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> 4. Interpolation andApproximation

Most functions cannot be evaluated exactly:

√x, ex, lnx, trigonometric functions

since by using a computer we are limited to the use of elementaryarithmetic operations

+,−,×,÷

With these operations we can only evaluate polynomials and rationalfunctions (polynomial divided by polynomials).

4. Interpolation Math 1070

> 4. Interpolation andApproximation

Interpolation

Given pointsx0, x1, . . . , xn

and corresponding values

y0, y1, . . . , yn

find a function f(x) such that

f(xi) = yi, i = 0, . . . , n.

The interpolation function f is usually taken from a restricted class offunctions: polynomials.


> 4. Interpolation andApproximation > 4.1 Polynomial Interpolation Theory

Interpolation of functions

f(x)

x0, x1, . . . , xn

f(x0), f(x1), . . . , f(xn)

Find a polynomial (or other special function) such that

p(xi) = f(xi), i = 0, . . . , n.

What is the error f(x) = p(x)?


> 4. Interpolation andApproximation > 4.1.1 Linear interpolation

Linear interpolation

Given two sets of points (x0, y0) and (x1, y1) with x0 6= x1, draw a linethrough them, i.e., the graph of the linear polynomial

x0 x1

y0 y1

`(x) =x− x1

x0 − x1y0 +

x− x0

x1 − x0y1

`(x) =(x1 − x)y0 + (x− x0)y1

x1 − x0(5.1)

We say that `(x) interpolates the value yi at the point xi, i = 0, 1, or`(xi) = yi, i = 0, 1

Figure: Linear interpolation4. Interpolation Math 1070


Example

Let the data points be (1, 1) and (4,2). The polynomialP1(x) is given by

P1(x) =(4− x) · 1 + (x− 1) · 2

3(5.2)

The graph y = P1(x) and y =√x, from which the data points were taken.

Figure: y =√x and its linear interpolating polynomial (5.2)



Example

Obtain an estimate of e0.826 using the function values

e0.82 ·= 2.270500, e0.83 ·= 2.293319

Denote x0 = 0.82, x1 = 0.83. The interpolating polynomial P1(x)interpolating ex at x0 and x1 is

P1(x) =(0.83− x) · 2.270500 + (x− 0.82) · 2.293319

0.01(5.3)

andP1(0.826) = 2.2841914

while the true value s

e0.826 ·= 2.2841638

to eight significant digits.


> 4. Interpolation andApproximation > 4.1.2 Quadratic Interpolation

Assume three data points (x0, y0), (x1, y1), (x2, y2), with x0, x1, x2 distinct.We construct the quadratic polynomial passing through these points usingLagrange’s folmula

P2(x) = y0L0(x) + y1L1(x) + y2L2(x) (5.4)

with Lagrange interpolation basis functions for quadratic interpolatingpolynomial

L0(x) = (x−x1)(x−x2)(x0−x1)(x0−x2)

L1(x) = (x−x0)(x−x2)(x1−x0)(x1−x2)

L2(x) = (x−x0)(x−x1)(x2−x0)(x2−x1)

(5.5)

Each Li(x) has degree 2 ⇒ P2(x) has degree ≤ 2. MoreoverLi(xj) = 0, j 6= iLi(xi) = 1 for 0 ≤ i, j ≤ 2 i.e., Li(xj) = δi,j =

{1, i = j0, i 6= j

the Kronecker delta function.

P2(x) interpolates the data

P2(x) = yi, i=0,1,2



Example

Construct P2(x) for the data points (0,−1), (1,−1), (2, 7). Then

P2(x)=(x−1)(x−2)

2· (−1)+

x(x−2)−1

· (−1)+x(x−1)

2· 7 (5.6)

Figure: The quadratic interpolating polynomial (5.6)



With linear interpolation: obvious that there is only one straight linepassing through two given data points.With three data points: only one quadratic interpolating polynomialwhose graph passes through the points.Indeed: assume ∃Q2(x), deg(Q2) ≤ 2 passing through(xi, yi), i = 0, 1, 2, then it is equal to P2(x). The polynomial

R(x) = P2(x)−Q2(x)

has deg(R) ≤ 2 and

R(xi) = P2(xi)−Q2(xi) = yi − yi = 0, for i = 0, 1, 2

So R(x) is a polynomial of degree ≤ 2 with three roots ⇒ R(x) ≡ 0



Example

Calculate a quadratic interpolate to e0.826 from the function values

e0.82 ·= 2.27050 e0.83 ·= 2.293319 e0.84 ·= 2.231637

With x0 = e0.82, x1 = e0.83, x2 = e0.84, we have

P2(0.826) ·= 2.2841639

to eight digits, while the true answer e0.826 ·= 2.2841638 andP1(0.826) ·= 2.2841914.


