MEI PowerPoint Template#MEIConf2019 Defining Recurrence Relations Any sequence in which each...

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First Order

Recurrence

Relations

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Recurrence Relations

Specification Topic Mapping

Discrete/Decision Mathematics Topics AQA Edexcel MEI OCR A

Modelling problems, e.g. population growth AS D2

Solution of first order linear homogeneous andnon-homogeneous relations

AS D2

Solution of second order linear homogeneous andnon-homogeneous relations

A Level D2

Particular solution, complementary function,auxiliary equation

AS D2

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In this session: Term-to-term rule; Position-to-term rule

Recurrence relation; Closed formula

Examples in context

Definitions, vocabulary; homogeneous, linear

Solving first order recurrence relations

Back substitution (induction, recursion)

Complementary function (C.F.)

Particular Solution (P.S.)

General Solution (G.S.)

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Defining Recurrence Relations Any sequence in which each subsequent term

is dependent upon one or more previous terms

is a recurrence relation.

Recurrence relations are also known as

difference equations.

A recurrence relation may be described by a

‘term-to-term’ or inductive rule.

Solving a recurrence relation requires finding a

‘position-to-term’ rule or closed formula, i.e.

an expression for the nth term.

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Easy examplesConsider these two sequences:

a) 5, 8, 11, 14, 17, …….

b) 2, 6, 18, 54, 162, …….

How do your students describe these sequences?

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a) 5, 8, 11, 14, 17, ……. “goes up in 3s”

b) 2, 6, 18, 54, 162, ……. “times by 3”

They are giving the ‘term-to-term’ or iterative rule.

You might have been hoping for the nth term of the

sequence which is the ‘position-to-term’ rule.

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a) 5, 8, 11, 14, 17, ……. “goes up in 3s”

b) 2, 6, 18, 54, 162, ……. “times by 3”

a) an+1 = an + 3, n ≥ 1, a1 = 5

or an = an-1 + 3, n ≥ 1, a0 = 5

b) an+1 = 3an, n ≥ 1, a1 = 2

or an = 3an-1 , n ≥ 1, a0 = 2

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a) 5, 8, 11, 14, 17, ……. “goes up in 3s”

b) 2, 6, 18, 54, 162, ……. “times by 3”

a) an+1 = an + 3, n ≥ 1, a1 = 5 an = 3n + 2

b) an+1 = 3an, n ≥ 1, a1 = 2 an = 2 x 3n-1

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Examples in contextPopulation growth

The number of seals on a

particular island was

estimated by a survey team

as 8500 at the start of 2011.

They also found that the

number of seals born in one year was equal to 40% of

the population at the start of that year.

The number of seals dying in a year is estimated to be

15% of the total population at the start of that year.

Express this as a recurrence relation.

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Examples in contextPopulation growth

Let P1 = 8500 (Jan 2011)

No. of births = 0.4 Pn

No. of deaths = 0.15 Pn

Pn+1 = Pn + 0.4 Pn - 0.15 Pn

Pn+1 = Pn + 0.25 Pn

Pn+1 = 1.25 Pn

How might this model be improved?

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Examples in contextCompound interest

Penny has invested £5000 in

a building society with a fixed

interest rate of 2.5%

compounded annually and

calculated at the end of December.

If she withdraws £100 on the 1st January every

year, find the recurrence relation that gives her

balance for the rest of the year.

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Examples in contextCompound interest

Let I0 = £5000 (initially)

Let I1 be the amount in the

account after 1 year

i.e. January of the next year.

In+1 = In + 0.025 In - 100

In+1 = 1.025 In - 100

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Examples in contextThe Tower of Hanoi

In this classic puzzle,

all the rings must be

moved from peg A to

peg C without a larger

ring being placed on top of a smaller ring.

How many moves does it take to transfer:

one ring? Two rings? Three rings? Ten rings?

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Examples in contextThe Tower of Hanoi

T1 = 1

T2 = 3

T3 = 7

What is the recurrence relation? Why?

Tn+1 = 2Tn + 1

The stack on C must be moved to B, the new disc

moved from A to C and the stack returned to C.

