Exercise 2.3.30(19).
The iteration equation for the secant method can be written in the simpler
form
p
n
=
f
(
p
n

1
)
p
n

2

f
(
p
n

2
)
p
n

1
f
(
p
n

1
)

f
(
p
n

2
)
.
Explain why, in general, this iteration equation is likely to be less accurate than the one given in
Algorithm 2.4.
Solution.
In the usual secant method,
p
n
=
p
n

1

f
(
p
n

1
)(
p
n

1

p
n

2
)
f
(
p
n

1
)

f
(
p
n

2
)
, roundoff errors in computing
f
(
p
n

1
), (
p
n

1

p
n

2
) and
f
(
p
n

1
)

f
(
p
n

2
) cause loss of relative accuracy in the
correction
to
p
n

1
. But this correction is likely to be much smaller than
p
n

1
, so having only, say, two correct
digits of the corrector can cause, say, the 11th and 12th digit to be fixed in updating
p
n

1
to obtain
p
n
. In the formula above, the new
p
n
will not have more correct digits than the denominator, which
may suffer severely from cancellation error (especially as the terms converge to the correct answer
and
f
(
p
n

1
) and
f
(
p
n

2
) are both close to zero.)
Exercise 2.4.1.
Use Newton’s method to find solutions accurate to within 10

5
for the following
problems.
b.
cos(
x
+
√
2) +
x
(
x/
2 +
√
2) = 0
,
(

2
≤
x
≤ 
1)
.
Solution.
Let
f
(
x
) = cos(
x
+
√
2) +
x
(
x/
2 +
√
2). Then
f
0
(
x
) =

sin(
x
+
√
2) + (
x/
2 +
√
2) +
x
·
1
2
=
x
+
√
2

sin(
x
+
√
2)
.
So to implement Newton’s method, we use fixedpoint iteration with the function
g
(
x
) =
x

f
(
x
)
f
0
(
x
)
=
x

cos(
x
+
√
2) +
x
(
x/
2 +
√
2)
x
+
√
2

sin(
x
+
√
2)
.
(
*
)
Since the interval is [

2
,

1], we will start with initial value

1
.
5. Then the sequence of approxi
mations generated by Newton’s method is as follows:

1
.
5
,

1
.
47855
,

1
.
46247
,

1
.
4504
,

1
.
44135
,

1
.
43457
,

1
.
42948
,

1
.
42566
,

1
.
4228
,

1
.
42065
,

1
.
41904
,

1
.
41784
,

1
.
41693
,

1
.
41625
,

1
.
41574
,

1
.
41536
,

1
.
41507
,

1
.
41486
,

1
.
4147
,

1
.
41459
,

1
.
41451
,

1
.
41449
,

1
.
41445
,

1
.
41445
.
The slow convergence rate is due to
f
(
x
) having a zero of multiplicity 4 at
x
=

√
2. To see this,
note that
f
(
s

√
2) = cos(
s
)

1+
s
2
/
2 =
O
(
s
4
), so
f
(
x
) =
O
(

x
+
√
2

4
). In fact, Newton’s method
fails to converge to the requested accuracy due to roundoff error in doubleprecision. (The answer
is

1
.
41421.)