MATH 375 Numerical Analysisbanach.millersville.edu/~bob/math375/Neville/main.pdf · Neville’s...

Neville’s MethodMATH 375 Numerical Analysis

J. Robert Buchanan

Department of Mathematics

Spring 2019

Motivation

I We have learned how to approximate a function usingLagrange polynomials and how to estimate the error insuch an approximation.

I Today we will learn how to interpolate between data valuesfound in a table without knowing the function thatgenerated the values.

I We will see that we can perform the interpolation withoutexplicitly writing out the interpolating polynomial.

Interpolation from Data

Suppose we had only the following table of data for f (x):

x f (x)10.1 0.1753722.2 0.3778432.0 0.5299241.6 0.6639350.5 0.63608

Approximate f (27.5) using polynomial interpolation.

Linear approximation:

P1(27.5) =(27.5− 32.0)(22.2− 32.0)

f (22.2)+(27.5− 22.2)(32.0− 22.2)

f (32.0) ≈ 0.46009

Interpolation from Data

Suppose we had only the following table of data for f (x):

x f (x)10.1 0.1753722.2 0.3778432.0 0.5299241.6 0.6639350.5 0.63608

Approximate f (27.5) using polynomial interpolation.Linear approximation:

P1(27.5) =(27.5− 32.0)(22.2− 32.0)

f (22.2)+(27.5− 22.2)(32.0− 22.2)

f (32.0) ≈ 0.46009

Higher Order ApproximationsQuadratic approximation: (two choices)

P2(27.5) =(27.5− 22.2)(27.5− 32.0)(10.1− 22.2)(10.1− 32.0)

f (10.1)

+(27.5− 10.1)(27.5− 32.0)(22.2− 10.1)(22.2− 32.0)

f (22.2)

+(27.5− 10.1)(27.5− 22.2)(32.0− 10.1)(32.0− 22.2)

f (32.0) ≈ 0.46141

P̂2(27.5) =(27.5− 32.0)(27.5− 41.6)(22.2− 32.0)(22.2− 41.6)

f (22.2)

+(27.5− 22.2)(27.5− 41.6)(32.0− 22.2)(32.0− 41.6)

f (32.0)

+(27.5− 22.2)(27.5− 32.0)(41.6− 22.2)(41.6− 32.0)

f (41.6) ≈ 0.46200

There are also two potential cubic interpolating polynomialsand a single quartic polynomial.

Remarks

I Without knowing f (x) we have no idea of the size of theerrors in these approximations.

I The highest degree polynomial does not necessarilydeliver the smallest error.

I Knowing the Lagrange Interpolating Polynomial of degreek does not help us find the one of degree k + 1.

I There is a method for using previous calculations to derivehigher order interpolations.

Remarks

I Without knowing f (x) we have no idea of the size of theerrors in these approximations.

I The highest degree polynomial does not necessarilydeliver the smallest error.

I Knowing the Lagrange Interpolating Polynomial of degreek does not help us find the one of degree k + 1.

I There is a method for using previous calculations to derivehigher order interpolations.

Lagrange Polynomials

DefinitionIf f is a function defined at x0, x1, . . . , xn distinct real numbersand m1, m2, . . . , mk are k distinct integers with 0 ≤ mi ≤ n,then the Lagrange polynomial that agrees with f at the k pointsxm1 , xm2 , . . . , xmk is denoted Pm1,m2,··· ,mk .

Example (1 of 2)

Let x0 = 1, x1 = 2, x2 = 4, x3 = 5, and x4 = 7 and f (x) = J0(x)(Bessel function of the first kind of order zero).

Determine P1,3,4(x).

P1,3,4(x) =(x − 5)(x − 7)(2− 5)(2− 7)

J0(2) +(x − 2)(x − 7)(5− 2)(5− 7)

J0(5)

+(x − 2)(x − 5)(7− 2)(7− 5)

J0(7)

=

[1

15x2 − 4

5x +

73

]J0(2) +

[−1

6x2 +

32

x +73

]J0(5)

+

[1

10x2 − 7

10x + 1

]J0(7)

Example (1 of 2)

Let x0 = 1, x1 = 2, x2 = 4, x3 = 5, and x4 = 7 and f (x) = J0(x)(Bessel function of the first kind of order zero).

Determine P1,3,4(x).

