MEI PowerPoint Template#MEIConf2019 Defining Recurrence Relations Any sequence in which each...
Transcript of 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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Easy examplesConsider these two sequences:
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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Easy examplesConsider these two sequences:
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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Easy examplesConsider these two sequences:
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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Solving 1st order recurrence relationsExample: an = 3an-1, a1 = 2
a2 = 3a1 = 3x2 = 31x2
a3 = 3a2 = 3x(3x2) = 32x2
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Solving 1st order recurrence relationsExample: an = 3an-1, a1 = 2
a2 = 3a1 = 3x2 = 31x2
a3 = 3a2 = 3x(3x2) = 32x2
a4 = 3a3 = 3x(3x(3x2)) = 33x2
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Solving 1st order recurrence relationsExample: an = 3an-1, a1 = 2
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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Solving 1st order recurrence relationsExample: an = 3an-1, a1 = 2
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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Recursion a.k.a. Induction or Back substitution
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)
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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
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)
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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
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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Recursion a.k.a. Induction or Back substitution
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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Recursion a.k.a. Induction or Back substitution
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
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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’?
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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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Remember this problem?Example: an = 2an-1 – 3n + 5, a0 = 2
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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Remember this problem?Example: an = 2an-1 – 3n + 5, a0 = 2
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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Remember this problem?Example: an = 2an-1 – 3n + 5, a0 = 2
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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Remember this problem?Example: an = 2an-1 – 3n + 5, a0 = 2
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
Next, try: an = Bn2 + Cn + D
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
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A harder exampleExample: an - 4an-1 = 3n2 - 5n + 1, n ≥ 1, a0 = 3
Next, try: an = Bn2 + Cn + D
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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A harder exampleExample: an - 4an-1 = 3n2 - 5n + 1, n ≥ 1, a0 = 3
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
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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, …
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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) !!
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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.
5. A sequence is given by the recurrence relation:
𝑎𝑛 = 𝑎𝑛−1 + 3𝑛
𝑎0 = 2
a. Write out the first 5 terms of this sequence.
b. Find a closed formula for 𝑎𝑛 in terms of 𝑛.
6. A sequence is given by the recurrence relation:
𝑎𝑛 = 5𝑎𝑛−1 − 3𝑛 + 1
𝑎0 = 2
a. Write out the first 5 terms of this sequence.
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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Teaching Discrete Mathematics
First Order Recurrence Relations consolidation exercise solutions
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Teaching Discrete © MEI 2018
Teaching Discrete Mathematics
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.
Discrete/Decision Mathematics Topics AQA Edexcel MEI OCR A
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
2 of 3
Discrete/Decision Mathematics Topics AQA Edexcel MEI OCR A
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
Discrete/Decision Mathematics Topics AQA Edexcel MEI OCR A
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