CONTINUED FRACTIONS Lecture notes, R. M. Dudley, Math Lecture Series, January 15, 2014

Transcription

1 CONTINUED FRACTIONS Lecture notes, R. M. Dudley, Math Lecture Series, January 5, 204. Basic definitions and facts A continued fraction is given by two sequences of numbers {b n } n 0 and {a n } n. One traditional way to write a continued fraction is: () Q = b 0 + b + b 2 + a a 2 a 3 b 3 + a 4 b 4 + Recall that an infinite sum i= a i means the limit as n (if it exists) of the finite sum n i= a i, and an infinite product Π i= a i means the limit, if it exists, of the finite products a a 2 a n (sometimes with the proviso that none of the factors a i is 0). Similarly, an infinite continued fraction will be the limit, if it exists, of the sequence of numbers Q 0 = b 0, Q = b 0 + a /b, Q 2 = b 0 + a /(b + (a 2 /b 2 )),.... To be more precise, let T j (z) := T j (z;a j,b j ) := a j /(b j + z) for any number z and j =,2,... (here := means equals by definition ). Then the nth convergent of the continued fraction is given by (2) Q n = b 0 + T (T 2 ( (T n (0)) )) if the expression is defined. Here 0/0 is undefined but we define a/0 := for a 0 and b/(c + ) := 0 for any finite b,c. To multiply the continued fraction by a constant c, one can multiply both b 0 and a by c. Clearly, Q n = Q n (b 0,...,b n ;a,...,a n ) is a rational function (quotient of polynomials) of its 2n + arguments. The continued fraction Q = Q({b k } k 0, {a k } k ) will be called convergent to a finite value Q if for n large enough, Q n is defined and finite and lim n Q n = Q. For example, if b 0 = a = b = 0 and a 2 = b 2 = then Q is not defined but Q 2 is well defined and equals 0. A convergent continued fraction is said to terminate at the nth term for the smallest positive integer n such that a n = 0 and Q k is defined for all k > n. Then Q k = Q n for all k > n. Since expressions like () can take up a lot of space, we will follow several other authors in writing the continued fraction () as b 0 + a b + a 2 b 2 +. The following formula for a sequence {X n } n is called the Wallis-Euler recurrence formula: (3) X n = b n X n + a n X n 2, n =,2,.... Theorem (Wallis Euler). For the continued fraction () and n = 0,, 2,..., we have Q n = A n /B n where each of the two sequences {A n } and {B n } satisfies (3) for A :=, B := 0, A 0 := b 0, and B 0 :=. Here Q n = A n /B n means that either both sides are defined and equal, or neither side is defined.

2 2 Proof. Define A n recursively for n by (3) with A n in place of X n, and likewise define B n. Clearly Q n = A n /B n for n = 0 or. To prove this for n 2 by induction, for general sequences and not just fixed sequences {a j }, {b j }, suppose it holds for a given n. By (2), Q n+ equals Q n with T n (0) = a n /b n replaced by T n (T n+ (0)), in other words with b n replaced by b n + (a n+ /b n+ ) (which may be infinite or undefined if b n+ = 0). Then by (3) for n and the induction hypothesis, if b n+ 0, ( ) b n + an+ b n+ A n + a n A n 2 Q n+ = ( b n + an+ = ( A n + a n+ b n+ A n b n+ )B n + a n B n 2 ) / ( B n + a n+ b n+ B n = (b n+ A n + a n+ A n )/(b n+ B n + a n+ B n ) = A n+ /B n+ by (3) for n +, finishing the proof if b n+ 0. Or if b n+ = 0 and a n+ 0 then T n+ (0) = and T n (T n+ (0)) = 0 so Q n+ = Q n and A n+ B n+ = a n+a n a n+ B n = A n B n = Q n = Q n+ as stated. Lastly if a n+ = b n+ = 0 then A n+ /B n+ = 0/0 undefined and T n+ (0) is undefined so Q n+ is also undefined, finishing the proof. Q.E.D. From (3) and Theorem 3, it s clear that A n and B n are polynomials with integer coefficients in the 2n + variables b 0,a,b,...,a n,b n. Actually B n doesn t depend on b 0 or a. For j = 0,,...,n let Q n,j := Q n,j (b 0,b,...,b n ;a,...,a n ) := Q n j (0,b j+,...,b n ;a j+,...,a n ). Then for j = 0,...,n, (4) Q n = Q n (b 0,b,...,b n ;a,...,a n ) = Q j (b 0,...,b j,b j + Q