Whence by integrating the given quantity, and the last result, con tinually n times, the nth order of integration will be indicated by

placing Abefore the given difference.

Problem.

To find the integral of a given difference, or several orders of integrals from a given difference.

Observe to which of the forms given in Problems 3, 4, 5, 6, 7, 8, the given difference belongs; then it is evident that in whatever manner the difference has been formed from its original function, or integral, we must reverse that operation, in order to find the integral from the given difference.

Case 1. Whence if the given difference be a factorial, we must proceed by the following

Rule.

Diminish the first factor, and increase the exponent, each by unity divide the result by the product of the exponent thus increased, and the common difference of the factors, and the quotient will indicate the factorial integral required.

If the given difference be a factorial reciprocal, proceed according to the following

Rule.

Change the reciprocal into a factorial with a negative exponent: ncrease the exponent by unity, and divide the result by the product of the exponent thus increased, and the common difference of the factors, and the quotient, is the integral required.

Case 3.

If the function or given difference be an exponential r" proceed according to the following

Rule.

Divide the given exponential by (r-1) whether r be a whole number or a fraction.

To perform indicative operations in the direct, and inverse methods, whether to indicate any order of differences of an indicative function or to indicate any order of integrals of an indicative difference.

n

Let A placed before a quantity, represent the nth order of differences to be taken of that quantity, and let A, when placed beforc a quantity, represent the inverse operation, of taking the integral of that quantity m times: then to find the mth integral or rth increment of any indicative quantity, the following

Rule.

Add the indices or exponents of the two operations together, then place the sum above the sign ▲, and the delta with its superior before the quantity; then if the resulting index is affirmative, the superior will indicate as many orders of differences as it contains units; but if the resulting index be negative, the superior will indicate as many orders of integrals as it contains units.

The difference of a product will be found by indicating, as directed by problem, page 40. See the following examples.

The integral of the product u v of two variable quantities will be

For by taking the differences of the respective terms, we shal have the following values, viz.

Whence, by adding the first sides of these equations, and the second sides together, the equation of their sums will be

To find the integral of a functional product consisting of two

variable factors.

Rule

Take the successive orders or differences of the one variable factor and the respective orders of the partial integrals of the consecutive succeeding values of the other variable factor :

Then multiply the first variable factor into the first integral of the other for the first term of the required integral:

Again multiply the first difference of the first factor, into the second integral of the first succeeding value of the other, for the second term of the required integral :

Again, multiply the second difference of the first factor into the third integral of the second succeeding value of the other, and so on; observing to give each product its proper sign.

Then change the signs of all the even terms, and the result will be the integral required to be formed.

According to the series u'Av— Au. ▲ v1+Au. Av2 demonstrated in theorem

&c. already

N. B. If one of the variable factors of the given functional product be a factorial, that factor ought to be made the first, viz. that from which the differences are taken, as the orders of differences cannot exceed the number of factors.

T

Au=

2411

&c. &c.

Where R=1—r;

12

2

&c.

Whence — ( u. Av —▲ u. Av‚ + Â u.Ãv ̧—&c.)=

T

(Rrair then will (–

Before we proceed to show the application of the inverse method of differences: we may observe, that as the differences of any variable quantities z, z+a, &c are each indicated by az, the constant part having no difference; so reciprocally the integral of ax, may be %, %, +a, z+b,&c. Therefore, when the integral of any difference is found it may want a constant part, whether affirmative or negative, in order to make it correct: this correction will be best known from the nature of the problem. In finding the sums of series if we make x=oin the integral, then if this particular value of the integral be zero, it wants no correction; but if this value be plus or minus a constant quantity, we must subtract or add that quantity to the integral function.

Application of the inverse method of Differences to the summation of Series.

Definitions.

1. The general sum of a finite series, consisting of any number of terms, r is such a function of x, that when the number indicating the number of terms is substituted for x, the value of the function will then be the sum of as many terms of the series as is expressed by this particular value of x.

Thus, suppose x=3, then the numeral value of the function will be the sum of three terms.

2. A number is said to be the limit of the sum of a series, when that number exceeds the sum of any number of the terms of the series, and when the difference between the said limits, and the said sum, is less than any assignable number, however small.