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CS 441 Discrete for CS Lecture 10

Sequences and

Milos Hauskrecht [email protected] 5329 Sennott Square

CS 441 for CS M. Hauskrecht

Sequences

Definition: A is a from a subset of the of (typically the set {0,1,2,...} or the set {1,2,3,...} to a set

S. We use the notation an to denote the of the n. We call an a of the sequence.

Notation: {an} is used to represent the sequence (note {} is the same notation used for sets, so be careful). {an} represents the ordered list a1, a2, a3, ... .

1 2 3 4 5 6 ….

a1 a2 a3 a4 a5 a6 ….

{an}

CS 441 Discrete mathematics for CS M. Hauskrecht

1 Sequences

Examples: 2 • (1) an = n , where n = 1,2,3... – What are the elements of the sequence? 1, 4, 9, 16, 25, ... n • (2) an = (-1) , where n=0,1,2,3,... – Elements of the sequence? 1, -1, 1, -1, 1, ... n • 3) an = 2 , where n=0,1,2,3,... – Elements of the sequence? 1, 2, 4, 8, 16, 32, ...

CS 441 Discrete mathematics for CS M. Hauskrecht

Arithmetic progression

Definition: An progression is a sequence of the form a, a+d,a+2d, …, a+nd where a is the initial term and d is common difference, such that both belong to R.

Example:

•sn= -1+4n for n=0,1,2,3, … • members: -1, 3, 7, 11, …

CS 441 Discrete mathematics for CS M. Hauskrecht

Definition A geometric progression is a sequence of the form: a, ar, ar2, ..., ark, where a is the initial term, and r is the common ratio. Both a and r belong to R.

Example: n •an = ( ½ ) for n = 0,1,2,3, … members: 1,½, ¼, 1/8, …..

CS 441 Discrete mathematics for CS M. Hauskrecht

Sequences

• Given a sequence finding a rule for generating the sequence is not always straightforward

Example: • Assume the sequence: 1,3,5,7,9, …. • What is the formula for the sequence? • Each term is obtained by adding 2 to the previous term. 1, 1+2=3, 3+2=5, 5+2=7 • What type of progression this suggest?

CS 441 Discrete mathematics for CS M. Hauskrecht

3 Sequences

• Given a sequence finding a rule for generating the sequence is not always straightforward

Example: • Assume the sequence: 1,3,5,7,9, …. • What is the formula for the sequence? • Each term is obtained by adding 2 to the previous term. • 1, 1+2=3, 3+2=5, 5+2=7 • It suggests an : a+nd with a=1 and d=2

•an=1+2n

CS 441 Discrete mathematics for CS M. Hauskrecht

Sequences

• Given a sequence finding a rule for generating the sequence is not always straightforward

Example 2: • Assume the sequence: 1, 1/3, 1/9, 1/27, … • What is the sequence? • The denominators are powers of 3. 1, 1/3= 1/3, (1/3)/3=1/(3*3)=1/9, (1/9)/3=1/27 • This suggests a geometric progression: ark with a=1 and r=1/3 • (1/3 )n

CS 441 Discrete mathematics for CS M. Hauskrecht

4 Recursively defined sequences

• The n-th of the sequence {an} is defined recursively in terms of the previous elements of the sequence and the initial elements of the sequence.

Example :

•an = an-1 + 2 assuming a0 = 1;

•a0 = 1;

•a1 = 3;

•a2 = 5;

•a3 = 7;

• Can you write an non-recursively using n?

•an = 1 + 2n

CS 441 Discrete mathematics for CS M. Hauskrecht

Fibonacci sequence

• Recursively defined sequence, where

•f0 = 0;

•f1 = 1;

•fn = fn-1 + fn-2 for n = 2,3, …

•f2 = 1

•f3 = 2

•f4 = 3

•f5 = 5

CS 441 Discrete mathematics for CS M. Hauskrecht

5 Summations

Summation of the terms of a sequence: n  a j  a m  a m 1  ...  a n j  m The j is referred to as the index of . • m is the lower and • n is the upper limit of the summation.

CS 441 Discrete mathematics for CS M. Hauskrecht

Summations

Example: • 1) Sum the first 7 terms of {n2} where n=1,2,3, ... .