n,j ;a,...,a j ). Theorem 2. Suppose that for given {a i } i and {b i } i 0, and a given nonnegative integer j, the vectors (A j,a j ) and (B j,b j ) are linearly independent in the plane. Then the set of all sequences {X i } i j satisfying (3) for n j + is two-dimensional and has a basis given by {A i } i j and {B i } i j. The linear independence is true if and only if the determinant D j := A j B j B j A j 0, and we have (5) D 0 = and D j = ( ) j a a 2 a j, j. Proof. D 0 = follows from the definitions, and D = b 0 B A = b 0 b b b 0 a = a as stated. To prove (5) for larger j by induction, suppose it holds for a given j. Then by (3) D j+ = A j B j+ B j A j+ = A j (b j+ B j + a j+ B j ) B j (b j+ A j + a j+ A j ) = a j+ D j and (5) follows. The set of all {X i } i j satisfying (3) for the given a i and b i is a vector space since the equations (3) are linear in the X i. Once X j and X j are given, the X i for i > j are uniquely and linearly determined by (3) applied for n = )

3 3 j +,j + 2,.... So the set of solutions is indeed two-dimensional and if the two vectors (A j,a j ),(B j,b j ) are linearly independent, then clearly so are the sequences {A i } i j, {B i } i j. The equivalence of linear independence of vectors and the given determinant not vanishing is well known from linear algebra, and the conclusion follows. Q.E.D. Now, some inequalities for continued fractions will be developed. A continued fraction () will be called fully positive if a n 0 and b n > 0 for all n. Theorem 3 (Euler). If a continued fraction () is fully positive, and if Q converges, then Q 0 Q 2 Q 4 Q Q 5 Q 3 Q. If also a n > 0 for all n then each can be replaced by the strict inequality <. If Q doesn t converge, the inequalities remain true if Q is omitted. Proof. By induction on n, from (2), Q n is a nondecreasing function of b n (increasing if all a j > 0) for n even, and a nonincreasing (resp. decreasing) function of it for n odd. Thus by (4), if j is even, then Q j Q n for all n j, while if j is odd, Q j Q n for all n j, and the Theorem follows. Q.E.D. Theorem 3 implies that for a fully positive convergent continued fraction Q, if two successive convergents Q n and Q n+ are close together, then since Q is between them we have good lower and upper bounds for it. If A is an approximation to Q, the relative error of the approximation is defined as (A/Q). So given ε > 0, to compute Q with a relative error < ε we can take n large enough so that (Q 2n+ /Q 2n ) < ε and let A = Q 2n+. A similar thing happens for continued fractions with terms a j alternating in sign, as follows. Definition. A continued fraction () will be called alternating if the following all hold: (i) b 0 0 and b j for all j. (ii) Let K := i + for the least i such that a i = 0, or K := + if there is no such i. Then for all positive integers j < K, a j = ( ) j+ c j where c j 0 and if j is even, c j <. A monotonicity argument like the proof of Theorem 3 also can be applied to alternating continued fractions. This was noted at least in special cases by A. A. Markov around 920. The following formulation is not proved here, see Dudley (987). As shown there, some functions (e.g. hypergeometric functions) can be evaluated more efficiently, in some ranges, via fully positive or alternating continued fractions than by summing their Taylor series. Theorem 4. For any alternating continued fraction Q, if Q converges, then Q Q 4 Q 5 Q 8 Q Q 7 Q 6 Q 3 Q 2. For a convergent alternating continued fraction Q, and any n, Q is between Q n and Q n+2, so if Q n and Q n+2 are close, then we have good upper and lower bounds for Q. To compute an alternating continued fraction Q to within a relative error < ε, one can find k large enough so that (Q 4k+2 /Q 4k+ ) < ε and approximate Q by Q 4k+2. Alternating continued fractions won t be mentioned further in this talk.