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Examples in contextThe Tower of Hanoi The Trilogic Game

The Doctor had to defeat the Celestial Toymaker

before the 1023rd move was played.

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Definitions and vocabularyThe order of a recurrence relation is the difference

between the highest and lowest subscripts in the relation.

an = 6an-1 + 16 First order

un+1 = 2un – 5(n-2) + 6(n-3) First order

xn = xn-1 + xn-2 + 2n3 -5n2 Second order

an+1 = 8an – 3an-1 + 2(an-2)2 Order three

xn+1 = 2xn-1 +3xn-2 + …. + 7xn-p Order p+1

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Definitions and vocabularyHomogeneous means the ‘same type’.

Homogeneous recurrence relations are those only

containing the same type of terms, e.g. an, an-1 or un+1, un.

an = 6an-1 + 16 Non-homogeneous

un+1 = 2un – 5(n-2) + 6(n-3) Non-homogeneous

xn = xn-1 + xn-2 + 2n3 -5n2 Non-homogeneous

an+1 = 8an – 3an-1 + 2(an-2)2 Homogeneous

xn+1 = 2xn-1 +3xn-2 + …. + 7xn-p Homogeneous

The only

non-linear

recurrence

relation

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Solving 1st order recurrence relationsExample: an = 3an-1, a1 = 2

a2 = 3a1 = 3x2 = 31x2

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a2 = 3a1 = 3x2 = 31x2

a3 = 3a2 = 3x(3x2) = 32x2

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a2 = 3a1 = 3x2 = 31x2

a3 = 3a2 = 3x(3x2) = 32x2

a4 = 3a3 = 3x(3x(3x2)) = 33x2

#MEIConf2019


a2 = 3a1 = 3x2 = 31x2

a3 = 3a2 = 3x(3x2) = 32x2

a4 = 3a3 = 3x(3x(3x2)) = 33x2

a5 = 3a4 = 3x(3x(3x(3x2))) = 34x2

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a2 = 3a1 = 3x2 = 31x2

a3 = 3a2 = 3x(3x2) = 32x2

a4 = 3a3 = 3x(3x(3x2)) = 33x2

a5 = 3a4 = 3x(3x(3x(3x2))) = 34x2

an = 3an-1 = 3x …..(3x(3x(3x2))) = 3n-1x2

an = 3n-1x2

What if the first term had been a0?

This

process is

called

recursion,

back-

substitution

or

induction.

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Recursion a.k.a. Induction or Back substitution

Example: un = 3un-1 + 8, u1 = 2

u2 = 3u1+8 = 3x2+8 = 31x2 + 8

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Example: un = 3un-1 + 8, u1 = 2

u2 = 3u1+8 = 3x2+8 = 31x2 + 8

u3 = 3u2+8 = 3x(3x2+8)+8 = 32x2 + 8(3+1)

#MEIConf2019


Example: un = 3un-1 + 8, u1 = 2

u2 = 3u1+8 = 3x2+8 = 31x2 + 8

u3 = 3u2+8 = 3x(3x2+8)+8 = 32x2 + 8(3+1)

u4 = 3u3+8 = 3x(3x(3x2+8)+8)+8 = 33x2 + 8(32+3+1)

#MEIConf2019


Example: un = 3un-1 + 8, u1 = 2

u2 = 3u1+8 = 3x2+8 = 31x2 + 8

u3 = 3u2+8 = 3x(3x2+8)+8 = 32x2 + 8(3+1)

u4 = 3u3+8 = 3x(3x(3x2+8)+8)+8 = 33x2 + 8(32+3+1)

u5 = 3u4+8 = 3x(3x(3x(3x2+8)+8)+8)+8 = 34x2 + 8(33+32+3+1)

