CHAPTER 3 INTERPOLATION AND POLYNOMIAL APPROXIMATION 31 INTERPOLATION

CHAPTER 11 OECD AVERAGE AND OECD TOTAL BOX
 CONTENTS PREFACE IX INTRODUCTION 1 REFERENCES 5 CHAPTER
 NRC INSPECTION MANUAL NMSSDWM MANUAL CHAPTER 2401 NEAR‑SURFACE

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32 STAKEHOLDER ANALYSIS IN THIS CHAPTER A STAKEHOLDER ANALYSIS
CHAPTER 13 MULTILEVEL ANALYSES BOX 132 STANDARDISATION OF
CHAPTER 6 COMPUTATION OF STANDARD ERRORS BOX 61

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Interpolation and Polynomial Approximation

Chapter 3

Interpolation and Polynomial Approximation

3.1 Interpolation and the Lagrange Polynomial

3.1.1 Lagrangian Form

Consider a polynomial of degree (n  1):

P(x) = a₁x ^n-1 + a₂x ^n-2 +... + a_n-1x + a_n

where the a_i are constants. The polynomial can be written in Lagrangian form:

P(x) = c₁(x  ₂) (x  ₃)... (x  _n) + c₂(x  ₁) (x  ₃)... (x  _n) + ...

c_i(x  ₁) (x  ₂) ... (x  _i-1) (x  _i+1) ... (x  _n) + ...

c_n(x  ₁) (x  ₂)... (x  _n-1)

where _i, i = 1, 2, ..., n are arbitrary scalars, while the constants c_i are related to the constants a_i.

Example 3.1-1 _____________________________________________________

Write the polynomial P(x) = x ²  4x + 3 in the Lagrangian form.

Solution

The Lagrangian form for P(x) = x ²  4x + 3 is

P(x) = c₁(x  ₂) (x  ₃) + c₂(x  ₁) (x  ₃) + c₃(x  ₁) (x  ₂)

where _i, i = 1, 2, 3 are arbitrary scalars. Let ₁ = 1, ₂ = 2, ₃ = 3, then

P(x) = c₁(x  2) (x  3) + c₂(x  1) (x  3) + c₃(x  1) (x  2)

The constants c₁ can be evaluated from the above relation by substituting x = ₁ = 1

P(x = 1) = 1  4 + 3 = c₁(1  2) (1  3)  c₁ = 0

For x = ₂ = 2

P(x = 2) = 4  8 + 3 = c₂(2  1) (2  3)  c₂ = 1

For x = ₃ = 3

P(x = 3) = 9  12 + 3 = c₃(3  1) (3  2)  c₃ = 0

The Lagrangian form for the polynomial is

P(x) = (x  1)(x  3)

Let ₁ =  2, ₂ =  1, ₃ = 2, then

P(x) = c₁(x + 1) (x  2) + c₂(x + 2) (x  2) + c₃(x + 2) (x + 1)

The constants c_i can be evaluated to obtain: c₁ = 3.7500, c₂ = -2.6667, and c₃ = -0.0833. The Lagrangian form for the polynomial is

P(x) = 3.7500 (x + 1) (x  2)  2.6667 (x + 2) (x  2)  0.0833 (x + 2) (x + 1)

A short form notation for P(x) is

P(x) =

where denotes product of all terms (x  _k), for k varying from 1 to n except i. Let x = _i then

P(_i) = c_i(_i  ₁) (_i  ₂) ... (_i  _i-1) (_i  _i+1) ... (_i  _n)

The constant c_i can be expressed as

c_i =

3.1.2 Polynomial Approximation

Consider a function f(x) that passes through the two distinct points (x₀, f(x₀)) and (x₁, f(x₁)) as shown in Figure 3.1-1. The first order polynomial that approximates the function between these two points can be expressed as

P(x) = a + bx

where a and b are constants. P(x) can also be written in Lagrangian form as

P(x) = c₀(x  x₁) + c₁(x  x₀)

Figure 3.1-1 First and second order polynomial approximation.

where

c_i =

or

c₀ = = , and c₁ = =

The approximating polynomial is finally

P(x) = f(x₀) + f(x₁)

The first order polynomial basis function L₀(x) is defined as

L₀(x) = =

Similarly, the first order polynomial basis function L₁(x) is defined as

L₁(x) = =

In terms of the basis function, P(x) can be written as

P(x) = L₀(x) f(x₀) + L₁(x) f(x₁)

If a second order polynomial is used to approximate the function using three points (x₀, f(x₀)), (x₁, f(x₁)), and (x₂, f(x₂)) then

P(x) = f(x₀) + f(x₁) + f(x₂)

P(x) can also be written in terms of the second order polynomial basis function L_2,k(x)

P(x) = L_2,0(x) f(x₀) + L_2,1(x)f(x₁) + L_2,2(x)f(x₂)

where L_2,0(x) = =

In general: L_2,k(x_k) = 1 at node k, L_2,k(x_i) = 0 at other nodes.

