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Chapter 3

SplinesDefinition (3.1) : Given a function f defined on [a , b] and a

set of numbers, a = x0 < x1 < x2 < ……. < xn = b, called nodes,

a cubic spline interpolant S, for f is a function that satisfies

the following condition:

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(3.1). S is a cubic polynomial, denoted Sj , on the subinterval

[ xj , xj + 1 ] for each j = 0, 1, ..., n-1.

(3.2). S(xj) = f(xj) for each j = 0, 1, …, n.

(The spline passes through each data point.)

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(3.3). Sj + 1 (xj + 1 ) = Sj (xj + 1) for each j = 0, 1, …, n-2.

( The spline forms a continuous function.)

(3.4). Sj + 1(xj + 1) = Sj (xj + 1) for each j = 0, 1, …, n-2.

( The spline forms a smooth function.)

(3.5). Sj + 1(xj + 1) = Sj (xj + 1) for each j = 0, 1, …, n-2.

( The second derivative is continuous.)

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(3.6). One of the following set of boundary conditions is satisfied:

(4.6.1) S(x0) = Sxn) = 0 ( free or natural boundary)(4.6.2) S(x0) = f(x0) and S(xn) = f(xn) ( Clamped boundary)

Although cubic splines are defined with other boundary

conditions, the conditions given above are sufficient for our

purposes. When the free boundary conditions occur, the spline

is called a natural cubic spline.

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and its graph approximates the shape, which a long flexible

rod would assume if forced to go through each of the data

points {( x0 , f(x0)) , ( x1 , f(x1)) , …, ( xn , f(xn))}.

Similarly, when the clamped boundary conditions occur, the

spline is called a clamped cubic spline.

In general, clamped boundary conditions lead to more accurate

approximations since they include more information about the

function.

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However, for this type of boundary conditions to hold, it is

necessary to have either the value of the derivative at the

endpoints or acceptable approximations to those values. Hence

we concentrate here only on natural cubic interpolants.

To construct the cubic spline interpolant for a given function

f, the conditions in the definition (3.1) are applied to the cubic

polynomials.

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Sj(x) = aj + bj (x – xj) + cj (x – xj)2 + dj (x – xj )3;

for each j = 0, 1, …, n-1.

Clearly, Sj(xj) = aj = f(xj),

and if conditions (3.3) are applied,

aj + 1 = Sj + 1(xj + 1) = Sj (xj + 1) = aj + bj (xj +1 – xj) + cj (xj + 1 – xj)2

+ dj (xj + 1 – xj )3

for each j = 0, 1, …, n-2.

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Since the term (xj+1 – xj) will be used repeatedly in this

development, it is convenient to introduce the notation

hj = xj+1 – xj ; for each j = 0, 1, …, n-1.

If we also define an = f(xn), then the equation

aj + 1 = aj + bjhj + cjhj2 + djhj

3; holds for each j = 0 , 1, …, n-1. (3.9)

In a similar manner, define bn = S(xn) and observe that Sj(x) = bj + 2cj( x – xj) + 3dj (x- xj)2

Implies Sj(xj) = bj for each j = 0, 1, …, n-1.

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Appling condition (3.4), we get

bj + 1 = bj + 2cjhj + 3djhj2 ; for each j = 0, 1, …, n-1. (3.10)

Another relationship between the coefficients of Sj is obtained by

defining cn = Sxn) /2 and applying condition (3.5) In this

case,

cj + 1 = cj + 3djhj ; for each j = 0, 1, …, n-1. (3.11)

Solving for dj in Eq. (3.11) and substituting this value into Eqs.

(3.9) and (3.10) gives the new equations

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aj + 1 = aj + bjhj + (hj2 / 3) (2cj + cj + 1 ) ;

for each j = 0, 1, …, n-1. (3.12)

and

bj + 1 = bj + hj (cj + cj + 1) ; for each j = 0, 1, …, n-1. (3.13)

The final relationship involving the coefficients is obtained by

solving the appropriate equation in the form of equation (3.12),

first for bj,

bj = (1/hj) (aj +1 – aj) – (hj / 3) (2cj + cj +1) (3.14)

and then, with a reduction of the index, for bj – 1,

bj –1 = (1 / hj – 1) (aj – aj –1) – (hj –1 / 3)(2cj –1 + cj ).

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Substituting these values into the equation derived from Eq. (3.13),

when the index is reduced by 1, gives the linear system of

equations

hj –1 cj – 1+ 2(hj – 1 + hj )cj+hjcj +1 = (3/hj)(aj +1 – aj) – (3/ hj – 1)(aj – aj –1 );

for each j = 0, 1, …, n-1. (3.15)

This system involves, only cj s as unknowns, since the value of

hj s and aj s are given by the spacing of the nodes xj s and the

value of f at the nodes.

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Note that once the value of cj s are known, it is a simpler matter

to find the remainder of the constants bj s from Eq. (3.14) and

dj s from Eq. (3.11) and to construct the cubic polynomials Sjs.

The natural boundary conditions in this case imply that

cn = S(xn) / 2 = 0 that

0 =S(x0) = 2c0 +6d0(x0 – x0)

So, c0 = 0.

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The two equations c0 = 0 and cn = 0 together with the equations in

(3.15) produce a linear system described by the vector equation

Ax = b, where A is ( n+1 )*( n +1) matrix given below.

1000

1nh)

1nh

2n2(h

2nh

0

2h)

2h

12(h

1h 0

1h)

1h

02(h

0h

0001

A =

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nc

1c0

c

With x =

0

)2n

a1n

)(a2n

(3/h)1n

an

)(a1n

(3/h

)0a

1)(a0

(3/h)1a

2)(a1

(3/h

0

and b =

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Note : Solution of this system is usually obtained using Crouts Algorithm.

End of Chapter 3

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