Chapter 3
Interpolation and Polynomial Approximation
3.3 Finite Difference Approximation
Consider a function f(x) over an interval [x0, x1] as shown in Figure 3.3-1. The first degree
Lagrange polynomial approximating f(x) is given by
P1(x) =
( x  x0 )
( x  x1 )
f(x0) +
f(x1)
( x 0  x1 )
( x1  x 0 )
Let h = (x1  x0), then
P1(x) =
1
[(x1  x) f(x0) + (x  x0) f(x1)]
h
f(x1)
E1(x)
f(x)
P1(x)
f(x0)
x0
x1
x
Figure 3.3-1 Approximating by first-degree polynomial with error E1(x).
The function f(x) can be expressed in terms of its approximating polynomial with an error as
f(x) = P1(x) + E1(x)
The slope of the function at x0 is then
dE1
dE1
df
1
dP
(x0) = 1 (x0) +
(x0) = [ f(x1)  f(x0)] +
(x0)
dx
dx
h
dx
dx
Forward difference approximation is obtained when the slope of the interpolating polynomial
estimates the derivative of the function at x0 as shown in Figure 3.3-2. In term of the forward
difference operator 
f ( x0 ) dE1
df
(x0) =
+
(x0)
dx
h
dx
3-11
where f(xi) = f(xi+1)  f(xi).
f(x1)
f(x)
df (x )
1
dx
P1(x)
f(x0)
df (x )
0
dx
x0
x1
x
Figure 3.3-2 Derivatives of the function at x0 and at x1.
The error for the derivative can be estimated by taking derivative of the error
W1 ( x ) d 2 f
E1(x) =
()
2! dx 2
dE1
d 1
d2 f
(x0) =
[ (x  x0)(x  x1) 2 () ] x
0
dx
dx 2
dx
dE1
1 d2 f
d
(x0) =
() [(x  x0)(x  x1) ] x
2
0
dx
2 dx
dx
dE1
1 d2 f
1 d2 f
]
(x0) =
(

)
[(x

x
)
+
(x

x
)
=

h
() = O(h)
1
0
x0
dx
2 dx 2
2 dx 2
Backward difference approximation is obtained when the slope of the interpolating
polynomial estimates the derivative of the function at x1 as shown in Figure 3.3-2.
dE1
dE1
df
1
dP
(x1) = 1 (x1) +
(x1) = [ f(x1)  f(x0)] +
(x1)
dx
dx
h
dx
dx
The error term has the form
dE1
1 d2 f
1 d2 f
]
(x1) =
(

)
[(x

x
)
+
(x

x
)
=
h
() = O(h)
1
0
x1
dx
2 dx 2
2 dx 2
The error in the backward difference approximation, while having the same form as that in
the forward difference approximation, has a different sign. In term of the backward
difference operator 
3-12
f ( x i )
df
(xi) =
+ O(h)
dx
h
where f(xi) = f(xi)  f(xi-1)
Central difference approximation is obtained when the slope of the interpolating polynomial
estimates the derivative of the function at the midpoint x1 as shown in Figure 3.3-3.
f(x1)
f(x)
P2(x)
f(x0)
x0
x1
x
Figure 3.3-3 Derivative of the function at x1 is estimated by a second-degree polynomial.
Let h = x1  x0 = x2  x1, the three points interpolating polynomial over this interval is
P2(x) = L2,0(x) f(x0) + L2,1(x)f(x1) + L2,2(x)f(x2)
P2(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 )
P2(x) =
1
[(x  x1)(x  x2)f(x0)  2(x  x0)(x  x2)f(x1) + (x  x0)(x  x1) f(x2)]
2h 2
The function f(x) can be expressed in terms of its approximating polynomial with an error as
f(x) = P2(x) + E2(x)
The slope of the function at x1 is then
dP
dE 2
dE 2
df
1
(x1) = 2 (x1) +
(x1) =
[ f(x2)  f(x0)] +
(x1)
dx
dx
2h
dx
dx
The error term in this difference approximation is
dE 2
d 1
d3 f
(x1) =
[ (x  x0)(x  x1)(x  x2) 3 () ] x
1
dx
dx 3!
dx
3-13
dE 2
1
d3 f
(x1) =  h2
() = O(h2)
dx
3!
dx 3
The three points approximation is accurate to O(h2). In term of the central difference operator

df
1
h
h
(x1) =
[f(x1 + ) + f(x1  )] + O(h2)
dx
2h
2
2
where the central difference operator  is defined as
f(xi) = f(xi +
f(x1 +
h
h
)  (xi  )
2
2
h
h
) + f(x1  ) = f(x1 + h)  f(x1) + f(x1)  f(x1  h) = f(x2)  f(x0)
2
2
Finite difference approximation of higher order derivatives can also be obtained
f(x) = P2(x) + E2(x)
The second derivative of the function at x1 is then
d 2 P2
d 2 E2
d2 f
(x
)
=
(x
)
+
(x1)
1
1
dx 2
dx 2
dx 2
After some algebra
d 2 E2
1
d2 f
(x1) = 2 [f(x0)  2f(x1) + f(x2)] +
(x1)
h
dx 2
dx 2
The error term can be evaluated to yield
d 2 E2
2 2 d4 f
(x
)
=

h
() = O(h2)
1
4
2
3!
dx
dx
The central difference can also be written in terms of the central difference operator 
 2 f ( x1 )
d2 f
(x
)
=
+ O(h2)
1
dx 2
h2
where
2f(x1) = [f(x1)] = [ f(x1 +
h
h
)  f(x1  )] = f(x2)  f(x1)  [f(x1)  f(x0)]
2
2
3-14