> 4. Interpolation andApproximation > 4.1.3 Higher-degree interpolation

Lagrange’s Formula

Given n+ 1 data points (x0, y0), (x1, y1), . . . , (xn, yn) with all xi’sdistinct, ∃ unique Pn, deg(Pn) ≤ n such that

Pn(xi) = yi, i = 0, . . . , ngiven by Lagrange’s Formula

Pn(x) =n∑i=0

yiLi(x) = y0L0(x) + y1L1(x) + · · · ynLn(x) (5.7)

where Li(x)=n∏

j=0,j 6=i

x−xjxi−xj

=(x−x0) · · · (x−xi−1)(x−xi+1) · · · (x−xn)(xi−x0) · · · (xi−xi−1)(xi−xi) · · · (xi−xn)

,

where Li(xj) = δij


> 4. Interpolation andApproximation > 4.1.4 Divided differences

Remark; The Lagrange’s formula (5.7) is suited for theoretical uses,but is impractical for computing the value of an interpolating polynomial:knowing P2(x) does not lead to a less expensive way to compute P3(x).But for this we need some preliminaries, and we start with a discrete versionof the derivative of a function f(x).

Definition (First-order divided difference)

Let x0 6= x1, we define the first-order divided difference of f(x) by

f [x0, x1] =f(x1)− f(x2)

x1 − x2(5.8)

If f(x) is differentiable on an interval containing x0 and x1, then the meanvalue theorem gives

f [x0, x1] = f ′(c), for c between x0 and x1.

Also if x0, x1 are close together, then

f [x0, x1] ≈ f ′(x0 + x1

2

)usually a very good approximation.



Example

Let f(x) = cos(x), x0 = 0.2, x1 = 0.3.

Then

f [x0, x1] =cos(0.3)− cos(0.2)

0.3− 0.2·= −0.2473009 (5.9)

while

f ′(x0 + x1

2

)= − sin(0.25) ·= −0.2474040

so f [x0, x1] is a very good approximation of this derivative.



Higher-order divided differences are defined recursively

Let x0, x1, x2 ∈ R distinct. The second-order divideddifference

f [x0, x1, x2] =f [x1, x2]− f [x0, x1]

x2 − x0(5.10)

Let x0, x1, x2, x3 ∈ R distinct. The third-order divideddifference

f [x0, x1, x2, x3] =f [x1, x2, x3]− f [x0, x1, x2]

x3 − x0(5.11)

In general, let x0, x1, . . . , xn ∈ R, n+ 1 distinct numbers. Thedivided difference of order n

f [x0, . . . , xn] =f [x1, . . . , xn]− f [x0, . . . , xn−1]

xn − x0(5.12)

or the Newton divided difference.



Theorem

Let n ≥ 1, f ∈ Cn[α, β] and x0, x1, . . . , xn n+ 1 distinct numbers in [α, β].Then

f [x0, x1, . . . , xn] = 1n!

f (n)(c) (5.13)

for some unknown point c between the maximum and the minimum ofx0, . . . , xn.

Example

Let f(x) = cos(x), x0 = 0.2, x1 = 0.3, x2 = 0.4.

The f [x0, x1] is given by (5.9), and

f [x1, x2] =cos(0.4)− cos(0.3)

0.4− 0.3·= −0.3427550

hence from (5.11)f [x0, x1, x2]

·=−0.3427550− (−0.2473009)

0.4− 0.2= −0.4772705 (5.14)

For n = 2, (5.13) becomes

f [x0, x1, x2] =12f ′′(c) = −1

2cos(c) ≈ −1

2cos(0.3) ·= −0.4776682

which is nearly equal to the result in (5.14).


