Recurrences, Part 1 Troy Vasiga Centre for Education in Mathematics - - PowerPoint PPT Presentation

Recurrences, Part 1

Troy Vasiga Centre for Education in Mathematics and Computing Faculty of Mathematics, University of Waterloo cemc.uwaterloo.ca

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Outline

• Short biography of the speaker
• Recursion
• Why recursion is good
• Work on problems

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Short biography of the speaker

• Academic background:
• B.Math (Computer Science and Combinatorics and

Optimization, Double Degree), UW, 19

• M.Math (Combinatorics and Optimization) [thesis topic:

“α-resolvable designs of block size 4”], UW, 19

• B.Ed, UBC, 19
• Ph.D (Computer Science) “Error detection and correction in

number-theoretic algorithms”, UW, 20

• Current Position:
• Lecturer, David R. Cheriton School of Computer Science
• Associate Director, Centre for Education in Mathematics and

Computing

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Short biography of the speaker

• Academic life
• teach (mostly) first- and second-year CS courses (CS 115, 116,

135, 241)

• coordinate the Canadian Computing Competition (organize

problem creation, publication of results, communication with teachers, select Stage 2 invitees, train and supervise the Canadian IOI team)

• do lots of school visits (elementary and secondary schools in

Ontario, Malaysia, Singapore, China, Hong Kong)

• reviewer and grader for other math competitions (Euclid,

CIMC/CSMC)

• sit on or chair various CS committees (Undergraduate

Recruiting, Outreach, Scholarship committees)

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Short biography of the speaker

• Non-academic life
• bicycle to work every day (yes, every day)
• enjoy working out at the gym regularly
• enjoy travel (through work and outside of work)
• love playing Uno, Sleeping Queens, and Clue and reading with

daughter Natalie

• love dinner and movies with lovely wife Krista
• enjoy hanging out with neighbourhood friends

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Recursion

• Definition: see Recursion
• Real Definition:

Defining a function/structure using the function/structure itself in the definition

• To repeat something, you must recurse
• A better definition of recursion:

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Recursion

• Definition: see Recursion
• Real Definition:

Defining a function/structure using the function/structure itself in the definition

• To repeat something, you must recurse
• A better definition of recursion: if you don’t understand

recursion, see recursion.

• Notice that recursion requires a base case as well as recursive

cases.

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Natural Recursion

Why is recursion good?

• Easier to determine what the function does
• A natural way of seeing the world

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Recursion Examples: Babushka Dolls

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Recursion Examples: Shells

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Recursion Examples: The spiral of life

• I was created by my parents
• My parents were created by their parents
• My grandparents were created by their parents
• ....
• Base case?

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Recursive space filling: M. C. Escher

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Recursive space filling: M. C. Escher

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Recurrence relations in sequences

A sequence a0, a1, a2, ... can be specified recursively by giving

• base cases (aka “initial conditions”)
• a recursive definition

For example, we could have: a0 = a1 = 5 as the base cases and an = 2an−1 − an−2 We would like to solve this: that is, find a closed form (ideally, a number or formula depending on n) for an.

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 =

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 =

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 =

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 = 2(5) − 0 = 10 a3 =

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 = 2(5) − 0 = 10 a3 = 2(10) − 5 = 15 a4 =

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 = 2(5) − 0 = 10 a3 = 2(10) − 5 = 15 a4 = 2(15) − 10 = 20 a5 =

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 = 2(5) − 0 = 10 a3 = 2(10) − 5 = 15 a4 = 2(15) − 10 = 20 a5 = 2(20) − 15 = 25

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 = 2(5) − 0 = 10 a3 = 2(10) − 5 = 15 a4 = 2(15) − 10 = 20 a5 = 2(20) − 15 = 25 It looks like the closed form for this recurrence is: an = 5n

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“Guess” and check

Write out the first few terms, and try to find a pattern. a0 = a1 = 5 a2 = 2(5) − 0 = 10 a3 = 2(10) − 5 = 15 a4 = 2(15) − 10 = 20 a5 = 2(20) − 15 = 25 It looks like the closed form for this recurrence is: an = 5n But we would like to prove it!

