2601ch3c

MATH2601 Chapter 3 Approximating Functions 118

3.3 Spline Interpolation

A spline function consists of polynomial pieces on subintervals

joined together with certain continuity conditions.

Formally, suppose that n + 1 points t0, t1, · · · , tn have been

specified and satisfy

t0 < t1 < · · · < tn

These points are called knots.

Suppose also that an integer k ≥ 0 has been prescribed. A spline

function of degree k having knots t0, t1, · · · , tn is a function

S such that:

1. On each interval [ti−1, ti], S is a polynomial of degree ≤ k.

2. S has a continuous (k − 1)st derivative on [t0, tn].

Hence, S is a piecewise polynomial of degree at most k having

continuous derivatives of all orders up to k − 1.


Figure 7: A spline of degree 0

Splines of degree 0 are piecewise constants. A typical spline of

degree 0 with six knots is shown in Figure 7.

A spline of degree 0 can be given explicitly in the form

S(x) =

S0(x) = c0 x ∈ [t0, t1)

S1(x) = c1 x ∈ [t1, t2)... ...

Sn−1(x) = cn−1 x ∈ [tn−1, tn]

The intervals [ti−1, ti) do not intersect each other, and so no

ambiguity arises in defining such a function at the knots.


Figure 8: A spline of degree 1

Figure 8 shows the graph of a typical spline function of degree 1

with nine knots. A function such as this can be defined explicitly

by

S(x) =

S0(x) = a0x + b0 x ∈ [t0, t1]

S1(x) = a1x + b1 x ∈ [t1, t2]... ...

Sn−1(x) = an−1x + bn−1 x ∈ [tn−1, tn]

The function S is continuous, and so the piecewise polynomials

match up at the knots; that is, Si(ti+1) = Si+1(ti+1).

◦ If the knots ti and the coefficients ai, bi are all prescribed,

then the value of S at x is obtained by first identifying the

subinterval [ti, ti+1) that contains x.


◦ The spline function can be defined on the entire real line.

For convenience, we can use the expression a0x + b0 on the

interval (−∞, t1) and the expression an−1x + bn−1 on the

interval [tn−1,∞).

Example 3.13 Determine whether the following function is a

quadratic spline.

f(x) =

x x ∈ (−∞, 1]

−1

2(2 − x)2 + 3

2x ∈ [1, 2]

3

2x ∈ [2,∞)

Solution It is easy to see that

f ′(x) =

1 x ∈ (−∞, 1]

2 − x x ∈ [1, 2]

0 x ∈ [2,∞)

Since

limx→1−

f(x) = 1 = limx→1+

f(x)

limx→2−

f(x) =3

2= lim

x→2+f(x)

limx→1−

f ′(x) = 1 = limx→1+

f ′(x)

limx→2−

f ′(x) = 0 = limx→2+

f ′(x)

we see that both f and f ′ are continuous on R. So f is a

quadratic spline.


Example 3.14 Is the function in Example 3.13 a cubic spline?

Solution Note that

f ′′(x) =

0 x ∈ (−∞, 1]

−1 x ∈ [1, 2]

0 x ∈ [2,∞)

Since

limx→1−

f ′′(x) = 0 6= −1 = limx→1+

f ′′(x)

By definition, f is not a cubic spline function since f ′′ is not

continuous.

Example 3.15 If S is a first degree spline function that inter-

polates f at a sequence of knots 0 = t0 < t1 < · · · < tn = 1,

what is∫

1

0S(x) dx?

Solution Suppose the given table of values is

x t0 t1 t2 · · · tn

y y0 y1 y2 · · · yn

and

S(x) =

S0(x) = a0x + b0 x ∈ [t0, t1)

S1(x) = a1x + b1 x ∈ [t1, t2)... ...

