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 +... + an-1x + an
where the ai are constants. The polynomial can be written in Lagrangian form:
P(x) = c1(x - a2) (x - a3)... (x - an) + c2(x - a1) (x - a3)... (x - an) + ...
ci(x - a1) (x - a2) ... (x - ai-1) (x - ai+1) ... (x - an) + ...
cn(x - a1) (x - a2)... (x - an-1)
where ai, i = 1, 2, ..., n are arbitrary scalars, while the constants ci are related to the constants ai.
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 - a2) (x - a3) + c2(x - a1) (x - a3) + c3(x - a1) (x - a2)
where ai, i = 1, 2, 3 are arbitrary scalars. Let a1 = 1, a2 = 2, a3 = 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 = a1 = 1
P(x = 1) = 1 - 4 + 3 = c1(1 - 2) (1 - 3) Þ c1 = 0
For x = a2 = 2
P(x = 2) = 4 - 8 + 3 = c2(2 - 1) (2 - 3) Þ c2 = 1
For x = a3 = 3
P(x = 3) = 9 - 12 + 3 = c3(3 - 1) (3 - 2) Þ c3 = 0
The Lagrangian form for the polynomial is
P(x) = (x - 1)(x - 3)
Let a1 = - 2, a2 = - 1, a3 = 2, then
P(x) = c1(x + 1) (x - 2) + c2(x + 2) (x - 2) + c3(x + 2) (x + 1)
The constants ci 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 - ak), for k varying from 1 to n except i. Let x = ai then
P(ai) = ci(ai - a1) (ai - a2) ... (ai - ai-1) (ai - ai+1) ... (ai - an)
The constant ci can be expressed as
ci =
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
ci =
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 L2,k(x)
P(x) = L2,0(x) f(x0) + L2,1(x)f(x1) + L2,2(x)f(x2)
where L2,0(x) = =
In general: L2,k(xk) = 1 at node k, L2,k(xi) = 0 at other nodes.
We now seek a polynomial P(x) of degree n that interpolates a given function f(x) between the node xi of the grid for which there are n+1 nodes x0, x1, ¼, xn and
P(xk) = f(xk) for each k = 1, 2, ¼, n
The polynomial is given by
P(x) = Ln,0(x) f(x0) + Ln,1(x) f(x1) + ¼ + Ln,n(x)f(xn) = f(xk)
where Ln,k(x) = ; Ln,k(xi) = 0 and Ln,k(xk) = 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 Ln,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 / 4f(xk) / 1 / 1 / 2 / 5
Solution
P(x) = f(xk)
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 / 5f(xk) / 0 / 16 / 48 / 88 / 0
Solution
Global interpolation by one polynomial: P(x) = f(xk)
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 En(x) associated with the interpolation of f(x) by Pn(x) over the interval [x0, xn] can be estimated as
En(x) = f(x) - Pn(x) = (z)
where z is some number lying in the open interval (x0, xn) and
Wn(x) = (x - x0)(x - x1) ¼ (x - xn)
When the spacial increments are uniform
xk+1 - xk = h, k = 0, 1, 2, ¼, n-1
Let x = x0 + ah, since
x1 = x0 + h Þ x - x1 = (a - 1)h
xn = x0 + nh Þ x - xn = (a - n)h
Wn(x) = (x - x0)(x - x1) ¼ (x - xn) = (ah)[(a - 1)h] ¼[(a - n)h]
The error associated with interpolation is then
En(x) = (z) = (ah)[(a - 1)h] ¼[(a - n)h] (z)
The only variable in the above expression is h the spacing of the nodes, therefore
En(x) = Chn+1, x0 < z < xn
where C is a coefficient independent of h.
We can therefore write En(x) = O(hn+1) meaning that the ratio En(x)/ hn+1 is bounded by a constant as h ® 0. As the increment h decreases, so also will the interpolation error En.
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 / 0.9ln(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: En(x) = (x - x0)(x - x1) ¼ (x - xn) (z)
E1(x) = |(x - x0)(x - x1)|
f(x) = ln(x + 1) Þ f’(x) = Þ 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
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