1 RECURRENCE 1. Sequence 2. Recursively defined sequence 3. Finding an explicit formula for...
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Transcript of 1 RECURRENCE 1. Sequence 2. Recursively defined sequence 3. Finding an explicit formula for...
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RECURRENCE
1. Sequence2. Recursively defined sequence3. Finding an explicit formula for
recurrence relation
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Learning Outcomes
You should be able to solve first-order and second-order linear
homogeneous recurrence relation with constant coefficients
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Sequences A sequence is an ordered list of objects (or events).
Like a set, it contains members (also called terms) Sequences can be finite or infinite.
2,4,6,8,… for i ≥ 1 ai = 2i (explicit formula)infinite sequence with infinite distinct values
-1,1,-1,1,… for i ≥ 1 bi = (-1)i infinite sequence with finite distinct values
For 1<=i<=6 ci = i+5 finite sequence (with finite distinct values)6,7,8,9,10,11
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Ways to define sequence
Write the first few terms: 3,5,7,… Use explicit formula for its nth term an = 2n for n ≥ 1 Use recursion How to define a sequence using a
recursion?
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Recursively defined sequencesRecursion can be used to defined a sequence.This requires:A recurrence relation: a formula that relates each term ak
to some previous terms ak-1, ak-2, … ak = ak-1 + 2ak-2
The initial conditions: the values of the first few terms a0, a1, …
Example: For all integers k ≥ 2, find the terms b2, b3 and b4: bk = bk-1 + bk-2 (recurrence relation) b0 = 1 and b1 = 3 (initial conditions)Solution:b2 = b1 + b0 = 3 + 1 = 4b3 = b2 + b1 = 4 + 3 = 7b4 = b3 + b2 = 7 + 4 = 11
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Explicit formula and recurrence relation Show that the sequence 1,-1!,2!,-3!,4!,…,(-1)nn!,
… for n≥0, satisfies the recurrence relation sk = (-k)sk-1 for all integers k≥1. The general term of the sequence: sn=(-1)nn! substitute k and k-1 for n to get sk=(-1)kk! sk-1=(-1)k-1(k-1)! Substitute sk-1 into recurrence relation: (-k)sk-1 = (-k)(-1)k-1(k-1)! = (-1)k(-1)k-1(k-1)! = (-1)(-1)k-1 k(k-1)! = (-1)k k! = sk
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Examples of recursively sequence Famous recurrences Arithmetic sequences: ak = ak-1 + de.g. 1,4,7,10,13,…geometric sequences: ak = ark-1
e.g. 1,3,9,27,… Factorial: f(n) = n . f(n-1) Fibonacci numbers: fk = fk-1+fk-2
1,1,2,3,5,8,…Tower of Hanoi problem
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Application of recurrence Analysis of algorithm containing recursive function such
as factorial function.
Algorithm f(n) /input: A nonnegative integer /output: The value of n! If n = 0 return 1 else return f(n-1)*n
No. of operations (multiplication) determines the efficiency of algo.
Recurrence relation is used to express the no. of operation in the algorithm.
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Solving Recurrence relation by Iteration It is helpful to know an explicit formula
for a sequence. An explicit formula is called a solution
to the recurrence relation Most basic method is iteration - start from the initial condition - calculate successive terms until a
pattern can be seen - guess an explicit formula
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Some examplesLet a0,a1,a2,… be the sequence defined recursively as
follows: For all integers k≥1,(1) ak = ak-1+2(2) a0 = 1Use iteration to guess an explicit formula for the sequence.a0=1a1=a0+2a2=a1+2=(1+2)+2 = 1+2.2a3=a2+2=(1+2.2)+2 = 1+3.2a4=a3+2=(1+3.2)+2 = 1+4.3….Guess: an=1+n.2=1+2nThe above sequence is an arithmetic sequence.
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Geometric SequenceLet r be a fixed nonzero constant, and suppose a sequence
a0,a1,a2,… is defined as follows: ak = rak-1 for all integers k ≥ 1 a0 = aUse iteration to guess an explicit formula for the sequence a0=a a1=ra0=ra a2=ra1=r(ra)=r2a a3=ra2=r(r2a)=r3a Guess: an=rna = arn for all integers n≥0The above sequence is geometric sequence and r is a
common ratio.
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Explicit formula for tower of Hanoi
mn = 2n – 1. (exponential order) To move 1 disk takes 1 second m64 = 264 –1 = 1.844674 * 1019
seconds = 5.84542 * 1011
years = 584.5 billion years.
