Chapter 8
243
(b) S
·
12
22
2
n(n
+
1)(2n
+
1)
mce
+
+
... +
n
=
''6''
see Problem 4 of Section 5 .
Ithe result follows
because a polynomial has the same order as its highest power, by Proposition 8.2.7.
8. (a) Letf(n)
=
3n
3
5n
2
+2n+1.
Bythetriangleinequality,forn
~
l,lf(n)1
~
3n
3
+5n
2
+2n+
1
~
3n
3
+5n
3
+2n
3
+n
3
=
11n
3
,
so
f(n)
=
O(n
3
)
with
no
=
1 andc
=
11 in Definition 8.2.1.
Conversely (and here we borrow an idea from the proof of Proposition 8.2.7),
3[ 5
2
1] 3[ 5
2
1]
f(n)
=
n
3  
+  + 
~
n
3    
 
.
n
n
2
n
3
n
n
2
n
3
Choose
no
such that
!,
~
and
;;ta
are all less than
~
for
n
~
no,
for example,
no
=
16. Then
5
1

>
n

3'
2
1
>
n
2

3
and
1
1
>
n
3

3
so that
f(n)
~
n
3
[3 
~

~

~]
=
2n
3
;
hence
n
3
~ ~f(n)
for
n
~
no.
So
n
3
=
O(f(n))
with c
=
~
in Definition 8.2.1.
(b) Let
f(n)
=
n
4

2n
3
+3n
2

5n+7.
By the triangle inequality, forn
~
l,lf(n)1
~
n
4
+2n
3
+
3n
2
+5n+7
~
n
4
+2n4 +3n
4
+5n
4
+7n
4
=
18n
4
,
so
f(n)
=
O(n4)
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 Summer '10
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 Graph Theory, Prime number, Sieve of Eratosthenes, 3n3 +5n 2 +2n+ 1 ~ 3 n3 +5n3

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