 + bx + c = 0 is a quadratic equation, and b2 − 4ac is not a perfect square Then we will say that the roots are a conjugate pair (α, α). Example. (a) If b2 − 4ac < 0, then α, α are complex numbers and they are conjugates in the ordinary sense. √ √ (b) The conjugate of the golden mean α = (1 + 5)/2 is (1 − 5)/2. √ √ (c) The conjugate of k (when k is not a perfect square) is − k. Theorem. The α > 1 with purely periodic continued fractions: (i) Are irrational numbers, which (ii) Are roots of a quadratic ax2 + bx + c = 0 with a, b, c ∈ Z and (iii) Have a conjugate root α that satisfies −1 < α < 0. Example. Any α of the form: √ α = k + k 2 + m with 0 < m ≤ 2k is irrational, a root of the quadratic equation: x2 − 2kx − m = 0 and √ has conjugate root α = k − k 2 + m < 0 satisfying −1 < α < 0. 9 4. Pell’s Equation. We seek a solution to an equation of the form: x2 − dy 2 = 1 where d > 0 is a natural number that is not itself a perfect square. √ √ Strategy. Such a solution satisfies (x − y d)(x + y d) = 1 hence: √ x √ 1 1 1 √ √ < 2 x−y d= and 0 < − d = y y x+y d y(x + y d) √ Such good approximations of an irrational ( d) by a rational (x/y) are precisely what continued fraction√ expansions produce. For example, the continued fraction expansion of 7 is 2, 1, 1, 1, 4, 1, 1, 1, 4, . . . and the associated sequence of rational numbers: 2 3 5 8 , , , ,... 1 1 2 3 satisfy, respectively: 22 − 7 · 12 = −3 32 − 7 · 12 = 2 52 − 7 · 22 = −3 82 − 7 · 32 = 1 and (8, 3) is a solution to Pell’s equation. √ A more involved example is 13, whose expansion is: 3, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, . . . producing the sequence: 3 4 7 11 18 , , , , , 1 1 2 3 5 which satisfy: 32 − 13 · 12 42 − 13 · 12 72 − 13 · 22 112 − 13 · 32 182 − 13 · 52 = −4 = 3 = −3 = 4 = −1 and although (18, 5) only solves x2 − 13y 2 = −1, we know how to convert that to a solution to Pell’s equation by squaring: √ √ √ (18 − 5 13)2 = (324 + 25 · 13) − 180 13 = 649 − 180 13 giving the solution (649, 180). 10 Proposition 4. Let: a0 − k, a1 , . . . , an , a0 , a1 , . . . , an , . . . √ be the continued fraction expansion of k 2 + m (m ≤ 2k). Then either: (a) n is even, and: A2n − (k 2 + m) · Bn2 = −1 or else (b) n is odd, and: A2n − (k 2 + m) · Bn2 = 1 In the latter case, Pells’ equation is solved, and in the former: √ √ (An − Bn k 2 + m)2 = (A2n + (k 2 + m)Bn2 ) − 2An Bn k 2 + m solves it with the pair (A2n + (k 2 + m)Bn2 , 2An Bn ). √ Proof. α = k + k 2 + m has periodic expansion: √ a0 , , a1 , . . . , an , a0 , . . . , an , . . . k 2 + m. Then: αAn + An−1 (β + k)An + An−1 β= = αBn + Bn−1 (β + k)Bn + Bn−1 from which it follows that β is a root of the polynomial: Let β = Bn x2 − (kBn + Bn−1 − An )x − (kAn + An−1 ) = 0 √ But β = k 2 + m, so β is a root of the polynomial: x2 − (k 2 + m) = 0 and it follows that: (i) kBn + Bn−1 − An = 0 and (ii) kAn + An−1 = (k 2 + m)Bn . Multiplying (i) through by An gives us: kAn Bn + An Bn−1 − A2n = 0 and Proposition 3(a) gives us An Bn−1 = An−1 Bn + (−1)n−1 hence: kAn Bn +An−1 Bn +(−1)n−1 −A2n = (kAn +An−1 )Bn −A2n +(−1)n−1 = 0 and then (ii) gives: (k 2 + m)Bn2 − A2n = (−1)n which is exactly what we needed to prove. 11 Final Remarks. Our examples from §3 give us the following expansions: √ k 2 + m (k, m) expansion √ 2 (1, 1) 1, 2 √ 3 (1, 2) 1, 1, 2 √ 5 (2, 1) 2, 4 √ 6 (2, 2) 2, 2, 4 √ 7 (2, 3) not from the examples √ 8 (2, 4) 2, 1, 4 10 (3, 1) 3, 6 11 (3, 2) 3, 3, 6 12 (3, 3) 3, 2, 6 13 (3, 4) not from the examples 14 (3, 5) not from the examples 15 (3, 6) 3, 1, 6 √ √ √ √ √ √ and while these expansions are predictable, the others are mysterious: √ (1) 7 has expansion 2, 1, 1, 1, 4 (2) √ 13 has expansion 3, 1, 1, 1, 1, 6 (3) √ 14 has expansion 3, 1, 2, 1, 6 and some really long periods appear for small numbers. For example: √ 46 has expansion 6, 1, 3, 1, 1, 2, 6, 2, 1, 1, 3, 1, 12 leading to an enormous smallest solution to Pell’s equation of (24335, 3588).