> 4. Interpolation andApproximation > 4.1.5 Properties of divided differences

The divided differences (5.12) have special properties that help simplify workwith them.(1) Let (i0, i1, . . . , in) be a permutation (rearrangement) of the integers(0, 1, . . . , n). It can be shown that

f [xi0 , xi1 , . . . , xin] = f [x0, x1, . . . , xn] (5.15)

The original definition (5.12) seems to imply that the order of x0, x1, . . . , xn

is important, but (5.15) asserts that it is not true.For n = 1

f [x0, x1] =f(x0)− f(x1)

x0 − x1=f(x1)− f(x0)

x1 − x0= f [x1, x0]

For n = 2 we can expand (5.11) to get

f [x0, x1, x2] =f(x0)

(x0 − x1)(x0 − x2)+

f(x1)(x1 − x0)(x1 − x2)

+f(x2)

(x2 − x0)(x2 − x1)

(2) The definitions (5.8), (5.11) and (5.12) extend to the case where some or

all of the xi coincide, provided that f(x) is sufficiently differentiable.



For example, define

f [x0, x0] = limx1→x0

f [x0, x1] = limx1→x0

f(x0)− f(x1)x1 − x0

f [x0, x0] = f ′(x0)

For an arbitrary n ≥ 1, let all xi → x0; this leads to the definition

f [x0, . . . , x0] =1n!f (n)(x0) (5.16)

For the cases where only some of nodes coincide: using (5.15), (5.16)we can extend the definition of the divided difference.For example

f [x0, x1, x0] = f [x0, x0, x1] =f [x0, x1]− f [x0, x0]

x1 − x0=f [x0, x1]− f ′(x0)

x1 − x0



MATLAB program: evaluating divided differences

Given a set of values f(x0), . . . , f(xn) we need to calculate the set ofdivided differences

f [x0, x1], f [x0, x1, x2], . . . , f [x0, x1, . . . , xn]

We can use the MATLAB function divdif using the function calldivdif y = divdif(x nodes, y values)Note that MATLAB does not allow zero subscripts, hence x nodes andy values have to be redefined as vectors containing n+ 1 components:

x nodes = [x0, x1, . . . , xn]x nodes(i) = xi−1, i = 1, . . . , n+ 1y values = [f(x0), f(x1), . . . , f(xn)]y values(i) = f(xi−1), i = 1, . . . , n+ 1



MATLAB program: evaluating divided differences

function divdif y = divdif(x nodes,y values) %% This is a function% divdif y = divdif(x nodes,y values)% It calculates the divided differences of the function% values given in the vector y values, which are the values of% some function f(x) at the nodes given in x nodes. On exit,% divdif y(i) = f[x 1,...,x i], i=1,...,m% with m the length of x nodes. The input values x nodes and% y values are not changed by this program.%divdif y = y values;m = length(x nodes);for i=2:mfor j=m:-1:idivdif y(j) = (divdif y(j)-divdif y(j-1)) ...

/(x nodes(j)-x nodes(j-i+1));end

end



This program is illustrated in this table.

i xi cos(xi) Di

0 0.0 1.000000 0.1000000E+1

1 0.1 0.980067 -0.9966711E-1

2 0.4 0.921061 -0.4884020E+0

3 0.6 0.825336 0.4900763E-1

4 0.8 0.696707 0.3812246E-1

5 1.0 0.540302 -0.3962047E-2

6 1.2 0.36358 -0.1134890E-2

Table: Values and divided differences for cos(x)


> 4. Interpolation andApproximation > 4.1.6 Newton’s Divided Differences

Newton interpolation formula

Interpolation of f at x0, x1, . . . , xnIdea: use Pn−1(x) in the definition of Pn(x):

Pn(x) = Pn−1(x) + C(x)

C(x) ∈ Pn correction termC(xi) = 0 i = 0, . . . , n− 1

}=⇒

=⇒ C(x) = a(x− x0)(x− x1) . . . (x− xn−1)Pn(x) = axn + . . .



Definition: The divided difference of f(x) at points x0, . . . , xn

is denoted by f [x0, x1, . . . , xn] and is defined to be the coefficient ofxn in Pn(x; f).

C(x) = f [x0, x1, . . . , xn](x− x0) . . . (x− xn−1)pn(x; f) = pn−1(x; f)+f [x0, x1, . . . , xn](x−x0) . . . (x−xn−1) (5.17)

p1(x; f) = f(x0) + f [x0, x1](x− x0) (5.18)

p2(x; f)=f(x0)+f [x0, x1](x−x0)+f [x0, x1, x2](x−x0)(x−x1) (5.19)

...

pn(x; f) = f(x0) + f [x0, x1](x− x0) + . . . (5.20)

+f [x0, . . . , xn](x− x0) . . . (x− xn−1)=⇒ Newton interpolation formula



For (5.18), consider p1(x0), p1(x1). Easily, p1(x0) = f(x0), and

p1(x1)=f(x0)+(x1−x0)[f(x1)−f(x0)x1 − x0

]=f(x0)+[f(x1)−f(x0)]=f(x1)

So: deg(p1) ≤ 1 and satisfies the interpolation conditions. Then theuniqueness of polynomial interpolation ⇒ (5.18) is the linearinterpolation polynomial to f(x) at x0, x1.

