Theory Practice References. Farey Sequences. Some practical consequences of the properties of the Farey sequence. Maximilian Christ.

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1 Some practical consequences of the properties of the Farey sequence January 12, 2014

2 Overview Generating of the sequence Applications Papers Books

3 Theorem 3 The Farey Sequence F n converges to [0, 1] Q.

4 Theorem 3 The Farey Sequence F n converges to [0, 1] Q. Proof of Theorem 3. This follows from lim sup F n = n lim inf n F n = n=1 m=n n=1 m=n F m = F m = n=1 F n n=1 m=1 F m = (1) F n n=1

5 Theorem 3 The Farey Sequence F n converges to [0, 1] Q. Proof of Theorem 3. This follows from lim sup F n = n lim inf n F n = So the limit exists and n=1 m=n n=1 m=n lim F n = n F m = F m = n=1 F n n=1 m=1 F m = (1) F n n=1 F n = [0, 1] Q. (2) n=1

6 Theorem 4 () Let a 1 F n. < a 2 < a 3 b 3 be three consecutive and reduced frictions from

7 Theorem 4 () Let a 1 F n. Then < a 2 < a 3 b 3 be three consecutive and reduced frictions from a 2 = a 1 + a 3 + b 3. (3) This means that a 2 is the mediant of a 1 and a 3 b 3

8 Theorem 4 () Let a 1 F n. Then < a 2 < a 3 b 3 be three consecutive and reduced frictions from a 2 = a 1 + a 3 + b 3. (3) This means that a 2 is the mediant of a 1 and a 3 b 3 Proof of Theorem 4. Proof by Induction, for n = 2 the claim is true. Let it be true for n, then we have to show that it is also true for n + 1.

9 Theorem 4 () Let a 1 F n. Then < a 2 < a 3 b 3 be three consecutive and reduced frictions from a 2 = a 1 + a 3 + b 3. (3) This means that a 2 is the mediant of a 1 and a 3 b 3 Proof of Theorem 4. Proof by Induction, for n = 2 the claim is true. Let it be true for n, then we have to show that it is also true for n + 1. If we want to create F n+1 from F n, we have to add all frictions n + 1 coprime. p n+1 with p and

10 Theorem 4 () Let a 1 F n. Then < a 2 < a 3 b 3 be three consecutive and reduced frictions from a 2 = a 1 + a 3 + b 3. (3) This means that a 2 is the mediant of a 1 and a 3 b 3 Proof of Theorem 4. Proof by Induction, for n = 2 the claim is true. Let it be true for n, then we have to show that it is also true for n + 1. If we want to p n+1 create F n+1 from F n, we have to add all frictions with p and n + 1 coprime. Precisely, { } r F n+1 = F n n + 1 r = 1,..., n ; r n + 1 = F n N n+1 (4)

11 Proof of Theorem 4. With the third Part 3.) of Theorem 1 we can write the set of new frictions N n+1 of F n+1 in another way

12 Proof of Theorem 4. With the third Part 3.) of Theorem 1 we can write the set of new frictions N n+1 of F n+1 in another way { } r N n+1 := n + 1 r = 1,..., n ; r n + 1 { a + c = a b + d b, c } (5) d F b + d n ; gcd(a + c, b + d) = n + 1.

13 Proof of Theorem 4. With the third Part 3.) of Theorem 1 we can write the set of new frictions N n+1 of F n+1 in another way { } r N n+1 := n + 1 r = 1,..., n ; r n + 1 { a + c = a b + d b, c } (5) d F b + d n ; gcd(a + c, b + d) = n + 1. This is because Theorem 1 3.) states that the mediant is the unique friction with the smallest denominator between 2 frictions.

14 Proof of Theorem 4. With the third Part 3.) of Theorem 1 we can write the set of new frictions N n+1 of F n+1 in another way { } r N n+1 := n + 1 r = 1,..., n ; r n + 1 { a + c = a b + d b, c } (5) d F b + d n ; gcd(a + c, b + d) = n + 1. This is because Theorem 1 3.) states that the mediant is the unique friction with the smallest denominator between 2 frictions.this means, that the frictions from N n+1 with the denominator n + 1 can be expressed as a mediant of two frictions from F n.

