151A HW 6 Solutions
Will Feldman
1. Let
f
∈
C
n
+1
([
a, b
]) and
p
be the interpolating polynomial at the
n
+ 1 equally spaced node points
a
=
x
0
, ..., x
n
=
b
with (
b

a
)
/n
=
h
=
x
j
+1

x
j
. Recall that we have the error bound

f
(
x
)

p
(
x
)
 ≤

f
(
n
+1)
(
ξ
)

(
n
+ 1)!
Π
n
j
=0

x

x
j

.
(a) Now for
x
∈
[
a, b
] arbitrary there is some
j
such that
x
j
≤
x
≤
x
j
+1
. We an upper bound for
g
(
x
) =

(
x

x
j
)(
x

x
j
+1
)

= (
x

x
j
)(
x
j
+1

x
). Just use calculus to find 0 =
g
0
(
x
) =
x
j
+
x
j
+1

2
x
so that the maximum value of
g
occurs when
x
= (
x
j
+
x
j
+1
)
/
2.
Plugging in we get that

(
x

x
j
)(
x

x
j
+1
)
 ≤
(
h/
2)
2
.
(b) Now for
i < j
use the bound (
x

x
i
)
≤
(
j

i
+ 1)
h
so that
Π
i<j
(
x

x
i
)
≤
h
j
Π
i<j
(
j

i
+ 1) = (
j
+ 1)!
h
j
.
Similarly for
i > j
+ 1 we have that (
x
i

x
)
≤
(
i

j
)
h
so that
Π
i>j
+1
(
x
i

x
)
≤
h
n

j

1
Π
i>j
+1
(
i

j
) = (
n

j
)!
h
n

j

1
.
Combining the two above bounds with the one from (a) gives the desired result.
(c) Note that (
n

j
)! = (
n

j
)
·
(
n

j

1)
· · ·
2 has
n

j

1 terms in the product and
n
!
(
j
+1)!
=
n
·
(
n

1)
· · ·
(
j
+ 2) also has
n

j

1 terms and
n

j

i
≤
n

i
for
i
= 0
, ..., n

j

2 so
(
n

j
)! = Π
n

j

2
i
=0
(
n

j

i
)
≤
Π
n

j

2
i
=0
(
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 Spring '08
 hitrik
 Math, Midpoint Formula, Polynomial interpolation, xj, Midpoint

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