1 Linear Recurrence Relations Part I Jorge Cobb The University of Texas at Dallas.
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Transcript of 1 Linear Recurrence Relations Part I Jorge Cobb The University of Texas at Dallas.
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1
Linear Recurrence Relations
Part I
Jorge Cobb
The University of Texas at Dallas
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Solving recurrence relations 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
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Linear recurrence relations of degree k
an = c1an-1 + c2an-2 + ... + ckan-k with c1,c2,...,ck real numbers and ck≠0
Linear: 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)
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an = c1an-1 + c2an-2 + ... + ckan-k with c1,c2,...,ck real numbers and ck≠0
Homogeneous: No terms appear on the right hand side that are not
multiples of a previous term
Of degree k: The recurrence goes back k terms, i.e., the earliest
previous term on the right hand side is an-k
Constant coefficients: The multipliers of the previous terms are all constants, not
functions that depend on n
Linear homogeneous recurrence relations of degree k with constant coefficients
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Classifying recurrences
an=2an-1 + an-22
an=an-1an-2
an=an-1+an-2
an=1.05an-1
an=nan-1
an=2an-1+1
an=an-1+an-4
Classification doesn’t depend on initial values
Not linear
Non-homogeneous
Not linearYes, degree 2
Coefficients are not constant
Yes, degree 1
Yes, degree 4
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First degree linear homogeneous recurrence relations with const. coef.
an = c1an-1
Recall the compound interest example Through iterative expansion, an = c1an-1 = c1(c1an-2) = c1
2an-2 = c12(c1an-3) =
c13an-3 = ... = c1
na0
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Approach for a general solution
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
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Second degree linear homogeneous recurrence relations
Assumptions
Recurrence relation: an = c1an-1 + c2an-2 Characteristic equation has two distinct roots r1, r2.
Theorem 1
{an} is a solution of the recurrence
if and only if
{an} is of the form an = b1r1n + b2r2
n for all n≥0, and for some constants b1, b2
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Proving Theorem 1 We need to prove both directions:
If {an} is a solution, then it must be of the form an= b1r1
n + b2r2n
for some appropriately chosen constants b1
and b2
If {an} is of the form an= b1r1
n + b2r2n
for all n≥0, then it must be a solution.
We prove the second one first (harder )
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Proving Theorem 1, first part Show an = b1r1
n + b2r2n, n≥2, is a solution
Because r2 – c1r – c2 = 0 and r1 and r2 are roots,
Therefore,
)()(
)()(
2212
222112
11
222
2112
122
1111
2211
crcrbcrcrb
rbrbcrbrbc
acaca
nn
nnnn
nnn
22221
21211 , rcrcrcrc
nn
nnn
rbrb
rrbrrba
2211
22
222
21
211
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Proving Theorem 1, first part
What about n=1 and n=0?
These are just the initial conditions of our solution
210
220
110
2211122
1111
bbrbrba
rbrbrbrba
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Proving Theorem 1, second part
Show that if {an} is a solution it must be of the form an = b1r1
n + b2r2n (we must find b1 and b2)
Note: a second-degree linear homogeneous recurrence has a unique solution {an} if a0 and a1 are specified
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 = b1r1
n + 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.
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Proving Theorem 1, second part
We first show that there are b1 and b2 such that b1r1
0 + b2r20 = a0 and b1r1
1 + b2r21 = a1
These simplify to b1 + b2 = a0
b1r1 + b2r2 = a1, (two linear equations, two unknowns)
12
1012101212122120 )()(
rr
raabraabrrarbrba
12
120
12
1011020
12
1010201 rr
ara
rr
raarara
rr
raaabab
201 bab
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Consider the sequence {a’n}, where a’n = b1r1
n + b2r2n, and a’0 = a0, a’1 = a1
From the first part of the theorem, we have shown that a’n = b1r1
n + 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.
Proving Theorem 1, second part
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Notes about the proof
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)
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Example solution
an = an-1 + 2an-2, a0=2 and a1=7 Characteristic equation
r2 – r – 2 = 0 The quadratic formula for ax2+bx+c = 0,
gives roots r1=(1-(3))/2=-1 and r2=(1+(3))/2 = 2
a
acbbx
2
42
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Example solution
Therefore,
33
9
)1(2
)1(27
12
1012
rr
raab
13
3
)1(2
722
12
1201
rr
arab
nnnnnnn rbrba )1(2323)1(12211