Sn−1(x) = an−1x + bn−1 x ∈ [tn−1, tn]


Then∫

1

0

S(x) dx =

n−1∑

i=0

∫ ti+1

ti

(aix + bi) dx

=n−1∑

i=0

[ai

2(t2i+1 − t2i ) + bi(ti+1 − ti)

]

=n−1∑

i=0

[ai

2(ti+1 + ti) + bi

]

(ti+1 − ti)

=n−1∑

i=0

[

1

2(aiti+1 + bi + aiti + bi)

]

(ti+1 − ti)

=n−1∑

i=0

1

2(yi+1 + yi)(ti+1 − ti)

Remark If f is nonnegative, then so is S. In this case,∫

1

0S(x) dx

is the sum of area of trapezoids over [ti, ti+1].


Example 3.16 Determine all the values of a, b, c, d, e for which

the following function is a cubic spline:

f(x) =

a(x − 2)2 + b(x − 1)3 x ∈ (−∞, 1]

c(x − 2)2 x ∈ [1, 3]

d(x − 2)2 + e(x − 3)3 x ∈ [3,∞)

Next, determine the value of parameters so that the cubic spline

interpolates this table

x 0 1 4

y 26 7 25

Solution From the definition of f , we deduce that

f ′(x) =

2a(x − 2) + 3b(x − 1)2 x ∈ (−∞, 1]

2c(x − 2) x ∈ [1, 3]

2d(x − 2) + 3e(x − 3)2 x ∈ [3,∞)

and

f ′′(x) =

2a + 6b(x − 1) x ∈ (−∞, 1]

2c x ∈ [1, 3]

2d + 6e(x − 3) x ∈ [3,∞)


The continuity of f , f ′ and f ′′ at x = 1 and x = 3 implies the

following, respectively:

f : a = c and c = d

f ′ : −2a = −2c and 2c = 2d

f ′′ : 2a = 2c and 2c = 2d

So a = c = d. Therefore f is a cubic spline iff a = c = d, and

b and e can take any values. (There are only three degrees of

freedom.)

According to the table, f(0) = 26, f(1) = 7, and f(4) = 25, so

4a − b = 26

c = 7

4d + e = 25

In addition, a = c = d. Therefore the values of these parameters

can be uniquely determined:

a = c = d = 7, b = 2 and e = −3


Cubic Splines

We shall develop more thoroughly the theory and construction of

the cubic splines (k = 3) since these are often used in practice.

We assume that a table of values has been given:

x t0 t1 t2 · · · tn

y y0 y1 y2 · · · yn

(3.22)

and that a cubic spline S it to be constructed to interpolate the

table.

On each interval [t0, t1], [t1, t2], · · · , [tn−1, tn], S is given by a

different cubic polynomial.

Let Si be the cubic polynomial that represents S on [ti, ti+1].

Thus,

S(x) =

S0(x) x ∈ [t0, t1]

S1(x) x ∈ [t1, t2]... ...

Sn−1(x) x ∈ [tn−1, tn]

(3.23)


The polynomials

Si−1 on [ti−1, ti]

and

Si on [ti, ti+1]

interpolate the same value at the point ti and therefore

Si−1(ti) = yi = Si(ti) (1 ≤ i ≤ n − 1)

Hence, S is automatically continuous.

Moreover, S ′ and S ′′ are assumed to be continuous, and these

conditions will be used in the derivation of the cubic spline func-

tion.


Does the continuity of S, S ′ and S ′′ provide enough conditions

to define a cubic spline?

There are 4n coefficients in the piecewise cubic polyno-

mial since there are four coefficients in each of the n cubic

polynomials.

◦ On each subinterval [ti, ti+1], there are two interpolation con-

ditions, S(ti) = yi and S(ti+1) = yi+1, giving 2n conditions.

◦ The continuity of S gives no additional conditions since it has

already been counted in the interpolation conditions.

◦ The continuity of S ′ gives one condition at each interior knot,

S ′i−1(ti) = S ′

i(ti)

accounting for n − 1 additional conditions.

◦ Similarly, the continuity of S ′′ gives another n− 1 conditions.

Thus there are altogether 4n − 2 conditions for determining 4n

coefficients. Two degrees of freedom remain, and there are

various ways of using them to advantage.


Now we derive the equation for Si(x) on the interval [ti, ti+1].

1. First we define the numbers zi = S ′′(ti). Clearly, zi exists for

0 ≤ i ≤ n an satisfy

limx→t+i

S ′′(x) = zi = limx→t−i

S ′′(x) (1 ≤ i ≤ n − 1)

because S ′′ is continuous at each interior knot.