P1,3,4(x) =(x − 5)(x − 7)(2− 5)(2− 7)

J0(2) +(x − 2)(x − 7)(5− 2)(5− 7)

J0(5)

+(x − 2)(x − 5)(7− 2)(7− 5)

J0(7)

=

[1

15x2 − 4

5x +

73

]J0(2) +

[−1

6x2 +

32

x +73

]J0(5)

+

[1

10x2 − 7

10x + 1

]J0(7)

Example (2 of 2)

Let x0 = 1, x1 = 2, x2 = 4, x3 = 5, and x4 = 7 and f (x) = J0(x).

Use P1,3,4(3) to approximate J0(3).

P1,3,4(x) =(x − 5)(x − 7)(2− 5)(2− 7)

J0(2) +(x − 2)(x − 7)(5− 2)(5− 7)

J0(5)

+(x − 2)(x − 5)(7− 2)(7− 5)

J0(7)

P1,3,4(3) =815

J0(2) +23

J0(5)−15

J0(7)

≈ −0.0590053

Actual value: J0(3) ≈ −0.260052

Example (2 of 2)

Let x0 = 1, x1 = 2, x2 = 4, x3 = 5, and x4 = 7 and f (x) = J0(x).

Use P1,3,4(3) to approximate J0(3).

P1,3,4(x) =(x − 5)(x − 7)(2− 5)(2− 7)

J0(2) +(x − 2)(x − 7)(5− 2)(5− 7)

J0(5)

+(x − 2)(x − 5)(7− 2)(7− 5)

J0(7)

P1,3,4(3) =815

J0(2) +23

J0(5)−15

J0(7)

≈ −0.0590053

Actual value: J0(3) ≈ −0.260052

Example (2 of 2)

Let x0 = 1, x1 = 2, x2 = 4, x3 = 5, and x4 = 7 and f (x) = J0(x).

Use P1,3,4(3) to approximate J0(3).

P1,3,4(x) =(x − 5)(x − 7)(2− 5)(2− 7)

J0(2) +(x − 2)(x − 7)(5− 2)(5− 7)

J0(5)

+(x − 2)(x − 5)(7− 2)(7− 5)

J0(7)

P1,3,4(3) =815

J0(2) +23

J0(5)−15

J0(7)

≈ −0.0590053

Actual value: J0(3) ≈ −0.260052

Theoretical Result

TheoremLet f be defined at x0, x1, . . . , xk and xi and xj be two distinctreal numbers in this set. Then the kth degree Lagrangepolynomial that interpolates f at x0, x1, . . . , xk is

P(x) =(x − xj)

(xi − xj)P0,1,...,j−1,j+1,...,k (x)−

(x − xi)

(xi − xj)P0,1,...,i−1,i+1,...,k (x).

This theorem allows us to build a Lagrange polynomial ofdegree k from two Lagrange polynomials each of degree k −1.

Theoretical Result

TheoremLet f be defined at x0, x1, . . . , xk and xi and xj be two distinctreal numbers in this set. Then the kth degree Lagrangepolynomial that interpolates f at x0, x1, . . . , xk is

P(x) =(x − xj)

(xi − xj)P0,1,...,j−1,j+1,...,k (x)−

(x − xi)

(xi − xj)P0,1,...,i−1,i+1,...,k (x).

This theorem allows us to build a Lagrange polynomial ofdegree k from two Lagrange polynomials each of degree k −1.

ProofLet Q(x) = P0,1,...,j−1,j+1,...,k (x) and Q̂(x) = P0,1,...,i−1,i+1,...,k (x),then

P(x) =(x − xj)Q(x)− (x − xi)Q̂(x)

(xi − xj).

If x = xi , then P(xi) = Q(xi) = f (xi). If x = xj , thenP(xj) = Q̂(xj) = f (xj). If l ∈ {0,1, . . . , k} with i 6= l 6= j then

P(xl) =(xl − xj)Q(xl)− (xl − xi)Q̂(xl)

(xi − xj)

=(xl − xj)f (xl)− (xl − xi)f (xl)

(xi − xj)= f (xl).

Thus P(x) is a polynomial of degree at most k whichinterpolates f (x) at xi for i = 0,1, . . . , k and thus

P(x) = P0,1,...,k (x).

ProofLet Q(x) = P0,1,...,j−1,j+1,...,k (x) and Q̂(x) = P0,1,...,i−1,i+1,...,k (x),then

P(x) =(x − xj)Q(x)− (x − xi)Q̂(x)

(xi − xj).