15 Proof of Theorem 4. Let now o denote a friction from F n and n a friction from N n+1. The o stands for old and the n for new friction.

16 Proof of Theorem 4. Let now o denote a friction from F n and n a friction from N n+1. The o stands for old and the n for new friction. If we choose 3 consecutive frictions from F n+1, we have 3 different possible settings: o o o, o n o, n o n

17 Proof of Theorem 4. Let now o denote a friction from F n and n a friction from N n+1. The o stands for old and the n for new friction. If we choose 3 consecutive frictions from F n+1, we have 3 different possible settings: o o o, o n o, n o n Only in the last case n o n we have to show the mediant property.

18 Proof of Theorem 4. Let now o denote a friction from F n and n a friction from N n+1. The o stands for old and the n for new friction. If we choose 3 consecutive frictions from F n+1, we have 3 different possible settings: o o o, o n o, n o n Only in the last case n o n we have to show the mediant property.in the case o o o it is clear because of the induction assumption. The case o n o is covered because n is, by construction of F n+1 from F n, the mediant of its neighbours.

19 Proof of Theorem 4. Let now o denote a friction from F n and n a friction from N n+1. The o stands for old and the n for new friction. If we choose 3 consecutive frictions from F n+1, we have 3 different possible settings: o o o, o n o, n o n Only in the last case n o n we have to show the mediant property.in the case o o o it is clear because of the induction assumption. The case o n o is covered because n is, by construction of F n+1 from F n, the mediant of its neighbours. But in the case n o n we take another two frictions o F n so that o n o n o and show that o fulfils the mediant property in respect to both n.

20 Proof of Theorem 4. So now be a 1, a 3 b 3 N and a 2 F n so that a 1 < a 2 < a 3 b 3

21 Proof of Theorem 4. So now be a 1, a 3 b 3 N and a 2 F n so that a 1 < a 2 < a 3 b 3. From a 2 F n and the induction assumption follows that there exists neighbours x y, z w F n such that x y < a 1 < a 2 < a 1 < z w and

22 Proof of Theorem 4. So now be a 1, a 3 b 3 N and a 2 F n so that a 1 < a 2 < a 3 b 3. From a 2 F n and the induction assumption follows that there exists neighbours x y, z w F n such that x y < a 1 < a 2 < a 1 < z w and a 2 = x + z y + w. (6)

23 Proof of Theorem 4. So now be a 1, a 3 b 3 N and a 2 F n so that a 1 < a 2 < a 3 b 3. From a 2 F n and the induction assumption follows that there exists neighbours x y, z w F n such that x y < a 1 < a 2 < a 1 < z w and Because of a 1, a 3 b 3 N follows that a 2 = x + z y + w. (6) a 1 = 2x + z 2y + w and a 3 = x + 2z b 3 y + 2w. (7)

24 Proof of Theorem 4. So now be a 1, a 3 b 3 N and a 2 F n so that a 1 < a 2 < a 3 b 3. From a 2 F n and the induction assumption follows that there exists neighbours x y, z w F n such that x y < a 1 < a 2 < a 1 < z w and Because of a 1, a 3 b 3 The mediant of a 1, a 3 b 3 N follows that a 2 = x + z y + w. (6) a 1 = 2x + z 2y + w and a 3 = x + 2z b 3 y + 2w. (7) is then a 1 + a 3 + b 3 = 3x + 3z 3y + 3w = a 2. (8)

25 Theorem 5 The Farey Sequence F n grows quadratically. This means: F n = O(n 2 ). (9)

26 Theorem 5 The Farey Sequence F n grows quadratically. This means: F n = O(n 2 ). (9) Proof of Theorem 5. We use Euler s totient function, which counts the number of coprime numbers of n. Precisely, ϕ(n) := { k N 1 k n, gcd(k, n) = 1 }. (10) Here gcd(k, n) describes the greatest common divisor of k and n.