• 7 7 2 a j  j 1  4  16  25  36  49  140 jj1 1

• 2) What is the value of 8 8 j a j  (  1) 1  (  1)  1  (  1)  1  1 kk4 4

CS 441 Discrete mathematics for CS M. Hauskrecht

6 Arithmetic

Definition: The sum of the terms of the arithmetic progression a, a+d,a+2d, …, a+nd is called an arithmetic series.

Theorem: The sum of the terms of the arithmetic progression a, a+d,a+2d, …, a+nd is n n n(n 1) S  (a  jd)  na  d j  na  d j11j 2

• Why?

CS 441 Discrete mathematics for CS M. Hauskrecht

Arithmetic series

Theorem: The sum of the terms of the arithmetic progression a, a+d,a+2d, …, a+nd is n n n(n 1) S  (a  jd)  na  d j  na  d j11j 2 Proof: n n n n S  (a  jd)  a  jd  na  d j j11j1 j j1

n  j 1 2  3 4 .... (n  2)  (n 1)  n j1

CS 441 Discrete mathematics for CS M. Hauskrecht

7 Arithmetic series

Theorem: The sum of the terms of the arithmetic progression a, a+d,a+2d, …, a+nd is

n n n(n 1) S  (a  jd)  na  d j  na  d j11j 2 Proof: n n n n S  (a  jd)  a  jd  na  d j j11j1 j j1

n  j 1 2  3 4 .... (n  2)  (n 1)  n j1

n+1 n+1 … n+1 n *(n 1) 2 CS 441 Discrete mathematics for CS M. Hauskrecht

Arithmetic series 5 Example: S  (2  j3)  j1 5 5  2  j3  jj1 1

5 5  21 3 j  jj1 1

5  2*5  3 j  j1 (5 1) 10  3 *5  2

10  45  55

CS 441 Discrete mathematics for CS M. Hauskrecht

8 Arithmetic series

5 Example 2: S  (2  j3)  j3  5   2   (2  j3)  (2  j3) Trick  j1   j1   5 5   2 2   21 3 j  21 3 j  jj1 1   jj1 1 

 5513  42

CS 441 Discrete mathematics for CS M. Hauskrecht

Double summations 4 2 Example: S  (2i  j)  ij1 1 4  2 2    2i  j  ij1  1 j 1  4  2 2   2i 1 j  ijj1  1 1  4  2   2i*2  j  ij1  1  4  2i*2  3  i1 4 4  4i  3  ii1 1 4 4  4i  3 1  4*10  3*4  28 ii1 1

CS 441 Discrete mathematics for CS M. Hauskrecht

Definition: The sum of the terms of a geometric progression a, ar, ar2, ..., ark is called a geometric series.

Theorem: The sum of the terms of a geometric progression a, ar, ar2, ..., arn is

n n n1 j j r 1 S  (ar )  a r a  j00j  r 1 

CS 441 Discrete mathematics for CS M. Hauskrecht

Geometric series

Theorem: The sum of the terms of a geometric progression a, ar, 2 n ar , ..., ar is n n n1 j j r 1 S  (ar )  a r a  j00j  r 1  Proof: n S  ar j  a  ar  ar 2  ar 3 ... ar n j0 • multiply S by r n rS  rar j  ar  ar 2  ar 3 ... ar n1 j0 • Substract rS  S  ar  ar 2  ar 3 ... ar n1 a  ar  ar 2.. ar n   ar n1  a ar n1  a r n1 1 S   a  r 1  r 1 

CS 441 Discrete mathematics for CS M. Hauskrecht

10 Geometric series

Example: 3 S  2(5) j  j0

General formula: n n n1 j j r 1 S  (ar )  a r a  j00j  r 1 

3 54 1 S  2(5) j 2*  j0 5 1 625 1 624  2*  2*  2*156  312 4 4

CS 441 Discrete mathematics for CS M. Hauskrecht

Infinite geometric series

• Infinite geometric series can be computed in the closed form for x<1 • How?

 k k 1 n n x 1 1 1  x  lim k  x  lim k    n0 n0 x 1 x 1 1  x

• Thus:  1  x n  n  0 1  x

CS 441 Discrete mathematics for CS M. Hauskrecht

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