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Chapter 7 Finite Differences and Interpolation

7.3 Antidifferences

We recall from elementary calculus that when we know the first derivative of a function, we can
integrate or antidifferentiate to find the function. By a similar method, we can find the antidifference

–
1
of a factorial polynomial. We denote the antidifference as Δ p x() . It is computed from
n
(
n +
1
x ()
–
1
Δ () n () = -------------------- ) (7.34)
x
( n + 1 )
Example 7.4
Compute the antidifference of the algebraic polynomial
4
3
px() = x – 5x + 3x + 4 (7.35)
Solution:
This is the same algebraic polynomial as that of the previous example, where we found that the
corresponding factorial polynomial is

p x() = 4 – x () 1 () – 8x() 2 () + x () 3 () + x () 4 () (7.36)
n
Then, by (7.34), its antidifference is

3 ()
2 ()
4 ()
5 ()
x ()
x ()
x ()
x ()
1
–
Δ p x() = ------------- + ------------- – 8------------- – ------------- + 4x() 1 () + C (7.37)
n
3
4
2
5
where C is an arbitrary constant.
Antidifferences are very useful in finding sums of series. Before we present an example, we need
to review the definite sum and the fundamental theorem of sum calculus. These are discussed
below.
In analogy with definite integrals for continuous functions, in finite differences we have the defi-
)
nite sum of p x() which for the interval a ≤ x ≤ a + ( n 1 h is denoted as
–
n
)
α + ( n – 1 h
∑ p x() = p α() + p α + h + p α + 2h + … + p α + ( n – 1 h ] (7.38)
(
)
[
)
(
)
n
n
n
n
n
x = α
Also, in analogy with the fundamental theorem of integral calculus which states that
7−12 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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