3.4 Hermite Interpolation 3.5 Cubic Spline Interpolation

3.4 Hermite Interpolation
3.5 Cubic Spline Interpolation

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Hermite Polynomial

Definition. Suppose ∈ 1[, ]. Let 0, … , be
distinct numbers in [, ], the Hermite polynomial
() approximating is that:

1. = , for = 0, … ,

2. = , for = 0, … ,


Remark: () and () agree not only function values but
also 1st derivative values at , = 0, … , .

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Theorem. If ∈ 1 , and 0, … , ∈ , distinct
numbers, the Hermite polynomial is:



2+1 = ,() + ′ ,()

=0 =0

Where , = [1 − 2( − )′,()]2,()

, = − 2,
Moreover, if ∈ 2+2 , , then
− 0 2 … − 2
= 2+1 + 2 + 2 ! 2+2 (())


for some ∈ , .

Remark:
1. 2+1 is a polynomial of degree at most 2 + 1.
2. ,() is jth Lagrange basis polynomial of degree .
−0 2… − 2
3. 2+2 ! 2+2 (()) is the error term.

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3rd Degree Hermite Polynomial

• Given distinct 0, 1 and values of and ′ at these
numbers.
2
3 = 1 + 2 − 0 1 − 0
1 − 0 1 − 0

+ − 0 1 − 2 0
1 − 0
′

+ 1 + 2 1 − 0 − 2 1
1 − 0 0 − 1


+ − 1 0 − 2 1
0 − 1
′

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Hermite Polynomial by Divided Differences

DSuepfipnoese0,…0, …, 2, +1abnyd , ′ are given at these numbers.
2 = 2+1 = , for = 0, … ,

Construct divided difference table, but use
′ 0 , ′ 1 , . . , ′

to set the following undefined divided difference:
0, 1 , 2, 3 , … , 2, 2+1 .

The Hermite polynomial is

2+1

2+1 = 0 + 0, … , − 0 … ( − −1)

=1

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Divided Difference Notation for Hermite

Interpolation

• Divided difference notation:

3 + ′ 0 − 0
= 0
+ 0, 0, 1 − 0 2
+ [0, 0, 1, 1] − 0 2( − 1)

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Problems with High Order Polynomial Interpolation

• 21 equal-spaced numbers to interpolate
1
= 1+162 . The interpolating polynomial

oscillates between interpolation points.

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Cubic Splines

• Idea: Use piecewise polynomial interpolation, i.e,
divide the interval into smaller sub-intervals, and
construct different low degree polynomial
approximations (with small oscillations) on the
sub-intervals.

• Challenge: If ′() are not known, can we still
generate interpolating polynomial with
continuous derivatives?

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Definition: Given a function on [, ] and nodes = 0 <
⋯ < = , a cubic spline interpolant for satisfies:
(a) () is a cubic polynomial () on [, +1], ∀ =

0,1, … , − 1.
(b) = () and +1 = +1 , ∀ = 0,1, … , −

1.
(c) +1 = +1 +1 , ∀ = 0,1, … , − 2.
(d) ′ +1 = ′+1 +1 , ∀ = 0,1, … , − 2.
(e) ′′ +1 = ′′+1 +1 , ∀ = 0,1, … , − 2.
(f) One of the following boundary conditions:

(i) ′′ 0 = ′′ = 0 (called free or natural boundary)
(ii) ′ 0 = ′(0) and ′ = ′() (called clamped

boundary)

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Things to match at interior point +1:

• The spline segment () is on , +1 .

• The spline segment +1() is on +1, +2 .

• Their function values: +1 = +1 +1 =
+1

• First derivative values: ′ +1 = ′+1 +1
• Second derivative values: ′′ +1 = ′′+1 +1

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Building Cubic Splines

• Define:
= + − + ( − )2+( − )3

and ℎ = +1 − , ∀ = 0,1, … , − 1.

Solve for coefficients , , , by:

1. = = ()
2. +1 +1 = +1 = + ℎ + (ℎ)2+(ℎ)3
3. ′ = , also +1 = + 2ℎ + 3(ℎ)2
4. ′′ = 2, also +1 = + 3ℎ

5. Natural or clamped boundary conditions

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Solving the Resulting Equations

∀ = 1, … , − 1

ℎ−1−1 + 2 ℎ−1 + ℎ + ℎ+1
3 3
= ℎ +1 − − ℎ−1 − −1

Remark: (n-1) equations for (n+1) unknowns {}=0.

Once compute , we then compute:

= +1− − ℎ 2++1
ℎ 3

= +1 −
3ℎ

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Completing the System

• Natural boundary condition:

1. 0 = ′′0 0 = 20 → 0 = 0
2. 0 = ′′ = 2 → = 0

• Clamped boundary condition:

a) ′0 0 = 0 = ′(0)
b) ′−1 = = −1 + ℎ−1(−1 + ) = ′()

Remark: a) and b) gives additional equations:
2+ℎℎ−011==ℎ3−0 (ℎ301
2ℎ00 − 0) − 3′(0)
ℎ−1−1 + − −1 + 3′()

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Error Bound

If ∈ 4[, ], let = max |4()|. If is the

≤≤
unique clamped cubic spline interpolant to with
respect to the nodes: = 0 < ⋯ < = , then

with

ℎ = max +1 −

0≤≤−1

max | − ()| ≤ 5ℎ4 .
384
≤≤

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