Sums. ITT9131 Konkreetne Matemaatika

Transcription

1 Sums ITT9131 Konkreetne Matemaatika Chapter Two Notation Sums and Recurrences Manipulation of Sums Multiple Sums General Methods Finite and Innite Calculus Innite Sums

2 Contents 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

3 Next section 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

4 Sequences Denition A sequence of elements of a set A is any function f : N A, where N is set of natural numbers. Notations used: f = {a n }, where a n = f (n) {a n } n N {a n } a n is called n-th term of a sequence f

5 Sequences Denition A sequence of elements of a set A is any function f : N A, where N is set of natural numbers. Notations used: f = {a n }, where a n = f (n) {a n } n N {a n } a n is called n-th term of a sequence f

6 Sequences Denition A sequence of elements of a set A is any function f : N A, where N is set of natural numbers. Notations used: f = {a n }, where a n = f (n) {a n } n N {a n } a n is called n-th term of a sequence f

7 Sequences Denition A sequence of elements of a set A is any function f : N A, where N is set of natural numbers. Notations used: f = {a n }, where a n = f (n) {a n } n N {a n } a n is called n-th term of a sequence f Example a 0 = 0, a 1 = 1 2 3, a 2 = 2 3 4, a 3 = 3 4 5, 0, or 1 6, 1 6, 3 20, 2 15,, n (n + 1)(n + 2),

8 Sequences Denition A sequence of elements of a set A is any function f : N A, where N is set of natural numbers. Notations used: f = {a n }, where a n = f (n) {a n } n N {a n } a n is called n-th term of a sequence f Notation f (n) = a n = n (n + 1)(n + 2) or n (n + 1)(n + 2)

9 Sets of indexes N set of indexes of the sequence f = {a n } n N Any countably innite set can be used for index. Examples of other frequently used indexes are: N + = N {0} N N K N, where K is any nite subset of N Z N {1, 3, 5, 7,...} = ODD N {0, 2, 4, 6,...} = EVEN N

10 Sets of indexes N set of indexes of the sequence f = {a n } n N Any countably innite set can be used for index. Examples of other frequently used indexes are: N + = N {0} N N K N, where K is any nite subset of N Z N {1, 3, 5, 7,...} = ODD N {0, 2, 4, 6,...} = EVEN N

11 Sets of indexes N set of indexes of the sequence f = {a n } n N Any countably innite set can be used for index. Examples of other frequently used indexes are: N + = N {0} N N K N, where K is any nite subset of N Z N {1, 3, 5, 7,...} = ODD N {0, 2, 4, 6,...} = EVEN N A B denotes that sets A and B are of the same cardinality, i.e. A = B.

12 Sets of indexes N set of indexes of the sequence f = {a n } n N Any countably innite set can be used for index. Examples of other frequently used indexes are: N + = N {0} N N K N, where K is any nite subset of N Z N {1, 3, 5, 7,...} = ODD N {0, 2, 4, 6,...} = EVEN N Two sets A and B have the same cardinality if there exists a bijection, that is, an injective and surjective function, from A to B. (See for detailed explanation)

13 Finite sequence Finite sequence of elements of a set A is a function f : K A, where K is set a nite subset of natural numbers For example: f : {1,2,3,4,,n}n A, n N Special case: n = 0, i.e. empty sequence: f (/0) = e

14 Domain of the sequence a n = f : T A n (n 2)(n 5) Domain of f is T = N {2,5}

15 Next section 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

16 Notation For a nite set K = {1,2,,m} and a given sequence f : K R with f (n) = a n we write m a k = a 1 + a a m k=1 Alternative notations m k=1 a k = 1 k m a k = k {1,,m} a k = a k K

17 Notation For a nite set K = {1,2,,m} and a given sequence f : K R with f (n) = a n we write m a k = a 1 + a a m k=1 Alternative notations m k=1 a k = 1 k m a k = k {1,,m} a k = a k K

18 Warmup: What does this notation mean? 0 q k k=4 Options: 1 0 k=4 q k = q 4 + q 3 + q 2 + q 1 + q 0 = k {4,3,2,1,0} q k = 4 k=0 q k. This seems the sensible thingbut: 2 4 k 0 q k = 0 also looks like a feasible interpretationbut: 3 If n q k = q k q k, k=m k n k<m (provided the two sums on the right-hand side exist nite) then 0 k=4 q k = k 0 q k k<4 q k = q 1 q 2 q 3.

