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Hermite Polynomial
Definition. Suppose ๐ โ ๐ถ1[๐, ๐]. Let ๐ฅ0, โฆ , ๐ฅ๐ be distinct numbers in [๐, ๐], the Hermite polynomial ๐(๐ฅ) approximating ๐ is that:
1. ๐ ๐ฅ๐ = ๐ ๐ฅ๐ , for ๐ = 0, โฆ , ๐
2.๐๐ ๐ฅ๐
๐๐ฅ=
๐๐ ๐ฅ๐
๐๐ฅ, for ๐ = 0,โฆ , ๐
Remark: ๐(๐ฅ) and ๐(๐ฅ) agree not only function values but also 1st derivative values at ๐ฅ๐, ๐ = 0,โฆ , ๐.
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Theorem. If ๐ โ ๐ถ1 ๐, ๐ and ๐ฅ0, โฆ , ๐ฅ๐ โ ๐, ๐ distinct numbers, the Hermite polynomial is:
๐ป2๐+1 ๐ฅ = ๐ ๐ฅ๐ ๐ป๐,๐(๐ฅ)
๐
๐=0
+ ๐โฒ ๐ฅ๐ ๐ป ๐,๐(๐ฅ)
๐
๐=0
Where
๐ป๐,๐ ๐ฅ = [1 โ 2(๐ฅ โ ๐ฅ๐)๐ฟโฒ๐,๐(๐ฅ๐)]๐ฟ๐,๐
2 (๐ฅ)
๐ป ๐,๐ ๐ฅ = ๐ฅ โ ๐ฅ๐ ๐ฟ๐,๐2 ๐ฅ
Moreover, if ๐ โ ๐ถ2๐+2 ๐, ๐ , then
๐ ๐ฅ = ๐ป2๐+1 ๐ฅ +๐ฅ โ ๐ฅ0
2โฆ ๐ฅ โ ๐ฅ๐2
2๐ + 2 !๐ 2๐+2 (๐(๐ฅ))
for some ๐ ๐ฅ โ ๐, ๐ . Remark: 1. ๐ป2๐+1 ๐ฅ is a polynomial of degree at most 2๐ + 1.
2. ๐ฟ๐,๐(๐ฅ) is jth Lagrange basis polynomial of degree ๐.
3.๐ฅโ๐ฅ0
2โฆ ๐ฅโ๐ฅ๐2
2๐+2 !๐ 2๐+2 (๐(๐ฅ)) is the error term.
3
3rd Degree Hermite Polynomial
โข Given distinct ๐ฅ0, ๐ฅ1 and values of ๐ and ๐โฒ at these numbers.
๐ป3 ๐ฅ = 1 + 2๐ฅ โ ๐ฅ0๐ฅ1 โ ๐ฅ0
๐ฅ1 โ ๐ฅ
๐ฅ1 โ ๐ฅ0
2
๐ ๐ฅ0
+ ๐ฅ โ ๐ฅ0๐ฅ1 โ ๐ฅ
๐ฅ1 โ ๐ฅ0
2
๐โฒ ๐ฅ0
+ 1 + 2๐ฅ1 โ ๐ฅ
๐ฅ1 โ ๐ฅ0
๐ฅ0 โ ๐ฅ
๐ฅ0 โ ๐ฅ1
2
๐ ๐ฅ1
+ ๐ฅ โ ๐ฅ1๐ฅ0 โ ๐ฅ
๐ฅ0 โ ๐ฅ1
2
๐โฒ ๐ฅ1
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Hermite Polynomial by Divided Differences
Suppose ๐ฅ0, โฆ , ๐ฅ๐ and ๐, ๐โฒ are given at these numbers. Define ๐ง0, โฆ , ๐ง2๐+1 by
๐ง2๐ = ๐ง2๐+1 = ๐ฅ๐ , for ๐ = 0,โฆ , ๐
Construct divided difference table, but use ๐โฒ ๐ฅ0 , ๐โฒ ๐ฅ1 , . . , ๐โฒ ๐ฅ๐
to set the following undefined divided difference: ๐ ๐ง0, ๐ง1 , ๐ ๐ง2, ๐ง3 , โฆ , ๐ ๐ง2๐, ๐ง2๐+1 .
The Hermite polynomial is
๐ป2๐+1 ๐ฅ = ๐ ๐ง0 + ๐ ๐ง0, โฆ , ๐ง๐ ๐ฅ โ ๐ง0 โฆ(๐ฅ โ ๐ง๐โ1)
2๐+1
๐=1
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Divided Difference Notation for Hermite Interpolation
โข Divided difference notation: ๐ป3 ๐ฅ
= ๐ ๐ฅ0 + ๐โฒ ๐ฅ0 ๐ฅ โ ๐ฅ0+ ๐ ๐ฅ0, ๐ฅ0, ๐ฅ1 ๐ฅ โ ๐ฅ0
2
+ ๐[๐ฅ0, ๐ฅ0, ๐ฅ1, ๐ฅ1] ๐ฅ โ ๐ฅ02(๐ฅ โ ๐ฅ1)
6
Problems with High Order Polynomial Interpolation
โข 21 equal-spaced numbers to interpolate
๐ ๐ฅ =1
1+16๐ฅ2. The interpolating polynomial
oscillates between interpolation points.
7
Cubic Splines
โข 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.
โข Challenge: If ๐โฒ(๐ฅ๐) are not known, can we still generate interpolating polynomial with continuous derivatives?
