3.4 Hermite Interpolation 3.5 Cubic Spline Interpolation

1 Illustration. Consider to interpolate tanh( ) using Lagrange polynomial and nodes = 1.5, = 0, = 1.5. 𝑥𝑥 𝑥𝑥0 − 𝑥𝑥1 𝑥𝑥2

Now interpolate tanh( ) using nodes = 1.5, = 0, = 1.5. Moreover, 𝑥𝑥Let 1st 0 1 derivative of𝑥𝑥 interpolating− 𝑥𝑥 2 polynomial𝑥𝑥 agree with derivative of tanh( ) at these nodes. Remark:This is called𝑥𝑥 Hermite interpolating polynomial.

2 Hermite Polynomial

Definition. Suppose [ , ]. Let , … , be distinct numbers in [ , ], 1the Hermite polynomial 0 𝑛𝑛 ( ) approximating 𝑓𝑓 is∈ that:𝐶𝐶 𝑎𝑎 𝑏𝑏 𝑥𝑥 𝑥𝑥 1. = , for𝑎𝑎 𝑏𝑏 = 0, … , 𝑃𝑃 𝑥𝑥 𝑓𝑓 2. 𝑃𝑃 𝑥𝑥𝑖𝑖 = 𝑓𝑓 𝑥𝑥𝑖𝑖 , for 𝑖𝑖 = 0, … ,𝑛𝑛 𝑑𝑑𝑃𝑃 𝑥𝑥𝑖𝑖 𝑑𝑑𝑑𝑑 𝑥𝑥𝑖𝑖 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑖𝑖 𝑛𝑛 Remark: ( ) and ( ) agree not only function values but also 1st derivative values at , = 0, … , . 𝑃𝑃 𝑥𝑥 𝑓𝑓 𝑥𝑥 𝑥𝑥𝑖𝑖 𝑖𝑖 𝑛𝑛 3 Osculating Polynomials

Definition 3.8 Let , … , be distinct numbers in , and for = 0, … , , let be a nonnegative integer. 0 𝑛𝑛 Suppose that 𝑥𝑥 , 𝑥𝑥 , where = max . The𝑎𝑎 𝑏𝑏 𝑖𝑖 𝑖𝑖 𝑛𝑛 𝑚𝑚 𝑚𝑚 osculating polynomial approximating is the polynomial𝑖𝑖 𝑓𝑓 ∈ 𝐶𝐶 𝑎𝑎 𝑏𝑏 𝑚𝑚 0≤𝑖𝑖≤𝑛𝑛 𝑚𝑚 ( ) of least degree such that = for each 𝑘𝑘 𝑓𝑓 𝑘𝑘 𝑑𝑑 𝑃𝑃 𝑥𝑥𝑖𝑖 𝑑𝑑 𝑓𝑓 𝑥𝑥𝑖𝑖 = 0, … , and k= 0, … , . 𝑘𝑘 𝑘𝑘 𝑃𝑃 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑖𝑖 𝑛𝑛 𝑚𝑚𝑖𝑖 Remark: the degree of ( ) is at most = + . 𝑛𝑛 𝑃𝑃 𝑥𝑥 𝑀𝑀 ∑𝑖𝑖=0 𝑚𝑚𝑖𝑖 𝑛𝑛

4 Theorem 3.9 If , and , … , , distinct numbers, the Hermite1 polynomial of degree at most + 1 is: 𝑓𝑓 ∈ 𝐶𝐶 𝑎𝑎 𝑏𝑏 𝑥𝑥0 𝑥𝑥𝑛𝑛 ∈ 𝑎𝑎 𝑏𝑏 = 𝑛𝑛 , ( ) + 𝑛𝑛 2𝑛𝑛, ( )

