where the interval [a,b] is partitioned into [x₀=a, x₁, x₂,...,x_n=b]
and Δx_i=x_i-x_i-1

The limiting process is over n and partitions of the interval such that the maximum increment goes to zero; i.e., max (Δx_i) → 0.

It would be extremely cumbersome and tedious to evaluate integrals by this definition. Fortunately there is an easier way.

If F(x) is a function such that its derivative is equal to f(x);

D(F(x)) = dF/dx = f(x),
 

then (The Fundamental Theorem of Calculus)

∫_a^bf(x)dx = F(b) - F(a)
 

The function F(x) is called the anti-derivative of f(x).

Ronald L. Graham, Donald E. Knuth and Oren Patashnik present an interest extension of this system for the summation of series. They call their system Finite Calculus. It is given in their book Concrete Mathematics (Addison-Wesley 1989).

Analogies exist in Finite Calculus for most concepts, operations and relations in the Differential Calculus. (Differential Calculus is a misnomer; the field should be called Derivative Calculus.)

In Finite Calculus the differencing operation

Δf = f(x+1)−f(x)
 

takes the place of the derivative. Note that it is a forward difference Then if F(x) is such that ΔF=f(x) then

Σ_a^bf(x)dx = F(b) - F(a)
 

F(x) is called the anti-difference of f(x).

While it may be easy to find the anti-difference of some functions, such as b^x, in general it is not easy to do so without some special tools. In the case of b^x

Δb^x = b^x+1−b^x = (b-1)b^x.
 

Therefore the anti-difference of b^x is b^x/(b-1). This means that for b=2

Δ2^x = 2^x.
 

The function 2^x serves the same role in the Finite Calculus as e^x serves in the Differential Calculus.

The power functions xⁿ do not have obvious anti-differences, but there is another type of function that provides a convenient basis for computing the anti-differences. This function is called the falling factorial power function. It is defined as

x^m = x(x-1)(x-2)··(x-m+1)
 

Note that there are m factors in this product.

Consider now

Δx^m = (x+1)x(x-1)(x-2)··(x+1-m+1)−x(x-1)(x-2)··(x-m+1)
x(x-1)(x-2)··(x-m+2)[(x+1)−(x-m+1)]
= x(x-1)(x-2)··(x-m+2)m
= mx^m-1.
 

This is analogous to d(x^m)/dx =mx^m-1.

Since Δx^m = mx^m-1 the anti-diffence of x^m is x^m+1/(m+1).

The definition of x^m may be extended beyond the positive integers. The appropriate definition of negative integers is

x^−m = 1/(x+1)(x+2)··(x+m)
 

This leaves x⁰ and x⁻⁰ to be defined. They are equal to each other and equal to 1. To justify this definition note that

x¹ = xx⁰
 

but x¹=x so x⁰ must be equal to 1. Likewise,

x⁻¹ = x⁻⁰/(x+1).
 

But x⁻¹=1/(x+1) so x⁻⁰=1.

For negative integers then

Δ(x^−m) = 1/[(x+2)··(x+m+1) − 1/(x+1)··(x+m)]
= 1/[(x+2)··(x+m)][1/(x+m+1) − 1/(x+1)]
= [1/(x+2)··(x+m)][(x+1)-(x+m+1)]/[(x+1)(x+m+1)]
= - m[1/(x+1)(x+2)··(x+m+1)]
= - mx^m+1
 

Therefore the anti-difference of x^−m is x^−(m-1)/(m-1) for m≠1.

For the special case of m equal to 1 consider the harmonic function defined for positive integers as:

H(x) = 1 + 1/2 + 1/3 + ·· 1/x.

The difference of ΔH is

H(x+1)−H(x) = 1/(x+1).
 

Therefore the anti-difference of x⁻¹ is H(x). The harmonic function is the analogue in the Finite Calculus of the logarithm function in the Differential Calculus.

Application of Anti-Differencing
to Series Summation

One of the simplest series is the geometric series, 1+a+a²+a³+…+aⁿ. The difference of a^x is (a−1)a^x-1 and thus the difference of a^x/(a−1) is equal to a^x-1. Hence S_n,the sum of a^x from 0 to n. is

a^n-1/(a−1)−a⁰)/(a−1) = (a^n-1−1)/(a−1)

In order to sum powers, such as squares and cubes, the powers must be represented in terms of the falling factorial power functions. For example,

x=x¹
This means that
the antidifference of x is
(1/2)x² = (1/2)x(x-1)

Therefore the sum of the integers from 1 to n is ½n(n-1).

Since x²=x²−x this means that

x² = x² + x¹
and thus the antidifference of x² is
(1/3)x³ + (1/2)x²

Therefore the sum of the first n squared integers (0, 1, 4, 9, 16, …)

(1/3)n(n-1)(n-2)+(1/2)n(n-1)
which can be represented as
n(n-1)(n/3-2/3+1/2)
which reduces to
n(n-1)(2n-1)/6

Thus when n=1, the sum is 0. When n=2 the sum is 1 and when n=3 the sum is 5, and so on.

Now consider the sum of the first n intergers having a remainder of j when divided by k. These numbers have the form kx+j for x=0 to (n-1). The antidifference of kx+j is k[½x(x-1)]+jx and thus the sum is

k[½n(n-1)]+jn

For example, the first 5 integers having a remainder of 0 when divided by 4 are (0, 4, 8, 12, 16). The consecutive sums are (0, 4, 12, 24, 40). For n=1 the formula gives 0. For n=2 the formula gives 4(1)=4. For n=3, the formula gives 4(3)=12. For n=4 the formula gives 4(6)=24 and for n=5 it is 4(10)=40. The success of the formula indicates that the numbering should start with 0.

For a remainder of 1 when divided by 4 the numbers are (1, 5, 9, 13, 17) and the successive sums are (1, 6, 15, 28, 45). For n=5 the formula gives 4(10)+5=45.

(To be continued.)