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

• 2. assume the result holds for all values ≤ k for some k which is

at least as big as the base case(s)

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

• 2. assume the result holds for all values ≤ k for some k which is

at least as big as the base case(s)

• this step is known as the inductive hypothesis

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

• 2. assume the result holds for all values ≤ k for some k which is

at least as big as the base case(s)

• this step is known as the inductive hypothesis
• when we make this hypothesis, we are using strong induction

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

• 2. assume the result holds for all values ≤ k for some k which is

at least as big as the base case(s)

• this step is known as the inductive hypothesis
• when we make this hypothesis, we are using strong induction
• there is a variant known as weak induction where we assume

the result holds only for one particular value of k

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

• 2. assume the result holds for all values ≤ k for some k which is

at least as big as the base case(s)

• this step is known as the inductive hypothesis
• when we make this hypothesis, we are using strong induction
• there is a variant known as weak induction where we assume

the result holds only for one particular value of k

• 3. show the result holds for k + 1

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Mathematical Induction

Mathematical induction is the mirror image of recursion. When we try to prove a result using induction, we need to do three things:

• 1. show the result holds for the base case(s): this is usually, but

not always, easy

• 2. assume the result holds for all values ≤ k for some k which is

at least as big as the base case(s)

• this step is known as the inductive hypothesis
• when we make this hypothesis, we are using strong induction
• there is a variant known as weak induction where we assume

the result holds only for one particular value of k

• 3. show the result holds for k + 1
• this step is called the inductive step

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0.

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1.

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases.

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n.

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1.

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1. We need to show that ak+1 = 5(k + 1). But, we know that ak+1 = 2ak − ak−1

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1. We need to show that ak+1 = 5(k + 1). But, we know that ak+1 = 2ak − ak−1 = 2(5k) − 5(k − 1)

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1. We need to show that ak+1 = 5(k + 1). But, we know that ak+1 = 2ak − ak−1 = 2(5k) − 5(k − 1) = 10k − 5k + 5

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1. We need to show that ak+1 = 5(k + 1). But, we know that ak+1 = 2ak − ak−1 = 2(5k) − 5(k − 1) = 10k − 5k + 5 = 5k + 5

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1. We need to show that ak+1 = 5(k + 1). But, we know that ak+1 = 2ak − ak−1 = 2(5k) − 5(k − 1) = 10k − 5k + 5 = 5k + 5 = 5(k + 1)

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Our first inductive proof

Prove: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n for all n ≥ 0. Proof: By (strong) mathematical induction. If n = 0, then 5n = 0 = a0. If n = 1, then 5n = 5 = a1. Thus, the result holds for the base cases. Assume that for all values of n such that n ≤ k, the result holds. That is, for any n ≤ k, we know that an = 5n. We now prove the result for n = k + 1. We need to show that ak+1 = 5(k + 1). But, we know that ak+1 = 2ak − ak−1 = 2(5k) − 5(k − 1) = 10k − 5k + 5 = 5k + 5 = 5(k + 1) which proves the result.

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Exercise

Suppose you have: a0 = 1 an = 2an−1 + 1 (This is sometimes known as the Tower of Hanoi recurrence). Find a closed form, and prove that it is correct using mathematical induction.

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Solution

Start with some values: a0 = 1

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Solution

Start with some values: a0 = 1 a1 =

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 =

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7 a3 =

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7 a3 = 2(7) + 1 = 15

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7 a3 = 2(7) + 1 = 15 a4 =

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7 a3 = 2(7) + 1 = 15 a4 = 2(15) + 1 = 31

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7 a3 = 2(7) + 1 = 15 a4 = 2(15) + 1 = 31

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Solution

Start with some values: a0 = 1 a1 = 2(1) + 1 = 3 a2 = 2(3) + 1 = 7 a3 = 2(7) + 1 = 15 a4 = 2(15) + 1 = 31 We conjecture that an = 2n+1 − 1.

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1.

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1. We need to show that the result holds when n = k + 1.

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1. We need to show that the result holds when n = k + 1. ak+1 = 2ak + 1

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1. We need to show that the result holds when n = k + 1. ak+1 = 2ak + 1 = 2(2k+1 − 1) + 1

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1. We need to show that the result holds when n = k + 1. ak+1 = 2ak + 1 = 2(2k+1 − 1) + 1 = 2k+2 − 2 + 1

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1. We need to show that the result holds when n = k + 1. ak+1 = 2ak + 1 = 2(2k+1 − 1) + 1 = 2k+2 − 2 + 1 = 2k+2 − 1

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Proof of closed form

We only have one base case: a0 = 1 = 2(0+1) − 1 Assume that for all n ≤ k (k ≥ 0), we have an = 2n+1 − 1. We need to show that the result holds when n = k + 1. ak+1 = 2ak + 1 = 2(2k+1 − 1) + 1 = 2k+2 − 2 + 1 = 2k+2 − 1 which proves the result.

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Formalizing a formula

Guessing a closed form seems like a not very extendible idea, even for very “simple” things like:

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Formalizing a formula

Guessing a closed form seems like a not very extendible idea, even for very “simple” things like: f0 = f1 = 1 fn = fn−1 + fn−2

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Formalizing a formula

Guessing a closed form seems like a not very extendible idea, even for very “simple” things like: f0 = f1 = 1 fn = fn−1 + fn−2 A recurrence is called linear if each term of a sequence defined by the recurrence is linear function (i.e., only multiplied by a constant

• r has a constant added) of earlier terms in the sequence.

