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1200
Fourth-Degree Interpolation — Editorial
algebra
polynomials
interpolation
Define
Q
(
x
)
=
P
(
x
)
−
x
.
Q(x) = P(x) - x.
Q
(
x
)
=
P
(
x
)
−
x
.
Then
Q
Q
Q
is a degree-
4
4
4
polynomial with
Q
(
1
)
=
Q
(
2
)
=
Q
(
3
)
=
Q
(
4
)
=
0
,
Q(1) = Q(2) = Q(3) = Q(4) = 0,
Q
(
1
)
=
Q
(
2
)
=
Q
(
3
)
=
Q
(
4
)
=
0
,
so
Q
(
x
)
=
k
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Q(x) = k(x-1)(x-2)(x-3)(x-4)
Q
(
x
)
=
k
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
for some constant
k
.
k.
k
.
From
P
(
5
)
=
15
P(5) = 15
P
(
5
)
=
15
we get
Q
(
5
)
=
10
,
Q(5) = 10,
Q
(
5
)
=
10
,
hence
10
=
k
⋅
4
⋅
3
⋅
2
⋅
1
=
24
k
⟹
k
=
5
12
.
10 = k \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 24k \implies k = \frac{5}{12}.
10
=
k
⋅
4
⋅
3
⋅
2
⋅
1
=
24
k
⟹
k
=
12
5
.
Therefore
P
(
6
)
=
6
+
Q
(
6
)
=
6
+
5
12
⋅
5
⋅
4
⋅
3
⋅
2
=
6
+
50
=
56.
P(6) = 6 + Q(6) = 6 + \frac{5}{12} \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 6 + 50 = 56.
P
(
6
)
=
6
+
Q
(
6
)
=
6
+
12
5
⋅
5
⋅
4
⋅
3
⋅
2
=
6
+
50
=
56.