4 4 2. Convergence conditions A continued fraction () and a series j 0 c j are called equivalent if for each n = 0,,2,..., Q n = n j=0 c j. In particular, all Q n must be defined. Clearly, for any continued fraction () with all Q n defined, there is a unique equivalent series, with c 0 = b 0 and c n = Q n Q n for all n. Thus by Theorems and 3, c n = D n /(B n B n ) = ( ) n+ a a 2 a n /(B n B n ). Since convergence of a series doesn t depend on its first term, it follows that: Theorem 5. A continued fraction () with all Q n defined is convergent if and only if, for D j as in (5), the following series converges: D j (6) B j B j. j= A continued fraction () will be called unary if a n = for all n. For such a continued fraction, if b n are small, approaching 0, putting in one more term makes a big difference: b n is small, but b n + (/b n+ ) is large, and so on. So for the continued fraction to converge, b n should not be too small. Theorem 6. A unary continued fraction () with n b n < does not converge. Proof. For a unary continued fraction we have B n Π n j= (+ b j ) for all n, as follows from B = 0, B 0 =, Theorem, which gives B n b n B n + B n 2, and by induction on n. Now j b j < implies Π j= (+ b j ) < since +x e x for x 0. Thus B n remain bounded, and since D j = ±, the terms of the series (6) don t approach 0, so it diverges, Q.E.D. Theorem 7 (Seidel and Stern). A unary, fully positive continued fraction () is convergent if and only if n= b n = +. Proof. Only if follows from Theorem 6. To prove if, suppose n= b n = +. By Theorems 2 and 5, we need to show that the series (6), j= ( )j /(B j B j ) in this case, converges. By Theorem, B j = b j B j + B j 2 > B j 2. Thus the terms of (6) are alternating in sign and decreasing in absolute value, so it is enough to show that their absolute values approach 0. It will be shown by induction on n that for n, (7) B 2n + b (b 2 + b b 2n ), B 2n+ > b + b b 2n+. We have B = b, B 2 = b 2 b +, and B 3 = b b 2 b 3 + b 3 + b > b + b 3, so (7) holds for n =. Assuming (7) for some n, we get from Theorem B 2n+2 = b 2n+2 B 2n+ + B 2n > b 2n+2 b + B 2n, B 2n+3 = b 2n+3 B 2n+2 + B 2n+ > b 2n+3 + B 2n+, which gives the induction step. Since n b n = + and b > 0, either B 2n + or B 2n+ +, so the terms of (6) approach 0 and it converges. Q.E.D. 3. Rational approximation of real numbers A continued fraction () will be called canonical if it is unary, b 0 Z := {0, ±, ±2,...}, and b n is a positive integer for all n. By Theorem 7, every canonical continued fraction is convergent.