For (5.19), note that

p2(x) = p1(x) + (x− x0)(x− x1)f [x0, x1, x2]

It satisfies: deg(P2) ≤ 2 and

p2(xi) = p1(xi) + 0 = f(xi), i = 0, 1p2(x2) = f(x0) + (x2 − x0)f [x0.x1] + (x2 − x0)(x2 − x1)f [x0, x1, x2]

= f(x0) + (x2 − x0)f [x0.x1] + (x2 − x1){f [x1, x2]− f [x0, x1]}= f(x0) + (x1 − x0)f [x0.x1] + (x2 − x1)f [x1, x2]= f(x0) + {f(x1)− f(x0)}+ {f(x2)− f(x1)} = f(x2)

By the uniqueness of polynomial interpolation, this is the quadraticinterpolating polynomial to f(x) at {x0, x1, x2}.



Example

Find p(x) ∈ P2 such that p(−1) = 0, p(0) = 1, p(1) = 4.

p(x) = p(x0) + p[x0, x1](x− x0) + p[x0, x1, x2](x− x0)(x− x1)

xi p(xi) p[xi, xi+1] p[xi, xi+1, xi+1]−1 0 1 10 1 31 4

p(x) = 0 + 1(x+ 1) + 1(x+ 1)(x− 0) = x2 + 2x+ 1



Example

Let f(x) = cos(x). The previous table contains a set of nodes xi, the valuesf(xi) and the divided differences computed with divdif

Di = f [x0, . . . , xi] i ≥ 0

pn(0.1) pn(0.3) pn(0.5)1 0.9900333 0.9700999 0.95016642 0.9949173 0.9554478 0.87690613 0.9950643 0.9553008 0.87764134 0.9950071 0.9553351 0.87758415 0.9950030 0.9553369 0.87758236 0.9950041 0.9553365 0.8775825

True 0.9950042 0.9553365 0.8775826

Table: Interpolation to cos(x) using (5.20)

This table contains the values of pn(x) for various values of n, computed

with interp, and the true values of f(x).



In general, the interpolation node points xi

need not to be evenly spaced,nor be arranged in any particular order

to use the divided difference interpolation formula (5.20)

To evaluate (5.20) efficiently we can use a nested multiplicationalgorithm

Pn(x) = D0 + (x− x0)D1 + (x− x0)(x− x1)D2

+ . . .+ (x− x0) · · · (x− xn−1)Dn

with D0 = f(x0), Di = f [x0, . . . , xi] for i ≥ 1Pn(x) = D0 + (x− x0)[D1 + (x− x1)[D2 + · · · (5.21)

+ (x− xn−2)[Dn−1 + (x− xn−1)Dn] · · · ]]For example

P3(x) = D0 + (x− x0)D1 + (x− x1)[D2 + (x− x2)D3]

(5.21) uses only n multiplications to evaluate Pn(x) and is more convenient

for a fixed degree n. To compute a sequence of interpolation polynomials of

increasing degree is more efficient to se the original form (5.20).



MATLAB: evaluating Newton divided difference for polynomial interpolation

function p eval = interp(x nodes,divdif y,x eval) %% This is a function% p eval = interp(x nodes,divdif y,x eval)% It calculates the Newton divided difference form of% the interpolation polynomial of degree m-1, where the% nodes are given in x nodes, m is the length of x nodes,% and the divided differences are given in divdif y. The% points at which the interpolation is to be carried out% are given in x eval; and on exit, p eval contains the% corresponding values of the interpolation polynomial.%m = length(x nodes);p eval = divdif y(m)*ones(size(x eval));for i=m-1:-1:1p eval = divdif y(i) + (x eval - x nodes(i)).*p eval;

end


> 4. Interpolation andApproximation > 4.2 Error in polynomial interpolation

Formula for error E(t) = f(t)− pn(t; f)

Pn(x) =n∑j=0

f(xj)Lj(x)

Theorem

Let n ≥ 0, f ∈ Cn+1[a, b], x0, x1, . . . , xn distinct points in [a, b]. Then

f(x)− Pn(x) = (x− x0) . . . (x− xn)f [x0, x1, . . . , xn, x] (5.22)

=(x− x0)(x− x1) · · · (x− xn)

(n+ 1)!f (n+1)(cx) (5.23)

for x ∈ [a, b], cx unknown between the minimum and maximum ofx0, x1, . . . , xn.



