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The Summation of Series Using
the Anti-Differencing Operation

In the integral calculus the Riemanian integral of a function f(x) over an interval [a,b] is defined as the limit of the


abf(x)dx = lim Σ1nf(xii
 

where the interval [a,b] is partitioned into [x0=a, x1, x2,...,xn=b]
and Δi=xi-xi-1

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


abf(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.

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. Then if F(x) is such that ΔF=f(x) then


Σabf(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 bx, in general it is not easy to do so without some special tools. In the case of bx


Δbx = bx+1−bx = (b-1)bx.
 

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


Δ2x = 2x.
 

The function 2x serves the same role in the Finite Calculus as ex 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


xm = x(x-1)(x-2)··(x-m+1)
 

Note that there are m factors in this product.

Consider now


Δxm = (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
= mxm-1.
 

This is analogous to d(xm)/dx =mxm-1.

Since Δxm = mxm-1 the anti-diffence of xm is xm+1/(m+1).

The definition of xm 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)]
= - mxm+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 ax is (a−1)ax-1 and thus the difference of ax/(a−1) is equal to ax-1. Hence Sn,the sum of ax from 0 to n. is

an-1/(a−1)−a0)/(a−1) = (an-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.)


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