19 Warmup: What does this notation mean? 0 q k k=4 Options: 1 0 k=4 q k = q 4 + q 3 + q 2 + q 1 + q 0 = k {4,3,2,1,0} q k = 4 k=0 q k. This seems the sensible thingbut: 2 4 k 0 q k = 0 also looks like a feasible interpretationbut: 3 If n q k = q k q k, k=m k n k<m (provided the two sums on the right-hand side exist nite) then 0 k=4 q k = k 0 q k k<4 q k = q 1 q 2 q 3.

20 Warmup: What does this notation mean? 0 q k k=4 Options: 1 0 k=4 q k = q 4 + q 3 + q 2 + q 1 + q 0 = k {4,3,2,1,0} q k = 4 k=0 q k. This seems the sensible thingbut: 2 4 k 0 q k = 0 also looks like a feasible interpretationbut: 3 If n q k = q k q k, k=m k n k<m (provided the two sums on the right-hand side exist nite) then 0 k=4 q k = k 0 q k k<4 q k = q 1 q 2 q 3.

21 Warmup: What does this notation mean? 0 q k k=4 Options: 1 0 k=4 q k = q 4 + q 3 + q 2 + q 1 + q 0 = k {4,3,2,1,0} q k = 4 k=0 q k. This seems the sensible thingbut: 2 4 k 0 q k = 0 also looks like a feasible interpretationbut: 3 If n q k = q k q k, k=m k n k<m (provided the two sums on the right-hand side exist nite) then 0 k=4 q k = k 0 q k k<4 q k = q 1 q 2 q 3.

22 Warmup: Interpreting the Σ-notation Compute {0 k 5} a k and {0 k2 5} a k 2. First sum {0 k 5} = {0,1,2,3,4,5} : thus, {0 k 5} a k = a 0 + a 1 + a 2 + a 3 + a 4 + a 5. Second sum {0 k 2 5} = {0,1,2, 1, 2} : thus, {0 k 5} a k 2 = a a a 2 2+a ( 1) 2 + a ( 2) 2 = a 0 + 2a 1 + 2a 2.

23 Warmup: Interpreting the Σ-notation Compute {0 k 5} a k and {0 k2 5} a k 2. First sum {0 k 5} = {0,1,2,3,4,5} : thus, {0 k 5} a k = a 0 + a 1 + a 2 + a 3 + a 4 + a 5. Second sum {0 k 2 5} = {0,1,2, 1, 2} : thus, {0 k 5} a k 2 = a a a 2 2+a ( 1) 2 + a ( 2) 2 = a 0 + 2a 1 + 2a 2.

24 Warmup: Interpreting the Σ-notation Compute {0 k 5} a k and {0 k2 5} a k 2. First sum {0 k 5} = {0,1,2,3,4,5} : thus, {0 k 5} a k = a 0 + a 1 + a 2 + a 3 + a 4 + a 5. Second sum {0 k 2 5} = {0,1,2, 1, 2} : thus, {0 k 5} a k 2 = a a a 2 2+a ( 1) 2 + a ( 2) 2 = a 0 + 2a 1 + 2a 2.

25 Warmup: Interpreting the Σ-notation Compute {0 k 5} a k and {0 k2 5} a k 2. First sum {0 k 5} = {0,1,2,3,4,5} : thus, {0 k 5} a k = a 0 + a 1 + a 2 + a 3 + a 4 + a 5. Second sum {0 k 2 5} = {0,1,2, 1, 2} : thus, {0 k 5} a k 2 = a a a 2 2+a ( 1) 2 + a ( 2) 2 = a 0 + 2a 1 + 2a 2.

