Post on 07-Jan-2020
Illustration. Consider to interpolate tanh(π₯π₯) using Lagrange polynomial and nodes π₯π₯0 = β1.5, π₯π₯1 = 0, π₯π₯2 = 1.5.
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Now interpolate tanh(π₯π₯)using nodes π₯π₯0 = β1.5, π₯π₯1 =0, π₯π₯2 = 1.5. Moreover, Let 1st
derivative of interpolating polynomial agree with derivative of tanh(π₯π₯) at these nodes.Remark:This is called Hermiteinterpolating polynomial.
Hermite Polynomial
Definition. Suppose ππ β πΆπΆ1[ππ, ππ]. Let π₯π₯0, β¦ , π₯π₯ππ be distinct numbers in [ππ, ππ], the Hermite polynomial ππ(π₯π₯) approximating ππ is that:1. ππ π₯π₯ππ = ππ π₯π₯ππ , for ππ = 0, β¦ ,ππ
2. ππππ π₯π₯πππππ₯π₯
= ππππ π₯π₯πππππ₯π₯
, for ππ = 0, β¦ ,ππ
Remark: ππ(π₯π₯) and ππ(π₯π₯) agree not only function values but also 1st derivative values at π₯π₯ππ, ππ = 0, β¦ ,ππ.
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Osculating Polynomials
Definition 3.8 Let π₯π₯0, β¦ , π₯π₯ππ be distinct numbers in ππ, ππand for ππ = 0, β¦ ,ππ, let ππππ be a nonnegative integer. Suppose that ππ β πΆπΆππ ππ, ππ , where ππ = max
0β€ππβ€ππππππ. The
osculating polynomial approximating ππ is the polynomial
ππ(π₯π₯) of least degree such that ππππππ π₯π₯πππππ₯π₯ππ
= ππππππ π₯π₯πππππ₯π₯ππ
for each ππ = 0, β¦ ,ππ and k= 0, β¦ ,ππππ .
Remark: the degree of ππ(π₯π₯) is at most ππ = βππ=0ππ ππππ + ππ.
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Theorem 3.9 If ππ β πΆπΆ1 ππ, ππ and π₯π₯0, β¦ , π₯π₯ππ β ππ, ππ distinct numbers, the Hermite polynomial of degree at most 2ππ + 1 is:
π»π»2ππ+1 π₯π₯ = οΏ½ππ=0
ππ
ππ π₯π₯ππ π»π»ππ,ππ(π₯π₯) + οΏ½ππ=0
ππ
ππβ² π₯π₯ππ οΏ½π»π»ππ,ππ(π₯π₯)
Where π»π»ππ,ππ π₯π₯ = [1 β 2(π₯π₯ β π₯π₯ππ)πΏπΏβ²ππ,ππ(π₯π₯ππ)]πΏπΏππ,ππ
2 (π₯π₯)οΏ½π»π»ππ,ππ π₯π₯ = π₯π₯ β π₯π₯ππ πΏπΏππ,ππ
2 π₯π₯Moreover, if ππ β πΆπΆ2ππ+2 ππ, ππ , then
ππ π₯π₯ = π»π»2ππ+1 π₯π₯ +π₯π₯ β π₯π₯0
2β¦ π₯π₯ β π₯π₯ππ
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2ππ + 2 !ππ 2ππ+2 (ππ(π₯π₯))
for some ππ π₯π₯ β ππ, ππ .Remark: 1. π»π»2ππ+1 π₯π₯ is a polynomial of degree at most 2ππ + 1.2. πΏπΏππ,ππ(π₯π₯) is jth Lagrange basis polynomial of degree ππ.
3.π₯π₯βπ₯π₯0
2β¦ π₯π₯βπ₯π₯ππ
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2ππ+2 !ππ 2ππ+2 (ππ(π₯π₯)) is the error term.
