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Discrete Mathematics University of Kentucky CS 275 Spring ... - MGNet

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Example: a n = 8a n41 + 10 n41 with a 0 = 1, which gives us a 1 = 9. Find a n in closed<br />

form. First multiply the recurrence relation by x n to give us<br />

a n x n + 8a n!1<br />

x n + 10 n-1 x n !<br />

. If f(x) = " a k<br />

x k , then<br />

f(x) 4 1 =<br />

!<br />

!<br />

!<br />

k=1<br />

!<br />

k=1<br />

k=0<br />

a k<br />

x k<br />

= (8a k-1<br />

x k +10 k-1 x k )<br />

= 8xf(x) + x/(1410x)<br />

Hence,<br />

f(x) = 1!9x<br />

(1!8x)(1!10x)<br />

= 1 "<br />

1<br />

2 1!8x + 1!10x<br />

1<br />

$<br />

$<br />

'<br />

#<br />

!<br />

k=0<br />

!<br />

%<br />

'<br />

'<br />

&<br />

= 1 8 k +10 k<br />

2<br />

#<br />

&x k<br />

or a n = .5(8 k +10 k ).<br />

"<br />

$<br />

%<br />

153<br />

Note: It is possible to prove many identities using generating functions.<br />

Exclusion-Inclusion Theorem: Given sets A i , 1;i;n, the number <strong>of</strong> elements in<br />

the union is<br />

n<br />

i=1<br />

A i<br />

n<br />

i=1<br />

"<br />

! = ! A i<br />

! 1!i

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