26 Next section 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

27 Sums and Recurrences Computation of any sum S n = n a k k=1 can be presented in the recursive form: S 0 = a 0 S n = S n 1 + a n Techniques from CHAPTER ONE can be used for nding closed formulas for evaluating sums.

28 Recalling repertoire method Given g(0) = α g(n) = Φ(g(n 1)) + Ψ(β,γ,...) for n > 0. where Φ and Ψ are linear, for example if g(n) = λ 1 g 1 (n) + λ 2 g 2 (n) then Φ(g(n)) = λ 1 Φ(g 1 (n)) + λ 2 Φ(g 2 (n)). Closed form is g(n) = αa(n) + βb(n) + γc(n) + (1) Functions A(n),B(n),C(n),... could be found from the system of equations α 1 A(n) + β 1 B(n) + γ 1 C(n) + = g 1 (n). =... α m A(n) + β m B(n) + γ m C(n) + = g m (n) where α i,β i,γ i are constants committing (1) and recurrence relationship for the repertoire case g i (n) and any n.

29 Recalling repertoire method Given g(0) = α g(n) = Φ(g(n 1)) + Ψ(β,γ,...) for n > 0. where Φ and Ψ are linear, for example if g(n) = λ 1 g 1 (n) + λ 2 g 2 (n) then Φ(g(n)) = λ 1 Φ(g 1 (n)) + λ 2 Φ(g 2 (n)). Closed form is g(n) = αa(n) + βb(n) + γc(n) + (1) Functions A(n),B(n),C(n),... could be found from the system of equations α 1 A(n) + β 1 B(n) + γ 1 C(n) + = g 1 (n). =... α m A(n) + β m B(n) + γ m C(n) + = g m (n) where α i,β i,γ i are constants committing (1) and recurrence relationship for the repertoire case g i (n) and any n.

30 Recalling repertoire method Given g(0) = α g(n) = Φ(g(n 1)) + Ψ(β,γ,...) for n > 0. where Φ and Ψ are linear, for example if g(n) = λ 1 g 1 (n) + λ 2 g 2 (n) then Φ(g(n)) = λ 1 Φ(g 1 (n)) + λ 2 Φ(g 2 (n)). Closed form is g(n) = αa(n) + βb(n) + γc(n) + (1) Functions A(n),B(n),C(n),... could be found from the system of equations α 1 A(n) + β 1 B(n) + γ 1 C(n) + = g 1 (n). =... α m A(n) + β m B(n) + γ m C(n) + = g m (n) where α i,β i,γ i are constants committing (1) and recurrence relationship for the repertoire case g i (n) and any n.

31 Next subsection 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

32 Example 1: arithmetic sequence Arithmetic sequence: a n = a + bn Recurrent equation for the sum S n = a 0 + a 1 + a a n : S 0 = a S n = S n 1 + (a + bn), for n > 0. Let's nd a closed form for a bit more general recurrent equation: R 0 = α R n = R n 1 + (β + γn), for n > 0.

33 Example 1: arithmetic sequence Arithmetic sequence: a n = a + bn Recurrent equation for the sum S n = a 0 + a 1 + a a n : S 0 = a S n = S n 1 + (a + bn), for n > 0. Let's nd a closed form for a bit more general recurrent equation: R 0 = α R n = R n 1 + (β + γn), for n > 0.

34 Evaluation of terms R n = R n 1 + (β + γn) R 0 = α R 1 = α + β + γ R 2 = α + β + γ + (β + 2γ) = α + 2β + 3γ R 3 = α + 2β + 3γ + (β + 3γ) = α + 3β + 6γ Observation R n = A(n)α + B(n)β + C(n)γ A(n), B(n), C(n) can be evaluated using repertoire method: we will consider three cases 1 R n = 1 for all n 2 R n = n for all n 3 R n = n 2 for all n.

35 Evaluation of terms R n = R n 1 + (β + γn) R 0 = α R 1 = α + β + γ R 2 = α + β + γ + (β + 2γ) = α + 2β + 3γ R 3 = α + 2β + 3γ + (β + 3γ) = α + 3β + 6γ Observation R n = A(n)α + B(n)β + C(n)γ A(n), B(n), C(n) can be evaluated using repertoire method: we will consider three cases 1 R n = 1 for all n 2 R n = n for all n 3 R n = n 2 for all n.