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Remark:
1. When ππ β ππ: π»π»ππ,ππ π₯π₯ππ = 0; οΏ½π»π»ππ,ππ π₯π₯ππ = 0.2. When ππ = ππ:
οΏ½π»π»ππ,ππ π₯π₯ππ = 1 β 2 π₯π₯ππ β π₯π₯ππ πΏπΏβ²ππ,ππ π₯π₯ππ πΏπΏππ,ππ
2 π₯π₯ππ = 1οΏ½π»π»ππ,ππ π₯π₯ππ = π₯π₯ππ β π₯π₯ππ πΏπΏππ,ππ
2 π₯π₯ππ = 0
βΉπ»π»2ππ+1 π₯π₯ππ = ππ(π₯π₯ππ).
3. π»π»β²ππ,ππ π₯π₯ = πΏπΏππ,ππ π₯π₯ β2πΏπΏππ,ππβ² π₯π₯ππ πΏπΏππ,ππ π₯π₯ + 1 β 2 π₯π₯ β π₯π₯ππ πΏπΏππ,ππ
β² π₯π₯ππ 2πΏπΏππ,ππβ² π₯π₯
βΉ When ππ β ππ:π»π»β²ππ,ππ π₯π₯ππ = 0; When ππ = ππ:π»π»β²
ππ,ππ π₯π₯ππ = 0.
4. οΏ½π»π»β²ππ,ππ π₯π₯ = πΏπΏππ,ππ2 π₯π₯ + 2 π₯π₯ β π₯π₯ππ πΏπΏππ,ππ(π₯π₯)πΏπΏβ²ππ,ππ π₯π₯
βΉ When ππ β ππ: οΏ½π»π»β²ππ,ππ π₯π₯ππ = 0; When ππ = ππ: οΏ½π»π»β²ππ,ππ π₯π₯ππ = 1.
Example 3.4.1 Use Hermite polynomial that agrees with the data in the table to find an approximation of ππ 1.5
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ππ π₯π₯ππ ππ(π₯π₯ππ) ππβ²(π₯π₯ππ)0 1.3 0.6200860 β0.52202321 1.6 0.4554022 β0.56989592 1.9 0.2818186 β0.5811571
3rd Degree Hermite Polynomial β’ Given distinct π₯π₯0, π₯π₯1 and values of ππ and ππβ² at these
numbers. π»π»3 π₯π₯
= 1 + 2π₯π₯ β π₯π₯0π₯π₯1 β π₯π₯0
π₯π₯1 β π₯π₯π₯π₯1 β π₯π₯0
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ππ π₯π₯0
+ π₯π₯ β π₯π₯0π₯π₯1 β π₯π₯π₯π₯1 β π₯π₯0
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ππβ² π₯π₯0
+ 1 + 2π₯π₯1 β π₯π₯π₯π₯1 β π₯π₯0
π₯π₯0 β π₯π₯π₯π₯0 β π₯π₯1
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ππ π₯π₯1
+ π₯π₯ β π₯π₯1π₯π₯0 β π₯π₯π₯π₯0 β π₯π₯1
2
ππβ² π₯π₯1
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Hermite Polynomial by Divided Differences
Suppose π₯π₯0, β¦ , π₯π₯ππ and ππ, ππβ² are given at these numbers. Define π§π§0, β¦ , π§π§2ππ+1 by
π§π§2ππ = π§π§2ππ+1 = π₯π₯ππ , for ππ = 0, β¦ ,ππ
Construct divided difference table, but use ππβ² π₯π₯0 ,ππβ² π₯π₯1 , . . ,ππβ² π₯π₯ππ
to set the following undefined divided difference: ππ π§π§0, π§π§1 ,ππ π§π§2, π§π§3 , β¦ , ππ π§π§2ππ, π§π§2ππ+1 .
Namely, ππ π§π§0, π§π§1 = ππβ² π₯π₯0 , ππ π§π§2, π§π§3 = ππβ² π₯π₯1 , β¦ππ π§π§2ππ, π§π§2ππ+1 = ππβ² π₯π₯ππ .
