5.1: Divided differences
Definition.
Let p ∈ P n be the polynomial, which interpolates a given function f ∈
C[a, b] at the distinct points {xi, i = 0, . . . , n} ⊂ [a, b].
The leading coefficient of p is denoted by f [x0, x1, . . . , xn]. It is called the
divided difference of order n.
The name becomes clear from:
f (x) =
n
X
k=0
f (xi)lk (x) =
n
X
n
f (xk ) Π
k=0
i=0
i6=k
x − xi
⇒
x k − xi
f [x0, x1, . . . , xn] =
n
X
k=0
f (xk )
Πni=0 (xk − xi)
i6=k
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5.2: Divided differences and higher derivatives
Theorem. [cf. Th. 5.1]
Let f ∈ C n[a, b] and let {xi, i = 0, . . . , n} ⊂ [a, b] be ordered distinct
points, i.e. xi ≤ xi+1.
Then there exists a point ξ ∈ [x0, xn] such that
f [x0, x1, . . . , xn] =
1 (n)
f (ξ)
n!
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5.3: Newton polynomials and interpolation
Theorem. [cf. Th. 5.1]
Let pn ∈ P n be the polynomial which interpolates f at {xi, i = 0, . . . , n}
and let pn+1 ∈ P n+1 interpolate f at the same points and additionally at
xn+1 ∈ [a, b].
Then
n
pn+1(x) = pn(x) + f [x0, x1, . . . , xn+1] Π (x − xi)
i=0
j
The polynomials πj (x) = Πj−1
i=0 (x − xi ) ∈ P , j = 1, . . . and π0 (x) ≡ 1 are
called Newton polynomials.
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5.4: Newton interpolation formula
A consequence of the theorem is the following representation of the interpolation polynomial:
n−1
pn(x) = f [x0] + f [x0, x1](x − x0) + · · · + f [x0, x1, . . . , xn] Π (x − xi)
i=0
with f [x0] := f (x0).
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5.5: Divided differences, recursion
x0
k=0
f [x0]
k=1
k=2
...
k=n
f [x0, x1]
x1
f [x1]
f [x0, x1, x2]
f [x1, x2]
x2
f [x2]
..
..
..
..
f [xn−2, xn−1, xn]
...
f [x0, . . . , xn]
f [xn−1, xn]
xn
f [xn]
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5.6: Divided differences, recursion
(Cont.)
Theorem. [cf. Th. 5.3]
f [xj , . . . , xj+k+1] =
f [xj+1, . . . , xj+k+1] − f [xj , . . . , xj+k ]
xj+k+1 − xj
For the proof consider a polynomial pk which interpolates f at xj , . . . , xj+k
and a polynomial qk which interpolates at xj+1, . . . , xj+k+1. Then discuss
pk+1(x) =
(x − xj )qk (x) − (xj+k+1 − x)pk (x)
xj+k+1 − xj
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5.7: Hermite interpolation
Hermite interpolation task:
For a given function f ∈ C m[a, b] and given points {xi, i = 0, . . . , m} find
a polynomial p ∈ P n such that for given points
p(j)(xi) = f (j)(xi)
j = 0, . . . , li
i = 0, . . . , m
n+1=
m
X
(li + 1).
i=0
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5.8: Hermite interpolation:
Solvability and uniqueness
Theorem. [cf. Th. 5.4]
The Hermite interpolation task has a unique solution, provided that the xi
are distinct.
The proof makes use of the fact that the functions xj , j = 0, . . . , n form a
basis of P.
Then it suffices to show that 0 data implies 0 coefficents.
li
solves the interpolation task but is
Note furthermore that c Πm
(x
−
x
)
i
i=0
a polynomial of degree n + 1 unless c = 0.
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5.9: Divided Differences: Extention to the Hermite case
We allow multiplicity of arguments corresponding to multiple input data at
the same interpolation point xi:
F[ x
. . , x}0 , x1, . . . , xi−1, x
. . , x}i , xi+1 . . . , xn].
| 0, .{z
| i, .{z
(l0 +1)−times
(li +1)−times
In the recurrence relation division by zero occurs. Consequently the defintion
of the devided differences is in this case slightly altered:
1 (li)
F[ x
. . , x}i ] := f (xi)
| i, .{z
li!
(li +1)−times
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5.10: Example
Let’s find the Hermite interpolation polynomial to the data
sin(0) = 0, sin0(0) = 1, sin00(0) = 0, sin000(0) = −1, sin(π/2) = 1
The divided difference scheme is
k=0 k=1
k=2
k=3
k=4
0
0
0
0
1
0
0
1
0
1
0
1
0
− 3!
0
π/2
1
2/π −0.231335 −0.147272 0.012346
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5.11: Example
(Cont.)
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