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x 0 ) n + 1
( n + 1 )!
= ( x
f [ n + 1 ] ( ψ )
for some ψ satisfying x 0
ψ
x .
2 2 n− 1 /
by applying it to the function f ( w )=
( 1 + w ) 1 on the interval [ 0, 1 ] . The Taylor expansion of this function is
( 1 + w ) 1 = 1
Taylor's theorem is used to expand
|
u
|
w 3 ( 1 + ψ ) 4
w + w 2
1. The magnitude of the last term is at most w 3 .
for some 0
ψ
Let n
12, k =
n/ 2
, l =
n/ 12
and restrict
|
u
|
as follows:
= 2 k +
|
u
|
|
a
|
where
= 2 l
|
a
|
|
b
|
+ 1and
2 l− 1
|
b
|≤
1
2 2 l− 1
2 l + 1 < 2 2 l− 1 for l
It follows that
|
a
|≤
1. Applying the Taylor series expansion
/ 2 k ) 1 ,wehave
2 2 n− 1
( 2 k +
to ( 1 +
|
a
|
= 2 2 n− 1 −k 1
( 1 + ψ ) 4
+ |
2 k 2
|
2 k 3
|
|
2 k
a
a
|
a
|
(2.14)
|
a
|
)
for some 0
both the sum of the first two terms
and the third term on the right-hand side have the following bounds:
2 2 n− 1 −k ( 1
ψ
1. For the given range of values for
|
u
|
/ 2 k ) > 2 2 n− 1 −k 1
2 2 l− 1 / 2 k
−|
a
|
/ 2 k ) 2 < 2 2 n− 1 −k 2 2 l− 1 / 2 k 2
Since 2 2 l− 1 / 2 k < 1 / 2, the value of the third term, 2 2 n− 1 −k (
2 2 n− 1 −k (
|
a
|
/ 2 k ) 2 , is an integer that does
not overlap in any bit positions with the sum of the first two terms.
The fourth term is negative; its magnitude has the following upper bound:
2 2 n− 1 4 k
|
a
|
3 ( 1 + ψ ) 4 < 2 3 ( 2 l− 1 )+ 2 n− 1 4 k
|
a
|
Expanding the third term, we have
2 2 n− 1 3 k (
) 2 = 2 2 n− 1 3 k ( 2 2 l
2 + 2 l + 1
|
a
|
|
b
|
|
b
|
+ 1 )
Because 3 ( 2 l
k , the third term on the right-hand side of this expansion has value
2 2 n− 1 3 k and is larger than the magnitude of the fourth term in ( 2.14 ). Consequently the
fourth term does not affect the value of the result in ( 2.14 ) in positions occupied by the binary
representation of 2 2 n− 1 3 k ( 2 2 l
1 )
2 + 2 l + 1
) .Inturn,2 l + 1
is less than 2 2 l ,whichmeans
|
b
|
|
b
|
|
b
|
that the binary representation of 2 2 n− 1 3 k ( 2 2 l
2 ) appears in the output shifted but otherwise
without modification. This provides the following result.
|
b
|
LEMMA 2.10.1 The reciprocal function f ( n )
recip
contains as a subfunction the squaring function
f ( m )
square for m =
1 .
Proof The value of the l -bit binary number denoted by b appears in the output if l =
n/ 12
n/ 12
1.
Lower bounds similar to those derived for the reciprocal function can be derived for special
fractional powers of binary numbers. (See Problem 2.35 .)
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