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Formalizing a formula

Guessing a closed form seems like a not very extendible idea, even for very “simple” things like: f0 = f1 = 1 fn = fn−1 + fn−2 A recurrence is called linear if each term of a sequence defined by the recurrence is linear function (i.e., only multiplied by a constant

• r has a constant added) of earlier terms in the sequence.

So, something like qn = (qn−1)2 + 4 is not a linear recurrence, but all the other recurrences we have seen so far are linear.

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More formalism

A linear homogeneous recurrence relation of degree k is a relation

• f the form

an = c1an−1 + c2an−2 + . . . + ckan−k where each ci is a real number and ck = 0.

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More formalism

A linear homogeneous recurrence relation of degree k is a relation

• f the form

an = c1an−1 + c2an−2 + . . . + ckan−k where each ci is a real number and ck = 0. Such a recurrence must include k initial conditions: a0 = V0 a1 = V1 . . . ak = Vk

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r.

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r. That would mean:

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r. That would mean: rn = c1rn−1 + c2rn−2 + · · · + ckrn−k

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r. That would mean: rn = c1rn−1 + c2rn−2 + · · · + ckrn−k which we could rewrite as:

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r. That would mean: rn = c1rn−1 + c2rn−2 + · · · + ckrn−k which we could rewrite as: rn − c1rn−1 − c2rn−2 − · · · − ckrn−k = 0

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r. That would mean: rn = c1rn−1 + c2rn−2 + · · · + ckrn−k which we could rewrite as: rn − c1rn−1 − c2rn−2 − · · · − ckrn−k = 0 and then we could divide both sides by rn−k to get:

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Characteristic equation

Let’s suppose that the closed form to a recurrence an = c1an−1 + c2an−2 + · · · + ckan−k looks like an = rn for some constant r. That would mean: rn = c1rn−1 + c2rn−2 + · · · + ckrn−k which we could rewrite as: rn − c1rn−1 − c2rn−2 − · · · − ckrn−k = 0 and then we could divide both sides by rn−k to get: rk − c1rk−1 − c2rk−2 − · · · − ck = 0 This is the characteristic equation of the recurrence.

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A helpful theorem

Suppose we have the characteristic equation rk − c1rk−1 − c2rk−2 − · · · − ck = 0 for the recurrence an = c1an−1 + c2an−2 + · · · + ckan−k and there are solutions r1, r2, . . . rm to the characteristic equation. Then, an = α1rn

1 + α2rn 2 + · · · + αmrn m satisfies the recurrence.

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Using this theorem

Recall the Fibonacci recurrence: f0 = f1 = 1 fn = fn−1 + fn−2

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Using this theorem

Recall the Fibonacci recurrence: f0 = f1 = 1 fn = fn−1 + fn−2 The characteristic equation is:

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Using this theorem

Recall the Fibonacci recurrence: f0 = f1 = 1 fn = fn−1 + fn−2 The characteristic equation is: r2 − r − 1 = 0. Using the quadratic formula we get: r = 1 ±

• (−1)2 − 4(1)(−1)

2 = 1 ± √ 5 2

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Solving the Fibonacci recurrence

Thus, we know that fn = α1

• 1 +

√ 5 2 n + α2

• 1 −

√ 5 2 n

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Solving the Fibonacci recurrence

Thus, we know that fn = α1

• 1 +

√ 5 2 n + α2

• 1 −

√ 5 2 n What to do next?

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Solving the Fibonacci recurrence

Thus, we know that fn = α1

• 1 +

√ 5 2 n + α2

• 1 −

√ 5 2 n What to do next? Notice that we haven’t used our initial conditions: those will help us solve for α1 and α2.

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Solving the Fibonacci recurrence

Thus, we know that fn = α1

• 1 +

√ 5 2 n + α2

• 1 −

√ 5 2 n What to do next? Notice that we haven’t used our initial conditions: those will help us solve for α1 and α2. f0 = α1 + α2 = 0 which means that α2 = −α1.

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Using the second base case

We know: f1 = α1

• 1 +

√ 5 2

• + α2
• 1 −

√ 5 2

• = 1

but we also know that α2 = −α1 from the previous page.

