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Homework 11 Solutions
Math 371, Fall 2011
(1)
(Richardson Extrapolation)
P. 454 #11
(a) Adding the Taylor series for
f
(
x
0
+
h
) and
f
(
x
0

h
) gives
f
(
x
0
+
h
) +
f
(
x
0

h
) =
2
f
(
x
0
) +
h
2
f
00
(
x
0
) +
h
4
12
f
(4)
(
x
0
) +
h
8
2880
f
(8)
(
x
0
) +
h
8
2880
[
f
(8)
(
ξ
)

f
(8)
(
x
0
)]
.
Thus
f
00
(
x
0
) =
f
(
x
0
+
h
)

2
f
(
x
0
) +
f
(
x
0

h
)
h
2

h
2
12
f
(4)
(
x
0
)

h
4
360
f
(6)
(
x
0
)

h
6
2880
f
(8)
(
x
0
) +
o
(
h
6
)
.
(b)
h
O
(
h
2
)
O
(
h
4
)
O
(
h
6
)
0.5
2.255252
0.25
2.062826 2.049998
0.125 2.015645 2.012500 2.010000
(2)
(Newton–Cotes).
Suppose that
f
is a function with four continuous derivatives on the
interval [
a,b
]. Recall that the error bound for the composite trapezoidal rule
T
(
h
) with
panel width
h
is
T
(
h
)

Z
b
a
f
(
x
) d
x
=
(
b

a
)
h
2
12
f
00
(
ξ
)
for some
ξ
∈
[
a,b
]. The error in the composite Simpson’s rule
S
(
h
) with panel width
h
is
S
(
h
)

Z
b
a
f
(
x
) d
x
=
(
b

a
)
h
4
180
f
(4)
(
ξ
)
for some (diﬀerent)
ξ
∈
[
a,b
].
(a) Let
f
(
x
) = e

x
sin
x
. For the composite trapezoidal rule and the composite Simpson’s
rule, ﬁnd the number of panels
n
required to integrate
f
on the interval [0
,
2
π
] with
error at most 10

4
. Recall that
h
= 2
π/n
. How many function evaluations are required
in each case?
The ﬁrst ﬁve derivatives of
f
are
1
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f
0
(
x
) = e

x
(cos
x

sin
x
)
f
00
(
x
) =

2e

x
cos
x
f
000
(
x
) = 2e

x
(sin
x
+ cos
x
)
f
(4)
(
x
) =

4e

x
sin
x
f
(5)
(
x
) = 4e

x
(sin
x

cos
x
)
To develop the bound for the composite trapezoidal rule, we need to calculate the
maximum value of

f
00
(
x
)

on the interval [0
,
2
π
]. The extremum must occur at
x
= 0,
x
= 2
π
, or at a point where
f
000
(
x
) = 0. A short calculation establishes that the
maximum occurs when
x
= 0, so
max
x
∈
[0
,
2
π
]

f
00
(
x
)

= 2
.
To ensure that the error is no more than 10

4
, the error bound requires
2
π h
2
12
max
x
∈
[0
,
2
π
]

f
00
(
x
)
 ≤
10

4
.
Solving this inequality for the panel width
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 Fall '08
 KRASNY
 Taylor Series

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