Proof of Theorem

Fix t ∈ R. Consider

pn+1(x; f)− interpolates f at x0, . . . , xn, t

pn(x; f)− interpolates f at x0, . . . , xn

From Newton

pn+1(x; f) = pn(x; f) + f [x0, . . . , xn, t](x− x0) . . . (x− xn)

Let x = t.

f(t) := pn+1(t, f) = pn(t; f) + f [x0, . . . , xn, t](t− x0) . . . (t− x)E(t) := f [x0, x1, . . . , xn, t](t− x0) . . . (t− xn) �



Examples

f(x) = xn, f [x0, . . . , xn] = n!n! = 1 or

pn(x; f) = 1 · xn ⇒ f [x0, . . . , xn] = 1f(x) ∈ Pn−1, f [x0, x1, . . . , xn] = 0f(x) = xn+1, f [x0, . . . , xn] = x0 + x1 + . . .+ xn

R(x) = xn+1 − pn(x; f) ∈ Pn+1

R(xi) = 0, i = 0, . . . , n

R(xi) = (x− x0) . . . (x− xn) = xn+1 − (x0 + . . .+ xn)xn + . . .︸︷︷︸Pn

=⇒ f [x0, . . . , xn] = x0 + x1 + . . .+ xn.

If f ∈ Pm,

f [x0, . . . , xm, x] ={

polynomial degree m− n− 1 if n ≤ m− 10 if n > m− 1



Example

Take f(x) = ex, x ∈ [0, 1] and consider the error in linear interpolationto f(x) using x0, x1 satisfying 0 ≤ x0 ≤ x1 ≤ 1.

From (5.23)

ex − P1(x) =(x− x0)(x− x1)

2ecx

for some cx between the max and min of x0, x1 and x. Assumingx0 < x < x1, the error is negative and approximately a quadraticpolynomial

ex − P1(x) = −(x1 − x)(x− x0)2

ecx

Since x0 ≤ cx ≤ x1

(x1 − x)(x− x0)2

ex0 ≤ |ex − P1(x)| =(x1 − x)(x− x0)

2ex1



For a bound independent of x

maxx0≤x≤x1

(x1 − x)(x− x0)2

=h2

8, h = x1 − x0

and ex1 ≤ e on [0, 1]

|ex − P1(x)| ≤h2

8e, 0 ≤ x0 ≤ x ≤ x1 ≤ 1

independent of x, x0, x1.Recall that we estimated e0.826 ·= 2.2841914 by e0.82 and e0.83, i.e.,h = 0.01.

|ex − P1(x)| ≤h2

8e = |ex − P1(x)| ≤

0.012

82.72 = 0.0000340,

The actual error being −0.0000276, it satisfies this bound.



Example

Again let f(x) = ex on [0, 1], but consider the quadratic interpolation.

ex − P2(x) =(x− x0)(x− x1)(x− x2)

6ecx

for some cx between the min and max of x0, x1, x2 and x.Assuming evenly spaced points, h = x1 − x0 = x2 − x1, and0 ≤ x0 < x < x2 ≤ 1, we have as before

|ex − P2(x)| ≤∣∣∣∣ (x− x0)(x− x1)(x− x2)

6

∣∣∣∣ e1while

maxx0≤x≤x2

(x− x0)(x− x1)(x− x2)6

=h3

9√

3(5.24)

hence

|ex − P2(x)| ≤h3e

9√

3≈ 0.174h3



For h = 0.01, 0 ≤ x ≤ 1

|ex − P2(x)| ≤ 1.74× 10−7



Let w2(x) =(x+ h)x(x− h)

6=x3 − xh

6

a translation along x-axis of polynomial in (5.24). Then x = ± h√3

satisfies

0 = w′2(x) =3x2 − h2

6and gives

∣∣∣∣w2

(± h√

3

)∣∣∣∣ = h3

9√

3

Figure: y = w2(x)4. Interpolation Math 1070

> 4. Interpolation andApproximation > 4.2.2 Behaviour of the error

When we consider the error formula (5.23) or (5.22), the polynomial

ψn(x) = (x− x0) · · · (x− xn)

is crucial in determining the behaviour of the error. Let assume thatx0, . . . , xn are evenly spaced and x0 ≤ x ≤ xn.

