Section 8.2
223
7. (a) [BB] Since
5n
:s;
n
3
for all
n
~
3, we can take C
=
1,
no
=
3, or C
=
5,
no
=
1.
(b) For
n
~
1, we have
17n
4
+
8n
3
+
5n
2
+
6n
+
1
:s;
17n
4
+
8n
n
+
5n
4
+
6n
4
+
n
4
=
37n4.
So
we can take C
=
37 and
no
=
1.
(c) [BB] For
n
~
1, we have
8n
3
+
4n
2
+
5n
+
1
:s;
9n
4
+
18n
2
+
24n
+
6
=
3(3n4
+
6n
2
+
8n
+
2),
so we can take C
=
3,
no
=
1.
(d) For
n
~
1,2
n
:s;
3
n
,
so take C
=
1,
no
=
1.
(e) For
n
~
1, we have
f(n)
=
2n
2
+
3n
+
1
:s;
2n2
+
3n
2
+
5
=
5n
2
+
5
=
5g(n),
so take
no
=
1
andc
=
5.
(f) We have
f(n)
=
2n
2

3n
+
5
:s;
2n
2
+
3n
+
5
:s;
2n
2
+
3n
2
+
5n
2
=
10n
2
for any
n
~
1.
Now
n
2
:s;
2(n
2

7n)
=
2n2
14n
=
2g(n)
for
n
~
14 since
(2n2 14n)

n
2
=
n
2
14n
=
n(n

14)
~
0 for
n
~
14. So, for
n
~
14,
f(n)
:s;
10n
2
:s;
20g(n).
So
f
=
O(g)
with
no
=
14
and c
=
20.
(g) Since 10g2
n
5
=
510g
2
n
and since 10g2
n
:s;
n
for any
n
~
1 we have 10g2
n
5
:s;
5n
for
n
~
1.
So take c
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 any
 Graph Theory, Trigraph, Existence, Proposition, Sufficiently large

Click to edit the document details