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Using the second base case

We know: f1 = α1

• 1 +

√ 5 2

• + α2
• 1 −

√ 5 2

• = 1

but we also know that α2 = −α1 from the previous page. So, we have α1

• 1 +

√ 5 2

• − α1
• 1 −

√ 5 2

• = 1

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Using the second base case

We know: f1 = α1

• 1 +

√ 5 2

• + α2
• 1 −

√ 5 2

• = 1

but we also know that α2 = −α1 from the previous page. So, we have α1

• 1 +

√ 5 2

• − α1
• 1 −

√ 5 2

• = 1

which implies

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Using the second base case

We know: f1 = α1

• 1 +

√ 5 2

• + α2
• 1 −

√ 5 2

• = 1

but we also know that α2 = −α1 from the previous page. So, we have α1

• 1 +

√ 5 2

• − α1
• 1 −

√ 5 2

• = 1

which implies α1

• 1 +

√ 5 2 − 1 − √ 5 2

• = 1

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Using the second base case

We know: f1 = α1

• 1 +

√ 5 2

• + α2
• 1 −

√ 5 2

• = 1

but we also know that α2 = −α1 from the previous page. So, we have α1

• 1 +

√ 5 2

• − α1
• 1 −

√ 5 2

• = 1

which implies α1

• 1 +

√ 5 2 − 1 − √ 5 2

• = 1

and so α

• 2

√ 5 2

• = 1, which implies α1 =

1 √ 5.

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Solution

Thus, the Fibonacci sequence has closed form: fn = 1 √ 5

• 1 +

√ 5 2 n − 1 √ 5

• 1 −

√ 5 2 n

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Golden ratio

Notice that if we have a and b satisfying: a + b b = a b

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Golden ratio

Notice that if we have a and b satisfying: a + b b = a b then a

b = ϕ, where

ϕ = 1 + √ 5 2 .

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5.

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is:

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0 which you can solve by factoring to get

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0 which you can solve by factoring to get (r − 1)2 = 0.

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0 which you can solve by factoring to get (r − 1)2 = 0. Thus, r = 1 is a solution.

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0 which you can solve by factoring to get (r − 1)2 = 0. Thus, r = 1 is a solution. Problem: an = α1(1)n + α2(1)n for FIXED α1 and α2 has no solution!

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0 which you can solve by factoring to get (r − 1)2 = 0. Thus, r = 1 is a solution. Problem: an = α1(1)n + α2(1)n for FIXED α1 and α2 has no solution! Fortunately, we can augment our theorem:

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Theorem isn’t exact enough yet

Recall: an = 2an−1 + an−2 with a0 = 0 and a1 = 5. The characteristic polynomial is: r2 − 2r − 1 = 0 which you can solve by factoring to get (r − 1)2 = 0. Thus, r = 1 is a solution. Problem: an = α1(1)n + α2(1)n for FIXED α1 and α2 has no solution! Fortunately, we can augment our theorem: If there is a root ri with multiplicity m + 1 (i.e., (r − ri)m+1 is a factor), then rn

i , nrn i , . . . nmrn i

are all solutions of the recurrence.

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Using this updated theorem

Since (r − 1)2 = r2 − 2r − 1, in our recurrence, r = 1 has multiplicity 2.

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Using this updated theorem

Since (r − 1)2 = r2 − 2r − 1, in our recurrence, r = 1 has multiplicity 2. Thus, we know that an = α101n + α11n1n which means an = α10 + α11n and we know a0 = 0, so α10 = 0.

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Using this updated theorem

Since (r − 1)2 = r2 − 2r − 1, in our recurrence, r = 1 has multiplicity 2. Thus, we know that an = α101n + α11n1n which means an = α10 + α11n and we know a0 = 0, so α10 = 0. Also, we know a1 = 5 and so 5 = α11. Thus, an = 5n, as we proved earlier.

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1.

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous!

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn.

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is r = 2.

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is r = 2. The constant term is 1 in an, so we guess that bn = c for some constant c.

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is r = 2. The constant term is 1 in an, so we guess that bn = c for some constant c. Therefore, bn = 2bn−1 + 1 and if bn = c, we get

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is r = 2. The constant term is 1 in an, so we guess that bn = c for some constant c. Therefore, bn = 2bn−1 + 1 and if bn = c, we get c = 2c + 1

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is r = 2. The constant term is 1 in an, so we guess that bn = c for some constant c. Therefore, bn = 2bn−1 + 1 and if bn = c, we get c = 2c + 1 which gives c = −1,

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A few words on non-homogeneous linear recurrences

We have seen an = 2an−1 + 1 with (a0 = 1) has closed form an = 2n+1 − 1. But, this is non-homogeneous! Trick: Solve the homogeneous equation, hn, and then try to figure

• ut what the additional term is, in terms of n, as bn, then

an = hn + bn. For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n since the characteristic equation is r = 2. The constant term is 1 in an, so we guess that bn = c for some constant c. Therefore, bn = 2bn−1 + 1 and if bn = c, we get c = 2c + 1 which gives c = −1, and thus (after verifying) our recurrence is an = 2n − 1.

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Work on problems!

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