2. Since Si is a polynomial of degree 3 on [ti, ti+1], S ′′i is a poly-

nomial of degree 1. We may write

S ′′i (x) = Aix + Bi, x ∈ [ti, ti+1]

Note that S ′′i satisfies

S ′′i (ti) = zi and S ′′

i (ti+1) = zi+1

Then

Aiti + Bi = zi

Aiti+1 + Bi = zi+1

Let hi = ti+1 − ti, we have

Ai =zi+1 − zi

ti+1 − ti=

zi+1 − zi

hi

Bi = zi − Aiti = zi −zi+1 − zi

hiti


Hence

S ′′i (x) = Aix + Bi

=zi+1 − zi

hix + zi −

zi+1 − zi

hiti

=zi+1

hix −

zi+1

hiti + zi −

zi

hix +

zi

hiti

=zi+1

hi(x − ti) + zi −

zi

hix +

zi

hi(ti+1 − hi)

=zi+1

hi(x − ti) + zi −

zi

hix +

zi

hiti+1 − zi

=zi+1

hi(x − ti) +

zi

hi(ti+1 − x)

So S ′′i (x) is the straight line between zi and zi+1:

S ′′i (x) =

zi

hi(ti+1 − x) +

zi+1

hi(x − ti) (3.24)

If this is integrated once,

S ′i(x) =

zi

hi

∫

(ti+1 − x) dx +zi+1

hi

∫

(x − ti) dx

= −1

2

zi

hi(ti+1 − x)2 +

1

2

zi+1

hi(x − ti)

2 + Ci

Integrating one more time, the result is Si itself:

Si(x) =zi

6hi(ti+1 − x)3 +

zi+1

6hi(x − ti)

3

+ Ci(x − ti) + Di (3.25)


3. The interpolation conditions Si(ti) = yi and Si(ti+1) = yi+1

can now be imposed on Si to determine Ci and Di.

First,

yi =zi

6hi(ti+1 − ti)

3 + Di =⇒ Di = yi −zih

2i

6

Then

yi+1 =zi+1

6hi(ti+1 − ti)

3 + Ci(ti+1 − ti) + Di

leads to

Ci =yi+1 − yi

hi+

hi

6(zi − zi+1)

Hence

Si(x) =zi

6hi(ti+1 − x)3 +

zi+1

6hi(x − ti)

3

+

[

yi+1 − yi

hi+

hi

6(zi − zi+1)

]

(x − ti)

+

(

yi −zih

2i

6

)

which can be rewritten as

Si(x) =zi

6hi(ti+1 − x)3 +

zi+1

6hi(x − ti)

3

+

(

yi+1

hi−

zi+1hi

6

)

(x − ti)

+

(

yi

hi−

zihi

6

)

(ti+1 − x) (3.26)


4. Once the values of z0, z1, · · · , zn have been determined,

(3.23) and (3.26) can be used to evaluate S(x) for any x in

the interval [t0, tn].

5. To determine z1, z2, · · · , zn−1, we use the continuity condi-

tions for S ′ at the interior knots ti:

S ′i−1(ti) = S ′

i(ti)

Since

S ′i(x) = −

1

2

zi

hi(ti+1 − x)2 +

1

2

zi+1

hi(x − ti)

2

+yi+1 − yi

hi+

hi

6(zi − zi+1)

Then substitution of x = ti and simplification lead to

S ′i(ti) = −

hi

3zi −

hi

6zi+1 +

yi+1 − yi

hi(3.27)

and

S ′i−1(ti) =

hi−1

6zi−1 +

hi−1

3zi +

yi − yi−1

hi−1

(3.28)

When the right-hand sides of (3.27) and (3.28) are set equal

to each other, the result can be written as

hi−1zi−1 + 2(hi + hi−1)zi + hizi+1

=6

hi(yi+1 − yi) −

6

hi−1

(yi − yi−1) (3.29)


If we define

ci =6

hi(yi+1 − yi)

then

6

hi(yi+1 − yi) −

6

hi−1

(yi − yi−1) = ci − ci−1

Let

bi = ci − ci−1

Then (3.29) can be written as

hi−1zi−1 + 2(hi + hi−1)zi + hizi+1 = bi (3.30)

which is valid only for i = 1, 2, · · · , n − 1. (Why?)