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Second-Order Linear Homogeneous with constant coefficients
A second-order linear homogeneous recur. relation with c.c. is a recur. relation of the form
ak = Aak-1 + Bak-2 for all integers k ≥ some fixed integer,
where A and B are fixed real numbers with B ≠ 0.
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Terminology ak = Aak-1 + Bak-2
Second order: ak contains the two previous terms
Linear: ak-1 and ak-2 appear in separate terms and to the first power
Homogeneous: total degree of each term is the same (no constant term)
Constant coefficients: A and B are fixed real numbers
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Examples Second-Order Linear Homogeneous with
constant coefficients ak = 3ak-1 + 2ak-2 - yes. bk = bk-1 + bk-2 + bk-3 - no dk = (dk-1)2 + dk-1dk-2 - no; not linear ek = 2ek-2 - yes; A = 0, B = 2. fk = 2fk-1 + 1 - no; not homogeneous
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Characteristic equationak = Aak-1 + Bak-2 for k>=2 ….. (1)Suppose that sequence 1, t, t2, t3,… satisfies relation (1) where t is a nonzero real number. General term for the sequence an=tn.Hence, ak-1=tk-1 and ak-2= tk-2
Substitute ak-1 and ak-2 into relation (1)tk = Atk-1 + Btk-2
Divide the equation by tk-2:t2 = At + B or t2 – At – B = 0 This equation is called the characteristic equation of therelation. Recurrence relation (1) is satisfied by the sequence 1,t,t2,t3,… iff t satisfies the characteristic equation.
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Using the characteristic equation to find sequences Example: Consider the following recurrence relationak = ak-1+2ak-2 for all k >= 2.Find sequences that satisfy the relation.Solution: For the given relation, A=1 and B=2.Relation is satisfied by the sequence 1,t,t2,t3,… iff t
satisfies the characteristic equation t2 – At – B = 0 ort2 – t – 2 = 0(t – 2)(t + 1) = 0.t = 2 or t = -1. Sequences: 1,2,22,23,… and 1,-1,(-1)2,(-1)3, … or 1,-1,1,-1, …,(-1)n, …
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Linear combination of sequences LemmaIf r0,r1,r2,…,rn,.. and s0,s1,s2,…,sn,… are sequences
that satisfy the same second-order linear homogeneous recurrence relation with c.c., and if C and D are any numbers, then the sequence a0,a1,a2,…
defined by the formula an = C.rn + D.sn for all integer n>=0 also satisfies the same recurrence relation. C and D can be calculated using initial
conditions.
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Two possible solutions
For the characteristic equation t2 – At – B = 0 there are two possible solutions: - Distinct-roots case - Single-root case
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Explicit formula for second-order relation – Distinct-roots case
Find a sequence (explicit formula) that satisfies the recurrence relation
ak=ak-1 + 2ak-2 for k>=2 and initial conditions a0=1 and a1=8Solution: A=1 and B=2. Characteristic equation : t2 – At – B =0 Substitute A and B, t2 – t – 2 = 0 Sequences: 1,2,22,23,… and 1,-1,1,-1, …,(-1)n, …
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Steps for finding explicit formula
1. Form the characteristic equation.2. Solve the equation – let r and s be
the roots. ( r ≠ s)3. Set up an explicit formula: ak = C.rk + D.sk
4. Find C and D using initial conditions.
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Single-Root Case
Two sequences that satisfy the relation ak = A.ak-1+B.ak-2
S1: 1,r,r2,r3,…,rn,…
S2: 0,r,2r2,3r3,…,nrn,…
where r is the root of t2 - A.t - B = 0. Explicit formula for the new sequence ak
= C.rk + D.nrn
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Example A sequence b0, b1, b2,… satisfies the rec. relation
bk = 4bk-1 – 4bk-2 for k>=2 with initial conditions b0=1 and b1=3.
Find explicit formula for the sequence.Solution: A=4 and B=-4Charac eq: t2 – 4t +4 =0 (t-2)2=0. t=2.Seq: 1,2,22, …, 2n,.. 0,2,2.22,3.23,…,n.2n,… Explicit formula: bn
= C.rn + D.nrn = C.2n + D.n2n
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Example bn = C.2n + D.n2n
b0 = 1 = C + D.0 = C b1 = 3 = C.2 + D.2 Hence D = ½. Therefore bn = 2n + (1/2).n.2n
= 2n (1+ n/2) for integer n>=0.
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Summary
How to express the sequence Find explicit formula for first-order
relation Find explicit formula for second-
order relation (distinct and single root)