2𝑛𝑛+1 𝑗𝑗 𝑛𝑛 𝑗𝑗 𝑗𝑗 𝑛𝑛 𝑗𝑗 Where 𝐻𝐻 𝑥𝑥 � 𝑓𝑓 𝑥𝑥 𝐻𝐻 𝑥𝑥 � 𝑓𝑓′ 𝑥𝑥 𝐻𝐻� 𝑥𝑥 𝑗𝑗=0 𝑗𝑗=0 , = [1 2( ) , ( )] , ( ) 2 , = ′ 𝐻𝐻𝑛𝑛 𝑗𝑗 𝑥𝑥 − 𝑥𝑥 − 𝑥𝑥𝑗𝑗 𝐿𝐿 𝑛𝑛 𝑗𝑗, 𝑥𝑥𝑗𝑗 𝐿𝐿𝑛𝑛 𝑗𝑗 𝑥𝑥 Moreover, if , , then 2 𝐻𝐻�𝑛𝑛 𝑗𝑗 𝑥𝑥 𝑥𝑥 − 𝑥𝑥𝑗𝑗 𝐿𝐿𝑛𝑛 𝑗𝑗 𝑥𝑥 2𝑛𝑛+2 … = 𝑓𝑓 ∈ 𝐶𝐶 +𝑎𝑎 𝑏𝑏 ( ( )) 2 + 2 ! 2 𝑥𝑥 − 𝑥𝑥0 𝑥𝑥 − 𝑥𝑥𝑛𝑛 2𝑛𝑛+2 for some𝑓𝑓 𝑥𝑥 𝐻𝐻2𝑛𝑛+1 𝑥𝑥, . 𝑓𝑓 𝜉𝜉 𝑥𝑥 Remark: 2𝑛𝑛 1. 𝜉𝜉 is𝑥𝑥 a polynomial∈ 𝑎𝑎 𝑏𝑏 of degree at most + 1. 2. , ( ) is jth Lagrange basis polynomial of degree . 2𝑛𝑛+1 𝐻𝐻 …𝑥𝑥 2𝑛𝑛 3. 𝐿𝐿𝑛𝑛 𝑗𝑗 𝑥𝑥2 2 ( ( )) is the error term. 𝑛𝑛 𝑥𝑥−𝑥𝑥0 𝑥𝑥−! 𝑥𝑥𝑛𝑛 2𝑛𝑛+2 2𝑛𝑛+2 𝑓𝑓 𝜉𝜉 𝑥𝑥 5 Remark:

1. When : , = 0; , = 0. 2. When = : 𝑖𝑖 ≠ 𝑗𝑗 𝐻𝐻𝑛𝑛 𝑗𝑗 𝑥𝑥𝑖𝑖 𝐻𝐻�𝑛𝑛 𝑗𝑗 𝑥𝑥𝑖𝑖 , 𝑖𝑖 = 𝑗𝑗1 2 , , = 1 = ′ = 0 2 𝐻𝐻𝑛𝑛,𝑗𝑗 𝑥𝑥𝑗𝑗 − 𝑥𝑥𝑗𝑗 −, 𝑥𝑥𝑗𝑗 𝐿𝐿 𝑛𝑛 𝑗𝑗 𝑥𝑥𝑗𝑗 𝐿𝐿𝑛𝑛 𝑗𝑗 𝑥𝑥𝑗𝑗 � 2 𝐻𝐻�𝑛𝑛 𝑗𝑗 𝑥𝑥𝑗𝑗 𝑥𝑥=𝑗𝑗 −(𝑥𝑥𝑗𝑗 )𝐿𝐿.𝑛𝑛 𝑗𝑗 𝑥𝑥𝑗𝑗

