Interpolation and Polynomial Approximation
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 ( (2) (x ( (3)... (x ( (n) + c2(x ( (1) (x ( (3)... (x ( (n) + ...
ci(x ( (1) (x ( (2) ... (x ( (i-1) (x ( (i+1) ... (x ( (n) + ...
cn(x ( (1) (x ( (2)... (x ( (n-1)
where (i, 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 ( (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 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) = [pic][pic]
where [pic] denotes product of all terms (x ( (k), for k varying from 1 to n except i. Let x = (i then
P((i) = ci((i ( (1) ((i ( (2) ... ((i ( (i-1) ((i ( (i+1) ... ((i ( (n)
The constant ci can be expressed as
ci = [pic]
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)
[pic]
Figure 3.1-1 First and second order polynomial approximation.
where
ci = [pic]
or
c0 = [pic] = [pic], and c1 = [pic] = [pic]
The approximating polynomial is finally
P(x) = [pic] f(x0) + [pic] f(x1)
The first order polynomial basis function L0(x) is defined as
L0(x) = [pic] = [pic]
Similarly, the first order polynomial basis function L1(x) is defined as
L1(x) = [pic] = [pic]
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) = [pic] f(x0) + [pic] f(x1) + [pic] 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) = [pic] = [pic]
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, (, xn and
P(xk) = f(xk) for each k = 1, 2, (, n
The polynomial is given by
P(x) = Ln,0(x) f(x0) + Ln,1(x) f(x1) + ( + Ln,n(x)f(xn) = [pic] f(xk)
where Ln,k(x) = [pic]; 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 |4 |
|f(xk) |1 |1 |2 |5 |
Solution
P(x) = [pic] f(xk)
P(x) = [pic](1) + [pic](1)
+ [pic](2) + [pic](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) = [pic] f(xk)
P(x) = [pic](0) + [pic](16)
+ [pic](48) + [pic](88) + 0
Piecewise polynomial interpolation with quadratic
P(x) = [pic](0) + [pic](16) + [pic](48); 0 ( x ( 2
P(x) = [pic](48) + [pic] (88) + [pic](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) = [pic][pic](()
where ( is some number lying in the open interval (x0, xn) and
Wn(x) = (x ( x0)(x ( x1) ( (x ( xn)
When the spacial increments are uniform
xk+1 ( xk = h, k = 0, 1, 2, (, n-1
Let x = x0 + (h, since
x1 = x0 + h ( x ( x1 = (( ( 1)h
xn = x0 + nh ( x ( xn = (( ( n)h
Wn(x) = (x ( x0)(x ( x1) ( (x ( xn) = ((h)[(( ( 1)h] ([(( ( n)h]
The error associated with interpolation is then
En(x) = [pic][pic](() = [pic]((h)[(( ( 1)h] ([(( ( n)h] [pic](()
The only variable in the above expression is h the spacing of the nodes, therefore
En(x) = Chn+1, x0 < ( < 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 |0.9 |
|ln(x + 1) |1 |0.47000 |0.64185 |
Solution
First degree polynomial
P1(x) = [pic](0) + [pic](0.47) = 0.78334x
P1(0.45) = 0.3525
Error bound: En(x) = [pic](x ( x0)(x ( x1) ( (x ( xn) [pic](()
E1(x) = |[pic](x ( x0)(x ( x1)|
f(x) = ln(x + 1) ( f’(x) = [pic] ( f”(x) = ([pic] f””(x) = [pic]
E1(x) = |[pic][pic](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) = [pic](0) + [pic](0.47)
+ [pic](0.64185)
P2(0.45) = 0.36829
Error bound: E2(x) = |[pic](x ( x0)(x ( x1)(x ( x2)|
E2(x) = |[pic][pic](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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