151A HW 5 Solutions
Will Feldman
1. See book for similar problems.
2. Using the Newton forward difference formula given in the problem statement
P
n
(
x
0
+
sh
) =
f
(
x
0
) +
n
X
k
=1
s
k
4
k
f
(
x
0
)
we want to calculate (
4
2
P
n
)(10) given values for (
4
k
P
n
)(0) for
k
= 2
,
3
,
4. A derivation of the above
formula can be found in the textbook. You can just think of it as a simplification of the formula for
the Newton polynomial
P
n
(
x
) =
f
[
x
0
] +
∑
n
k
=1
f
[
x
0
, ..., x
k
]Π
k

1
i
=0
(
x

x
i
) in the case that the
x
j
are
evenly spaced. We can derive a formula for (
4
2
P
n
)(
x
0
+
sh
) similar to the Newton forward difference
formula above. Note that
(
4
P
n
)(
x
0
+
sh
) =
P
n
((
x
0
+
h
) +
sh
)

P
n
(
x
0
+
sh
)
=
P
n
(
x
0
+
h
) +
n
X
k
=1
s
k
(
4
k
P
n
)(
x
0
+
h
)

"
P
n
(
x
0
+
h
) +
n
X
k
=1
s
k
(
4
k
P
n
)
#
=
P
n
(
x
0
+
h
)

P
n
(
x
0
+
h
) +
n
X
k
=1
s
k
(
4
k
P
n
)(
x
0
+
h
)

(
4
k
P
n
)(
x
0
)
=
4
P
n
(
x
0
) +
n
X
k
=1
s
k
(
4
k
+1
P
n
)(
x
0
)
.
By applying the same reasoning again we get that,
(
4
2
P
n
)(
x
0
+
sh
) = (
4
2
P
n
)(
x
0
) +
n
X
k
=1
s
k
(
4
k
+2
P
n
)(
x
0
)
.
This is almost what we want since it is a formula for
4
2
P
n
which depends only on
4
k
P
n
(0) for
k
≥
2.
To finish the problem you just need to note that
4
P
n
(
x
) =
Q
n

1
(
x
) where
Q
n

1
is a polynomial of
degree
≤
n

1. This will imply that Δ
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 Spring '08
 hitrik
 Math, Numerical Analysis, pn, Polynomial interpolation, Newton polynomial

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