The document discusses recurrence relations and provides examples of how they can be used to model real-world scenarios like loan payments. It explains that a recurrence relation is a function that includes itself, and provides examples of simple and more complex recurrence relations. It also gives an example of how a recurrence relation can model the remaining debt on a loan each month over the life of the loan based on the initial amount, monthly payment, and interest rate.

1. RECURRENCE RELATIONS
RECURRENCE RELATIONS
RECURRENCE RELATIONS
RECURRENCE RELATIONS
RECURRENCE RELATIONS
Keaton Brandt 2010 MST Gateway Project

2. What are Recurrence Relations?
• Functions that Include Themselves

3. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=f(n-1)+2

4. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=f(n-1)+2
n f(n)
0 1
1 3
2 5
3 7
4 9
5 11
6 13

5. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=f(n-1)+2
n f(n)
0 1 Starting Point
1 3
2 5
3 7
4 9
5 11
6 13

6. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=f(n-1)+2
n f(n)
0 1 Starting Point
1 3 Starting Point + 2
2 5
3 7
4 9
5 11
6 13

7. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=f(n-1)+2
n f(n)
0 1 Starting Point
1 3 Starting Point + 2
2 5 (Starting Point + 2) + 2
3 7
4 9
5 11
6 13

8. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=f(n-1)+2
n f(n)
0 1 f(0)
1 3 f(1-1)+ 2
2 5 f(2-1) + 2
3 7
4 9
5 11
6 13

9. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=7f(n-1)-12f(n-2)
n f(n)
0 2
1 7
2 25
3 91
4 337
5 1267
6 4825

10. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=7f(n-1)-12f(n-2)
n f(n)
0 2 Starting Point 1
1 7 Starting Point 2
2 25
3 91
4 337
5 1267
6 4825

11. What are Recurrence Relations?
• Functions that Include Themselves
f(n)=7f(n-1)-12f(n-2)
n f(n)
0 2 Starting Point 1
1 7 Starting Point 2
2 25 7(7)-12(2)
3 91
4 337
5 1267
6 4825