Figure: y = Ψ6(x)4. Interpolation Math 1070


Remark that the interpolation error

is relatively larger in [x0, x1] and [x5, x6], and

is likely to be smaller near the middle of the node points

⇒ In practical interpolation problems, high-degree polynomial interpolationwith evenly spaced nodes is seldom used.But when the set of nodes is suitable chosed, high-degree polynomials can bevery useful in obtaining polynomial approximations to functions.

Example

Let f(x) = cos(x), h = 0.2, n = 8 and then interpolate at x = 0.9.

Case (i) x0 = 0.8, x8 = 2.4⇒ x = 0.9 ∈ [x0, x1]. By direct calculation ofP8(0.9),

cos(0.9)− P8(0.9) ·= −5.51× 10−9

Case (ii) x0 = 0.2, x8 = 1.8⇒ x = 0.9 ∈ [x3, x4], where x4 is the midpoint.

cos(0.9)− P8(0.9) ·= 2.26× 10−10,

a factor of 24 smaller than the first case.



Example

Le f(x) = 11+x2 , x ∈ [−5, 5], n > 0 an even integer, h = 10

n

xj = −5 + jh, j = 0, 1, 2, . . . , n

It can be shown that for many points x in [−5, 5], the sequence of {Pn(x)}does not converge to f(x) as n→∞.

Figure: The interpolation to 11+x2


> 4. Interpolation andApproximation > 4.3 Interpolation using spline functions

x 0 1 2 2.5 3 3.5 4

y 2.5 0.5 0.5 1.5 1.5 1.125 0The simplest method of interpolating data in a table: connecting the nodepoints by straight lines: the curve y = `(x) is not very smooth.

Figure: y = `(x): piecewise linear interpolation

We want to construct a smooth curve that interpolates the given data point

that follows the shape of y = `(x).4. Interpolation Math 1070


Next choice: use polynomial interpolation. With seven data point, weconsider the interpolating polynomial P6(x) of degree 6.

Figure: y = P6(x): polynomial interpolation

Smooth, but quite different from y = `(x)!!!



A third choice: connect the data in the table using a succession ofquadratic interpolating polynomials: on each [0, 2], [2, 3], [3, 4].

Figure: y = q(x): piecewise quadratic interpolation

Is smoother than y = `(x), follows it more closely than y = P6(x), butat x = 2 and x = 3 the graph has corners, i.e., q′(x) is discontinuous.


> 4. Interpolation andApproximation > 4.3.1 Spline interpolation

Cubic spline interpolation

Suppose n data points (xi, yi), i = 1, . . . , n are given and

a = x1 < . . . < xn = b

We seek a function S(x) defined on [a, b] that interpolates the data

S(xi) = yi, i = 1, . . . , n

Find S(x) ∈ C2[a, b], a natural cubic spline such that

S(x)is a polynomial of degree ≤ 3on each subinterval [xj−1, xj ], j = 2, 3, . . . ,n;S(xi) = yi, i = 1, . . . , n,S′′(a) = S′′(b).

On [xi−1, xi]: 4 degrees of freedom (cubic spline) ⇒ 4(n− 1) DOF.


> 4. Interpolation andApproximation > 4.3.1 Spline interpolation

S(xi) = yi : n constraints

S′(xi − 0) = S′(xi + 0) continuity 2 = 1, . . . , n− 1S′′(xi − 0) = S′′(xi + 0) i = 2, . . . , n− 1S(xi − 0) = S(xi + 0) i = 2, . . . , n− 1

3n− 6

n+ (3n− 6) = 4n− 6 constraints

Two more constraints needed to be added, Boundary Conditions:

S′′ = 0 at a = x1, xn = b,

for a natural cubic spline.(Linear system, with symmetric, positive definite, diagonally dominantmatrix.)