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Example: un = 3un-1 + 8, u1 = 2

u2 = 3u1+8 = 3x2+8 = 31x2 + 8

u3 = 3u2+8 = 3x(3x2+8)+8 = 32x2 + 8(3+1)

u4 = 3u3+8 = 3x(3x(3x2+8)+8)+8 = 33x2 + 8(32+3+1)

u5 = 3u4+8 = 3x(3x(3x(3x2+8)+8)+8)+8 = 34x2 + 8(33+32+3+1)

un = 3un-1+8 = 3x(………………...….)+8 = 3n-1x2 + 8(3n-2+…+1)

N.B. (3n-2+3n-3+ …+32+31+1) is a G.P.; a=1 , ‘n’ = n-1, r=3

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Example: un = 3un-1 + 8, u1 = 2

u2 = 3u1+8 = 3x2+8 = 31x2 + 8

u3 = 3u2+8 = 3x(3x2+8)+8 = 32x2 + 8(3+1)

u4 = 3u3+8 = 3x(3x(3x2+8)+8)+8 = 33x2 + 8(32+3+1)

u5 = 3u4+8 = 3x(3x(3x(3x2+8)+8)+8)+8 = 34x2 + 8(33+32+3+1)

un = 3un-1+8 = 3x(………………...….)+8 = 3n-1x2 + 8(3n-2+…+1)

un = 3n-1x2 + 8[½(3n-1-1)]

un = 3n-1x2 + 4x3n-1- 4

un = 6x3n-1 - 4

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A general solutionExample: an = pan-1 + q, first term a1

a2 = pa1+q = pxa1+q = p1xa1 + q

a3 = pa2+q = px(pxa1+q)+q = p2xa1 + q(p+1)

a4 = pa3+q = px(px(pxa1+q)+q)+q = p3xa1 + q(p2+p+1)

a5 = pa4+q = px(px(px(pxa1+q)+q)+q)+q = p4xa1 + q(p3+p2+p+1)

an = pan-1+q = px(………………...….)+q = pn-1xa1 + q(pn-2+…+1)

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A general solutionExample: an = pan-1 + q, first term a1

an = pan-1+q = px(………………...….)+q = pn-1xa1 + q(pn-2+…+1)

an = pn-1xa1 + q

an = Axpn-1 + B

Why not just use this form?

1 1

1

np

p

1 1

1( 1)

1

n n

n

p p a q p qa

p

1

1(( 1) )

1 1

n

n

p a q p qa

p p

#MEIConf2019

Using an = Axpn-1 + BExample: an = 3an-1 + 2, u1 = 4

This sequence is 4, 14, 44, 134, …

We can write down: an = A x 3n-1 + B

From the first term: a1 = A + B = 4

From the second term: a2 = 3A + B = 14

⇒ 2A = 10 ⇒ A = 5 and B = -1

Solution: an = 5 x 3n-1 -1

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A harder problemExample: an = 2an-1 – 3n + 5, a0 = 2

This sequence is 2, 6, 11, 18, 29, …

We have seen that for recurrence relations of the type

an = pan-1 + q

we can simply use the structure

an = Axpn + B (N.B. first term a0)

and substitute to find the missing constants.

What do we do about that ‘extra bit’?

#MEIConf2019

A harder problemExample: an = 2an-1 – 3n + 5, a0 = 2

This sequence is 2, 6, 11, 18, 29, …

Try instead: an = A x 2n + Bn + C

From the first term: a0 = A + C = 2

From the second term: a1 = 2A + B + C = 6

From the third term: a2 = 4A + 2B + C = 11

⇒ A = 1, B =3, C = 1

Solution: an = 2n + 3n + 1

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The extra bit …Example: an = 2an-1 – 3n + 5, a0 = 2

This sequence is 2, 6, 11, 18, 29, …

Try instead: an = A x 2n + Bn + C

Essentially what we did was to solve the homogeneous

recurrence relation first and then try a possibility that would

solve the ‘other bit’.

This is leading towards a more general method …

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A general methodSolve first the homogeneous recurrence relation.

This gives the Complementary Function (C.F.).

Next by using a well-chosen trial function, find a

Particular Solution (P.S.) of the full relation.

Combine these to produce a General Solution

(G.S.) and then evaluate the missing constants.

G.S. = C.F. + P.S.

What does this remind you of?