27 Theorem 5 The Farey Sequence F n grows quadratically. This means: F n = O(n 2 ). (9) Proof of Theorem 5. We use Euler s totient function, which counts the number of coprime numbers of n. Precisely, ϕ(n) := { k N 1 k n, gcd(k, n) = 1 }. (10) Here gcd(k, n) describes the greatest common divisor of k and n.its clear that ϕ(k) < c k (11) for 0 c 1.

28 Proof of Theorem 5. The Farey Sequence F n contains every reduced friction with denominators smaller or equal n. If we want to create F n+1 we have to add all reduced frictions with a denominator that is coprime to n + 1.

29 Proof of Theorem 5. The Farey Sequence F n contains every reduced friction with denominators smaller or equal n. If we want to create F n+1 we have to add all reduced frictions with a denominator that is coprime to n + 1.Therefore F n+1 = F n + ϕ(n + 1) (12) for n 2. For n = 1 we get F 1 = 1 + ϕ(1).

30 Proof of Theorem 5. The Farey Sequence F n contains every reduced friction with denominators smaller or equal n. If we want to create F n+1 we have to add all reduced frictions with a denominator that is coprime to n + 1.Therefore F n+1 = F n + ϕ(n + 1) (12) for n 2. For n = 1 we get F 1 = 1 + ϕ(1).we can now show the assertion: F n = 1 + n ϕ(k) c Equation (11) k=1 n k = c 2 n (n + 1) = O(n2 ). k=1 (13)

31 Generating of the sequence Applications Because of the mediant property exists a simple Algorithm to calculate every F n efficiently.

32 Generating of the sequence Applications Because of the mediant property exists a simple Algorithm to calculate every F n efficiently. If we know two of three consecutive neighbours somewhere in the Farey Sequence, we can calculate the third one with the mediant property.

33 Generating of the sequence Applications Because of the mediant property exists a simple Algorithm to calculate every F n efficiently. If we know two of three consecutive neighbours somewhere in the Farey Sequence, we can calculate the third one with the mediant property. Besides this, the first two frictions in the Farey Sequence F n are always 0 n and 1 n, so beginning from this two frictions we calculate the third one, then the fourth one and so on.

34 Generating of the sequence Applications For the problem of finding the k-th element of F n, there exists an algorithm that runs in time O (n log(n)) and uses space O(n).

35 Generating of the sequence Applications For the problem of finding the k-th element of F n, there exists an algorithm that runs in time O (n log(n)) and uses space O(n). The same bounds hold for the problem of determining the rank in the Farey Sequence of a given fraction.

36 Generating of the sequence Applications For the problem of finding the k-th element of F n, there exists an algorithm that runs in time O (n log(n)) and uses space O(n). The same bounds hold for the problem of determining the rank in the Farey Sequence of a given fraction. A more complicated solution can reduce the space to O ( n 1 3 log(log(n)) 2 3 ), and, for the problem of determining the rank of a fraction, reduce the time to O(n).

37 Generating of the sequence Applications For the problem of finding the k-th element of F n, there exists an algorithm that runs in time O (n log(n)) and uses space O(n). The same bounds hold for the problem of determining the rank in the Farey Sequence of a given fraction. A more complicated solution can reduce the space to O ( n 1 3 log(log(n)) 2 3 ), and, for the problem of determining the rank of a fraction, reduce the time to O(n). For more information see the paper of C. E. Patrascu and M. Patrascu, Computing Order Statistics in the Farey Sequence.

38 Applications I Generating of the sequence Applications There are various applications for the Farey-Sequence, which justify a deeper analysis:

39 Applications I Generating of the sequence Applications There are various applications for the Farey-Sequence, which justify a deeper analysis: The net resistance of n resistors with resistances R 1, R 2..., R n connected in series is given by R series = R 1 + R R n (14) whereas the net resistance of these resistors connected in parallel is given by 1 R parallel = 1 R R (15) R n

40 Applications I Generating of the sequence Applications There are various applications for the Farey-Sequence, which justify a deeper analysis: The net resistance of n resistors with resistances R 1, R 2..., R n connected in series is given by R series = R 1 + R R n (14) whereas the net resistance of these resistors connected in parallel is given by 1 R parallel = 1 R R (15) R n We can employ the Farey Sequence to establish strict upper and lower bounds for the order of the set of equivalent resistances for a circuit constructed from equal resistors combined in series and in parallel.