36 Repertoire method: case 1 Lemma 1: A(n) = 1 for all n Hence 1 = R 0 = α From R n = R n 1 + (β + γn) follows that 1 = 1 + (β + γn). This is possible only when β = γ = 0 1 = A(n) 1 + B(n) 0 + C(n) 0

37 Repertoire method: case 2 Lemma 2: B(n) = n for all n Hence α = R 0 = 0 From R n = R n 1 +(β +γn) follows that n = (n 1)+(β +γn). I.e. 1 = β + γn. This gives that β = 1 and γ = 0 n = A(n) 0 + B(n) 1 + C(n) 0

38 Repertoire method: case 3 Lemma 3: C(n) = n2 +n 2 for all n Hence α = R 0 = 0 2 = 0 Equation R n = R n 1 + (β + γn) can be transformed as n 2 = (n 1) 2 + β + γn n 2 = n 2 2n β + γn 0 = (1 + β) + n(γ 2) This is valid i 1 + β = 0 and γ 2 = 0 Due to Lemma 2 we get n 2 = A(n) 0 + B(n) ( 1) + C(n)γ 2 n 2 = n + 2C(n)

39 Repertoire method: summing up According to Lemma 1, 2, 3, we get 1 R n = 1 for all n = A(n) = 1 2 R n = n for all n = B(n) = n 3 R n = n 2 for all n = C(n) = (n 2 + n)/2.

40 Repertoire method: summing up According to Lemma 1, 2, 3, we get 1 R n = 1 for all n = A(n) = 1 2 R n = n for all n = B(n) = n 3 R n = n 2 for all n = C(n) = (n 2 + n)/2. That means that ( n 2 ) + n R n = α + nβ + γ 2

41 Repertoire method: summing up According to Lemma 1, 2, 3, we get 1 R n = 1 for all n = A(n) = 1 2 R n = n for all n = B(n) = n 3 R n = n 2 for all n = C(n) = (n 2 + n)/2. That means that ( n 2 ) + n R n = α + nβ + γ 2 The sum for arithmetic sequence we obtain taking α = β = a and γ = b: n S n = (a + bk) = (n + 1)a + k=0 n(n + 1) b 2

42 Next subsection 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

43 Perturbation method Finding the closed form for S n = 0 k n a k : Rewrite S n+1 by splitting o rst and last term: S n + a n+1 = a 0 + a k = 1 k n+1 = a 0 + a k+1 = 1 k+1 n+1 = a 0 + a k+1 0 k n Work on last sum and express in terms of S n. Finally, solve for S n.

44 Example 2: geometric sequence Geometric sequence: a n = ax n Recurrent equation for the sum S n = a 0 + a 1 + a a n = 0 k n ax k : S 0 = a S n = S n 1 + ax n, for n > 0.

45 Example 2: geometric sequence Geometric sequence: a n = ax n Recurrent equation for the sum S n = a 0 + a 1 + a a n = 0 k n ax k : S 0 = a S n = S n 1 + ax n, for n > 0. Splitting o the rst term gives S n + a n+1 = a 0 + a k+1 = 0 k n = a + ax k+1 = 0 k n = a + x ax k = 0 k n Hence, we have the equation = a + xs n

46 Example 2: geometric sequence Geometric sequence: a n = ax n Recurrent equation for the sum S n = a 0 + a 1 + a a n = 0 k n ax k : S 0 = a S n = S n 1 + ax n, for n > 0. Hence, we have the equation S n + ax n+1 = a + xs n Solution: S n = a axn+1 1 x

47 Example 2: geometric sequence Geometric sequence: a n = ax n Recurrent equation for the sum S n = a 0 + a 1 + a a n = 0 k n ax k : S 0 = a S n = S n 1 + ax n, for n > 0. Hence, we have the equation S n + ax n+1 = a + xs n Solution: S n = a axn+1 1 x