The Hermite polynomial is
π»π»2ππ+1 π₯π₯ = ππ π§π§0 + οΏ½ππ=1
2ππ+1
ππ π§π§0, β¦ , π§π§ππ π₯π₯ β π§π§0 β¦ (π₯π₯ β π§π§ππβ1)9
Example 3.4.2 Use divided difference method to determine the Hermite polynomial that agrees with the data in the table to find an approximation of ππ 1.5
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ππ π₯π₯ππ ππ(π₯π₯ππ) ππβ²(π₯π₯ππ)0 1.3 0.6200860 β0.52202321 1.6 0.4554022 β0.56989592 1.9 0.2818186 β0.5811571
Divided Difference Notation for HermiteInterpolation
β’ Divided difference notation for Hermitepolynomial interpolating 2 nodes: π₯π₯0, π₯π₯1.
π»π»3 π₯π₯= ππ π₯π₯0 + ππβ² π₯π₯0 π₯π₯ β π₯π₯0 + ππ π₯π₯0, π₯π₯0, π₯π₯1 π₯π₯ β π₯π₯0 2
+ ππ[π₯π₯0, π₯π₯0, π₯π₯1, π₯π₯1] π₯π₯ β π₯π₯0 2(π₯π₯ β π₯π₯1)
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Problems with High Order Polynomial Interpolation
β’ 21 equal-spaced numbers to interpolate ππ π₯π₯ = 1
1+25π₯π₯2. The interpolating polynomial
oscillates between interpolation points.
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3.5 Cubic Splinesβ’ 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.
Example. Piecewise-linear interpolation
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β’ Challenge: If ππβ²(π₯π₯ππ) are not known, can we still generate interpolating polynomial with continuous derivatives?
β’ Spline: A spline consists of a long strip of wood (a lath) fixed in position at a number of points. The lath will take the shape which minimizes the energy required for bending it between the fixed points, and thus adopt the smoothest possible shape.
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β’ In mathematics, a spline is a function that is piecewise-defined by polynomial functions, and which possesses a high degree of smoothness at the places where the polynomial pieces connect.
β’ Example. Irwin-Hall distribution. Nodes are -2, -1, 0, 1, 2.
ππ π₯π₯ =
14π₯π₯ + 2 3 β 2 β€ π₯π₯ β€ β1
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3 π₯π₯ 3 β 6π₯π₯2 + 4 β 1 β€ π₯π₯ β€ 114
2 β π₯π₯ 3 1 β€ π₯π₯ β€ 2
Notice: ππβ² β1 = 34
, ππβ² 1 = β34
,ππβ²β² β1 = 64
,ππβ²β² 1 = 64
.
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β’ Piecewise-polynomial approximation using cubic polynomials between each successive pair of nodes is called cubic spline interpolation.
Definition 3.10 Given a function ππ on [ππ, ππ] and nodes ππ = π₯π₯0 <β― < π₯π₯ππ = ππ, a cubic spline interpolant ππ for ππ satisfies:(a) ππ(π₯π₯) is a cubic polynomial ππππ(π₯π₯) on [π₯π₯ππ , π₯π₯ππ+1] with:
ππππ π₯π₯ = ππππ + ππππ π₯π₯ β π₯π₯ππ + ππππ π₯π₯ β π₯π₯ππ2 + ππππ π₯π₯ β π₯π₯ππ
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βππ = 0,1, β¦ ,ππ β 1.(b) ππππ π₯π₯ππ = ππ(π₯π₯ππ) and ππππ π₯π₯ππ+1 = ππ π₯π₯ππ+1 , βππ = 0,1, β¦ ,ππ β 1.(c) ππππ π₯π₯ππ+1 = ππππ+1 π₯π₯ππ+1 , βππ = 0,1, β¦ ,ππ β 2.
