Download Linear Recurrence Relations, Part I

Linear Recurrence
Relations
Part I
Jorge Cobb
The University of Texas at Dallas
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Solving recurrence relations
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Setting up a recurrence relation is important –
it corresponds to modeling a problem
Solving it (providing a sequence that satisfies
the recurrence) is also needed
Sometimes, this can be done through
iteration/induction
For certain types of recurrence relations,
there are systematic methods for solving
them
Linear recurrence relations of degree k
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an = c1an-1 + c2an-2 + ... + ckan-k with c1,c2,...,ck
real numbers and ck≠0
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Linear:
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The right-hand side is a sum of weighted previous
terms of the sequence – the weights do not
depend on the sequence (but not necessarily
constant, may be a function of n)
Linear homogeneous recurrence
relations of degree k with constant
coefficients
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an = c1an-1 + c2an-2 + ... + ckan-k with c1,c2,...,ck real
numbers and ck≠0
Homogeneous:
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Of degree k:
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The recurrence goes back k terms, i.e., the earliest
previous term on the right hand side is an-k
Constant coefficients:
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No terms appear on the right hand side that are not
multiples of a previous term
The multipliers of the previous terms are all constants, not
functions that depend on n
Classifying recurrences
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Not linear
an=2an-1 + an-22
Not linear
an=an-1an-2
Yes, degree 2
an=an-1+an-2
an=1.05an-1
Yes, degree 1
an=nan-1
Coefficients are not constant
an=2an-1+1
Non-homogeneous
an=an-1+an-4
Yes, degree 4
Classification doesn’t depend on initial values
First degree linear homogeneous
recurrence relations with const. coef.
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an = c1an-1
Recall the compound interest example
Through iterative expansion,
an = c1an-1 = c1(c1an-2) = c12an-2 = c12(c1an-3) =
c13an-3 = ... = c1na0
Approach for a general
solution
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an = c1an-1 + c2an-2 + ... + ckan-k
Pretend that there is a solution of the form
an=rn for some constant r
rn = c1rn-1 + c2rn-2 + ... + ckrn-k
rk – c1rk-1 – c2rk-2 - ... – ck-1r – ck = 0
This is the characteristic equation of the
recurrence relation; the numbers r that satisfy
it are the characteristic roots of the
recurrence relation
Second degree linear homogeneous
recurrence relations
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Assumptions
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Recurrence relation: an = c1an-1 + c2an-2
 Characteristic equation has two distinct roots r1, r2.
Theorem 1
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{an} is a solution of the recurrence
if and only if
{an} is of the form an = b1r1n + b2r2n
for all n≥0, and for some constants b1, b2
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Proving Theorem 1
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We need to prove both directions:
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If {an} is a solution, then it must be of the form
an= b1r1n + b2r2n
for some appropriately chosen constants
b1 and b2
If {an} is of the form
an= b1r1n + b2r2n
for all n≥0, then it must be a solution.
We prove the second one first (harder )
Proving Theorem 1, first part
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Show an = b1r1n + b2r2n, n≥2, is a solution
an  c1an 1  c2 an 2
 c1 (b1r1n 1  b2 r2n 1 )  c2 (b1r1n 2  b2 r2n2 )
n2
1 1
br
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n2
2 2
(c1r1  c2 )  b r
(c1r2  c2 )
Because r2 – c1r – c2 = 0 and r1 and r2 are roots,
c1r1  c2  r , c1r2  c2  r
2
1
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Therefore,
an  b1r1n2 r12  b2 r2n2 r22
 b1r1n  b2 r2n
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2
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Proving Theorem 1, first part
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What about n=1 and n=0?
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These are just the initial conditions of our solution
a1  b r  b r  b1r1  b2 r2
1
1 1
1
2 2
a0  b1r10  b2 r20  b1  b2
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Proving Theorem 1, second part
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Show that if {an} is a solution it must be of the form
an = b1r1n + b2r2n (we must find b1 and b2)
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Note: a second-degree linear homogeneous
recurrence has a unique solution {an} if a0 and a1 are
specified
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Because there is a single way to generate the terms an
from the two initial values
Therefore, if we show that there is a solution of the
form an = b1r1n + b2r2n, n ≥0, that satisfies the initial
values a0 and a1, this must be the solution that was
given to us, and we are done.
Proving Theorem 1, second part
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We first show that there are b1 and b2 such that
b1r10 + b2r20 = a0 and b1r11 + b2r21 = a1
These simplify to
b 1 + b2 = a0
b1r1 + b2r2 = a1, (two linear equations, two unknowns)
b1  a0  b2
a1  a0 r1
(a0  b2 )r1  b2 r2  a1  ( r2  r1 )b2  a1  a0 r1  b2 
r2  r1
a1  a0 r1 a0 r2  a0 r1  a1  a0 r1 a0 r2  a1
b1  a0  b2  a0 


r2  r1
r2  r1
r2  r1
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Proving Theorem 1, second part
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Consider the sequence {a’n}, where
a’n = b1r1n + b2r2n, and a’0 = a0, a’1 = a1
From the first part of the theorem, we have shown that
a’n = b1r1n + b2r2n is a solution for any b1 and b2
We have shown (previous slide) that b1 and b2 can be
chosen so that the satisfy a’0 = a0, a’1 = a1
Thus, {a’n} is a solution, just like {an}, and with the same
initial conditions.
However, the initial conditions determine the rest of the
sequence.
Hence, {a’n} = {an} and {an} is in the desired form.
Notes about the proof
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Our proof of the second part also serves as
the means to find the solution
It depended on r1 ≠ r2
The characteristic roots r1 and r2 may be
complex numbers (the proof and the solution
are still valid)
Example solution
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an = an-1 + 2an-2, a0=2 and a1=7
Characteristic equation
r2 – r – 2 = 0
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The quadratic formula for ax2+bx+c = 0,
 b  b2  4ac
x
2a
gives roots r1=(1-(3))/2=-1 and r2=(1+(3))/2 =
2
Example solution
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Therefore,
a1  a0 r1 7  2  ( 1) 9
b2 

 3
r2  r1
2  ( 1)
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a0 r2  a1 2  2  7  3
b1 


 1
r2  r1
2  ( 1)
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an  b1r1n  b2 r2n  1  ( 1)n  3  2n  3  2n  ( 1)n
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