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2 SOLVING RECURRENCE RELATIONS
EXAMPLE 1. Consider the recurrence relation with initial conditions given by an = 2anโ 1 โ anโ 2 , a 0 = 0, a 1 = 3
a. Use Excel to try to find a (nonrecursive) solution.
Formulas A B 1 n a_n 2 recursively 3 0 0 4 =A3+1 3 (^5) โ =2*B4โB (^6) โ โ (^7) โ โ (^8) โ โ (^9) โ โ (^10) โ โ
Values A B 1 n a_n 2 recursively 3 0 0 4 1 3 5 2 6 6 3 9 7 4 12 8 5 15 9 6 18 10 7 21
From the table of values, it appears that a solution of this recurrence relation with initial conditions is
an = 3n. We can check to see if our answer is reasonable by inserting a third column containing the values of the nonrecursive sequence. Formulas A B C 1 n a_n a_n 2 recursively nonrecursively 3 0 0 =3A 4 =A3+1 (^3) โ (^5) โ =2B4โB3 โ (^6) โ โ โ (^7) โ โ โ (^8) โ โ โ (^9) โ โ โ (^10) โ โ โ
Values A B C 1 n a_n a_n 2 recursively nonrecursively 3 0 0 0 4 1 3 3 5 2 6 6 6 3 9 9 7 4 12 12 8 5 15 15 9 6 18 18 10 7 21 21
b. Verify that an = 3n is indeed a solution of this recurrence relation with initial conditions.
Note that an = 3n is a solution of this recurrence relation since 2anโ 1 โ anโ 2 = 2โ
3(n โ 1) โ 3(n โ 2) = 6(n โ 1) โ 3(n โ 2) = 6n โ 6 โ 3n + 6 = 3n = an. Also we have that a 0 = 3โ
0 = 0 and a 1 = 3โ
1 = 3. Thus, the initial conditions are also satisfied.
EXAMPLE 2. Find the solution of the following recurrence relation with initial condition: an = 2anโ 1 , a 0 = 4
a. List the first few terms and make a guess at the formula.
b. Use an iterative approach to develop the formula.
EXAMPLE 4. Find the solution of the following recurrence relation with initial condition: an = anโ 1 + 5, a 1 = 13
EXAMPLE 5. For the quadratic sequence fn = an^2 + bn + c, complete the following table.
n fn = an^2 + bn + c First differences Second differences
1 2 3 4 5 6 7
a. What is true about the first differences?
b. What is true about the second differences?
c. Regression commands on TI-83 Plus calculator can be used to find linear, quadratic, cubic, and quartic equations that fit given data.
Store the data. Enter command at home screen.
Result of command.# Thus, we have that
1 + 2 + 3 + โฆ + n = 0.5n^2 + 0.5n + 0 = 0.5n^2 + 0.5n = 0.5n(n + 1)
=
n(n 1) 2
Set up window for plot. Set up STAT PLOT.
Result of STAT PLOT. Store regression equation.
Result with reg. equation.
(^) If R (^2) value does not show up, use to the command DiagonisticOn from the Catalog menu to activate the
R^2 value. An R^2 value of 1 indicates that the equation fits the data perfectly.
The following formulas will be useful in solving certain recurrence relations:
1 + 2 + 3 + โฆ + n =
n(n 1) 2
12 + 2^2 + 3^2 + โฆ + n^2 =
n(n 1)(2n 1) 6
13 + 2^3 + 3^3 + โฆ + n^3 =
n (n 1) 4
1 + r + r^2 + โฆ + rn^ =
r 1 r 1
n + (^1) โ
โ
for r โ 1
EXAMPLE 7. Find the solution of the following recurrence relation with initial condition: an = anโ 1 + 6n, a 0 = 5
ANSWERS
- S = 1 + r + r^2 + โฆ + rn
rS = r + r^2 + r^3 + โฆ + rn+
rS โ S = (r + r^2 + r^3 + โฆ + rn+1) โ (1 + r + r^2 + โฆ + rn) rS โ S = r + r^2 + r^3 + โฆ + rn+1^ โ 1 โ r โ r^2 โ โฆ โ rn) rS โ S = rn+1^ โ 1 (r โ 1)S = rn+1^ โ 1
S = r 1 r 1
n + (^1) โ
โ
4a. an = โ7n + 52
4b. bn = 5n+
4c. cn = 8n^3 + 12n^2 + 4n + 60
4d. en = n!
4e. fn = (โ1)n
4f. gn = 3n^4 + 6n^3 + 3n^2