Remark: (c) is derived from (b).(d) ππβ²ππ π₯π₯ππ+1 = ππβ²ππ+1 π₯π₯ππ+1 , βππ = 0,1, β¦ ,ππ β 2.(e) ππβ²β²ππ π₯π₯ππ+1 = ππβ²β²ππ+1 π₯π₯ππ+1 , βππ = 0,1, β¦ ,ππ β 2.(f) One of the following boundary conditions:
(i) ππβ²β² π₯π₯0 = ππβ²β² π₯π₯ππ = 0 (called free or natural boundary)(ii) ππβ² π₯π₯0 = ππβ²(π₯π₯0) and ππβ² π₯π₯ππ = ππβ²(π₯π₯ππ) (called clamped boundary)
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The spline segment ππππ(π₯π₯) is on π₯π₯ππ , π₯π₯ππ+1 . The spline segment ππππ+1(π₯π₯) is on π₯π₯ππ+1, π₯π₯ππ+2 . Things to match at interior point π₯π₯ππ+1:β’ Their function values: ππππ π₯π₯ππ+1 = ππππ+1 π₯π₯ππ+1 =ππ π₯π₯ππ+1
β’ First derivative values: ππβ²ππ π₯π₯ππ+1 = ππβ²ππ+1 π₯π₯ππ+1β’ Second derivative values: ππβ²β²ππ π₯π₯ππ+1 = ππβ²β²ππ+1 π₯π₯ππ+1
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Building Cubic Splinesβ’ Define: ππππ π₯π₯ = ππππ + ππππ π₯π₯ β π₯π₯ππ + ππππ(π₯π₯ β π₯π₯ππ)2+ππππ(π₯π₯ β π₯π₯ππ)3
and βππ = π₯π₯ππ+1 β π₯π₯ππ, βππ = 0,1, β¦ , (ππ β 1).β’ Also define ππππ = ππ π₯π₯ππ ; ππππ = ππβ² π₯π₯ππ ; ππππ = ππβ²β²(π₯π₯ππ)/2.From Definition 3.10:1) ππππ π₯π₯ππ = ππππ = ππ π₯π₯ππ for ππ = 0,1, β¦ , (ππ β 1).2) ππππ+1 π₯π₯ππ+1 = ππππ+1 = ππππ + ππππβππ + ππππ(βππ )2+ππππ(βππ )3
for ππ = 0,1, β¦ , (ππ β 1).ππππππππ: ππππ = ππππβ1 + ππππβ1βππβ1 + ππππβ1(βππβ1)2+ππππβ1(βππβ1)3
3) ππβ²ππ π₯π₯ππ = ππππ, also ππππ+1 = ππππ + 2ππππβππ + 3ππππ(βππ )2for ππ = 0,1, β¦ , (ππ β 1).
4) ππβ²β²ππ π₯π₯ππ = 2ππππ, also ππππ+1 = ππππ + 3ππππβππfor ππ = 0,1, β¦ , ππ β 1 .
5) Natural or clamped boundary conditions19
Solve ππππ , ππππ , ππππ ,ππππ by substitution:1. Solve Eq. 4) for ππππ =
ππππ+1βππππ3βππ
, and substitute into Eqs. 2) and 3) to get:
2. ππππ+1 = ππππ + ππππβππ +βππ2
32ππππ + ππππ+1 ; (3.18)
ππππ+1 = ππππ + βππ ππππ + ππππ+1 . (3.19)3. Solve for ππππ in Eq. (3.18) to get:
ππππ = 1βππ
ππππ+1 β ππππ ββππ3
2ππππ + ππππ+1 . (3.20)Reduce the index by 1 to get:
ππππβ1 = 1βππβ1
ππππ β ππππβ1 ββππβ13
2ππππβ1 + ππππ .
