Recurrences, Part 1 - CEMC€¦ · Recurrences, Part 1 Troy Vasiga Centre for Education in...

Recurrences, Part 1

Troy Vasiga

Centre for Education in Mathematics and ComputingFaculty of Mathematics, University of Waterloo

cemc.uwaterloo.ca

WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING

Outline

• Short biography of the speaker

• Recursion

• Why recursion is good

• Work on problems


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



• 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 withteachers, select Stage 2 invitees, train and supervise theCanadian IOI team)

• do lots of school visits (elementary and secondary schools inOntario, Malaysia, Singapore, China, Hong Kong)

• reviewer and grader for other math competitions (Euclid,CIMC/CSMC)

• sit on or chair various CS committees (UndergraduateRecruiting, Outreach, Scholarship committees)



• 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


Recursion

• Definition: see Recursion

• Real Definition:

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

• To repeat something, you must recurse

• A better definition of recursion:

if you don’t understandrecursion, see recursion.

• Notice that recursion requires a base case as well as recursivecases.


Recursion

• Definition: see Recursion

• Real Definition:

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

• To repeat something, you must recurse

• A better definition of recursion: if you don’t understandrecursion, see recursion.

• Notice that recursion requires a base case as well as recursivecases.


Natural Recursion

Why is recursion good?

• Easier to determine what the function does

• A natural way of seeing the world


Recursion Examples: Babushka Dolls


Recursion Examples: Shells


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?


Recursive space filling: M. C. Escher


Recurrence relations in sequencesA sequence a0, a1, a2, ... can be specified recursively by giving

• base cases (aka “initial conditions”)

• a recursive definition

For example, we could have:

a0 = 0

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, anumber or formula depending on n) for an.


“Guess” and checkWrite out the first few terms, and try to find a pattern.

a0 =

0

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!



a0 = 0

a1 =

5

a2 = 2(5)− 0 = 10

a3 = 2(10)− 5 = 15

a4 = 2(15)− 10 = 20

a5 = 2(20)− 15 = 25


an = 5n




a0 = 0

a1 = 5

a2 =

2(5)− 0 = 10

a3 = 2(10)− 5 = 15

a4 = 2(15)− 10 = 20

a5 = 2(20)− 15 = 25


an = 5n




a0 = 0

a1 = 5

a2 = 2(5)− 0 = 10

a3 =

2(10)− 5 = 15

a4 = 2(15)− 10 = 20

a5 = 2(20)− 15 = 25


an = 5n




a0 = 0

a1 = 5

a2 = 2(5)− 0 = 10

a3 = 2(10)− 5 = 15

a4 =

2(15)− 10 = 20

a5 = 2(20)− 15 = 25


an = 5n




a0 = 0

a1 = 5

a2 = 2(5)− 0 = 10

a3 = 2(10)− 5 = 15

a4 = 2(15)− 10 = 20

a5 =

2(20)− 15 = 25


an = 5n




a0 = 0

a1 = 5

a2 = 2(5)− 0 = 10

a3 = 2(10)− 5 = 15

a4 = 2(15)− 10 = 20

a5 = 2(20)− 15 = 25


an = 5n




a0 = 0

a1 = 5

a2 = 2(5)− 0 = 10

a3 = 2(10)− 5 = 15

a4 = 2(15)− 10 = 20

a5 = 2(20)− 15 = 25


an = 5n

But we would like to prove it!WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING

Mathematical InductionMathematical induction is the mirror image of recursion.

When we try to prove a result using induction, we need to do threethings:

1. show the result holds for the base case(s): this is usually, butnot always, easy

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






• 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








• 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 assumethe result holds only for one particular value of k









3. show the result holds for k + 1

• this step is called the inductive step


Our first inductive proofProve: If a0 = 0, a1 = 5 and an = 2an−1 − an−2, then an = 5n forall 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.