48 Example 2: geometric sequence Geometric sequence: a n = ax n Recurrent equation for the sum S n = a 0 + a 1 + a a n = 0 k n ax k : S 0 = a S n = S n 1 + ax n, for n > 0. Closed formula for geometric sum: S n = a(xn+1 1) x 1

49 Next subsection 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

50 Example 3: Hanoi sequence The Tower of Hanoi recurrence: T 0 = 0 T n = 2T n 1 + 1

51 Example 3: Hanoi sequence The Tower of Hanoi recurrence: T 0 = 0 T n = 2T n This sequence can be transformed into geometric sum using following manipulations: Divide equations by 2 n : T 0 /2 0 = 0 T n /2 n = T n 1 /2 n 1 + 1/2 n Set S n = T n /2 n to have: S 0 = 0 S n = S n n (This is geometric sum with the parameters a = 1 and x = 1/2.)

52 Example 3: Hanoi sequence The Tower of Hanoi recurrence: T 0 = 0 T n = 2T n This sequence can be transformed into geometric sum using following manipulations: Divide equations by 2 n : T 0 /2 0 = 0 T n /2 n = T n 1 /2 n 1 + 1/2 n Set S n = T n /2 n to have: S 0 = 0 S n = S n n (This is geometric sum with the parameters a = 1 and x = 1/2.)

53 Example 3: Hanoi sequence The Tower of Hanoi recurrence: T 0 = 0 T n = 2T n Hence, S n = 0.5(0.5n 1) = 1 2 n (a 0 = 0 has been left out of the sum) T n = 2 n S n = 2 n 1

54 Example 3: Hanoi sequence The Tower of Hanoi recurrence: T 0 = 0 T n = 2T n Hence, S n = 0.5(0.5n 1) = 1 2 n (a 0 = 0 has been left out of the sum) T n = 2 n S n = 2 n 1 Just the same result we have proven by means of induction! :))

55 Next subsection 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

56 Linear recurrence in form a n T n = b n T n 1 + c n Here {a n }, {b n } and {c n } are any sequences and initial value T 0 is a constant. The idea: Find a summation factor s n satisfying the property s n b n = s n 1 a n 1 for any n If such a factor exists, one can do following transformations: s n a n T n = s n b n T n 1 + s n c n = s n 1 a n 1 T n 1 + s n c n Setting S n = s n a n T n, to rewrite the equation as Closed formula (!) for solution: S 0 = s 0 a 0 T 0 S n = S n 1 + s n c n T n = 1 n (s 0 a 0 T 0 + s n a n s k c k ) = 1 n (s 1 b 1 T 0 + s k=1 n a n s k c k ) k=1

57 Linear recurrence in form a n T n = b n T n 1 + c n Here {a n }, {b n } and {c n } are any sequences and initial value T 0 is a constant. The idea: Find a summation factor s n satisfying the property s n b n = s n 1 a n 1 for any n If such a factor exists, one can do following transformations: s n a n T n = s n b n T n 1 + s n c n = s n 1 a n 1 T n 1 + s n c n Setting S n = s n a n T n, to rewrite the equation as Closed formula (!) for solution: S 0 = s 0 a 0 T 0 S n = S n 1 + s n c n T n = 1 n (s 0 a 0 T 0 + s n a n s k c k ) = 1 n (s 1 b 1 T 0 + s k=1 n a n s k c k ) k=1

58 Finding summation factor Assuming that b n 0 for all n: Set s 0 = 1 Compute next elements using the property s n b n = s n 1 a n 1 : s 1 = a 0 b 1 s 2 = s 1a 1 = a 0a 1 b 2 b 1 b 2 s 3 = s 2a 2 = a 0a 1 a 2 b 3 b 1 b 2 b 3... s n = s n 1a n 1 b n (To be proved by induction!) = a 0a 1...a n 1 b 1 b 2...b n

