Infinite Continued Fractions

Transcription

1 Infinite Continued Fractions The value of an infinite continued fraction [a 0 ; a, a 2, ] is lim c k, where c k is the k-th convergent k If [a 0 ; a, a 2, ] is an infinite continued fraction with positive terms, then lim k c k exists The convergents of an infinite continued fraction with positive terms converge by oscillation to its value If [a 0 ; a, a 2, ] is an infinite continued fraction with positive terms, then its value is an irrational number Every irrational number x has a unique infinite continued fraction expansion [a 0 ; a, a 2, ] whose terms are given recursively by x 0 x, and a k [x k ], x k+ for k If [a 0 ; a, a 2, ] is an infinite continued fraction, I want to define its value to be the limit of the convergents: lim k c k, where c k is the k-th convergent For this to make sense, I need to show that this limit exists In what follows, take as given an infinite continued fraction [a 0 ; a, a 2, ] Let p 0 a 0, q 0 p a a 0 +, q a p k a k p k + p k 2, a k + 2, k 2, Lemma c k p k (a) c k c k ( )k (b) c k c k 2 a k( ) k 2 Proof For (a), For (b), c k c k p k p k p k p k ( )k c k c k 2 p k p k 2 2 p k 2 p k 2 2 (a kp k + p k 2 ) 2 p k 2 (a k + 2 ) 2 a k (p k 2 p k 2 ) 2 a k ( ) k 2 2 a k( ) k 2

2 c 3 c 5 Lemma c > c 3 > c 5 > > c 4 > c 2 > c 0 That is, the odd convergents get smaller, the even convergents get bigger, and any odd convergent is bigger than any even convergent c c 2 c 4 c 0 Proof If k is even, then This shows that the even terms get bigger If k is odd, then c k c k 2 a k( ) k c k c k 2 a k( ) k 2 > 0, so c k > c k 2 2 < 0, so c k < c k 2 This shows that the odd terms get smaller Finally, I have to show any odd term is bigger than any even term Note that it s true if the terms are adjacent: c 2n+ c 2n ( )(2n+) > 0, so c 2n+ > c 2n Next, do the general case Let c 2n+ be an odd term and let c 2m be an even term Then c 2n+ > c 2n+2m+ > c 2n+2m > c 2m The first inequality is true because odd terms decrease The second inequality is true because I just observed that an odd term is bigger than an adjacent even Finally, the last inequality is true because even terms increase I ve shown that any odd term is bigger than any even term, and that completes the proof Example The first ten convergents for the golden ratio are: 2 > > 625 > 6905 > 688 c c 3 c 5 c 7 c > 6538 > 6 > 5 > 2 c 8 c 6 c 4 c 2 c 0

3 The odd convergents get smaller, the even convergents get bigger, and any odd convergent is bigger than any even convergent Lemma k for all k Proof I ll induct on k q a, so the result holds for k Take k >, and assume it holds for numbers k I ll prove that it holds for k + + a k+ + a k+ k + (k ) k + (k ) 2k k + (k ) k + (The last inequality used k 2) This completes the induction step Theorem Let [a 0 ; a, a 2, ] be an infinite continued fraction with a k > 0 for k, and let c k be the k-th convergent Then lim c k exists k Proof Consider the sequence of odd convergents c > c 3 > c 5 > This is a decreasing sequence of numbers, and it s bounded below by any even convergent, for example A standard result from analysis (see, for example, Theorem 34 of []) asserts that such a sequence must have a limit, so lim c 2k+ exists k Likewise, consider the sequence of even convergents: > c 4 > c 2 > c 0 This is an increasing sequence of numbers that s bounded above by any odd convergent, for example The result from analysis mentioned above says that the sequence has a limit: lim k c 2k exists I have to show that the two limits agree The previous lemma implies that q 2k 2k and q 2k+ 2k +, so 0 c 2k+ c 2k ( )2k+ q 2k q 2k+ (2k)(2k + ) Now let k 0, so by the Squeezing Theorem of calculus, (2k)(2k + ) lim (c 2k+ c 2k ) 0, ie lim c 2k+ lim c 2k k k k Since the odd and even terms approach the same limit, lim k c k exists Knowing this, I m justified in defining What can I say about its value? [a 0 ; a, a 2, ] lim k c k 3

4 Theorem Let [a 0 ; a, a 2, ] be an infinite continued fraction with a k > 0 for k Then [a 0 ; a, a 2, ] is irrational Proof Write x [a 0 ; a, a 2, ] for short I want to show that x is irrational Suppose on the contrary that x p, where p and q are integers I will show this leads to a contradiction q Since the odd convergents are bigger than x and the even convergents are smaller than x, c 2k+ > x > c 2k Then c 2k+ c 2k > x c 2k > 0, ( ) 2k q 2k q 2k+ > x c 2k > 0, q 2k q 2k+ > x c 2k > 0, q 2k q 2k+ > x p 2k q 2k > 0, q 2k+ > xq 2k p 2k > 0, > pq 2k p 2k > 0, q 2k+ q q q 2k+ > pq 2k p 2k q > 0 Notice that this inequality is true for all k, and that the junk in the middle is an integer But q is fixed, q and q 2k+ 2k +, so if I make k sufficiently large eventually q 2k+ will become bigger than q Then q 2k+ will be a fraction less than, and I have an integer pq 2k p 2k q caught between 0 and a fraction less than Since this is impossible, x can t be rational Now I know that every infinite continued fraction made of positive integers represents an irrational number The converse is also true true, and the next result gives an algorithm for computing the continued fraction expansion Theorem Let x R be irrational Let x 0 x, and a k [x k ], x k+ for k 0 Then x [a 0 ; a, a 2, ] Proof Step x k is irrational for k 0 Since x is irrational and x 0 x, the result is true for k 0 Assume that k > 0 and that the result is true for k I want to show that x k is irrational Suppose on the contrary that x k s, where s, t Z Then t s t so x k a k + t x k a k s 4