12. Closed Forms
Closed Forms of Recursive Equations produce
the same values, but do not reference themselves.

13. Finding Closed Forms
This is where it gets hard.

14. A Simple Example
b(n) = { 1
b(n-1)*2
if n = 0
if n > 0

15. A Simple Example
b(n) = { 1
b(n-1)*2
if n = 0
if n > 0
1,2,4,8,16,32,64,128

16. A Simple Example
b(n) = { 1
b(n-1)*2
if n = 0
if n > 0
150
112.5
75
37.5
0
0 1 2 3 4 5 6 7

17. A Simple Example
b(n) = { 1
b(n-1)*2
if n = 0
if n > 0
150
112.5
75
Exponential
37.5
0
0 1 2 3 4 5 6 7

18. A Simple Example
b(n) = { 1
b(n-1)*2
if n = 0
if n > 0
Exponential
f(n) = abn

19. A Simple Example
f(n) = f(n-1) * 2

20. A Simple Example
f(n) = 2 f(n-1)

21. A Simple Example
n-1
f(n) = 2

22. A Simple Example
n-1
f(n) = a 2

23. A Simple Example
n-1
f(n) = a 2
Exponential
f(n) = abn

24. A Simple Example
n-1
f(n) = a 2
What is a?

25. A Simple Example
n-1
f(n) = a 2

26. A Simple Example
n-1
f(n) = a 2
f(0) = 1

27. A Simple Example
n-1
f(n) = a 2
f(0) = 1
1 = a2 0-1

28. A Simple Example
1= a2 0-1

29. A Simple Example
1= a2 0-1
1 = a2 -1

30. A Simple Example
1= a2 0-1
1 = a2 -1
1 = a*.5

31. A Simple Example
1= a2 0-1
1 = a2 -1
1 = a*.5
.5 .5

32. A Simple Example
1 = a*.5
.5 .5

33. A Simple Example
1 = a*.5
.5 .5
2=a

34. A Simple Example
1 = a*.5
.5 .5
2=a
a=2

35. A Simple Example
b(n) = 2*2n-1

36. A Simple Example
b(n) = 2*2n-1
b(n) = 2n

37. A Simple Example
b(n) = 2n
1,2,4,8,16,32,64,128

38. A Harder Example
b(n) = { 2
5
5b(n-1)+-6b(n-2)
if n = 0
if n = 1
if n > 1

39. A Harder Example
b(n) = { 2
5
5b(n-1)+-6b(n-2)
if n = 0
if n = 1
if n > 1
2,5,13,35,97,275,793

40. A Harder Example
b(n) = { 2
5
5b(n-1)+-6b(n-2)
if n = 0
if n = 1
if n > 1
Exponential + Exponential
f(n) = abn+ cdn

41. A Harder Example
b(n ) = 5b(n-1)+-6b(n-2)

42. A Harder Example
b n= 5bn-1 - 6bn-2
bn-2 bn-2 bn-2

43. A Harder Example
b 2= 5b1-6

44. A Harder Example
0 = 1b 2+-5b +6
a b c

45. A Harder Example
-b ± √(b2-4(
a*c ))
2a

46. A Harder Example
5 ± √(25-4(1 *6 ))
2 *1

47. A Harder Example
5 ± √(25-24)
2

48. A Harder Example
5±√1
2

49. A Harder Example
5±1
2

50. A Harder Example
5±1
2
6 4
&
2 2

51. A Harder Example
5±1
2
Roots: (3,2)

52. A Harder Example
b(n) = a*3n+ c*2 n

53. A Harder Example
b(n) = a*3n+ c*2 n

54. A Harder Example
b(n) = a*3n+ c*2 n
2= a*3 0+ c*2 0

55. A Harder Example
b(n) = a*3n+ c*2 n
2= a*3 0+ c*2 0
2 = a+ c

56. A Harder Example
2 = a+ c
5= a*3 1+ c*2 1

57. A Harder Example
2 = a+ c
5= a*3 1+ c*2 1
5 = 3a + 2c

58. A Harder Example
5 = 3a + 2c
2 = a+ c

59. A Harder Example
5 = 3a + 2c
2 = a+ c
*3 *3 *3

60. A Harder Example
5 = 3a + 2c
2 = a+ c
*3 *3 *3
6 = 3a + 3c

61. A Harder Example
5 = 3a + 2c
6 = 3a + 3c

62. A Harder Example
5 = 3a + 2c
6 = 3a + 3c
1 = 0a + 1c

63. A Harder Example
1 = 1c

64. A Harder Example
1 = 1c
c=1

65. A Harder Example
c=1
a+c=2
-c -c

66. A Harder Example
c=1
a+c=2
-c -c
a=2-c

67. A Harder Example
c=1
a+c=2
-c -c

68. A Harder Example
c=1
a+c=2
-c -c
a=2-1

69. A Harder Example
c=1

70. A Harder Example
c=1
a=1

71. A Harder Example
a=1 c=1
b=3 d=2

72. A Harder Example
a=1 c=1
b=3 d=2
b(n) = 1*3n + 1*2 n

73. A Harder Example
b(n) = 1*3 n + 1*2 n

74. A Harder Example
b(n) = 1*3 n + 1*2 n
2,5,13,35,97,275,793

75. The Fibonacci Sequence
f(n)=f(n-1)+f(n-2)
Starting Points: (0,1)

76. The Fibonacci Sequence
f(n)=f(n-1)+f(n-2)
Starting Points: (0,1)
0,1,1,2,3,5,8,13,21,34,55

77. The Fibonacci Sequence
1,1,2,3,5,8,13,21,34,55

78. The Fibonacci Sequence
1,1,2,3,5,8,13,21,34,55

79. An Aside
The golden ratio

80. The golden ratio
1+
ϕ=
2

81. The fibonacci solution
1+
ϕ=
2
n n
ϕ -(1-ϕ)
F(n) =

82. The fibonacci solution
n n
ϕ -(1-ϕ)
F(n) =
√(5

83. The fibonacci solution
n n
ϕ -(1-ϕ)
F(n) =
√(5
0,1,1,2,3,5,8,13,21,34,55

84. Φ = 1.6180339887498

85. The Golden Ratio
13
8

86. The Golden Ratio
13/ 8 = 1.625
ϕ = 1.6180339887498

89. Calculating Loans

90. Calculating Loans
The Total Cost, and monthly payments, of a
loan can calculated with a recurrence relation.

91. EXAMPLE

92. EXAMPLE
•A car costs $10,000

93. EXAMPLE
•A car costs $10,000
• The buyer can pay a $1000 down payment

94. EXAMPLE
•A car costs $10,000
• The buyer can pay a $1000 down payment
• The buyer can then pay $350 per month

95. EXAMPLE
•A car costs $10,000
• The buyer can pay a $1000 down payment
• The buyer can then pay $350 per month
• The yearly interest rate on the loan is 5%

96. EXAMPLE
•A car costs $10,000
• The buyer can pay a $1000 down payment
• The buyer can then pay $350 per month
• The yearly interest rate on the loan is 5%
• The seller wants the loan paid off in 24 months

97. Remaining debt per month
n = The number of months since the start of the loan
m = The monthly payment
i = The yearly interest rate (decimal)
b(n,m,i) = { 9000
(1+(i/12))*b(n-1,m,i)-m
if n = 0
if n > 0

98. Remaining debt per month
n = The number of months since the start of the loan
b(n) = { 9000 if n = 0
(1+(.05/12))*b(n-1)-350 if n > 0