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Remember this problem?Example: an = 2an-1 – 3n + 5, a0 = 2

First, consider: an = 2an-1 (the homogeneous relation)

This gives us: an = A x 2n Complementary Function

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Next, consider: an = 2an-1 – 3n + 5 (the full relation)

Trial function: an = Pn + Q (general linear fn.)

Substituting in:

Pn + Q = 2(P(n – 1) + Q) – 3n + 5

Pn + Q = 2Pn – 2P + 2Q – 3n + 5

Pn + Q = 2Pn – 3n – 2P + 2Q + 5

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Pn + Q = 2Pn – 3n – 2P + 2Q + 5

Equating similar terms:

Pn = 2Pn – 3n Q = – 2P + 2Q + 5

P = 2P – 3

3 = P

Q = – 6 + 2Q + 5

1 = Q

Trial function: an = Pn + Q

Particular Solution: an = 3n + 1

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There are two steps left to complete the solution; write down

the General Solution and evaluate the missing constant.

G.S. = C.F. + P.S.

C.F.: an = A x 2n

P.S.: an = 3n + 1

G.S.: an = A x 2n + 3n + 1

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G.S.: an = A x 2n + 3n + 1

Since a0 = 2

A x 20 + 3 x 0 + 1 = 2

A + 1 = 2

A = 1

Solution: an = 2n + 3n + 1

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A little bit of theorySuppose we have a first order linear homogeneous

recurrence relation of the form:

un – pun-1 = 0, n ≥ 1

Consider the substitution: un = Amn

Amn – pAmn-1 = 0 (the auxiliary equation)

Amn-1(m – p) = 0 Why is A ≠ 0?

So m = 0 or m = p Why is m ≠ 0?

Hence un = A x pn (C.F.)

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A harder exampleExample: an = 4an-1 + 3n2 – 5n + 1, n ≥ 1, a0 = 3

Write as: an - 4an-1 = 3n2 - 5n + 1

Solve: an - 4an-1 = 0

C.F.: an = A x 4n

Next, try: an = Bn2 + Cn + D

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A harder exampleExample: an - 4an-1 = 3n2 - 5n + 1, n ≥ 1, a0 = 3


Bn2 + Cn + D – 4[B(n – 1)2 + C(n – 1) + D] = 3n2 – 5n + 1

Bn2 + Cn + D – 4B(n – 1)2 – 4C(n – 1) – 4D = 3n2 – 5n + 1

Bn2 + Cn + D – 4B(n2 – 2n + 1) – 4Cn + 4C – 4D = 3n2 – 5n + 1

Bn2 + Cn + D – 4Bn2 + 8Bn - 4B – 4Cn + 4C – 4D = 3n2 – 5n + 1

#MEIConf2019



Bn2 + Cn + D – 4Bn2 + 8Bn – 4B – 4Cn + 4C – 4D = 3n2 – 5n + 1

n2: B – 4B = 3 B = -1

n1: C + 8B – 4C = – 5 C = -1

n0: D – 4B + 4C – 4D = 1 D = -⅓

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C.F.: an = A x 4n

P.S.: an = - n2 - n - ⅓

G.S. = C.F. + P.S.: an = A x 4n - n2 - n - ⅓

Since a0 = 2: a0 = A x 40 - 02 - 0 - ⅓ = 3

A = 3⅓

So: an = x 4n - n2 - n - ⅓10

3

#MEIConf2019

A special caseThis method, which is a subset of the method required to

solve second order recurrence relations, is very effective

but has its limitations. Consider the following case:

Example: an - an-1 = 3n – 4, n ≥ 1, a0 = 2

LHS suggests C.F.: an = A x 1n

RHS suggests trying: an = Bn + C

Could the solution be of the form: an = A x 1n + Bn + C ?

This sequence is: 2, 1, 3, 8, 16, …

#MEIConf2019

A special caseExample: an – an-1 = 3n - 4, n ≥ 1, a0 = 2

This sequence is: 2, 1, 3, 8, 16, …

The solution can’t be of the form: an = A x 1n + Bn + C

The sequence is quadratic and the proposed G.S. is linear.