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Definition: Given a function ๐ on [๐, ๐] and nodes ๐ = ๐ฅ0 <โฏ < ๐ฅ๐ = ๐, a cubic spline interpolant ๐ for ๐ satisfies:
(a) ๐(๐ฅ) is a cubic polynomial ๐๐(๐ฅ) on [๐ฅ๐ , ๐ฅ๐+1], โ๐ =0,1, โฆ , ๐ โ 1.
(b) ๐๐ ๐ฅ๐ = ๐(๐ฅ๐) and ๐๐ ๐ฅ๐+1 = ๐ ๐ฅ๐+1 , โ๐ = 0,1, โฆ , ๐ โ1.
(c) ๐๐ ๐ฅ๐+1 = ๐๐+1 ๐ฅ๐+1 , โ๐ = 0,1, โฆ , ๐ โ 2.
(d) ๐โฒ๐ ๐ฅ๐+1 = ๐โฒ๐+1 ๐ฅ๐+1 , โ๐ = 0,1,โฆ , ๐ โ 2.
(e) ๐โฒโฒ๐ ๐ฅ๐+1 = ๐โฒโฒ๐+1 ๐ฅ๐+1 , โ๐ = 0,1, โฆ , ๐ โ 2.
(f) One of the following boundary conditions: (i) ๐โฒโฒ ๐ฅ0 = ๐โฒโฒ ๐ฅ๐ = 0 (called free or natural boundary)
(ii) ๐โฒ ๐ฅ0 = ๐โฒ(๐ฅ0) and ๐โฒ ๐ฅ๐ = ๐โฒ(๐ฅ๐) (called clamped boundary)
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Things to match at interior point ๐ฅ๐+1:
โข The spline segment ๐๐(๐ฅ) is on ๐ฅ๐ , ๐ฅ๐+1 .
โข The spline segment ๐๐+1(๐ฅ) is on ๐ฅ๐+1, ๐ฅ๐+2 .
โข Their function values: ๐๐ ๐ฅ๐+1 = ๐๐+1 ๐ฅ๐+1 =๐ ๐ฅ๐+1
โข First derivative values: ๐โฒ๐ ๐ฅ๐+1 = ๐โฒ๐+1 ๐ฅ๐+1
โข Second derivative values: ๐โฒโฒ๐ ๐ฅ๐+1 = ๐โฒโฒ๐+1 ๐ฅ๐+1
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Building Cubic Splines โข Define:
๐๐ ๐ฅ = ๐๐ + ๐๐ ๐ฅ โ ๐ฅ๐ + ๐๐(๐ฅ โ ๐ฅ๐)2+๐๐(๐ฅ โ ๐ฅ๐)
3
and โ๐ = ๐ฅ๐+1 โ ๐ฅ๐, โ๐ = 0,1,โฆ , ๐ โ 1.
Solve for coefficients ๐๐ , ๐๐ , ๐๐ , ๐๐ by:
1. ๐๐ ๐ฅ๐ = ๐๐ = ๐(๐ฅ๐)
2. ๐๐+1 ๐ฅ๐+1 = ๐๐+1 = ๐๐ + ๐๐โ๐ + ๐๐(โ๐)2+๐๐(โ๐)
3
3. ๐โฒ๐ ๐ฅ๐ = ๐๐, also ๐๐+1 = ๐๐ + 2๐๐โ๐ + 3๐๐(โ๐)2
4. ๐โฒโฒ๐ ๐ฅ๐ = 2๐๐, also ๐๐+1 = ๐๐ + 3๐๐โ๐
5. Natural or clamped boundary conditions
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Solving the Resulting Equations
โ๐ = 1,โฆ , ๐ โ 1
โ๐โ1๐๐โ1 + 2 โ๐โ1 + โ๐ ๐๐ + โ๐๐๐+1
=3
โ๐๐๐+1 โ ๐๐ โ
3
โ๐โ1๐๐ โ ๐๐โ1
Remark: (n-1) equations for (n+1) unknowns {๐๐}๐=0๐ .
Once compute ๐๐, we then compute:
๐๐ = ๐๐+1โ๐๐
โ๐โ
โ๐ 2๐๐+๐๐+1
3
๐๐ =๐๐+1 โ ๐๐
3โ๐
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Completing the System
โข Natural boundary condition:
1. 0 = ๐โฒโฒ0 ๐ฅ0 = 2๐0 โ ๐0 = 0
2. 0 = ๐โฒโฒ๐ ๐ฅ๐ = 2๐๐ โ ๐๐ = 0
โข Clamped boundary condition:
a) ๐โฒ0 ๐ฅ0 = ๐0 = ๐โฒ(๐ฅ0)
b) ๐โฒ๐โ1 ๐ฅ๐ = ๐๐ = ๐๐โ1 + โ๐โ1(๐๐โ1 + ๐๐) = ๐โฒ(๐ฅ๐)
Remark: a) and b) gives additional equations:
2โ0๐0 + โ0๐1 =3
โ0(๐1 โ ๐0) โ 3๐โฒ(๐ฅ0)
โ๐โ1๐๐โ1 + 2โ๐โ1๐๐ = โ3
โ0๐๐ โ ๐๐โ1 + 3๐โฒ(๐ฅ๐)
13
Error Bound
If ๐ โ ๐ถ4[๐, ๐], let ๐ = max๐โค๐ฅโค๐
|๐4(๐ฅ)|. If ๐ is the
unique clamped cubic spline interpolant to ๐ with respect to the nodes: ๐ = ๐ฅ0 < โฏ < ๐ฅ๐ = ๐, then with
โ = max0โค๐โค๐โ1
๐ฅ๐+1 โ ๐ฅ๐
max๐โค๐ฅโค๐
|๐ ๐ฅ โ ๐(๐ฅ)| โค5๐โ4
384.
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