59 Example: application of summation factor a n = c n = 1 and b n = 2 gives Hanoi Tower sequence: Evaluate summation factor s n = s n 1a n 1 b n = a 0a 1...a n 1 b 1 b 2...b n = 1 2 n Solution is T n = 1 (s 1 b 1 T 0 + s n a n n k=1 s k c k ) = 2 n n k=1 1 2 k = 2n (1 2 n ) = 2 n 1

60 YAE: constant coecients Equation Z n = az n 1 + b Taking a n = 1,b n = a and c n = b : Evaluate summation factor s n = s n 1a n 1 b n = a 0a 1...a n 1 b 1 b 2...b n = 1 a n Solution is ( ) ) Z n = 1 n s 1 b 1 Z 0 + s n a n s k c k = a (Z n n b a k k=1 k=1 = a n Z 0 + b(1 + a + a a n 1 ) = a n Z 0 + an 1 a 1 b

61 YAE : check up on results Equation Z n = az n 1 + b Z n = az n 1 + b = = a 2 Z n 2 + ab + b = = a 3 Z n 3 + a 2 b + ab + b = = a k Z n k + (a k 1 + a k )b = = a k Z n k + ak 1 b (assuming a 1) a 1 Continuing until k = n: Z n = a n Z n n + an 1 a 1 b = = a n Z 0 + an 1 a 1 b

62 YAE : check up on results Equation Z n = az n 1 + b Z n = az n 1 + b = = a 2 Z n 2 + ab + b = = a 3 Z n 3 + a 2 b + ab + b = = a k Z n k + (a k 1 + a k )b = = a k Z n k + ak 1 b (assuming a 1) a 1 Continuing until k = n: Z n = a n Z n n + an 1 a 1 b = = a n Z 0 + an 1 a 1 b

63 Eciency of quick sort Average number of comparisons: C n = n n n 1 k=0 C k

64 Eciency of quick sort (2) The average number of comparison steps when it is applied to n items C 0 = 0 C n = n n n 1 C k k=0 The following transformations reduce this equation nc n = n 2 n 2 + n + 2 C k + 2C n 1 k=0 Write the last equation for n 1 and subtract to eliminate the sum: (n 1)C n 1 = (n 1) 2 n 2 + (n 1) + 2 C k k=0

65 Eciency of quick sort (2) The average number of comparison steps when it is applied to n items C 0 = 0 C n = n n n 1 C k k=0 The following transformations reduce this equation nc n = n 2 n 2 + n + 2 C k + 2C n 1 k=0 Write the last equation for n 1 and subtract to eliminate the sum: (n 1)C n 1 = (n 1) 2 n 2 + (n 1) + 2 C k k=0 nc n (n 1)C n 1 = n 2 + n + 2C n 1 (n 1) 2 (n 1)

66 Eciency of quick sort (2) The average number of comparison steps when it is applied to n items C 0 = 0 C n = n n n 1 C k k=0 The following transformations reduce this equation nc n = n 2 n 2 + n + 2 C k + 2C n 1 k=0 Write the last equation for n 1 and subtract to eliminate the sum: (n 1)C n 1 = (n 1) 2 n 2 + (n 1) + 2 C k k=0 nc n nc n 1 + C n 1 = n 2 + n + 2C n 1 n 2 + 2n 1 n + 1

67 Eciency of quick sort (2) The average number of comparison steps when it is applied to n items C 0 = 0 C n = n n n 1 C k k=0 The following transformations reduce this equation nc n = n 2 n 2 + n + 2 C k + 2C n 1 k=0 Write the last equation for n 1 and subtract to eliminate the sum: (n 1)C n 1 = (n 1) 2 n 2 + (n 1) + 2 C k k=0 nc n nc n 1 = C n 1 + 2n

68 Eciency of quick sort (2) The average number of comparison steps when it is applied to n items C 0 = 0 C n = n n n 1 C k k=0 The following transformations reduce this equation nc n = n 2 n 2 + n + 2 C k + 2C n 1 k=0 Write the last equation for n 1 and subtract to eliminate the sum: (n 1)C n 1 = (n 1) 2 n 2 + (n 1) + 2 C k k=0 nc n = (n + 1)C n 1 + 2n

69 Eciency of quick sort (3) Equation nc n = (n + 1)C n 1 + 2n Assuming a n = n,b n = n + 1 and c n = 2n evaluate summation factor s n = a 1a 2...a n (n 1) = b 2 b 3...b n (n + 1) = 2 n(n + 1) Solution is ( C n = 1 s n a n = n s 1 b 1 C 0 + n k=1 4k k(k + 1) ) n s k c k k=1 n 1 = 2(n + 1) = 2(n + 1) k + 1 k=1 = 2(n + 1)H n 2n ( n 1 k + 1 k=1 n ) where H n = n is n-th harmonic number.

