Linear Recurrence Relations Limits

1. Block 1
Linear Recurrence Relations
Limits

2. What is to be learned?
• When linear recurrence relations have a
limit
• How to calculate the limit
• How to interpret the results

3. Consider Un+1 = 0.5Un + 8 U0 = 10
U1 = 0.5(10) + 8 = 13
U2 = 0.5(13) + 8 = 14.5
U3 = 15.25
U4 = 15.625
U5 = 15.813
U6 = 15.906
U7 = 15.953
U8 = 15.977
U9 = 15.988
UU1111 = 15.997= 15.997
UU1212 = 15.999= 15.999
UU1313 = 15.999= 15.999
UU1414 = 16.000= 16.000
UU1515 = 16.000= 16.000
““Reaches” aReaches” a
limit of 16limit of 16

4. Un+1 = 0.5Un + 8 U14 = 16, U15 = 16
Limit = 16
U15 = 0.5(16) + 8
= 16
U16 = 0.5(16) + 8
= 16
etc.
Limit of 16 means does not go over 16

5. When is there a Limit?
for recurrence relation Un+1 = mUn + c
goes to limit if -1 < m < 1

6. Finding the Limit
Un+1 = 0.5Un + 8
-1 < m < 1
so as n∞
Equation becomes
L = 0.5L + 8
-0.5L -0.5L
0.5L = 8
L = 8
/0.5 = 16
UUnnUUn+1n+1 Limit LLimit L

7. When is there a Limit?
for recurrence relation Un+1 = mUn + c
goes to limit if -1 < m < 1

8. Finding the Limit
Un+1 = 0.8Un + 10
-1 < m < 1
so as n∞
Equation becomes
L = 0.8L + 10
-0.8L -0.8L
0.2L = 10
L = 10
/0.2 = 50
UUnnUUn+1n+1 Limit LLimit L

9. Key Question
Un+1 = 0.2Un + 20
-1 < m < 1
so as n∞
Equation becomes
L = 0.2L + 20
-0.2L -0.2L
0.8L = 20
L = 20
/0.8 = 25
UUnnUUn+1n+1 Limit LLimit L

10. Ex Paul Uter’s factory dumps 200kg of yuk
at the end of each day into the nearby
loch.
The council install a filter scheme
that removes 30% of the yuk each day
If the yuk exceeds 700kg, Tony trout
and his pals will perish horribly
Is Tony safe?
loch yukloch yuk outout
–– 30%30%
InIn
+ 200+ 200
YYn+1n+1 = 0.7Y= 0.7Ynn + 200+ 200
X 0.7X 0.7
Yn
X 0.7X 0.7 + 200+ 200
Yn+1

11. -1 < m < 1
so as n∞
Equation becomes
L = 0.7L + 200
-0.7L -0.7L
0.3L = 200
L = 200
/0.3 = 666.7Kg of yuk
After a period of time there will be a limit of
666.7 Kg at the start of each day.
Tony and pals are safe!
YYnnYYn+1n+1 Limit LLimit L
YYn+1n+1 = 0.7Y= 0.7Ynn + 200+ 200

12. Ex Bertie’s balloon has sprung a leak.
It is losing 20% of its air each hour.
To compensate Bertie pumps 10m3
into
it at the end of each hour.
If volume of air drops below 45m3
, the
balloon will crash.
Is Bertie doomed?
air volumeair volume outout
–– 20%20%
InIn
+ 10+ 10
VVn+1n+1 = 0.8V= 0.8Vnn + 10+ 10
(X 0.8)(X 0.8)
Vn
X 0.8X 0.8 + 10+ 10
Vn+1

13. -1 < m < 1
so as n∞
Equation becomes
L = 0.8L + 10
-0.8L -0.8L
0.2L = 10
L = 10
/0.2 = 50m3
After a period of time there will be a limit of
50m3
at the start of each hour.
Bertie is safe!
VVnnVVn+1n+1 Limit LLimit L
VVn+1n+1 = 0.8V= 0.8Vnn + 10+ 10
or is he?or is he?

14. -1 < m < 1
so as n∞
Equation becomes
L = 0.8L + 10
-0.8L -0.8L
0.2L = 10
L = 10
/0.2 = 50m3
After a period of time there will be a limit of
50m3
at the start of each hour.
Bertie is safe!
VVnnVVn+1n+1 Limit LLimit L
VVn+1n+1 = 0.8V= 0.8Vnn + 10+ 10
or is he?or is he?
will not gowill not go
aboveabove 5050

15. Volume at start of hour = 50m3
During next hour
V = 0.8(50) + 10
= 40
VVn+1n+1 = 0.8V= 0.8Vnn + 10+ 10
+ 10

16. Volume at start of hour = 50m3
During next hour
V = 0.8(50) + 10
= 40 + 10
Will drop below 45 during the hour
Nae luck Bertie
VVn+1n+1 = 0.8V= 0.8Vnn + 10+ 10

17. Limit Sneaky Formula
for Un+1 = mUn + c
Limit = c
1 – m1 – m

18. Henry’s hedge is causing a bit of annoyance to his
neighbor Brutal Boris. His hedge is increasing is
height by 80cm each year. To compensate Henry
cuts 30% from the height if the hedge each June.
Boris has said that if the height exceeds 2m then he
will do Henry some serious mischief.
Should Henry be worried?
Hn+1 = 0.7Hn + 80
-1 < m < 1 so as n∞
L = 0.7L + 80
0.3L = 80
L = 266 2
/3 cm
Not good for Henry!
HHnnHn+1Hn+1  Limit LLimit L
Key Question