We now seek a polynomial P(x) of degree n that interpolates a given function f(x) between the node x_i of the grid for which there are n+1 nodes x₀, x₁, , x_n and

P(x_k) = f(x_k) for each k = 1, 2, , n

The polynomial is given by

P(x) = L_n,0(x) f(x₀) + L_n,1(x) f(x₁) +  + L_n,n(x)f(x_n) = f(x_k)

where L_n,k(x) = ; L_n,k(x_i) = 0 and L_n,k(x_k) = 1

Polynomial approximation constitutes the foundation upon which we shall build the various numerical methods. The approximation P(x) to f(x) is known as a Lagrange interpolation polynomial, and the function L_n,k(x) is called a Lagrange basis polynomial.

Example 3.1-2 _____________________________________________________

Find the Lagrange interpolation polynomial that takes the values prescribed below

xk	0	1	2	4
f(xk)	1	1	2	5

Solution

P(x) = f(x_k)

P(x) = (1) + (1)

+ (2) + (5)

When working with grids having large numbers of intervals one typically assigns a set of low degree (n = 1, 2, or 3) basis functions to each adjacent set of n+1 = 2, 3, or 4 nodes.

Example 3.1-3 _____________________________________________________

Use global interpolation by one polynomial and piecewise polynomial interpolation with quadratic for the following nodes.

xk	0	1	2	4	5
f(xk)	0	16	48	88	0

Solution

Global interpolation by one polynomial: P(x) = f(x_k)

P(x) = (0) + (16)

+ (48) + (88) + 0

Piecewise polynomial interpolation with quadratic

P(x) = (0) + (16) + (48); 0  x  2

P(x) = (48) + (88) + (0); 2  x  5

The error E_n(x) associated with the interpolation of f(x) by P_n(x) over the interval [x₀, x_n] can be estimated as

E_n(x) = f(x)  P_n(x) = ()

where  is some number lying in the open interval (x₀, x_n) and

W_n(x) = (x  x₀)(x  x₁)  (x  x_n)

When the spacial increments are uniform

x_k+1  x_k = h, k = 0, 1, 2, , n-1

Let x = x₀ + h, since

x₁ = x₀ + h  x  x₁ = (  1)h

x_n = x₀ + nh  x  x_n = (  n)h

W_n(x) = (x  x₀)(x  x₁)  (x  x_n) = (h)[(  1)h] [(  n)h]

The error associated with interpolation is then

E_n(x) = () = (h)[(  1)h] [(  n)h] ()

The only variable in the above expression is h the spacing of the nodes, therefore

E_n(x) = Chⁿ⁺¹, x₀ <  < x_n

where C is a coefficient independent of h.

We can therefore write E_n(x) = O(hⁿ⁺¹) meaning that the ratio E_n(x)/ hⁿ⁺¹ is bounded by a constant as h  0. As the increment h decreases, so also will the interpolation error E_n.

Example 3.1-4 _____________________________________________________

For the function f(x) = ln(x + 1), construct interpolation polynomials of degree one and two to approximate f(0.45) from the given nodes. Find the error bound and the actual error.

xk	0	0.6	0.9
ln(x + 1)	1	0.47000	0.64185

Solution

First degree polynomial

P₁(x) = (0) + (0.47) = 0.78334x

P₁(0.45) = 0.3525

Error bound: E_n(x) = (x  x₀)(x  x₁)  (x  x_n) ()

E₁(x) = | (x  x₀)(x  x₁)|

f(x) = ln(x + 1)  f’(x) =  f”(x) =  f””(x) =

E₁(x) = | (0.45  0)(0.45  0.6)| = 3.37510^-2

Actual error = |ln(1 + 0.45)  P₁(0.45)| = 1.90610^-2

Second degree polynomial

P₂(x) = (0) + (0.47)

+ (0.64185)

P₂(0.45) = 0.36829

Error bound: E₂(x) = | (x  x₀)(x  x₁)(x  x₂)|

E₂(x) = | (0.45  0)(0.45  0.6)(0.45  0.9)| = 1.012510^-2

Actual error = |ln(1 + 0.45)  P₂(0.45)| = 3.272910^-3

CONFIGURING USER STATE MANAGEMENT FEATURES 73 CHAPTER 7 IMPLEMENTING
INTERPOLATION 41 CHAPTER 5 INTERPOLATION THIS CHAPTER SUMMARIZES POLYNOMIAL
PREPARING FOR PRODUCTION DEPLOYMENT 219 CHAPTER 4 DESIGNING A

Tags: interpolation and, construct interpolation, interpolation, approximation, polynomial, chapter

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