3. ⟹ 𝐻𝐻, 2𝑛𝑛+=1 𝑥𝑥,𝑗𝑗 𝑓𝑓2 𝑥𝑥,𝑗𝑗 , + 1 2 , 2 , ′ ′ ′ 𝑛𝑛 𝑗𝑗 𝑛𝑛 𝑗𝑗 𝑛𝑛 𝑗𝑗 𝑗𝑗 𝑛𝑛 𝑗𝑗 𝑗𝑗 𝑛𝑛 𝑗𝑗 𝑗𝑗 𝑛𝑛 𝑗𝑗 𝐻𝐻� When𝑥𝑥 𝐿𝐿 𝑥𝑥 : − 𝐿𝐿, 𝑥𝑥 𝐿𝐿= 0𝑥𝑥; When− 𝑥𝑥=− 𝑥𝑥: 𝐿𝐿 , 𝑥𝑥 =𝐿𝐿 0𝑥𝑥. ′ ′ 𝑛𝑛 𝑗𝑗 𝑖𝑖 𝑛𝑛 𝑗𝑗 𝑗𝑗 4. ⟹ , 𝑖𝑖 =≠ 𝑗𝑗 𝐻𝐻, 𝑥𝑥+ 2 𝑖𝑖 , 𝑗𝑗( 𝐻𝐻) ,𝑥𝑥 :2 = 0; = : ′ = 1. 𝐻𝐻�′When𝑛𝑛 𝑗𝑗 𝑥𝑥 𝐿𝐿𝑛𝑛 𝑗𝑗 ,𝑥𝑥 𝑥𝑥 −When𝑥𝑥𝑗𝑗 𝐿𝐿𝑛𝑛 𝑗𝑗 𝑥𝑥 𝐿𝐿, 𝑛𝑛 𝑗𝑗 𝑥𝑥 6 ⟹ 𝑖𝑖 ≠ 𝑗𝑗 𝐻𝐻�′𝑛𝑛 𝑗𝑗 𝑥𝑥𝑖𝑖 𝑖𝑖 𝑗𝑗 𝐻𝐻�′𝑛𝑛 𝑗𝑗 𝑥𝑥𝑗𝑗 Example 3.4.1 Use Hermite polynomial that agrees with the data in the table to find an approximation of 1.5

𝑓𝑓 ( ) ( ) 0 1.3 0.6200860 0.5220232 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑘𝑘1 1𝑥𝑥.6 0.4554022𝑓𝑓 𝑥𝑥 0.𝑓𝑓5698959� 𝑥𝑥 2 1.9 0.2818186 −0.5811571 − −

7 3rd Degree Hermite Polynomial • Given distinct , and values of and at these numbers. 𝑥𝑥0 𝑥𝑥1 𝑓𝑓 𝑓𝑓�

3 =𝐻𝐻 𝑥𝑥1 + 2 2 0 1 𝑥𝑥 − 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 0 1 0 1 0 𝑓𝑓 𝑥𝑥 + 𝑥𝑥 − 𝑥𝑥 𝑥𝑥2 − 𝑥𝑥 1 ′ 0 𝑥𝑥 − 𝑥𝑥 0 𝑥𝑥 − 𝑥𝑥 1 0 𝑓𝑓 𝑥𝑥 + 1 + 2 𝑥𝑥 − 𝑥𝑥 2 1 0 𝑥𝑥 − 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 1 1 0 0 1 𝑓𝑓 𝑥𝑥 + 𝑥𝑥 − 𝑥𝑥 𝑥𝑥2 − 𝑥𝑥 𝑥𝑥0 − 𝑥𝑥 ′ 𝑥𝑥 − 𝑥𝑥1 𝑓𝑓 𝑥𝑥1 𝑥𝑥0 − 𝑥𝑥1 8 Hermite Polynomial by Divided Differences

Suppose , … , and , are given at these numbers. Define , … , by 𝑥𝑥0 𝑥𝑥𝑛𝑛= 𝑓𝑓 =𝑓𝑓′ , for = 0, … , 𝑧𝑧0 𝑧𝑧2𝑛𝑛+1 2𝑖𝑖 2𝑖𝑖+1 𝑖𝑖 Construct divided𝑧𝑧 difference𝑧𝑧 𝑥𝑥 table, but use𝑖𝑖 𝑛𝑛 , , . . , to set the following undefined′ ′ divided difference:′ , 𝑓𝑓 ,𝑥𝑥0 𝑓𝑓, 𝑥𝑥1, … , 𝑓𝑓 𝑥𝑥,𝑛𝑛 . Namely, , = , , = , … 0 1 2 3 2𝑛𝑛 2𝑛𝑛+1 𝑓𝑓 𝑧𝑧 𝑧𝑧 ′ 𝑓𝑓 𝑧𝑧, 𝑧𝑧 =𝑓𝑓 𝑧𝑧 ′ 𝑧𝑧. 0 1 0 2 3 1 The Hermite𝑓𝑓 𝑧𝑧 polynomial𝑧𝑧 𝑓𝑓 𝑥𝑥 is 𝑓𝑓 𝑧𝑧 𝑧𝑧 ′ 𝑓𝑓 𝑥𝑥 𝑓𝑓 𝑧𝑧2𝑛𝑛 𝑧𝑧2𝑛𝑛+1 𝑓𝑓 𝑥𝑥𝑛𝑛 = + 2𝑛𝑛+1 , … , … ( )