5 5 Theorem 8. A number q is rational, q Q, if and only if there is a canonical continued fraction Q such that Q n = q for some n. Each rational number q has exactly two such representations. If it has one with n = 0, or n and b n 2, then the other has n replaced by n +, b n by b n and b n+ :=. Or if q = Q n with n and b n =, the other representation has n replaced by n and b n by b n +. Proof. In a canonical (or any unary) continued fraction, Q n only depends on b 0,b,...,b n, so once we have defined them, the values of b k for k > n won t change Q n, e.g. we can set b k = for k > n. If q Z the theorem holds, with either n = 0 and b 0 = q, or n =, b 0 = q, and b =. If q / Z, let b 0 be the largest integer < q. So we can reduce to the case b 0 = 0 and 0 < q <. Let q = k /m where 0 < k < m are integers and k /m is in lowest terms. There is a unique positive integer b such that /(b + ) < k /m /b. Thus b m /k < b +. If b = m /k (so k = ) let n = and we get Q = q as desired. Otherwise iterate the process and take the unique positive integer b 2 such that /(b 2 + ) < k 2 /m 2 := (m /k ) b /b 2, where k 2 /m 2 is in lowest terms, thus m 2 = k < m. We get a decreasing sequence m j of positive integers, so after finitely many steps, the process ends and gives q = Q n for some n m. It s clear that given one representation q = Q n, q has another representation as described. For any positive integers k and m j for j =,...,k, 0 < Q k (0,m,...,m k ;,,...,), with equality if and only if k = = m. This fact, applied to fractions Q n,j as in (4), implies that there are exactly two representations of q of the given form. Q.E.D. For irrational numbers we have: Theorem 9. There is a one-to-one correspondence between irrational real numbers x and canonical continued fractions Q = x. Proof. Let x 0 be irrational. Let b 0 be the largest integer < x 0. Then b 0 < x 0 < b 0 +. For n a sequence of positive integers b n and irrational numbers x n > will be defined recursively. Let x := /(x 0 b 0 ) >. Then x is irrational. Given x n > irrational there is a unique positive integer b n such that b n < x n < b n +. Let x n+ := /(x n b n ), which is irrational and >. Then x 0 = b 0 + x = b 0 + b + By Theorem 7, the continued fraction Q = b 0 + b + x 2 =... = b 0 + b + b 2 + b 2 + x n. converges. By Theorem 3 and its proof, Q 2n < x 0 < Q 2n+ for all n. So Q = x 0. The positive integers b j are functions of x 0. Let x 0 equal a canonical continued fraction β 0 + β + β. Then it s easily seen that 0 < x 2+ 0 β 0 < so β 0 = b 0. Considering /(x 0 β 0 ) we likewise get β = b, and iterating we get β n = b n for all n. So the irrational numbers are in correspondence with the canonical continued fractions, Q.E.D.

6 6 So, here is an iteration to find a canonical continued fraction for any real number x 0 : for each j = 0,,2,..., let b j be the largest integer x j. If x j b j = 0, then x 0 = Q j, which eventually happens if and only if x 0 is rational, as in Theorem 8. Otherwise, 0 < x j b j <. Let x j+ = /(x j b j ). So x j = b j + (/x j+ ). Then for i, b i, if defined, will be a positive integer, b i.. Example. Let x 0 = π = = Then b 0 = 3, x = /(π 3) = , b = 7, x 2 = /(x b ) =. /( ) = , b 2 = 5, x 3 = / =.00347, b3 =, x 4 = / = 292.6, and b 4 = 292. For a calculator working to some fixed number of digits of accuracy, some digits are lost at each stage, and eventually b k would become incorrect. The canonical continued fraction for π, for which we just found the first few terms, is π = , with no discernible pattern. This continued fraction was found (as