Section 8.2
227
also involves the multiplication
of
a number with perhaps
n
+ 2 digits by 2, requires at most
3n
+
[3(n
+
2)

2]
=
6n
+ 4 operations. The next step requires at most
6n
+ 7 operations, and
so on. Assuming
k
steps, the number
of
multiplications and divisions by 2 is at most
(6n

2) +
(6n
+
1)
+
(6n
+ 4) +
(6n
+
7)
+
...
+
(6n
+
3k

5)
=
k(6n)
+
[2
+
1
+ 4 +
...
+
(3k

5)]
=
k(6n)
+
~[2(
2)
+
(k

1)3]
=
6kn
+
~(3k

7)
=
~k2

~k
+
6kn
~
~k2
+
6kn,
using the fact that the sum
of
the first
k
terms
of
the arithmetic sequence
a,
a
+
d, a
+
2d,
...
is
~[2a
+
(k

l)d]see
(7)
on
p.
234. By part (a),
k
<
1
+ nlog21O, so the number
of
multiplications and divisions by 2 is at most
(c) We must add at most
k
integers, each with at most
n
+
k
digits. Using Problem 16, we see that
this requires at most
k[2(n
+
k)
+
k]
~
(1
+
nlog21O)((2n
+ 3 +
3nlog2
1O
)
=
0(n
2
)
basic operations.
(d)
We
use the result
of
Exercise 9. Since the number
of
operations required by each
of
the two stages
of
the algorithm is
0
(
n
2
), the total number
of
operations is
0
(
n
2
), as desired.
27. (a) For real numbers
a
and
b,
we have
(a+b)2
=
a
2
+2ab+b
2
~
lal
2
+2lallbl
+
Ibl
2
=
(Ial +
Ib1)2.
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 Summer '10
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 Graph Theory, Multiplication, Mathematical Induction, Natural number, ixi, lal + Ibl, im n+ oo

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