41 Applications II Generating of the sequence Applications Image processing and shape analysis: In order to describe an object boundary we use a sequence of straight line segments, which gives a convenient polygonal representation. For an efficient approximation, the successive edges, which are almost collinear, are merged. The are used to discretizise these edge slopes. This is done by finding the fraction of F n closest to a given p q efficiently.

42 Applications III Generating of the sequence Applications There exists a link between the Farey Sequence and the Riemann hypothesis ( The real part of every non-trivial zero of the Riemann zeta function ζ(s) = 1 n is 1 s 2 ). n=1

43 Applications III Generating of the sequence Applications There exists a link between the Farey Sequence and the Riemann hypothesis ( The real part of every non-trivial zero of the Riemann zeta function ζ(s) = n=1 1 n s is 1 2 ). Farey Sequences are used in two equivalent formulations of the Riemann hypothesis, which are and m n k=1 d 2 k,n = O(nr ) r > 1 (16)

44 Applications III Generating of the sequence Applications There exists a link between the Farey Sequence and the Riemann hypothesis ( The real part of every non-trivial zero of the Riemann zeta function ζ(s) = n=1 1 n s is 1 2 ). Farey Sequences are used in two equivalent formulations of the Riemann hypothesis, which are and m n k=1 m n k=1 Here d k,n = f k,n k m n and m n = F n. d 2 k,n = O(nr ) r > 1 (16) d k,n = O(n r ) r > 1/2. (17)

45 Applications IV Generating of the sequence Applications Ford Circles are a geometric representation of fractions. Therefore exists a strong connection between these circles and frictions from the Farey Sequence. In Addition it is even possible to prove most Theorems about with geometrical methods by using Ford circles.

46 Applications IV Generating of the sequence Applications Ford Circles are a geometric representation of fractions. Therefore exists a strong connection between these circles and frictions from the Farey Sequence. In Addition it is even possible to prove most Theorems about with geometrical methods by using Ford circles. Definition 4 For every rational number p q in lowest terms, the Ford circle C(p, q) is the circle with center ( p q, 1 1 ) and radius. This 2q 2 2q 2 means that C(p, q) is the circle tangent to the x-axis at x = p q 1 with radius. 2q 2

47 Applications V Generating of the sequence Applications Example 4 The following image shows some fractions with their responding Ford circles:

48 I Papers Books S. Das, K. Halder, S. Pratihar, P. Bhowmick Properties of Farey Sequence and their Applications to Digital Image Processing KInternational Journal of Pattern Recognition and Artificial Intelligence, Vol. 27(7), S.Kanemitsu and M. Yoshimoto Farey Series and the Riemann hypothesis Acta Arithmetica, Vol.75(4), 1996 N. Routledge Computing Farey Series The Mathematical Gazette, Vol. 92 (No. 523), 2008

49 II Papers Books J. Pawlewicz Order Statistics in the in Sublinear Time Lecture Notes in Computer Science, Vol. 4698, 2007 J. Pawlewicz, M. Patrascu Order Statistics in the in Sublinear Time and Counting Primitive Lattice Points in Polygons Algorithmica, Vol. 55(2), 2009 C. E. Patrascu, M. Patrascu Computing Order Statistics in the Farey Sequence Lecture Notes in Computer Science, Vol. 3076, 2004

50 III Papers Books S. A. Khan Farey sequences and resistor networks Proceedings - Mathematical Sciences, Vol. 122(2), 2012

51 IV Papers Books S. Guthery A Motif of Mathematics: History and Application of the Mediant and the Farey Sequence. CreateSpace Independent Publishing Platform, G.H. Hardy, E.M. Wright An Introduction to the of Numbers (Fifth Edition) Oxford University Press, 1979

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