far as given) by J. Wallis in his book Tractatus de Algebra in 685 and is stated in Perron, p As rational approximations to π the continued fraction gives Q 0 = 3, Q = 22/7 (over 2200 years ago Archimedes proved 223/7 < π < 22/7), Q 2 = 333/06, Q 3 = 355/3, an approximation found in China over 500 years ago. After that the denominators get much larger: Q 4 = 03993/3302. Note that for canonical continued fractions Q, the quantities A n and B n as defined in Theorem are all integers and B n > 0 for n 0 by (7). Theorem 0. For any canonical continued fraction Q and any n, the fraction Q n = A n /B n is in lowest terms, i.e. A n and B n are relatively prime. Proof. By (5), A n B n A n B n = ( ) n a a n = ( ) n. A common factor of A n and B n would be a factor of ( ) n, which is impossible, Q.E.D. Any real number x can be approximated by a rational number r/s for any positive integer s and some r Z with an error x (r/s) /(2s). It turns out that for suitable s, it s possible to make the error less than /s 2, and that continued fractions will give us such rational approximations, as the next fact shows. Conversely, if the error is less than /(2s 2 ), r/s must equal some Q n from the canonical continued fraction Q (Corollary ). Theorem (Lagrange). Let x 0 be an irrational real number with canonical continued fraction () from Theorem 9 x 0 = Q = Q({b k } k 0 ) := Q({b k } k 0, {a k } k ) where a k = for all k. Then for B k as usual (Theorem ), for all n 0, x 0 Q n < /(B n B n+ ) /B 2 n. Proof. Let x (n) := Q({b k+n } k 0 ) for n = 0,,..., and for any β 0,...,β n, Q n ({β k } n k=0) := Q n (β 0,...,β n ;a,...,a n ) where a = a 2 = = a n =. Then x 0 = Q n ({β k } n k=0 ) where β k = b k for k = 0,,...,n and β n = x (n). It follows from Theorem that for any n =,2,..., (8) x 0 = [x (n) A n + A n 2 ]/[x (n) B n + B n 2 ].

7 7 Therefore x 0 Q n = A n 2B n A n B n 2 B n (x (n) B n + B n 2 ). We have B k 0 for all k, B k > 0 for k 0, and x (n) > b n. By Theorem and (5) we then have x 0 Q n < /(B n B n ). Replacing n by n+ gives the first conclusion. Then noting that for a canonical continued fraction, B 0 B < B 2 <, we get the second bound. Q.E.D. In the example of the canonical continued fraction of π given before Theorem 0 we got Q 3 = 355/3 = where the given digits equal those of π except for the last one. From Theorem, we see that π = θ 3 2 for some θ with θ <. In fact in this case θ = , which is small, as is connected with the unusually large number 292 occurring in the canonical continued fraction expansion. So 355/3 is a remarkably good rational approximation of π in relation to its denominator, which is not very large. According to some websites, this approximation was first discovered in China by Zu or Zhu Chongzhi (spelled Tsu Ch ung Chi on another site), who lived from about 430 to 500 A.D., and a son. It was not improved until about,000 years later. Let x 0 (r/s) = θ/s 2 where r,s Z, s > 0, and r/s is in lowest terms. Then for r/s to equal some Q n for the canonical continued fraction of x 0, we have just seen that θ < is necessary, and it will be shown that θ < /2 is sufficient. The precise necessary and sufficient condition is as follows. Theorem 2. Let x 0 be a real irrational number. Let r,s Z with s > 0 where r/s is in lowest terms. Represent r/s = Q n ({γ k } n k=0 ) by Theorem 8 for integers γ k with γ k > 0 for k and n such that ( ) n (x 0 (r/s)) > 0 (by Theorem 8, we can choose n even or odd). Define [ θ := s 2 ( ) n x 0 r ] > 0. s Let A k and B k for k =,0,...,n be defined as A k and B k for Q n ({γ k } n k=0 ) by