In cases where: an – an-1 = f(n) i.e. an = an-1 + f(n)

a more powerful trial function of n needs to be used.

In this case the P.S. is of the form: an = Bn2 + Cn + D

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A special caseExample: an – an-1 = 3n - 4, n ≥ 1, a0 = 2

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Summary𝑎𝑛 = 𝑝𝑎𝑛−1 ⇔ 𝑎𝑛 = 𝐴⨯𝑝𝑛 (𝑛 ≥ 1)

𝑎𝑛 = 𝑝𝑎𝑛−1 + 𝑞 ⇔ 𝑎𝑛 = 𝐴⨯𝑝𝑛 + 𝐵 (𝑛 ≥ 1)

Methods

Back substitution (induction or recursion)

Trial solution, using initial term(s) to find constants

Solve Homogeneous form → Complementary Function

Use Trial function for RHS → Particular Solution

G.S. = C.F. + P.S.

and use initial term(s) to find constants

Take care if 𝑎𝑛 = 𝑎𝑛−1 + 𝑓(𝑛) !!

#MEIConf2019

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Teaching Discrete Mathematics

First Order Recurrence Relations consolidation exercise

1. A sequence is defined by the first-order recurrence relation:

𝑎𝑛 = 5𝑎𝑛−1 + 3

𝑎0 = 4

a. Write out the first 5 terms of this sequence.

b. Given that:

𝑎𝑛 = 𝐴 × 5𝑛 + 𝐵

show that 𝐴 =19

4 and 𝐵 = −

3

4 .

c. Use mathematical induction to prove that

𝑎𝑛 =19

4× 5𝑛 −

3

4

2. A sequence is defined by the formula:

𝑎𝑛 = 4 × 3𝑛 + 7

a. Write down the first 5 terms from 𝑎0 to 𝑎4

b. Given that the sequence also satisfies a recurrence relation

𝑎𝑛 = 𝑝𝑎𝑛−1 + 𝑞

𝑎0 = 𝑟

find the constants 𝑝, 𝑞, and 𝑟.

3. The following question links recurrence relations with proof by induction:

a. Find a closed formula for 𝑎𝑛 in terms of 𝑛 for the sequence given by the recurrence

relation:

𝑎𝑛 = 5𝑎𝑛−1 − 7

𝑎0 = 3

b. Use mathematical induction to prove that your formula is correct for all integers 𝑛 ≥ 0.


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4. A sequence is given by the recurrence relation:

𝑎𝑛 = 4𝑎𝑛−1 − 1

𝑎2 = 43

a. Find 𝑎1 and 𝑎0.

b. Find a closed formula for 𝑎𝑛 in terms of 𝑛 for the sequence.


𝑎𝑛 = 𝑎𝑛−1 + 3𝑛

𝑎0 = 2


b. Find a closed formula for 𝑎𝑛 in terms of 𝑛.


𝑎𝑛 = 5𝑎𝑛−1 − 3𝑛 + 1

𝑎0 = 2


b. Find a closed formula for 𝑎𝑛 in terms of 𝑛.

7. Consider this financial situation:

a. You decide to invest £500 on January 1st every year in a savings account that pays 3%

interest on December 31st each year. If 𝑎𝑛 is the amount you have on December 31st of

the 𝑛th year of the plan, write down a recurrence relation for 𝑎𝑛.

b. Use your recurrence relation to find a closed formula for 𝑎𝑛 in terms of 𝑛.