◦ (3.29) gives a system of n−1 linear equations for the n+1

unknowns z0, z1, · · · , zn.

◦ Recall that we have 2 degrees of freedom.

We can select z0 and zn arbitrarily and solve the resulting system

of equations to obtain z1, z2, · · · , zn−1.

One widely used choice is z0 = zn = 0. The resulting spline

function is called natural cubic spline. (This choice of spline

is supported by a theorem to be proved later.)


Equations (3.30) for 1 ≤ i ≤ n− 1 with z0 = 0 and zn = 0 leads

to a linear system of the following form.

u1 h1

h1 u2 h2

h2 u3 h3

. . . . . . . . .

hn−3 un−2 hn−2

hn−2 un−1

z1

z2

z3

...

zn−2

zn−1

=

b1

b2

b3

...

bn−2

bn−1

where

ui = 2(hi + hi−1) hi = ti+1 − ti

bi = ci − ci−1 ci =6

hi(yi+1 − yi)

◦ The coefficient matrix is symmetric, tridiagonal and diagonally

dominant.

◦ Hence Gaussian elimination without pivoting can be used to

solve the system for z1, z2, · · · , zn−1.

◦ Once zi are computed, Si(x) can be determined.


Example 3.17 Find the natural cubic spline function whose

knots are −1, 0 and 1 and that takes the value S(−1) = 13,

S(0) = 7 and S(1) = 9.

Solution We need z0 = z2 = 0.

Now

t0 = −1, t1 = 0, t2 = 1,

y0 = 13, y1 = 7, y2 = 9

and

h0 = h1 = 1

So (3.29) yields

h0z0 + 2(h1 + h0)z1 + h1z2 =6

h1

(y2 − y1) −6

h0

(y1 − y0)

2(2)z1 = 6(2) − 6(−6)

which implies z1 = 12.

By (3.26), we have

S0(x) =z0

6h0

(t1 − x)3 +z1

6h0

(x − t0)3 +

(

y1

h0

−z1h0

6

)

(x − t0)

+

(

y0

h0

−z0h0

6

)

(t1 − x)

=z1

6h0

(x − t0)3 +

(

y1

h0

−z1h0

6

)

(x − t0) +y0

h0

(t1 − x)

= 2(x + 1)3 + 5(x + 1) − 13x


and

S1(x) =z1

6h1

(t2 − x)3 +z2

6h1

(x − t1)3 +

(

y2

h1

−z2h1

6

)

(x − t1)

+

(

y1

h1

−z1h1

6

)

(t2 − x)

=z1

6h1

(t2 − x)3 +y2

h1

(x − t1) +

(

y1

h1

−z1h1

6

)

(t2 − x)

= 2(1 − x)3 + 9x + 5(1 − x)

So the natural cubic spline function is

S(x) =

{

2(x + 1)3 + 5(x + 1) − 13x x ∈ [−1, 0]

2(1 − x)3 + 9x + 5(1 − x) x ∈ [0, 1]

We now present a theorem to the effect that the natural spline

produces the smoothest possible interpolating function. The

word smooth is given a technical meaning in the theorem.

Theorem 3.6. (Theorem on Optimality of Natural Cubic

Splines) Let f ′′ be continuous on [a, b] and let a = t0 < t1 <

· · · < tn = b. If S is the natural cubic spline interpolating f at

the knots ti for 0 ≤ i ≤ n, then∫ b

a

[S ′′(x)]2 dx ≤

∫ b

a

[f ′′(x)]2 dx


Proof We need to show∫ b

a

[S ′′(x)]2 dx ≤

∫ b

a

[f ′′(x)]2 dx

Let g ≡ f − S. Then

g(ti) = 0 for 0 ≤ i ≤ n

and (f ′′)2 = (S ′′ + g′′)2 = (S ′′)2 + (g′′)2 + 2S ′′g′′ leads to∫ b

a

(f ′′)2 dx =

∫ b

a

(S ′′)2 dx +

∫ b

a

(g′′)2 dx + 2

∫ b

a

S ′′g′′ dx

The proof will be complete if we can show that∫ b

a S ′′g′′ dx ≥ 0.