5 Now all the a k s are clearly integers (since a k [x k ] means they re outputs of the greatest integer function), so a k + t s is the sum of an integer and a rational number Therefore, it s rational, so x k is rational, contrary to the induction hypothesis It follows that x k is irrational By induction, x k is irrational for all k 0 Step 2 The a k s are positive integers for k I already observed that the a k s are integers Let k 0 Since a k [x k ], the definition of the greatest integer function gives But x k is irrational, so a k x k Hence, a k x k < a k + a k < x k < a k +, 0 < <, x k+ >, a k+ [x k+ ] Since k 0, this proves that the a k s are positive integers for k Step 3 lim c k lim [a 0; a,, a k ] x k k First, I ll get a formula for x in terms of the p s, q s, and a s Then I ll find x p k and show that it s less than something which goes to 0 To get the formula for x, start with x k+ Do some algebra to get Write out this equation for a few values of k: x k a k + x k+ x 0 a 0 + x x a + x 2 x 2 a 2 + x 3 x 3 a 3 + x 4 Substituting the second equation of the set into the first gives x 0 a a + x 2

6 Substituting x 2 a 2 + x 3 into this equation gives x 0 a 0 + a + a 2 + x 3 Substituting x 3 a 3 + x 4 into this equation gives x 0 a 0 + a + a 2 + a 3 + x 4 You get the idea In general, x x 0 a 0 + a + a 2 + [a 0 ; a, a 2,, a k, x k+ ] + a k + x k+ In other words, the x k s are the infinite tails of the continued fraction Recall the recursion formulas for convergents: p k a k p k + p k 2 and a k + 2 The right sides only involve terms up to a k (and p s and q s of smaller indices) Therefore, the fractions [a 0 ; a, a 2,, a k, x k+ ] and [a 0 ; a, a 2,, a k, a k+, ] have the same p s and q s through index k Using the recursion formula for convergents, I get Therefore, Take absolute values: Now x x 0 [a 0 ; a, a 2,, a k, x k+ ] x k+p k + p k x k+ + x p k x k+p k + p k x k+ + p k x k+p k + p k x k+ p k p k (x k+ + ) p k p k (x k+ + ) ( ) k (x k+ + ) x p k (x k+ + ) x k+ > [x k+ ] a k+, so x k+ + > a k

7 Therefore, x k+ + < +, <, (x k+ + ) + x p k < + By an earlier lemma, k and + k +, so x p k < + Now lim 0, so by the Squeezing Theorem k k(k + ) lim k x p k 0 k(k + ) This implies that p k lim x k Example I ll compute the continued fraction expansion of π Here are the first two steps: x Continuing in this way, I obtain: x 0 π, a 0 [x 0 ] [π] 3 x 0 a , a [x ] 7 x k a k p k c k π Example I ll compute the continued fraction expansion of 5 x x 0 5 a 0 [ 5] [223607] 2 5 2, a [ ] [423607]

8 Here are the first 5 terms: x a In fact, the continued fraction expansion for 5 is [2; 4, 4, 4, ] Theorem The continued fraction expansion of an irrational number is unique Proof Suppose [a 0 ; a, a 2, ] x [b 0 ; b, b 2, ] are two continued fractions for the irrational number x, where a k, b k Z and a k, b k for k I want to show that a k b k for all k Recall that: The even convergents are smaller than x The odd convergents are greater than x Therefore, Now a 0 < x < a 0 + a a a a 0 + a a 0 + x < a 0 + Thus, a 0 is an integer less than x, and the next larger integer a 0 + is greater than x This means that a 0 [x] The same reasoning applies to the b s Therefore, b 0 [x], so a 0 b 0 Hence, [a 0 ; a, a 2, ] [b 0 ; b, b 2, ] a 0 + a + a 2 + a + a 2 + b 0 + b + b + b 2 + b 2 + [a ; a 2, a 3, ] [b ; b 2, b 3, ] 8

9 I can continue in the same way to show that a k b k for all k Here s a summary of some of the important results on infinite continued fractions: An irrational number has a unique infinite continued fraction expansion 2 The algorithm for computing the continued fraction expansion of an irrational number x is: Then x 0 x, and a k [x k ], x k+ x [a 0 ; a, a 2, ] for k 3 If [a 0 ; a, a 2, ] is the continued fraction expansion of an irrational number, then a k is a positive integer for k 4 If x [a 0 ; a, a 2, ] is the continued fraction expansion of an irrational number and {p k } and { } are defined by the recursion formulas for convergents, then x p k < + Example Here is the continued fraction expansion for e + π Now Thus, in this case, x k a k p k c k e + π while e + π < [] Walter Rudin, Principles of Mathematical Analysis (3 rd edition) New York: McGraw-Hill Book Company, 976 c 2008 by Bruce Ikenaga 9