If x = xi , then P(xi) = Q(xi) = f (xi). If x = xj , thenP(xj) = Q̂(xj) = f (xj). If l ∈ {0,1, . . . , k} with i 6= l 6= j then

P(xl) =(xl − xj)Q(xl)− (xl − xi)Q̂(xl)

(xi − xj)

=(xl − xj)f (xl)− (xl − xi)f (xl)

(xi − xj)= f (xl).

Thus P(x) is a polynomial of degree at most k whichinterpolates f (x) at xi for i = 0,1, . . . , k and thus

P(x) = P0,1,...,k (x).

ProofLet Q(x) = P0,1,...,j−1,j+1,...,k (x) and Q̂(x) = P0,1,...,i−1,i+1,...,k (x),then

P(x) =(x − xj)Q(x)− (x − xi)Q̂(x)

(xi − xj).

If x = xi , then P(xi) = Q(xi) = f (xi). If x = xj , thenP(xj) = Q̂(xj) = f (xj). If l ∈ {0,1, . . . , k} with i 6= l 6= j then

P(xl) =(xl − xj)Q(xl)− (xl − xi)Q̂(xl)

(xi − xj)

=(xl − xj)f (xl)− (xl − xi)f (xl)

(xi − xj)= f (xl).

Thus P(x) is a polynomial of degree at most k whichinterpolates f (x) at xi for i = 0,1, . . . , k and thus

P(x) = P0,1,...,k (x).

Recursive Generation of Polynomials

P0,1(x) =1

x1 − x0[(x − x0)P1 − (x − x1)P0]

P1,2(x) =1

x2 − x1[(x − x1)P2 − (x − x2)P1]

P0,1,2(x) =1

x2 − x0

[(x − x0)P1,2 − (x − x2)P0,1

]...

Neville’s Method

Define polynomial Qi,j(x) = Pi−j,i−j+1,...,i(x) where 0 ≤ j ≤ i .I j represents the degree of the polynomial.I i determines the consecutive values of x used.I Qi,j(x) can be found recursively.

Qi,j(x) =(x − xi−j)Qi,j−1(x)− (x − xi)Qi−1,j−1(x)

xi − xi−j

Tabular Calculation

Given x0, x1, . . . , xn and f (x0), f (x1), . . . f (xn) we can arrangethe Lagrange interpolating polynomials in the following table.

x0 f (x0) = P0 = Q0,0x1 f (x1) = P1 = Q1,0 P0,1 = Q1,1x2 f (x2) = P2 = Q2,0 P1,2 = Q2,1 P0,1,2 = Q2,2...

......

...xn f (xn) = Pn = Qn,0 Pn−1,n = Qn,1 Pn−2,n−1,n = Qn,2 · · · P0,1,...,n = Qn,n

Example

Approximate f (27.5) from the following data.

xi f (xi)

10.1 0.1753722.2 0.3778432.0 0.5299241.6 0.6639350.5 0.63608

Example

Order points by increasing distance from x = 27.5.

xi Qi,032.0 0.5299222.2 0.3778441.6 0.6639310.1 0.1753750.5 0.63608

Example

xi Qi,0 Qi,132.0 0.5299222.2 0.37784 0.46009†

41.6 0.66393 0.45600‡

10.1 0.17537 0.44524∗

50.5 0.63608 0.37380∗∗

†0.46009 =(27.5− 32.0)(0.37784)− (27.5− 22.2)(0.52992)

22.2− 32.0‡0.45600 =

(27.5− 22.2)(0.66393)− (27.5− 41.6)(0.37784)41.6− 22.2

∗0.44524 =(27.5− 41.6)(0.17537)− (27.5− 10.1)(0.66393)

10.1− 41.6∗∗0.37380 =

(27.5− 10.1)(0.63608)− (27.5− 50.5)(0.17537)50.5− 10.1

Example

xi Qi,0 Qi,1 Qi,232.0 0.5299222.2 0.37784 0.4600941.6 0.66393 0.45600 0.46200†

10.1 0.17537 0.44524 0.46071‡

50.5 0.63608 0.37380 0.55843∗

†0.46200 =(27.5− 32.0)(0.45600)− (27.5− 41.6)(0.46009)

41.6− 32.0‡0.46071 =

(27.5− 22.2)(0.44524)− (27.5− 10.1)(0.45600)10.1− 22.2

∗0.55843 =(27.5− 41.6)(0.37380)− (27.5− 50.5)(0.44524)

