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  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
3-1
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
n
n
 c  (x  
P(x) =
i 1
n
where
 (x  
k 1,#i
k
i
k 1,#i
k
)
) 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 =
P( i )
n
 (
k 1,#i
i
 k )
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)
3-2
f(x2)
f(x1)
f(x)
f(x)
f(x1)
f(x0)
f(x0)
x2
x1
x0
x0 x1
x
Figure 3.1-1 First and second order polynomial approximation.
x
where
ci =
P ( xi )
n
 (x
k  0 ,#i
i
 xk )
or
c0 =
P ( x0 )
f ( x0 )
=
, and
( x 0  x1 )
( x 0  x1 )
c1 =
P ( x1 )
f ( x1 )
=
( x1  x0 )
( x1  x0 )
The approximating polynomial is finally
P(x) =
( x  x0 )
( x  x1 )
f(x0) +
f(x1)
( x 0  x1 )
( x1  x 0 )
The first order polynomial basis function L0(x) is defined as
L0(x) =
1 at x  x0
( x  x1 )
= 
( x 0  x1 ) 0 at x  x1
Similarly, the first order polynomial basis function L1(x) is defined as
L1(x) =
1 at
( x  x0 )
= 
( x1  x 0 ) 0 at
x  x1
x  x0
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
3-3
P(x) =
( x  x0 )( x  x1 )
( x  x0 )( x  x 2 )
( x  x1 )( x  x 2 )
f(x0) +
f(x1) +
f(x2)
( x0  x1 )( x0  x 2 )
( x 2  x0 )( x 2  x1 )
( x1  x0 )( x1  x 2 )
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) =

( x  x1 )( x  x 2 )
= 
( x0  x1 )( x0  x 2 ) 0 at
1 at
x  x0
x  x1 and
x  x2
at
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) =
n
L
n ,k
( x ) f(xk)
k 0
n
where
Ln,k(x) =
( x  xi )
; Ln,k(xi) = 0 and Ln,k(xk) = 1
k  xi )
 (x
i 0,  k
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
f(xk)
1
1
2
2
Solution
3
P(x) =
L
k 0
3, k
( x ) f(xk)
P(x) =
( x  1)( x  2)( x  4)
( x  0)( x  2)( x  4)
(1) +
(1)
(0  1)( 0  2)( 0  4)
(1  0)(1  2)(1  4)
+
( x  0)( x  1)( x  2)
( x  0)( x  1)( x  4)
(2) +
(5)
( 2  0)( 2  1)( 2  4)
( 4  0)( 4  1)( 4  2)
3-4
4
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
0
f(xk)
1
16
2
48
4
88
5
0
Solution
4
Global interpolation by one polynomial: P(x) =
L
4 ,k
( x ) f(xk)
k 0
P(x) =
( x  1)( x  2)( x  4)( x  5)
( x  0)( x  2)( x  4)( x  5)
(0) +
(16)
(0  1)( 0  2)( 0  4)( 0  5)
(1  0)(1  2)(1  4)(1  5)
+
( x  0)( x  1)( x  4)( x  5)
( x  0)( x  1)( x  2)( x  5)
(48) +
(88) + 0
(2  0)( 2  1)( 2  4)( 2  5)
(4  0)( 4  1)( 4  3)( 4  5)
Piecewise polynomial interpolation with quadratic
P(x) =
( x  0)( x  2)
( x  0)( x  1)
( x  1)( x  2)
(0) +
(16) +
(48);
( 2  0)( 2  1)
(0  1)( 0  2)
(1  0)(1  2)
0x2
P(x) =
( x  4)( x  5)
( x  2)( x  5)
( x  2)( x  4)
(48) +
(88) +
(0);
(2  4)( 2  5)
(4  2)( 4  5)
(5  2)(5  4)
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) =
Wn ( x ) d n 1 f
()
(n  1)! dx n 1
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
3-5
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) =
Wn ( x ) d n 1 f
d n 1 f
1
(

)
=
(

h)[(


1)h]
[(


n)h]
()
(n  1)! dx n 1
(n  1)!
dx n 1
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
ln(x + 1)
0
1
0.6
0.47000
0.9
0.64185
Solution
First degree polynomial
P1(x) =
x  0 .6
x0
(0) +
(0.47) = 0.78334x
0  0 .6
0.6  0
P1(0.45) = 0.3525
Error bound: En(x) =
E1(x) = |
d n 1 f
1
(x  x0)(x  x1)  (x  xn)
()
dx n 1
(n  1)!
f " ( )
(x  x0)(x  x1)|
2!
3-6
f(x) = ln(x + 1)  f’(x) =
E1(x) = |
1
1
1
 f”(x) = 
f””(x) =
2
x 1
( x  1) 3
( x  1)
1
1
(0.45  0)(0.45  0.6)| = 3.37510-2
2
(0  1) 2
Actual error = |ln(1 + 0.45)  P1(0.45)| = 1.90610-2
Second degree polynomial
( x  0.6)( x  0.9)
( x  0)( x  0.9)
(0) +
(0.47)
(0  0.6)( 0  0.9)
(0.6  0)( 0.6  0.9)
( x  0)( x  0.6)
+
(0.64185)
(0.9  0.6)( 0.9  0.6)
P2(x) =
P2(0.45) = 0.36829
Error bound: E2(x) = |
E2(x) = |
f " ' ( )
(x  x0)(x  x1)(x  x2)|
3!
1
1
(0.45  0)(0.45  0.6)(0.45  0.9)| = 1.012510-2
2
(0  1) 6
Actual error = |ln(1 + 0.45)  P2(0.45)| = 3.272910-3
3-7