9 𝐻𝐻2𝑛𝑛+1 𝑥𝑥 𝑓𝑓 𝑧𝑧0 � 𝑓𝑓 𝑧𝑧0 𝑧𝑧𝑘𝑘 𝑥𝑥 − 𝑧𝑧0 𝑥𝑥 − 𝑧𝑧𝑘𝑘−1 𝑘𝑘=1 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 1.5 ( ) ( ) 0 1.3 0.6200860 0.5220232 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑓𝑓 𝑘𝑘1 1𝑥𝑥.6 0.4554022𝑓𝑓 𝑥𝑥 0.𝑓𝑓5698959� 𝑥𝑥 2 1.9 0.2818186 −0.5811571 − −

10 Divided Difference Notation for Hermite Interpolation • Divided difference notation for Hermite polynomial interpolating 2 nodes: , .

= + + , 𝑥𝑥0, 𝑥𝑥1 3 +𝐻𝐻 𝑥𝑥[ , , ′ , ] ( ) 2 0 0 0 0 0 1 0 𝑓𝑓 𝑥𝑥 𝑓𝑓 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 2 𝑓𝑓 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 𝑓𝑓 𝑥𝑥0 𝑥𝑥0 𝑥𝑥1 𝑥𝑥1 𝑥𝑥 − 𝑥𝑥0 𝑥𝑥 − 𝑥𝑥1

11 Problems with High Order Polynomial Interpolation

• 21 equal-spaced numbers to interpolate = . The interpolating polynomial 1 oscillates between2 interpolation points. 𝑓𝑓 𝑥𝑥 1+25𝑥𝑥

12 3.5 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. Example. Piecewise-linear interpolation

13 • Challenge: If ( ) are not known, can we still generate interpolating polynomial with continuous derivatives? 𝑓𝑓� 𝑥𝑥𝑖𝑖 • 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.

14 • 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. • Example. Irwin-Hall distribution. Nodes are -2, -1, 0, 1, 2. 1 + 2 2 1 4 1 3 = 3 𝑥𝑥 6 + 4 − ≤ 𝑥𝑥1≤ − 1 4 1 3 2 𝑓𝑓 𝑥𝑥 𝑥𝑥2 − 𝑥𝑥 1− ≤ 𝑥𝑥 2≤ 4 3 Notice: 1 =− 𝑥𝑥, 1 = , ≤ 𝑥𝑥1≤ = , 1 = . ′ 3 ′ 3 ′′ 6 ′′ 6 𝑓𝑓 − 4 𝑓𝑓 − 4 𝑓𝑓 − 4 𝑓𝑓 4 15 • Piecewise-polynomial approximation using cubic polynomials between each successive pair of nodes is called cubic spline interpolation.