We establish this in the following analysis, using integration by

parts, the condition S ′′(t0) = S ′′(tn) = 0, and the fact that S ′′′

is a constant (say ci) on [ti−1, ti].∫ b

a

S ′′g′′ dx =n

∑

i=1

∫ ti

ti−1

S ′′g′′ dx

=

n∑

i=1

{

(S ′′g′)(ti) − (S ′′g′)(ti−1) −

∫ ti

ti−1

S ′′′g′ dx

}

= −n

∑

i=1

∫ ti

ti−1

S ′′′g′ dx = −n

∑

i=1

ci

∫ ti

ti−1

g′ dx

= −n

∑

i=1

ci[g(ti) − g(ti − 1)] = 0


Theorem 3.7. Let a = t0 < t1 < · · · < tn = b. If g′′ is

continuous on [a, b] and g interpolates a given function f at the

knots ti for 0 ≤ i ≤ n, then∫ b

a

[S ′′(x)]2 dx ≤

∫ b

a

[g′′(x)]2 dx

where S is the natural cubic spline interpolating f at the knots

ti for 0 ≤ i ≤ n.

Proof Since S also interpolates g at ti for 0 ≤ i ≤ n, the

statement can be decuded directly from Theorem (3.6) with g in

place of f over there.

Remark For a curve defined by the equation y = g(x), the

curvature at a point x is

k(x) =|g′′(x)|

(1 + |g′(x)|2)3/2

which is commonly linearized to

k(x) ≈ |g′′(x)|

The quantity∫ b

a [g′′(x)]2 dx can therefore be viewed as a crude

measure of the total curvature over an interval.

In this measure, Theorem (3.7) asserts that any smooth interpo-

lating function must have a total curvature at least as large as

that of the natural cubic spline. That is why the natural spline

produces the smoothest possible interpolating function.


Example 3.18 Determine the values of a, b, c so that the fol-

lowing function is a cubic spline having knots 0, 1 and 2:

f(x) =

{

3 + x − 9x2 x ∈ [0, 1]

a + b(x − 1) + c(x − 1)2 + d(x − 1)3 x ∈ [1, 2]

Next, determine d so that∫

2

0[f ′′(x)]2 dx is minimized. Finally,

find the value of d that makes f ′′(2) = 0 and explain why this

value is different from the one previously determined.

Solution Observe that

f ′(x) =

{

1 − 18x x ∈ [0, 1]

b + 2c(x − 1) + 3d(x − 1)2 x ∈ [1, 2]

and

f ′′(x) =

{

−18 x ∈ [0, 1]

2c + 6d(x − 1) x ∈ [1, 2]

The continuity of f , f ′ and f ′′ at x = 1 implies the following,

respectively:

f : −5 = a

f ′ : −17 = b

f ′′ : −18 = 2c

So a = −5, b = −17 and c = −9.


Since∫

2

0

[f ′′(x)]2 dx

=

∫

1

0

[f ′′(x)]2 dx +

∫

2

1

[f ′′(x)]2 dx

=

∫

1

0

[−18]2 dx +

∫

2

1

[2c + 6d(x − 1)]2 dx

= 2(18)2 − 108d + 12d2 (as c = −9)

it can be seen that∫

2

0[f ′′(x)]2 dx is minimized when the first

derivative −108 + 24d = 0. Thus d = 9/2.

Setting f ′′(2) = 0 yields 2c + 6d = 0, so d = 3.

Since f ′′(0) = −18, f(x) is not a natural cubic spline for any d.

So when∫

2

0[f ′′(x)]2 dx is minimized, we need not have f ′′(2) = 0.

That is why the above values of d can be different.

Date post:	20-Feb-2015
Category:	Documents
Upload:	surender-reddy
View:	29 times
Download:	0 times

2601ch3c

Documents