### ppt - People Server at UNCW

```Chapter 8
Recursion
8.3
More Recurrence
Second-Order Recurrence
• Definition
– A second-order linear homogeneous recurrence relation
with constant coefficients is a recurrence relation of the
form,
ak = Aak-1 + Bak-2 for all ints k ≥ some fixed int, where A and
B are fixed real numbers and B≠0.
– This is called a second order because expression is based
on the previous two terms (ak-1 , ak-2).
– Its linear ak-1 , ak-2 are separate terms of the first order.
– Homogenous because are terms are of the same order.
– Constant coefficient b/c A and B are fixed real numbers
that do not depend on k.
Example
• State whether each is a second-order linear
homogenous recurrence relation.
•
•
•
•
•
ak = 3ak-1 + 2ak-2
bk = bk-1 + bk-2 + bk-3
dk = d2k-1 + dk-1 * dk-2
ek = 2ek-2
f k = 2f k-1 + 1
Distinct Roots
• Lemma 8.3.1
– Let A and B be real numbers. A recurrence relation
of the form ak = Aak-1 + Bak-2 is satisfied by the
sequence 1, t, t2, t3, … , tn, … where t is non-zero
real number if, and only if, t satisfies the equation
t2 – At – B = 0.
Characteristic Equation
• Definition
– Given a second-order linear homogenous
recurrence relation with constant coefficients:
ak = Aak-1 + Bak-2 for all ints k≥2,
the characteristic equation of the relation is
t2 – At – B = 0
Example
• Use characteristic eq to find solutions to a recurrence
problem.
– Consider recurrence relation where kth term of a sequence
equals the sum of the (k-1)st term plus twice the (k-2) term.
ak = ak-1 + 2ak-2
– Find all sequences that satisfy 1, t, t2, t3, … , tn, …
– Solution
• t2 – t - 2 = 0
• t2 – t - 2 = (t – 2)(t + 1), thus the values of t = 2, -1
• t can be: 1, 2, 22, 23, … , 2n OR 1, -12, -13, … -1n
• This example demonstrates how to find two distinct
sequences that satisfy a given second-order homogenous
recurrence relation with constant coefficients.
Single Root Case
• Consider ak = Aak-1 + Bak-2 for ints k≥2, but consider that the
characteristic equation t2 – At – B = 0 has a single root.
• By Lemma 8.3.1 one sequence that satisfies the relation is:
1, r, r2, r3, … rn, … however another is: 0, r, 2r2, 3r3, … , nrn
• To see this observe:
t2 – At – B = (t – r)2 = t2 – 2rt – r2, A=2r B=-r2
• sn = nrn for all ints n≥0
• Ask-1 + Bsk-2 = A(k -1)rk-1 + B(k -2)rk-2
= 2r(k -1)rk-1 + -r2 (k -2)rk-2
= 2(k -1)rk - (k -2)rk
= (2k – 2 – k + 2) rk
= krk = sk
Single Root
• Lemma 8.3.4
– Let A and B be real numbers and suppose the
characteristic eq t2 – At – B = 0
has a single root r. Then the sequence 1, r1, r2, … ,
rn, … and 0, r, 2r2, … nrn, … both satisfy the
recurrence relation
ak = Aak-1 + Bak-2 for all ints k≥2
Single Root
• Theorem 8.3.5
– Suppose a sequence satisfies a recurrence relation
ak = Aak-1 + Bak-2
for some real numbers A and B with B≠0 and for all
ints k≥2. If the characteristic eq t2 – At – B = 0
has a single (real) root r then the sequence a0, a1, a2,
… satisfies the explicit formula
an = Crn + Dnrn
where C and D are the real numbers whose values are
determined by the values of a0 and any other known
value of the sequence.
Example
• Suppose b0, b1, b2 … satisfies the recurrence relation bk = 4bk-1 – 4bk2 for all ints k ≥ 2 with initial conditions b0 = 1 and b1 = 3.
Find an explicit formula for the sequence.
• Solution
– sequences is of second-order linear homogenous recurrence relation
with constant coefficients (A=4 and B=-4). The single-root condition is
also met because the characteristic equation t2 – 4t + 4 = 0 has a single
root r = 2 ( (t-2)(t-2) )
– bn = C 2n + Dn2n
– to find C and D use initial conditions
• b0 = 1 = C 20 +D(0)20 => C = 1
• b1 = 3 = C 21 +D(1)21 => 2C + 2D = 3 (sub C = 1 from above)
• 3 = 2(1) + 2D => D = ½
– Hence, bn = 2n + ½ n2n for all ints n≥2
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