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)




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)




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)




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)




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)





ak+1 = 2ak − ak−1

= 2(5k)− 5(k − 1)

= 10k − 5k + 5

= 5k + 5

= 5(k + 1)





ak+1 = 2ak − ak−1

= 2(5k)− 5(k − 1)

= 10k − 5k + 5

= 5k + 5 = 5(k + 1)





ak+1 = 2ak − ak−1

= 2(5k)− 5(k − 1)

= 10k − 5k + 5

= 5k + 5 = 5(k + 1)

which proves the result.WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING

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 mathematicalinduction.


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.


Solution


a0 = 1

a1 =

2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 =

2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 =

2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 =

2(15) + 1 = 31



Solution


a0 = 1

a1 = 2(1) + 1 = 3

a2 = 2(3) + 1 = 7

a3 = 2(7) + 1 = 15

a4 = 2(15) + 1 = 31



Proof of closed formWe 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




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




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



Formalizing a formula

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

f0 = 0

f1 = 1

fn = fn−1 + fn−2

A recurrence is called linear if each term of a sequence defined bythe recurrence is linear function (i.e., only multiplied by a constantor has a constant added) of earlier terms in the sequence.So, something like qn = (qn−1)2 + 4 is not a linear recurrence, butall the other recurrences we have seen so far are linear.




f0 = 0

f1 = 1

fn = fn−1 + fn−2

A recurrence is called linear if each term of a sequence defined bythe recurrence is linear function (i.e., only multiplied by a constantor has a constant added) of earlier terms in the sequence.

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




f0 = 0

f1 = 1

fn = fn−1 + fn−2

A recurrence is called linear if each term of a sequence defined bythe recurrence is linear function (i.e., only multiplied by a constantor has a constant added) of earlier terms in the sequence.So, something like qn = (qn−1)2 + 4 is not a linear recurrence, butall the other recurrences we have seen so far are linear.


More formalism

A linear homogeneous recurrence relation of degree k is a relationof the form

an = c1an−1 + c2an−2 + . . .+ ckan−k

where each ci is a real number and ck 6= 0.

Such a recurrence must include k initial conditions:

a0 = V0 a1 = V1 . . . ak = Vk


More formalism

A linear homogeneous recurrence relation of degree k is a relationof the form

an = c1an−1 + c2an−2 + . . .+ ckan−k

where each ci is a real number and ck 6= 0.Such a recurrence must include k initial conditions:

a0 = V0 a1 = V1 . . . ak = Vk


Characteristic equationLet’s suppose that the closed form to a recurrence

an = c1an−1 + c2an−2 + · · ·+ ckan−k

looks likean = rn

for some constant r .

That would mean:

rn = c1rn−1 + c2rn−2 + · · ·+ ck rn−k

which we could rewrite as:

rn − c1rn−1 − c2rn−2 − · · · − ck rn−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.


Characteristic equationLet’s suppose that the closed form to a recurrence

an = c1an−1 + c2an−2 + · · ·+ ckan−k

looks likean = rn

for some constant r .That would mean:

rn = c1rn−1 + c2rn−2 + · · ·+ ck rn−k

which we could rewrite as:

rn − c1rn−1 − c2rn−2 − · · · − ck rn−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.


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 = α1rn1 + α2rn2 + · · ·+ αmrnm satisfies the recurrence.


Using this theoremRecall the Fibonacci recurrence:

f0 = 0

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


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 helpus solve for α1 and α2.

f0 = α1 + α2 = 0

which means that α2 = −α1.



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 helpus solve for α1 and α2.

f0 = α1 + α2 = 0




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 helpus solve for α1 and α2.

f0 = α1 + α2 = 0



Using the second base caseWe 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.


Using the second base caseWe 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.


Solution

Thus, the Fibonacci sequence has closed form:

fn =1√5

(1 +√

5

2

)n

− 1√5

(1−√

5

2

)n


Golden ratio

Notice that if we have a and b satisfying:

a + b

b=

a

b

then ab = ϕ, where

ϕ =1 +√

5

2.


