Review of recurrence relations
The sequence $10,15,20,25,30$10,15,20,25,30... is an arithmetic progression as the sequence recurs by a constant addition. It can be defined using the recursive rule:
$T_{n+1}=T_n+5,T_1=10$Tn+1=Tn+5,T1=10 or $T_n=T_{n-1}+5,T_1=10$Tn=Tn−1+5,T1=10
The sequence $2,6,18,54,162$2,6,18,54,162... is a geometric progression as the sequence recurs by a constant multiplication. It can be defined using the recursive rule:
$T_{n+1}=3T_n,T_1=2$Tn+1=3Tn,T1=2 or $T_n=3T_{n-1},T_1=2$Tn=3Tn−1,T1=2
Now consider the progression, $3,10,31,94,283,\dots$3,10,31,94,283,… This sequence recurs by multiplying the previous term by 3 and then adding 1 to the result.
Neither arithmetic nor geometric
The first order linear recurrence equation for the sequence mentioned above is $T_{n+1}=3T_n+1$Tn+1=3Tn+1 with $T_1=3$T1=3. This progression is neither geometric nor arithmetic, but a combination of both.
$T_{n+1}=kT_n+d$Tn+1=kTn+d with $T_1=a$T1=a or alternatively $T_n=kT_{n-1}+d$Tn=kTn−1+d with $T_1=a$T1=a
Approaching a steady state
For recurrence relations of the form $t_{n+1}=kt_n+d$tn+1=ktn+d, the long term behaviour of the sequence is dependent on the value of $k$k. Let's consider two different sequences and use our calculator to observe the long term behaviour using both a table and graph.
Sequence 1. $t_{n+1}=0.5t_n-2,\ t_1=12$tn+1=0.5tn−2, t1=12. In this example $k$k is between $-1$−1and $1$1.
Sequence 2. $t_{n+1}=1.5t_n-2,\ t_1=12$tn+1=1.5tn−2, t1=12. In this example $k$k is greater than $1$1.
Select your calculator brand below to view instructions for observing the long-term behaviour of a sequence.
Casio Classpad
How to use the CASIO Classpad to complete the following tasks regarding recursive rules
Generate the terms of the sequence with the recursive relationship $t_n=0.5t_{n-1}-2,t_1=12$tn=0.5tn−1−2,t1=12 using a table to observe the long-term behaviour.
Generate a graph of the sequence and describe the long-term behaviour.
Generate the terms of the sequence with the recursive relationship $t_n=1.5t_{n-1}-2,t_1=12$tn=1.5tn−1−2,t1=12 using a table to observe the long-term behaviour.
Generate a graph of the sequence and describe the long-term behaviour.
TI Nspire
How to use the TI Nspire to complete the following tasks regarding recursive rules
Generate the terms of the sequence with the recursive relationship $t_n=0.5t_{n-1}-2,t_1=12$tn=0.5tn−1−2,t1=12 using a table to observe the long-term behaviour.
Generate a graph of the sequence and describe the long-term behaviour.
Generate the terms of the sequence with the recursive relationship $t_n=1.5t_{n-1}-2,t_1=12$tn=1.5tn−1−2,t1=12 using a table to observe the long-term behaviour.
Generate a graph of the sequence and describe the long-term behaviour.
We can say that a recurrence relation $t_{n+1}=kt_n+d$tn+1=ktn+d converges to a steady state when $k$k is strictly between $-1$−1 and $1$1. Otherwise, we say the recurrence relation diverges.
A steady state is reached when $T_{n+1}=T_n$Tn+1=Tn, that is successive terms are not changing. We can solve algebraically for the steady state by assuming the terms on both sides of the recurrence formula are equal to a constant value, say $x$x, and then solve for $x$x.
Worked example
For the first order linear recurrence relation defined by $t_{n+1}=0.5t_n-2$tn+1=0.5tn−2, $t_1=12$t1=12 find the steady state solution.
Think: The steady state solution is the constant value that the terms of the sequence approach. We can find the steady-state solution by setting $t_{n+1}=t_n$tn+1=tn. To make the working out easier, we can let both these terms be $x$x, and then we solve for $x$x.
Do: Let $x$x be the steady state solution, then for the sequence $t_{n+1}=0.5t_n-2$tn+1=0.5tn−2, at the steady state we have:
| $x$x | $=$= | $0.5x-2$0.5x−2 |
Substitute $x$x for $t_{n+1}$tn+1 and $t_n$tn. |
| $0.5x$0.5x | $=$= | $-2$−2 |
Take $0.5x$0.5x from both sides. |
| $\therefore\ x$∴ x | $=$= | $-4$−4 |
Multiply both sides by $2$2. |
Hence, the steady state for this sequence is $-4$−4.
Reflect: This is the sequence we inspected in parts (a) and (b) of the calculator instructions given above. Does this solution agree with our previous observations? Also notice the starting value does not influence the long term behaviour.
When we are dealing with a steady state solution we set up an equation with $T_{n+1}=T_n$Tn+1=Tn.
Practice questions
Question 1
Consider the following sequence.
$4$4, $-28$−28, $224$224, $-1372$−1372, ...
Is the sequence arithmetic, geometric or neither?
arithmetic
Aneither
Bgeometric
C
Question 2
Consider the sequence defined by $a_1=5$a1=5 and $a_n=-2a_{n-1}-3$an=−2an−1−3 for $n\ge2$n≥2.
What is the first term of the sequence?
What is the second term of the sequence?
What is the third term of the sequence?
What is the fourth term of the sequence?
What is the fifth term of the sequence?
Question 3
Consider the recurrence relation $u_{n+1}=2u_n+2$un+1=2un+2, $u_0=2$u0=2.
Find $u_1$u1.
Find $u_2$u2.
Find $u_3$u3.
Find $u_4$u4.
Complete the table of values.
$n$n $1$1 $2$2 $3$3 $4$4 $u_n$un $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Graph the relation.
Loading Graph...Find the largest value of $n$n for which $u_n$un $<$< $22$22.
Question 4
The sequence defined by the recurrence relation $u_{n+1}=ku_n-16$un+1=kun−16, where $u_1=3$u1=3, approaches a long-term steady-state value of $-40$−40. Find the value of $k$k.