Definition 3.10 Given a function on [ , ] and nodes = < < = , a cubic spline interpolant for satisfies: (a) ( ) is a cubic polynomial 𝑓𝑓( ) on𝑎𝑎 [𝑏𝑏 , ] with:𝑎𝑎 𝑥𝑥0 ⋯ 𝑥𝑥𝑛𝑛 𝑏𝑏 𝑆𝑆 𝑓𝑓 = + + + 𝑆𝑆 𝑥𝑥 𝑆𝑆𝑗𝑗 𝑥𝑥 𝑥𝑥𝑗𝑗 𝑥𝑥𝑗𝑗+1 = 0,1, … , 1.2 3 𝑆𝑆𝑗𝑗 𝑥𝑥 𝑎𝑎𝑗𝑗 𝑏𝑏𝑗𝑗 𝑥𝑥 − 𝑥𝑥𝑗𝑗 𝑐𝑐𝑗𝑗 𝑥𝑥 − 𝑥𝑥𝑗𝑗 𝑑𝑑𝑗𝑗 𝑥𝑥 − 𝑥𝑥𝑗𝑗 (b) = ( ) and = , = 0,1, … , 1. ∀𝑗𝑗 𝑛𝑛 − = , = 0,1, … , 2. (c) 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗+1 𝑗𝑗+1 Remark:𝑆𝑆 𝑥𝑥 (c) is𝑓𝑓 derived𝑥𝑥 from𝑆𝑆 (b).𝑥𝑥 𝑓𝑓 𝑥𝑥 ∀𝑗𝑗 𝑛𝑛 − 𝑗𝑗 𝑗𝑗+1 𝑗𝑗+1 𝑗𝑗+1 (d) 𝑆𝑆 𝑥𝑥 =𝑆𝑆 𝑥𝑥 ∀, 𝑗𝑗 = 0,1, …𝑛𝑛, − 2. (e) ′ = ′ , = 0,1, … , 2. 𝑆𝑆 𝑗𝑗 𝑥𝑥𝑗𝑗+1 𝑆𝑆 𝑗𝑗+1 𝑥𝑥𝑗𝑗+1 ∀𝑗𝑗 𝑛𝑛 − (f) One′′ of the following′′ boundary conditions: 𝑗𝑗 𝑗𝑗+1 𝑗𝑗+1 𝑗𝑗+1 (i)𝑆𝑆 𝑥𝑥 = 𝑆𝑆 =𝑥𝑥0 (called∀𝑗𝑗 free or natural𝑛𝑛 − boundary) (ii) ′′ = (′′ ) and = ( ) (called clamped boundary) 0 𝑛𝑛 16 𝑆𝑆′ 𝑥𝑥 𝑆𝑆 𝑥𝑥 ′ 𝑆𝑆 𝑥𝑥0 𝑓𝑓� 𝑥𝑥0 𝑆𝑆 𝑥𝑥𝑛𝑛 𝑓𝑓� 𝑥𝑥𝑛𝑛 The spline segment ( ) is on , . The spline segment ( ) is on , . Things to match at 𝑗𝑗 𝑗𝑗 𝑗𝑗+1 interior point : 𝑆𝑆 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑆𝑆𝑗𝑗+1 𝑥𝑥 𝑥𝑥𝑗𝑗+1 𝑥𝑥𝑗𝑗+2 • Their function values: = = 𝑥𝑥𝑗𝑗+1 𝑆𝑆𝑗𝑗 𝑥𝑥𝑗𝑗+1 𝑆𝑆𝑗𝑗+1 𝑥𝑥𝑗𝑗+1 • First derivative values: = 𝑓𝑓 𝑥𝑥𝑗𝑗+1 • Second derivative values:′ =′ 𝑆𝑆 𝑗𝑗 𝑥𝑥𝑗𝑗+1 𝑆𝑆 𝑗𝑗+1 𝑥𝑥𝑗𝑗+1 ′′ ′′ 17 𝑆𝑆 𝑗𝑗 𝑥𝑥𝑗𝑗+1 𝑆𝑆 𝑗𝑗+1 𝑥𝑥𝑗𝑗+1 Example 3.5.1 Construct a natural spline through (1,2), (2,3) and (3.5). 𝑆𝑆 𝑥𝑥