> 4. Interpolation andApproximation > 4.3.2 Construction of the cubic spline

Introduce the variables M1, . . . ,Mn with

Mi ≡ S′′(xi), i = 1, 2, . . . , n

Since S(x) is cubic on each [xj−1, xj ]⇒ S′′(x) is linear, hence determinedby its values at two points:

S′′(xj−1) = Mj−1, S′′(xj) = Mj

⇒S′′(x) =(xj − x)Mj−1 + (x− xj−1)Mj

xj − xj−1, xj−1 ≤ x ≤ xj (5.25)

Form the second antiderivative of S′′(x) on [xj−1, xj ] and apply theinterpolating conditions

S(xj−1) = yj−1, S(xj) = yj

we get

S(x) =(xj − x)3Mj−1 + (x− xj−1)3Mj

6(xj − xj−1)+

(xj − x)yj−1 + (x− xj−1)yj

6(xj − xj−1)

− 16(xj − xj−1) [(xj − x)Mj−1 + (x− xj−1)Mj ] (5.26)

for x ∈ [xj−1, xj ], j = 1, . . . , n.



Formula (5.26) implies that S(x) is continuous over all [a, b] andsatisfies the interpolating conditions S(xi) = yi.Similarly, formula (5.25) for S′′(x) implies that is continuous on [a, b].To ensure the continuity of S′(x) over [a, b]: we require S′′(x) on[xj−1, xj ] and [xj , xj+1] have to give the same value at their commonpointxj , j = 2, 3, . . . , n− 1, leading to the tridiagonal linear system

xj − xj−1

6Mj−1 +

xj+1 − xj−1

3Mj +

xj+1 − xj6

Mj+1 (5.27)

=yj+1 − yjxj+1 − xj

− yj − yj−1

xj − xj−1, j = 2, 3, . . . , n− 1

These n− 2 equations together with the assumptionS′′(a) = S′′(b) = 0:

M1 = Mn = 0 (5.28)

leads to the values M1, . . . ,Mn, hence the function S(x).



Example

Calculate the natural cubic spline interpolating the data{(1, 1), (2,

12), (3,

13), (4,

14)}

n = 4, and all xj−1 − xj = 1. The system (5.27) becomes

16M1+ 2

3M2 + 16M3 = 1

316M2 + 2

3M3 + 16M4 = 1

12

and with (5.28) this yields

M2 =12, M3 = 0

which by (5.26) gives

S(x) =

112x

3 − 14x

2 − 13x+ 3

2 , 1 ≤ x ≤ 2− 1

12x3 + 3

4x2 − 7

3x+ 176 , 2 ≤ x ≤ 3

− 112x+ 7

12 , 3 ≤ x ≤ 4

Are S′(x) and S′′(x) continuous?



Example

Calculate the natural cubic spline interpolating the datax 0 1 2 2.5 3 3.5 4

y 2.5 0.5 0.5 1.5 1.5 1.125 0

n = 7, system (5.27) has 5 equations

Figure: Natural cubic spline interpolation y = S(x)Compared to the graph of linear and quadratic interpolation y = `(x) and

y = q(x), the cubic spline S(x) no longer contains corners.


> 4. Interpolation andApproximation > 4.3.3 Other interpolation spline functions

So far we only interpolated data points, wanting a smooth curve.

When we seek a spline to interpolate a known function, we are interested also

in the accuracy.

Let f(x) be given on [a, b], that we want to interpolate on evenlyspaced values of x. For n > 1, let

h =b− an− 1

, xj = a+ (j − 1)h, j = 1, 2, . . . , n

and Sn(x) be the natural cubic spline interpolating f(x) at x1, . . . , xn.It can be shown that

maxa≤x≤b

|f(x)− Sn(x)| ≤ ch2 (5.29)

where c depends on f ′′(a), f ′′(b) and maxa≤x≤b |f (4)(x)|. The reasonwhy Sn(x) doesn’t converge more rapidly (have an error bound with ahigher power of h) is that f ′′(a) 6= 0 6= f ′′(b), while by definitionS′′n(a) = S′′n(b) = 0. For functions f(x) with f ′′(a) = f ′′(b) = 0, theRHS of (5.29) can be replaced with ch4.



To improve on Sn(x), we look for other interpolating functions S(x) thatinterpolate f(x) on

a = x1 < x2 < . . . < xn = b

Recall the definition of natural cubic spline:

1 S(x) cubic on each subinterval [xj−1, xj ];2 S(x), S′(x) and S′′(x) are continuous on [a, b];3 S′′(x1) = S′′(xn) = 0.

We say that S(x) is a cubic spline on [a, b] if

1 S(x) cubic on each subinterval [xj−1, xj ];2 S(x), S′(x) and S′′(x) are continuous on [a, b];

With the interpolating conditions

S(xi) = yi, i = 1, . . . , n

the representation formula (5.26) and the tridiagonal system (5.27) arestill valid.