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First Order Recurrence Relations consolidation exercise solutions


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First Order Recurrence Relations key points

A recurrence relation is an equation defining a term in an iterative sequence based on one or more of its previous terms. This gives a ‘term-to-term’ rule for the sequence. A recurrence relation only defines a sequence uniquely if the value of one term (usually the first term) is also given. Such a sequence can be described as recursive and an alternative name for the connecting relationship is a difference equation. Examples:

a) a1 = 4, an = 5an-1, n ≥ 2

b) x0 = 2, x1 = 3, xn = 2xn-1 + 3xn-2, n ≥ 2

c) u1 = 0, u2 = 1, un = 2un-1 – 4un-2 + 6n2 + 3n, n ≥ 3

Terminology

The order of a recurrence relation is the difference between the highest and lowest subscripts on the general terms in the recurrence relation. A homogeneous recurrence relation contains only terms of the type un, un-1, etc., (homogeneous means ‘same type’). Linear recurrence relations are only based upon linear multiples of previous terms and do not involve powers or more complicated functions of these terms. Non-homogeneous recurrence relations will include an additional function of n or a constant term. The purpose of solving a recurrence relation is to find a closed formula which defines the terms in the sequence solely in terms of n, in other words a ‘position-to-term’ rule. The Complementary Function is the solution to the homogeneous recurrence relation. The Particular Solution is the function based on the non-homogeneous part, with constants found by trial, that satisfies the complete recurrence relation. Methods of solution

There are several methods for solving first order recurrence relations:

• Induction, also called back substitution or recursion, which involves repeatedly applying the recurrence relation and looking for a general pattern

• Inspection or recognition based on a substituting a few terms into a known general result

• General Solution equals Complementary Function plus Particular Solution, or G.S. = C.F. + P.S., a method which applies also to second order recurrence relations.

General results

un = pun-1 ⇔ un = A x pn, n ≥ 2 (In fact: un = pn-1u1 or un = pnu0)

un = pun-1 + q ⇔ un = A x pn + B, n ≥ 2 In the special case where p = 1, the result above does not apply. Instead:

un = un-1 + c ⇔ un = u1 + (n-1)c, n ≥ 2 [Additional extensive notes with alternative methods and a range of worked examples can be found in the document First Order Recurrence Relations: notes and examples.]

JT 27/06/19

Version 3.4

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Topic mapping for Discrete/Decision Mathematics courses from 2017

The following tables give a side-by-side comparison of the content of the various Discrete/Decision Maths units. It is a summary only and more detail can be found in the Further Mathematics specification documents for each awarding organisation.

Topics shown with a green background are covered in TD1. Topics shown with a purple background are covered in TD2.