18 Building Cubic Splines • Define: = + + ( ) + ( ) and = , = 0,1, … , ( 1). 2 3 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 • Also define𝑆𝑆 𝑥𝑥 𝑎𝑎= 𝑏𝑏 𝑥𝑥 ;− 𝑥𝑥 = 𝑐𝑐 𝑥𝑥 −;𝑥𝑥 =𝑑𝑑 𝑥𝑥(− 𝑥𝑥)/2. ℎ𝑗𝑗 𝑥𝑥𝑗𝑗+1 − 𝑥𝑥𝑗𝑗 ∀𝑗𝑗 𝑛𝑛 − From Definition 3.10: ′ ′′ 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 1) = =𝑎𝑎 𝑓𝑓for𝑥𝑥 =𝑏𝑏0,1, …𝑆𝑆, ( 𝑥𝑥 1)𝑐𝑐. 𝑆𝑆 𝑥𝑥 2) = = + + ( ) + ( ) 𝑆𝑆𝑗𝑗 𝑥𝑥𝑗𝑗 𝑎𝑎𝑗𝑗 𝑓𝑓 𝑥𝑥𝑗𝑗 𝑗𝑗 𝑛𝑛 − for = 0,1, … , ( 1). 2 3 𝑗𝑗+1 𝑗𝑗+1 𝑗𝑗+1 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑆𝑆 𝑥𝑥: = 𝑎𝑎 + 𝑎𝑎 𝑏𝑏 +ℎ 𝑐𝑐( ℎ ) + 𝑑𝑑 ℎ( ) 𝑗𝑗 𝑛𝑛 − 3) = = + 2 + 3 2 ( ) 3 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑎𝑎𝑛𝑛 ,𝑎𝑎 also𝑛𝑛−1 𝑏𝑏𝑛𝑛−1ℎ𝑛𝑛−1 𝑐𝑐𝑛𝑛−1 ℎ𝑛𝑛−1 𝑑𝑑𝑛𝑛−1 ℎ𝑛𝑛−1 ′ for = 0,1, … , ( 1). 2 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗+1 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 4) 𝑆𝑆 𝑥𝑥 =𝑏𝑏 , also𝑏𝑏 =𝑏𝑏 + 𝑐𝑐3 ℎ 𝑑𝑑 ℎ 𝑗𝑗 𝑛𝑛 − ′′ for = 0,1, … , 1 . 𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗+1 𝑗𝑗 𝑗𝑗 𝑗𝑗 5) Natural𝑆𝑆 𝑥𝑥 or clamped2𝑐𝑐 boundary𝑐𝑐 𝑐𝑐 conditions𝑑𝑑 ℎ 𝑗𝑗 𝑛𝑛 − 19 Solve , , , by substitution:

1. Solve𝑗𝑗 Eq.𝑗𝑗 𝑗𝑗 4) 𝑗𝑗for = , and substitute into Eqs. 2) and 3) to 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑑𝑑 𝑗𝑗+1 𝑗𝑗 get: 𝑐𝑐 −𝑐𝑐 𝑑𝑑𝑗𝑗 3ℎ𝑗𝑗

2. = + + 2 2 + ; (3.18) ℎ𝑗𝑗 = + + . (3.19) 𝑎𝑎𝑗𝑗+1 𝑎𝑎𝑗𝑗 𝑏𝑏𝑗𝑗ℎ𝑗𝑗 3 𝑐𝑐𝑗𝑗 𝑐𝑐𝑗𝑗+1 3. Solve for in Eq. (3.18) to get: 𝑏𝑏𝑗𝑗+1 𝑏𝑏𝑗𝑗 ℎ𝑗𝑗 𝑐𝑐𝑗𝑗 𝑐𝑐𝑗𝑗+1 = 2 + . (3.20) 𝑏𝑏𝑗𝑗 1 ℎ𝑗𝑗 Reduce the index by 1 to get: 𝑏𝑏𝑗𝑗 ℎ𝑗𝑗 𝑎𝑎𝑗𝑗+1 − 𝑎𝑎𝑗𝑗 − 3 𝑐𝑐𝑗𝑗 𝑐𝑐𝑗𝑗+1 = 2 + . 1 ℎ𝑗𝑗−1 4. Substitute𝑗𝑗−1 and 𝑗𝑗 𝑗𝑗into−1 Eq. (3.19):𝑗𝑗−1 𝑗𝑗 𝑏𝑏 ℎ𝑗𝑗−1 𝑎𝑎 − 𝑎𝑎 − 3 𝑐𝑐 𝑐𝑐 + 2 + + = (3.21) 𝑗𝑗 𝑗𝑗−1 𝑏𝑏 3 𝑏𝑏 3 ℎ𝑗𝑗−1𝑐𝑐𝑗𝑗−1 ℎ𝑗𝑗−1 ℎ𝑗𝑗 𝑐𝑐𝑗𝑗 ℎ𝑗𝑗𝑐𝑐𝑗𝑗(+1 ) 𝑗𝑗+1 =𝑗𝑗1,2, … , 𝑗𝑗 1 . 𝑗𝑗−1 𝑎𝑎for − 𝑎𝑎 − 𝑎𝑎 − 𝑎𝑎 20 ℎ𝑗𝑗 ℎ𝑗𝑗−1 𝑗𝑗 𝑛𝑛 − Solving the Resulting Equations = 1,2, … , ( 1) + 2 + + ∀𝑗𝑗 3 𝑛𝑛 − 3 ℎ=𝑗𝑗−1𝑐𝑐𝑗𝑗−1 ℎ𝑗𝑗−1 ℎ𝑗𝑗 𝑐𝑐𝑗𝑗 ℎ𝑗𝑗𝑐𝑐𝑗𝑗+1 (3.21)