70 Eciency of quick sort (3) Equation nc n = (n + 1)C n 1 + 2n Assuming a n = n,b n = n + 1 and c n = 2n evaluate summation factor s n = a 1a 2...a n (n 1) = b 2 b 3...b n (n + 1) = 2 n(n + 1) Solution is ( C n = 1 s n a n = n s 1 b 1 C 0 + n k=1 4k k(k + 1) ) n s k c k k=1 n 1 = 2(n + 1) = 2(n + 1) k + 1 k=1 = 2(n + 1)H n 2n ( n 1 k + 1 k=1 n ) where H n = n is n-th harmonic number. (k-th harmonic produced by a violin string is the fundamental tone produced by a string that is 1/k times as long.)

71 Eciency of quick sort (3) Equation nc n = (n + 1)C n 1 + 2n Assuming a n = n,b n = n + 1 and c n = 2n evaluate summation factor s n = a 1a 2...a n (n 1) = b 2 b 3...b n (n + 1) = 2 n(n + 1) Solution is ( C n = 1 s n a n = n s 1 b 1 C 0 + n k=1 4k k(k + 1) ) n s k c k k=1 n 1 = 2(n + 1) = 2(n + 1) k + 1 k=1 = 2(n + 1)H n 2n ( n 1 k + 1 k=1 n ) where H n = n lnn. (k-th harmonic produced by a violin string is the fundamental tone produced by a string that is 1/k times as long.)

72 Next section 1 Sequences 2 Notations for sums 3 Sums and Recurrences Repertoire method Perturbation method Reduction to the known solutions Summation factors 4 Manipulation of Sums

73 Manipulation of Sums Some properties of sums: For K being a nite set and p(k) is any permutation of the set of all integers. Distributive law Associative law Commutative law ca k = c a k k K k K (a k + b k ) = a k + b k k K k K k K a k = a p(k) k K p(k) K Application of these laws for S = 0 k n (a + bk) S = (a + b(n k)) = (a + bn bk) (commutativity) 0 n k n 0 k n 2S = ((a + bk) + (a + bn bk)) = (2a + bn) (associativity) 0 k n 0 k n 2S = (2a + bn) 1 = (2a + bn)(n + 1) (distributivity) 0 k n

74 Manipulation of Sums Some properties of sums: For K being a nite set and p(k) is any permutation of the set of all integers. Distributive law Associative law Commutative law ca k = c a k k K k K (a k + b k ) = a k + b k k K k K k K a k = a p(k) k K p(k) K Application of these laws for S = 0 k n (a + bk) S = (a + b(n k)) = (a + bn bk) (commutativity) 0 n k n 0 k n 2S = ((a + bk) + (a + bn bk)) = (2a + bn) (associativity) 0 k n 0 k n 2S = (2a + bn) 1 = (2a + bn)(n + 1) (distributivity) 0 k n

75 Yet another useful equality a k + k K a k = a k + k K k K K k K K a k Special cases: a) for 1 m n b) for n 0 c) for n 0 m k=1 n a k + a k = a m + k=m n k=1 a k = a 0 + a k 0 k n 1 k n a k S n + a n+1 = a 0 + a k+1 0 k n

76 Example: S n = n k=0 kxk For x 1: S n + (n + 1)x n+1 = 0 k n (k + 1)x k+1 = kx k k n = xs n + x(1 x n+1 ) 1 x x k+1 0 k n n kx k = x (n + 1)x n+1 + nx n+2 k=0 (x 1) 2