Theorem. Then for the canonical continued fraction Q of x 0, r/s = Q m for some m if and only if (9) θ < B n /(B n + B n ), and then n = m. Proof. The case m = 0 can occur if and only if s =. Then r/s = Q 0 if and only if r = b 0. Since B 0 = and B = 0 by definition, B 0 /(B 0 + B ) = and the equivalence holds in this case. So we can assume m and s 2. Then also n (n = 0 is only possible for s = ). To prove if, let w := (B n θb n )/(θb n ). Since 0 < θ < and B n B n > 0, we have w > 0. Solving for θ gives (0) θ = B n /(wb n + B n ). By Theorem 0, r = A n and s = B n. Thus by definition of θ, x 0 = r s + ( )n θ s 2 = r s + ( )n B 2 n B n (wb n + B n ) = A n(wb n + B n ) + ( ) n. B n (wb n + B n )

8 8 Now A n B n + ( ) n = A n B n by (5), so () x 0 = (A n w + A n )/(B n w + B n ). Since θ is irrational, so is w by (0). We have w > if and only if (9) holds, by (0). If w >, then by Theorem 9, we have the canonical continued fraction w = Q({ζ k } k 0 ) for some positive integers ζ k (ζ 0 since w > ). Let γ k := ζ k n for k n +. Then Q({γ k } k 0 ) is a canonical continued fraction, so it converges to some ξ, with ξ (n+) = w from the definitions. Then by (8) applied to ξ and to n+ in place of n, ξ = (A n w+ A n )/(B n w+ B n ) = x 0 by (3.4). Thus A k = A k and B k = B k for k =,0,,...,n, and r/s = Q n. So if holds, with n = m as stated. By Theorem 3, since a j > 0, m is uniquely determined. [To confirm that the definitions of n and θ in Theorem 2 give n = m as opposed to n = m ±, if r/s < x 0 then n is even, and if r/s = Q m then m is also even by Theorem 3. Likewise if Q m = r/s > x 0 then n and m are both odd.] To prove only if, still with m, n, and s 2, let w <, i.e. (9) fails. Since w > 0 we then have γ n + (/w) > γ n +. The canonical continued fraction expansion of γ n + /w is Q({c i } i 0 ) where c 0 = γ n + u with u. I claim that x 0 = Q n ({γ k }n k=0 ) where γ k = γ k for k = 0,,...,n and γ n = γ n + (/w). For this let A k and B k be defined like A k and B k respectively except with γ j in place of γ j, and let Q k = A k /B k for k =,0,...,n. Then A k = A k and B k = B k for k =,0,...,n, and we have by Theorem Q n = wa n wb n = (wγ n + )A n + wa n 2 (wγ n + )B n + wb n 2 = wa n + A n wb n + B n = x 0 by (3.4), proving the claim. It follows that in the canonical continued fraction for x 0, b k = γ k for k = 0,,...,n and b n = c 0 = γ n + u. Then Q n = A n /B n = A n /B n and Q n = (A n + ua n )/(B n + ub n ), while r/s = A n /B n. Now B n = B n B n < B n + ub n = B n. If n 2, or n = and γ >, then B n < B n and r/s cannot be any Q k, since s = B n is not equal to any B k, using Theorem 0. If n = γ =, then s =, a case treated at the beginning of the proof. Thus r/s is no Q k of x 0 in any case. This completes the proof of Theorem 2, Q.E.D. Corollary. If x 0 is irrational and x 0 (r/s) < /(2s 2 ), where r and s are integers and s > 0, then r/s = Q n for some n where Q is the canonical continued fraction of x 0. Proof. Since B n B n, θ < /2 implies θ < B n /(B n + B n ) (9) for any n, Q.E.D. 4. Musical intervals A vibrating string (of a piano, violin, etc.) has a basic frequency b and overtones 2b, 3b,... If basic frequencies b and b have ratio b/b = m/n for small integers m and n they will have overtones in common and sound consonant. The Pythagoreans over 2500 years ago noticed the consonance of octaves b /b = 2/ and fifths b /b = 3/2. Combining these they got a scale, but to keep the scale finite, an approximation is needed somewhere because no power of 3/2 exactly equals a power of 2, for integer powers not both 0.