4. Substitute ππππ and ππππβ1 into Eq. (3.19):βππβ1ππππβ1 + 2 βππβ1 + βππ ππππ + βππππππ+1 = (3.21)
3βππ
ππππ+1 β ππππ β3
βππβ1(ππππ β ππππβ1)
for ππ = 1,2, β¦ , ππ β 1 .20
Solving the Resulting Equationsβππ = 1,2, β¦ , (ππ β 1)
βππβ1ππππβ1 + 2 βππβ1 + βππ ππππ + βππππππ+1
=3βππ
ππππ+1 β ππππ β3
βππβ1ππππ β ππππβ1 (3.21)
Remark: (n-1) equations for (n+1) unknowns {ππππ}ππ=0ππ . Eq. (3.21) is solved with boundary conditions. β’ Once compute ππππ, we then compute:
ππππ = ππππ+1βππππβππ
β βππ 2ππππ+ππππ+13
(3.20)
and
ππππ = ππππ+1βππππ3βππ
(3.17) for ππ = 0,1,2, β¦ , (ππ β 1)21
Building Natural Cubic Spline β’ Natural boundary condition:
1. 0 = ππβ²β²0 π₯π₯0 = 2ππ0 β ππ0 = 02. 0 = ππβ²β²ππ π₯π₯ππ = 2ππππ β ππππ = 0
Step 1. Solve Eq. (3.21) together with ππ0 = 0 and ππππ = 0 to obtain {ππππ}ππ=0ππ .
Step 2. Solve Eq. (3.20) to obtain {ππππ}ππ=0ππβ1.
Step 3. Solve Eq. (3.17) to obtain {ππππ}ππ=0ππβ1.
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Building Clamped Cubic Splineβ’ Clamped boundary condition:
a) ππβ²0 π₯π₯0 = ππ0 = ππβ²(π₯π₯0)b) ππβ²ππβ1 π₯π₯ππ = ππππ = ππππβ1 + βππβ1(ππππβ1 + ππππ) = ππβ²(π₯π₯ππ)Remark: a) and b) gives additional equations:2β0ππ0 + β0ππ1 = 3
β0ππ1 β ππ0 β 3ππβ² π₯π₯0 (ππ)
βππβ1ππππβ1 + 2βππβ1ππππ = β3
βππβ1ππππ β ππππβ1 + 3ππβ² π₯π₯ππ (ππ)
Step 1. Solve Eq. (3.21) together with Eqs. (a) and (b) to obtain {ππππ}ππ=0ππ .Step 2. Solve Eq. (3.20) to obtain {ππππ}ππ=0ππβ1.Step 3. Solve Eq. (3.17) to obtain {ππππ}ππ=0ππβ1.
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Example 3.5.4 Let (π₯π₯0,ππ(π₯π₯0)) =(0,1), (π₯π₯1, ππ(π₯π₯1)) = (1, ππ), (π₯π₯2, ππ(π₯π₯2)) = (2, ππ2),(π₯π₯3, ππ(π₯π₯3)) = (3, ππ3). And ππβ² π₯π₯0 = ππ, ππβ² π₯π₯3 = ππ3.Determine the clamped spline ππ π₯π₯ .
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Theorem 3.11 If ππ is defined at the nodes: ππ =π₯π₯0 < β― < π₯π₯ππ = ππ, then ππ has a unique natural spline interpolant ππ on the nodes; that is a spline interpolant that satisfied the natural boundary conditions ππβ²β² ππ = 0, ππβ²β² ππ = 0.
Theorem 3.12 If ππ is defined at the nodes: ππ =π₯π₯0 < β― < π₯π₯ππ = ππ and differentiable at ππ and ππ, then ππ has a unique clamped spline interpolant ππon the nodes; that is a spline interpolant that satisfied the clamped boundary conditions ππβ² ππ = ππβ²(ππ), ππβ² ππ = ππβ²(ππ).
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Error BoundTheorem 3.13 If ππ β πΆπΆ4[ππ, ππ], let ππ =maxππβ€π₯π₯β€ππ
|ππ4(π₯π₯)|. If ππ is the unique clamped cubic spline interpolant to ππ with respect to the nodes: ππ = π₯π₯0 < β― < π₯π₯ππ = ππ, then with
β = max0β€ππβ€ππβ1
π₯π₯ππ+1 β π₯π₯ππ
maxππβ€π₯π₯β€ππ
|ππ π₯π₯ β ππ(π₯π₯)| β€ 5ππβ4
384.
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