Example. The sequence defined by the function. a : Z 0 Ñ Q defined by n fiñ 1{n

Transcription

1 Sequences A sequence is a function a from a subset of the set of integers (usually Z 0 or Z 0 )toasets, a : Z 0 Ñ S or a : Z 0 Ñ S. We write a n apnq, andcalla n the nth term of the sequence. Example The sequence defined by the function a : Z 0 Ñ Q defined by n fiñ 1{n is the sequence 1, 1{2, 1{3, 1{4,... We write a n 1{n. We can also write such a sequence like tapnqu n 1,2,... or tapnqu npz 0. For example, the sequence above is t1{nu npz 0. Some di erent kinds of sequences A geometric sequence (or progression is a sequence of the form c, cr, cr 2,cr 3,..., i.e. a : Z 0 Ñ S by n fiñ cr n, for some constants c and r. (This is a discrete version of the exponential function fpxq cr x.) An arithmetic progression is a sequence of the form b, b ` m, b ` 2m, b ` 3m,..., i.e. a : Z 0 Ñ S by n fiñ b ` mn, for some constants b and m. (This is a discrete version of the linear function fpxq b ` mx.) Notice, with a geometric sequence, the ratio is constant: if a n cr n, then a n {a n 1 r for all n. And with an arithmetic sequence the di erence is constant: if a n b ` mn, then a n a n 1 m for all n. (This is how we test to see if a sequence is geometric or arithmetic!)

2 Recurrence relations A recurrence relation for a sequence is an equation that expresses a n in terms of one of more of the previous terms of the sequence. For example: a n a n 1 2; a n a n 2 ` 1; a n a n 1 ` a n 2. A sequence is called a solution to a recurrence relation if its terms satisfy the recurrence relation. For example, a n 3 2 n is a solution to the recurrence relation a n a n 1 2; a n 2 n is also a solution to the recurrence relation a n a n 1 2; a n c 2 n is also a solution to the recurrence relation a n a n 1 2, for any c P R. An initial condition is a specified value for some fixed a i (usually a 0 and/or a 1 ). Without initial conditions, there are usually many solutions to a recurrence relation. For example, a n 3 2 n is the only solution to the r. rel. a n a n 1 2,a 0 3. Recurrence relations A recurrence relation for a sequence is an equation that expresses a n in terms of one of more of the previous terms of the sequence. A sequence is called a solution to a recurrence relation if its terms satisfy the recurrence relation. For example, a n 3 2 n is a solution to the recurrence relation a n a n 1 2; a n 2 n is also a solution to the recurrence relation a n a n 1 2; a n c 2 n is also a solution to the recurrence relation a n a n 1 2, for any c P R. And initial condition is a specified value for some fixed a i (usually a 0 and/or a 1 ). Without initial conditions, there are usually many solutions to a recurrence relation. For example, a n 3 2 n is the only solution to the r. rel. a n a n 1 2,a 0 3. A closed formula for a recurrence relation is a formula generating the sequence. We call a closed formula that satisfies a recurrence relation a solution to that relation. (Ex: a n c 2 n )

3 Going from a recurrence relation to a closed form is like calculating integrals it is not always even possible, let alone deterministic. We learn to recognize familiar types, and look for patterns. Geometric: Ifa n ra n 1,then a 1 r a 0, a 2 ra 1 rpra 0 q r 2 a 0, a 3 ra 1 rpr 2 a 0 q r 3 a 0... Claim: In general, a n a 0 r n for whatever constant a 0 is. Arithmetic: a n m ` a n 1,then a 1 m ` a 0, a 2 m ` a 1 m `pm ` a 0 q 2m ` a 0, a 3 m ` a 1 m `p2m ` a 0 q 3m ` a 0... Claim: In general, a n nm ` a 0 for whatever constant a 0 is. Going from a recurrence relation to a closed form is like calculating integrals it is not always even possible, let alone deterministic. We learn to recognize familiar types, and look for patterns. Factorial: Forn P Z 0,wedefinen factorial, denoted n!, by n! npn 1qpn 2q 2 1. For example, 4! For convenience, we define 0! 1. Then if a n na n 1,wehave a 1 1 a 0, a 2 2a 1 2p1 a 0 q p2 1qa 0, a 3 3a 1 3pp2 1qa 0 q p3 2 1qa 0... Claim: In general, a n n!a 0 for whatever constant a 0 is. You try: ICE 9

4 Summations Recall, for a sequence ta n u,thesummation notation ` a i a k ` a k`1 ` `a` i k and 8 a i a k ` a k`1 ` lim i k `Ñ8 i k ` a i. For example, let a i i. DefineS n n a i. Then S 1 1, S 2 1 ` 2 3, S 3 1 ` 2 ` 3 6,... In this example, 8 a i is not defined (the series does not converge). On the other hand, let a i p1{2q i,anddefines n n a i.then S 1 1{2, S 2 1{2 ` 1{4 3{4, S 3 1{2 ` 1{4 ` 1{8 7{8 1 1{8, S 4 1{2 ` 1{4 ` 1{8 ` 1{16 15{16 1 1{16... Claim: In general, S n n. So 8 a i lim nñ8 ˆ n 1.

5 Solving using partial sums The finite sum n S n a i a 0 ` a 1 ` `a n i 0 is called the partial sum for the series S 8 i 0 a i. We define S lim nñ8 S n. So to solve for S, it would be very helpful to get a closed form for S n. Example Show n cr i i 0 # cr n`1 c r 1 if r 1 pn ` 1qc if r 1 using partial sums. Namely, show rs n S n `pcr n`1 cq and solve for S n.thencalculate 8 i 0 cri. Identities: a i ` b i ips ips c a i c ips a i ` ips a i ips b i (addition is commutative) (distributive property) Set summations. a means add up everything in A. apa fpaq means add up fpaq for everything in A. apa Example: i ` 4 2 ` 6 2. ipt2,4,6u

6 More notation Double summations. Do the inside first! For example, ij ij j 1 j pi 1 ` i 2 ` i 3q ip1 ` 2 ` 3q 4 6i 6p1 ` 2 ` 3 ` 4q 60. More special summations. Theorem We have the following special summation identities: Notice n i npn ` 1q{2, and 8 ar i 8 What can we conclude? You try: ICE 10 a 1 r i x i 1 d dx for r Pp 1, 1q. 8 x i.