Remark: (n-1) 𝑎𝑎equations𝑗𝑗+1 − 𝑎𝑎𝑗𝑗 for− (n+1) 𝑎𝑎unknowns𝑗𝑗 − 𝑎𝑎𝑗𝑗−1 { } . Eq. ℎ𝑗𝑗 ℎ𝑗𝑗−1 (3.21) is solved with boundary conditions. 𝑛𝑛 𝑐𝑐𝑗𝑗 𝑗𝑗=0 • Once compute , we then compute:

=𝑐𝑐𝑗𝑗 (3.20) 𝑎𝑎𝑗𝑗+1−𝑎𝑎𝑗𝑗 ℎ𝑗𝑗 2𝑐𝑐𝑗𝑗+𝑐𝑐𝑗𝑗+1 𝑏𝑏𝑗𝑗 ℎ𝑗𝑗 −and 3 = (3.17) for = 0,1,2, … , ( 1) 𝑐𝑐𝑗𝑗+1−𝑐𝑐𝑗𝑗 21 𝑑𝑑𝑗𝑗 3ℎ𝑗𝑗 𝑗𝑗 𝑛𝑛 − Building Natural Cubic Spline • Natural boundary condition: 1. 0 = = 2 = 0 2. 0 = ′′ = 2 = 0 𝑆𝑆 0 𝑥𝑥0 𝑐𝑐0 → 𝑐𝑐0 Step 1. Solve′′ Eq. (3.21) together with = 0 and 𝑆𝑆 𝑛𝑛 𝑥𝑥𝑛𝑛 𝑐𝑐𝑛𝑛 → 𝑐𝑐𝑛𝑛 = 0 to obtain { } . 𝑐𝑐0 𝑛𝑛 { } 𝑐𝑐Step𝑛𝑛 2. Solve Eq. (3.20)𝑐𝑐𝑗𝑗 𝑗𝑗=0to obtain . { }𝑛𝑛−1 Step 3. Solve Eq. (3.17) to obtain 𝑏𝑏𝑗𝑗 𝑗𝑗=0 . 𝑛𝑛−1 𝑑𝑑𝑗𝑗 𝑗𝑗=0