9 9 To divide an octave into m notes so that the ratio of frequencies of successive notes is a constant, the constant must be 2 /m. To get a good approximation of a fifth, from each note to the kth note above it, we then need to have 2 k/m approximately 3/2. In other words, we need a good rational approximation k/m to λ defined as λ = log 2 (3/2). = The canonical continued fraction expansion of λ has denominators B n forming an increasing sequence of possible m s:,2,5,2,4,... Choosing among these, we can see that m = 5 or less would give a coarse scale of too few notes. Whereas, m = 4 gives too many notes: that many notes crowded into a single octave of a piano would give a fine approximation of fifths, but it wouldn t be worth it. So m = 2 is the adopted solution: in each octave there are 7 white keys and 5 black keys, counting only one of the two ends of the octave. Pianos, for about the past century, have equal temperament where the octave is divided into 2 half-tones, with ratio of successive frequencies 2 /2. This results in approximations et q of the so-called just consonances q = m/n as follows. Here d represents the absolute error d = et q q and d/q is the corresponding relative error. CF(et q ) is the canonical continued fraction of et q and the last column gives a convergent equal to q. Interval q = m n et q d d/q Q k CF(et q ) Octave Q 0 2 Fifth 3/2 2 7/ Q + 2+ Fourth 4/3 2 5/ Q Major third 5/4 2 / Q Minor third 6/5 2 / Q For the fifth and fourth, the approximations by equal temperament are close enough so that the human auditory system generally accepts them as consonances. For the thirds, the approximations are generally accepted by the broad musical community, but some specialists dislike them, e.g. Sethares (998). Only the first two of the following problems are assigned. PROBLEMS. (a) Find the canonical continued fraction of the number e far enough to evaluate q = Q 5 = m/n for positive integers m = A 5 and n = B 5. (b) Evaluate n 2 e q and check that it s less than, as it should be by Theorem. Also check if it s less than /2, so that if we had been given q in advance, we would have known q must equal Q k for some k (Corollary )., a sequence of rational numbers (coming from the Taylor series of e x around 0, evaluated at x = ) which converges to e rather fast. Find S 4 = µ/ν in lowest terms for positive integers µ and ν. (d) Find ν 2 e S 4. In terms of this, is the approximation of e by S 4 as good as those given by canonical continued fractions (as in part (b))? (c) Let S k = k j=0 j! 2. In the table for musical intervals, consider the next interval beyond a fifth, which would have an approximation et q = 2 8/2 = 2 2/3.

10 0 (a) Find q for this case (a ratio of single-digit integers). Hint: note that 2 2/3 = 2/2 /3 and look at the octave and major third lines of the table. (b) Find the canonical continued fraction of 2 2/3 to enough terms and with enough accuracy to do the next part. (c) For what k does Q k = q, as in the last column of the table? (d) Let q = m/n with m = A k and n = B k integers. Find n 2 et q q and verify it s less than. UNASSIGNED PROBLEMS You can do Problem 3 for extra credit. 3. In the canonical continued fraction expansion of 2, (a) Show that b k = 2 for all k. (b) Show that A k+ 5A k + 2A k2 and B k+ 5B k + 2B k 2 for all k For the same continued fraction: (a) Find constants K >, C,D,E,F such that for all n, A n = CK n + D( K) n and B n = EK n + F( K) n. (b) Show that B 2 n (A n /B n ) 2 is never 0 and converges to a non-zero limit ζ as n +. Find ζ. (c) Prove that for any sequences m k and n k of positive integers and q k = m k /n k, n 2 k q k 2 cannot approach 0 as k +. So, the order of approximation of irrational numbers by rationals given by Theorem can t be improved except by some constant factor. 5. Find the smallest possible value of n 2 (m/n) 2 for any positive integers m and n. Notes. Authors mentioned, whose works don t appear in the following short list, are cited in one or more of the references given. Sections 2 and 3 were based mainly on Perron (929). REFERENCES Dudley, R. M. (987). Some inequalities for continued fractions. Math. Computation 49, Jones, W. B., and Thron, W. J. (980). Continued Fractions: Analytic Theory and Applications. Addison-Wesley, Reading, MA. Perron, Oskar (929). Die Lehre von den Kettenbrüchen. Teubner, Leipzig. Sethares, W. A. (998). Tuning, Timbre, Spectrum, Scale. Springer, London. Wall, H. S. (948). Analytic Theory of Continued Fractions. Van Nostrand, Princeton, NJ.