50.5− 41.6

Example

xi Qi,0 Qi,1 Qi,2 Qi,332.0 0.5299222.2 0.37784 0.4600941.6 0.66393 0.45600 0.4620010.1 0.17537 0.44524 0.46071 0.46174†

50.5 0.63608 0.37380 0.55843 0.47901‡

†0.46174 =(27.5− 32.0)(0.46071)− (27.5− 10.1)(0.46200)

10.1− 32.0‡0.47901 =

(27.5− 22.2)(0.55843)− (27.5− 50.5)(0.46071)50.5− 22.2

Example

xi Qi,0 Qi,1 Qi,2 Qi,3 Qi,432.0 0.5299222.2 0.37784 0.4600941.6 0.66393 0.45600 0.4620010.1 0.17537 0.44524 0.46071 0.4617450.5 0.63608 0.37380 0.55843 0.47901 0.45754†

†0.45754 =(27.5− 32.0)(0.47901)− (27.5− 50.5)(0.46174)

50.5− 32.0

Neville’s Iterated Interpolation Algorithm

Goal: evaluate the interpolating polynomial P on n + 1 distinctnumbers x0, . . . , xn at x to approximate f (x).

INPUT values x , x0, . . . , xn, f (x0), . . . , f (xn)

STEP 1 For i = 0,1, . . . ,n set Qi,0 = f (xi).STEP 2 For i = 1,2, . . . ,n

for j = 1,2, . . . , iset Qi,j =

1xi − xj

[(x − xi−j)Qi,j−1 − (x − xi)Qi−1,j−1

].

STEP 3 OUTPUT Q. STOP.

Example (1 of 2)

Use Neville’s Iterated Interpolation to obtain the approximationsfor Lagrange interpolating polynomials of degrees one, two, andthree for f (0).

xi f (xi)

−0.50 1.93750−0.25 1.332030.25 0.800780.50 0.68750

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203

0.72656

2 0.25 0.80078

1.06641 0.95312

3 0.50 0.68750

0.91406 1.01562 0.98437

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203 0.726562 0.25 0.80078

1.06641 0.95312

3 0.50 0.68750

0.91406 1.01562 0.98437

Q1,1 =(0− x0)Q1,0 − (0− x1)Q0,0

x1 − x0

=(0.50)(1.33203) + (−0.25)(1.93750)

−0.25− (−0.50)= 0.72656

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203 0.726562 0.25 0.80078 1.06641

0.95312

3 0.50 0.68750

0.91406 1.01562 0.98437

Q2,1 =(0− x1)Q2,0 − (0− x2)Q1,0

x2 − x1

=(0.25)(0.80078) + (0.25)(1.33203)

0.25− (−0.25)= 1.06641

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203 0.726562 0.25 0.80078 1.06641

0.95312

3 0.50 0.68750 0.91406

1.01562 0.98437

Q3,1 =(0− x2)Q3,0 − (0− x3)Q2,0

x3 − x2

=(−0.25)(0.68750) + (0.50)(0.80078)

0.50− 0.25= 0.91406

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203 0.726562 0.25 0.80078 1.06641 0.953123 0.50 0.68750 0.91406

1.01562 0.98437

Q2,2 =(0− x0)Q2,1 − (0− x2)Q1,1

x2 − x0

=(0.50)(1.06641) + (0.25)(0.72656)

0.25− (−0.50)= 0.95312

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203 0.726562 0.25 0.80078 1.06641 0.953123 0.50 0.68750 0.91406 1.01562

0.98437

Q3,2 =(0− x1)Q3,1 − (0− x3)Q2,1

x3 − x1

=(0.25)(0.91406) + (0.50)(1.06641)

0.50− (−0.25)= 1.01562

Example (2 of 2)

Qi,j(0) =(0− xi−j)Qi,j−1(0)− (0− xi)Qi−1,j−1(0)

xi − xi−j

i xi Qi,0 Qi,1 Qi,2 Qi,30 −0.50 1.937501 −0.25 1.33203 0.726562 0.25 0.80078 1.06641 0.953123 0.50 0.68750 0.91406 1.01562 0.98437

Q3,3 =(0− x0)Q3,2 − (0− x3)Q2,2

x3 − x0

=(0.50)(1.01562) + (0.50)(0.95312)

0.50− (−0.50)= 0.98437

Homework

I Read Section 3.2.I Exercises: 1a, 3, 5, 7