Algorithms

Finite algorithms, definition AS D1 MwA AS

Trace and Interpret and apply algorithms AS D1 MwA AS

Repair, develop and adapt algorithms MwA AS

Complexity/Order of an algorithm (Big O notation) AS D1 MwA AS

Complexity and Efficiency AS

Proof, disproof, counter example MwA

Heuristics MwA AS

Sorting algorithms, comparisons and swaps AS D1 MwA AS

- Bubble sort AS D1 AS

- Quick sort AS D1 MwA A Level

- Shuttle sort AS

Packing algorithms, bin-packing AS D1 MwA AS

Extend knowledge of packing methods, e.g. 2D/3D, Knapsack A Level

Graph Theory

Nodes/vertices, degree/order AS AS D1 MwA AS

Arcs/edges, simple, connected, related vocabulary AS AS D1 MwA AS

Trees AS MwA AS

Euler’s relation: V - E + F = 2 AS A Level

Bipartite graphs, Km,n AS MwA A Level

Walk, trail, path, cycle AS AS

Eulerian, semi-Eulerian graphs AS AS D1 AS

Hamiltonian cycles AS A Level D1 A Level

Complete graphs, Kn AS AS D1 AS

Isomorphic Equivalence A Level AS D1 AS

Planar graphs AS AS D1 A Level

Subdivision and contraction A Level A Level

Planarity Algorithm A Level D1 Kuratowski’s Theorem A Level A Level

Thickness A Level

Complement of a graph A Level

Ore’s theorem A Level

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Networks

Modelling problems using weighted graphs AS AS D1 MwA AS

Adjacency/Incidence matrix AS AS D1 MwA AS

Minimum connector problem, Kruskal and Prim AS AS D1 MwA AS

Shortest path, Dijkstra AS D1 MwA AS

Complexity of Kruskal’s, Prim’s and Dijkstra’s algorithms MwA AS

Floyd’s algorithm A Level D1

Route inspection (Chinese Postman) AS AS D1 A Level

Travelling salesperson AS A Level D1 A Level

Solving network problems using technology MwA

Network Flows

Digraph representation AS AS D2 MwA AS

Evaluating/interpreting a cut AS AS D2 MwA

Max flow – min cut theorem AS AS D2 MwA

Labelling procedure for flow augmentation A Level AS D2

Supersource, supersink AS A Level D2 MwA

Upper and lower capacities A Level A Level D2

Restriction at a vertex A Level A Level D2

Linear Programming

Formulation of constrained problems into Linear Programs AS AS D1 MwA AS

Graphical solution using an objective function AS AS D1 MwA AS

Integer solution AS D1 MwA A Level

Slack variables A Level A Level D1 MwA A Level

Simplex Method A Level A Level D1 MwA A Level

Interpretation of Simplex A Level A Level D1 MwA A Level

Big M method A Level D1 MwA

Integer programming, branch-and-bound method A Level

Post-optimal analysis MwA A Level

Formulate a range of network problems as LPs MwA

Use of software and interpretation of output MwA

Critical Path Analysis

Precedence table for a set of sub-tasks AS AS D1 MwA AS

Construct an activity network (AQA: on node; others: on arc) AS AS D1 MwA AS

Forward/backward pass AS AS D1 MwA AS

Float times, critical activities, critical path AS AS D1 MwA AS

Refine model due to changes in context of problem AS

Interfering float MwA A Level

Gantt/Cascade chart A Level AS D1 MwA A Level

Resource histogram A Level A Level D1

Resource levelling A Level A Level D1

Scheduling A Level A Level D1 MwA A Level

Game Theory

Two-player, zero-sum games AS AS D2 AS

Pay-off matrix AS AS D2 AS

Play-safe strategies AS AS D2 AS

Stable solutions, value of the game AS AS D2 AS

Reduction of Pay-off matrix using dominance AS A Level D2 AS

Optimal mixed strategies AS AS D2 AS

Graphical solution AS AS D2 AS

Nash equilibrium solution A Level

Conversion of higher order games to LP problems A Level A Level D2 A Level

3 of 3


Allocation

Cost matrix reduction AS D2

Hungarian algorithm AS D2

Maximisation problems AS D2

Formulation as a Linear Programming problem A Level D2

Transportation problems

North-west corner method for an initial feasible solution A Level D2

Stepping-stone method for an improved solution A Level D2

Improvement indices A Level D2

Formulation as a Linear Programming problem A Level D2

Dynamic programming

Bellman’s principle A Level D2

Stage and State variables A Level D2

Maximum, minimum, maximin, minimax problems A Level D2

Network and table formats A Level D2

Recurrence relations

Modelling problems, e.g. population growth AS D2

Solution of first order linear homogeneous and non-homogeneous relations

AS D2

Solution of second order linear homogeneous and non-homogeneous relations

A Level D2

Particular solution, complementary function, auxiliary equation

AS D2

Decision Analysis

Construct and interpret decision trees A Level D2

Decision nodes, chance nodes, end pay-offs A Level D2

Expected monetary values (EMVs) and utility to compare courses of action

A Level D2

Mathematical Preliminaries

Classifying Problems AS

Set notation, Venn diagrams AS

Pigeonhole principle AS

Arrangement and selection, Permutations & Combinations AS

Inclusion-exclusion principle for two sets AS

Inclusion-exclusion principle for more than two sets A Level

Derangements A Level

Binary Operations

Modular Arithmetic and Matrix Multiplication AS

Commutativity, associativity AS

Cayley tables AS

Identity, inverse AS

Group Theory

Language: order, period, subgroup, proper, trivial, non-trivial A Level

Group axioms; closure, associativity, identity, inverse A Level

Finite/infinite groups A Level

Symmetry groups, abstract groups, modulo-n groups A Level

Cyclic groups and abelian groups A Level

Subgroups, order of group and element, Lagrange’s theorem A Level

Generators of groups A Level

Isomorphism between groups of finite order A Level

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