Theorem isn’t exact enough yetRecall: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 nosolution!Fortunately, we can augment our theorem:If there is a root ri with multiplicity m + 1 (i.e., (r − ri )

m+1 is afactor), then

rni , nrni , . . . nmrni

are all solutions of the recurrence.


Theorem isn’t exact enough yetRecall:an = 2an−1 + an−2 with a0 = 0 and a1 = 5.The characteristic polynomial is:

r2 − 2r − 1 = 0







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 nosolution!Fortunately, we can augment our theorem:If there is a root ri with multiplicity m + 1 (i.e., (r − ri )






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 nosolution!Fortunately, we can augment our theorem:If there is a root ri with multiplicity m + 1 (i.e., (r − ri )






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 nosolution!Fortunately, we can augment our theorem:If there is a root ri with multiplicity m + 1 (i.e., (r − ri )






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 nosolution!

Fortunately, we can augment our theorem:If there is a root ri with multiplicity m + 1 (i.e., (r − ri )






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 nosolution!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), thenrni , nrni , . . . n

mrni




r2 − 2r − 1 = 0






Using this updated theorem

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

Thus, we know that

an = α101n + α11n1n

which meansan = 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.



Since (r − 1)2 = r2 − 2r − 1, in our recurrence, r = 1 hasmultiplicity 2.Thus, we know that

an = α101n + α11n1n


and we know a0 = 0, so α10 = 0.

Also, we know a1 = 5 and so 5 = α11.Thus, an = 5n, as we proved earlier.



Since (r − 1)2 = r2 − 2r − 1, in our recurrence, r = 1 hasmultiplicity 2.Thus, we know that

an = α101n + α11n1n


and we know a0 = 0, so α10 = 0.Also, we know a1 = 5 and so 5 = α11.Thus, an = 5n, as we proved earlier.


A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1.

But, this is non-homogeneous!Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n sincethe characteristic equation is r = 2.The constant term is 1 in an, so we guess that bn = c for someconstant c. Therefore,

bn = 2bn−1 + 1

and if bn = c , we getc = 2c + 1

which gives c = −1, and thus (after verifying) our recurrence isan = 2n − 1.


A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1. But, this is non-homogeneous!

Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n sincethe characteristic equation is r = 2.The constant term is 1 in an, so we guess that bn = c for someconstant c. Therefore,

bn = 2bn−1 + 1




A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1. But, this is non-homogeneous!Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.

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

bn = 2bn−1 + 1




A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1. But, this is non-homogeneous!Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n sincethe characteristic equation is

r = 2.The constant term is 1 in an, so we guess that bn = c for someconstant c. Therefore,

bn = 2bn−1 + 1




A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1. But, this is non-homogeneous!Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n sincethe characteristic equation is r = 2.

The constant term is 1 in an, so we guess that bn = c for someconstant c. Therefore,

bn = 2bn−1 + 1




A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1. But, this is non-homogeneous!Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n sincethe characteristic equation is r = 2.The constant term is 1 in an, so we guess that bn = c for someconstant c.

Therefore,

bn = 2bn−1 + 1




A few words on non-homogeneous linear recurrencesWe have seen an = 2an−1 + 1 with (a0 = 1) has closed forman = 2n+1 − 1. But, this is non-homogeneous!Trick: Solve the homogeneous equation, hn, and then try to figureout what the additional term is, in terms of n, as bn, thenan = hn + bn.For this recurrence, hn = 2hn−1 is easily seen to be hn = 2n sincethe characteristic equation is r = 2.The constant term is 1 in an, so we guess that bn = c for someconstant c. Therefore,

bn = 2bn−1 + 1

and if bn = c , we get

c = 2c + 1




bn = 2bn−1 + 1





bn = 2bn−1 + 1


which gives c = −1,

and thus (after verifying) our recurrence isan = 2n − 1.



bn = 2bn−1 + 1




Work on problems!


Recurrences, Part 1 - CEMC€¦ · Recurrences, Part 1 Troy Vasiga Centre for Education in...

Documents

Transcript of Recurrences, Part 1 - CEMC€¦ · Recurrences, Part 1 Troy Vasiga Centre for Education in...