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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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