22 Building Clamped Cubic Spline • Clamped boundary condition: a) = = ( ) b) ′ = = + ( + ) = ( ) 𝑆𝑆 0 𝑥𝑥0 𝑏𝑏0 𝑓𝑓� 𝑥𝑥0 Remark:′ a) and b) gives additional equations: 𝑆𝑆 𝑛𝑛−1 𝑥𝑥𝑛𝑛 𝑏𝑏𝑛𝑛 𝑏𝑏𝑛𝑛−1 ℎ𝑛𝑛−1 𝑐𝑐𝑛𝑛−1 𝑐𝑐𝑛𝑛 𝑓𝑓� 𝑥𝑥𝑛𝑛 2 + = 3 ( ) 3 3 ′ ℎ0𝑐𝑐0 ℎ+0𝑐𝑐12 ℎ0 𝑎𝑎1=− 𝑎𝑎0 − 𝑓𝑓 𝑥𝑥0 +𝑎𝑎 3 ( ) ′ Stepℎ𝑛𝑛 −11.𝑐𝑐 𝑛𝑛Solve−1 Eq.ℎ𝑛𝑛− 1(3.21)𝑐𝑐𝑛𝑛 − together𝑎𝑎𝑛𝑛 − with𝑎𝑎𝑛𝑛−1 Eqs. 𝑓𝑓(a)𝑥𝑥and𝑛𝑛 (b)𝑏𝑏 to ℎ𝑛𝑛−1 obtain { } . Step 2. Solve𝑛𝑛 Eq. (3.20) to obtain { } . 𝑐𝑐𝑗𝑗 𝑗𝑗=0 Step 3. Solve Eq. (3.17) to obtain { }𝑛𝑛−1 . 𝑏𝑏𝑗𝑗 𝑗𝑗=0 𝑛𝑛−1 𝑑𝑑𝑗𝑗 𝑗𝑗=0 23 Example 3.5.4 Let ( , ( )) = (0,1), ( , ( )) = (1, ), ( , ( )) = (2, ), 0 0 ( , ( )) = (3, 𝑥𝑥). And𝑓𝑓 𝑥𝑥 = , =2 . 1 1 2 2 Determine𝑥𝑥 𝑓𝑓 the𝑥𝑥 clamped3 𝑒𝑒spline𝑥𝑥 𝑓𝑓 𝑥𝑥. ′ 𝑥𝑥3 𝑒𝑒 3 𝑥𝑥3 𝑓𝑓 𝑥𝑥3 𝑒𝑒 𝑓𝑓′ 𝑥𝑥0 𝑒𝑒 𝑓𝑓 𝑒𝑒 𝑆𝑆 𝑥𝑥

24 Theorem 3.11 If is defined at the nodes: = < < = , then has a unique natural spline interpolant𝑓𝑓 on the nodes; that is a 𝑎𝑎spline 𝑥𝑥interpolant0 ⋯ 𝑥𝑥that𝑛𝑛 satisfied𝑏𝑏 𝑓𝑓 the natural boundary conditions =𝑆𝑆 0, = 0. ′′ ′′ Theorem 3.12𝑆𝑆 If𝑎𝑎 is defined𝑆𝑆 𝑏𝑏 at the nodes: = < < = and differentiable at and , then has a unique𝑓𝑓 clamped spline interpolant𝑎𝑎 𝑥𝑥on0 the⋯ nodes;𝑥𝑥𝑛𝑛 that𝑏𝑏 is a spline interpolant𝑎𝑎that 𝑏𝑏 satisfied𝑓𝑓 the clamped boundary conditions 𝑆𝑆 = ( ), = ( ). ′ ′ 𝑆𝑆 𝑎𝑎 𝑓𝑓� 𝑎𝑎 𝑆𝑆 𝑏𝑏 𝑓𝑓� 𝑏𝑏 25 Error Bound Theorem 3.13 If [ , ], let = max | ( )|. If is the4 unique clamped cubic 𝑓𝑓 ∈ 𝐶𝐶 𝑎𝑎 𝑏𝑏 𝑀𝑀 spline interpolant4 to with respect to the nodes: 𝑎𝑎≤𝑥𝑥≤𝑏𝑏 𝑓𝑓 𝑥𝑥 𝑆𝑆 = < < = , then with = max𝑓𝑓 𝑎𝑎 𝑥𝑥0 ⋯ 𝑥𝑥𝑛𝑛 𝑏𝑏 𝑗𝑗+1 𝑗𝑗 maxℎ | 0≤𝑗𝑗≤𝑛𝑛−1 (𝑥𝑥 )| − 𝑥𝑥 . 4 5𝑀𝑀ℎ 𝑎𝑎≤𝑥𝑥≤𝑏𝑏 𝑓𝑓 𝑥𝑥 − 𝑆𝑆 𝑥𝑥 ≤ 384

26