This system has n− 2 equations and n unknowns: M1, . . . ,Mn. Byreplacing the end conditions (5.28) S′′(x1) = S′′(xn) = 0, we canobtain other interpolating cubic splines.

If the data (xi, yi) is obtained by evaluating a function f(x)

yi = f(xi), i = 1, . . . , n

then we choose endpoints (boundary conditions) for S(x) that wouldyield a better approximation to f(x). We require

S′(x1) = f ′(x1), S′(xn) = f ′(xn)

or

S′′(x1) = f ′′(x1), S′′(xn) = f ′′(xn)

When combined with(5.26-5.27), either of these conditions leads to aunique interpolating spline S(x), dependent on which of theseconditions is used. In both cases, the RHS of (5.29) can be replacedby ch4, where c depends on maxx∈[a,b]

∣∣f (4)(x)∣∣.



If the derivatives of f(x) are not known, then extra interpolatingconditions can be used to ensure that the error bond of (5.29) isproportional to h4. In particular, suppose that

x1 < z1 < x2, xn−1 < z2 < xn

and f(z1), f(z2) are known. Then use the formula for S(x) in (5.26)and

s(z1) = f(z1), s(z2) = f(z2) (5.30)

This adds two new equations to the system (5.27), one for M1 andM2, and another equation for Mn−1,Mn. This form is preferable tothe interpolating natural cubic spline, and is almost equally easy toproduce. This is the default form of spline interpolation that isimplemented in MATLAB. The form of spline formed in this way issaid to satisfy the not-a-knot interpolation boundary conditions.



Interpolating cubic spline functions are a popular way to represent dataanalytically because

1 are relatively smooth: C2,

2 do not have the rapid oscillation that sometime occurs with thehigh-degree polynomial interpolation,

3 are relatively easy to work with on a computer.

They do not replace polynomials, but are a very useful extension ofthem.


> 4. Interpolation andApproximation > 4.3.4 The MATLAB program spline

The standard package MATLAB contains the functionrm spline.The standard calling sequence is

y = spline(x nodes, y nodes, x)

which produces the cubic spline function S(x) whose graph passesthrough the points {(ξi, ηi) : i = 1, . . . , n} with

(ξi, ηi) = (x nodes(i), y nodes (i))

and n the length of x nodes (and y nodes). The not-a-knotinterpolation conditions of (5.30) are used. The point (ξ2, η2) is thepoint (z1, f(z1)) of (5.30) and (ξn−1, ηn−1) is the point (z2, f(z2)).



Example

Approximate the function f(x) = ex on the interval [a, b] = [0, 1]. Forn > 0, define h = 1/n and interpolating nodes

x1 = 0, x2 = h, x3 = 2h, . . . , xn+1 = nh = 1

Using spline, we produce the cubic interpolating spline interpolantSn,1 to f(x). With the not-a-knot interpolation conditions, the nodesx2 and xn are the points z1 and z2 in (5.30). For a general smoothfunction f(x), it turns out that te magnitude of the errorf(x)− Sn,1(x) is largest around the endpoints of the interval ofapproximation.



Example

Two interpolating nodes are inserted, the midpoints of the subintervals [0, h]and [1− h, 1]:

x1 =0, x2 =12h, x3 =h, x4 =2h, . . . , xn+1 =(n−1)h, xn+2 =1− 1

2h, xn+3 =1

Using spline results in a cubic spline function Sn,2(x); with the not-a-knotinterpolation conditions conditions, the nodes x2 and xn+2 are the points z1and z2 of (5.30). Generally, Sn,2 is a more accurate approximation than isSn,1(x).The cubic polynomials produced for Sn,2(x) by spline for the intervals[x1, x2] and [x2, x3] are the same, thus we can use the polynomial for [0, 1

2h]for the entire interval [0, h]. Similarly for [1− h, h].

n E(1)n Ratio E

(2)n Ratio

5 1.01E-4 1.11E-510 6.92E-6 14.6 7.88E-7 14.120 4.56E-7 15.2 5.26E-8 15.040 2.92E-8 15.6 3.39E-9 15.5

Table: Cubic spline approximation to f(x) = ex


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