99. Remaining debt per month
b(n) = { 9000 if n = 0
(1+(.05/12))*b(n-1)-350 if n > 0
n b(n) n b(n) n b(n)
0 9,000.00 11 5,489.98 22 1,815.70
1 8,687.50 12 5,162.86
23 1,473.27
2 8,373.70 13 4,834.37
24 1,129.40
3 8,058.59 14 4,504.51
25 784.11
4 7,742.17 15 4,173.28
26 437.38
5 7,424.42 16 3,840.67
6 7,105.36 17 3,506.67 27 89.20
7 6,784.97 18 3,171.28 28 -260.43
8 6,463.24 19 2,834.50
9 6,140.17 20 2,496.31
10 5,815.75 21 2,156.71

100. Remaining debt per month

101. Remaining debt per month
It takes 28 months to pay off the loan
The seller wants it payed off in 24 months
The total cost, down payment, and interest rate can’t be changed
The monthly payment must be increased

102. How much greater?

103. How much greater?
Need to Solve the Equation for m (Monthly Payment)
b(n,m,i) = { 9000
(1+(i/12))*b(n-1,m,i)-m
if n = 0
if n > 0

104. Remaining debt per month
n = The number of months since the start of the loan
b(n) = { 9000 if n = 0
(1+(.05/12)) *b(n-1)-350 if n > 0

105. Remaining debt per month
n = The number of months since the start of the loan
b(n) = { 9000 if n = 0
(1+(.05/12))*b(n-1)-350 if n > 0
a c
Formula
c-(1-a n)
b(n) = x*a n-1 -
a-1

106. Remaining debt per month
n = The number of months since the start of the loan
350-(1-(1+(.05/12))n)
b(n) = x * (1+(.05/12))n-1 -
.05/12

107. Remaining debt per month
n = The number of months since the start of the loan
350-(1-(1+(.05/12))0)
9000 = x * (1+(.05/12))0-1 -
.05/12

108. Remaining debt per month
n = The number of months since the start of the loan
350-(1-1)
9000 = x * 1/(1+(.05/12)) -
.05/12

109. Remaining debt per month
n = The number of months since the start of the loan
9000 = x * 0.99585 - 84000
+84000 +84000

110. Remaining debt per month
n = The number of months since the start of the loan
93000 = x * 0.99585
0.99585 0.99585

111. Remaining debt per month
n = The number of months since the start of the loan
93387 = x

112. Remaining debt per month
n = The number of months since the start of the loan
350-(1-(1+(.05/12))n)
bn = 93387 * (1+(.05/12))n-1 -
.05/12

113. Remaining debt per month
n = The number of months since the start of the loan
m-(1-(1+(.05/12))24)
0 = 93387 * (1+(.05/12))24-1 -
.05/12

114. Remaining debt per month
n = The number of months since the start of the loan
m+0.104941
0 = 93387 * 1.100357 -
.05/12

115. Remaining debt per month
n = The number of months since the start of the loan
0 = 102374.419 - (1/(.05/12))(m+0.104941)

116. Remaining debt per month
n = The number of months since the start of the loan
102374.419 = (1/(.05/12))(m+0.104941)
/(1/(.05/12)) /(1/(.05/12))

117. Remaining debt per month
n = The number of months since the start of the loan
406.56 = (m+0.104941)
-0.104941 -0.104941

118. Remaining debt per month
n = The number of months since the start of the loan
404.46 = m

119. Remaining debt per month
b(n) = { 9000 if n = 0
(1+(.05/12))*b(n-1)-404.56 if n > 0
n b(n) n b(n) n b(n)
0 9,000.00 11 4,854.70 22 515.39
1 8,630.94 12 4,468.37
23 110.98
2 8,260.34 13 4,080.43
24 -295.12
3 7,888.20 14 3,690.87
25 -702.91
4 7,514.51 15 3,299.69
26 -1,112.40
5 7,139.26 16 2,906.88
6 6,762.45 17 2,512.43 27 -1,523.59
7 6,384.06 18 2,116.34 28 -1,936.50
8 6,004.10 19 1,718.59
9 5,622.56 20 1,319.19
10 5,239.43 21 918.13

120. Loan Conclusion
The buyer will have to pay somewhere around $406.46 per
month in order to pay of his loan in time.

121. Recurrence Relations Conclusion
Any function that references itself.
A common example is the fibonacci sequence.
Can be solved for a non self-referential ‘closed form’
Can be used to calculate loan payment schedules

122. RECURRENCE RELATIONS
RECURRENCE RELATIONS
RECURRENCE RELATIONS
RECURRENCE RELATIONS
RECURRENCE RELATIONS
Keaton Brandt 2010 MST Gateway Project