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4. Find the fourth integral of Au.

By rule Au=AU.

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

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Theorem. The integral of the product u v of two variable quantities will be 4. AV-A 4.AVI + A 2. AU2

- Å u.

AU.Avg+&c.

2

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

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2

+ACA 4. Avi)=

+ A u. AV 2+ &c. -&c.

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

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A(U. Av-Au.Av, + AU. A V,— &c.)=wv;

therefore u.Av— A4. Av, + Au. Avg. -&c.=A(uv.) Whence the proposition is truc.

Problem. 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. A vs+au. Ava &c. already demonstrated in theorem

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.

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Where R=1-;

2

2.3

Whence

-(z.

u. Dv-S 2.Av,+2. v,&c.)= - + Rörsla+Rozur +&c.) Put

1then will ä(t)=(Psali

-Poi -&c.)

+ &c. Put I-r=-P

Scholium. 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+a, &c are each indicated by az, the constant part having no difference; so reciprocally the integral of Az, may be %, %, +a, 3+1,&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 I=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 r=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 sun, is less than any assignable number, however small.

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then unity is the limit of the sum, since the sum of any number of its terms is less than unity ; and since no number can be added to the sum of a certain number of its terms, but what must inake the sum more than a unit.

3. The sum of an infinite series is the limit of the sum of that series.

4. A series is said to be summable when a finele general expression can be found, that will be the value of a given number of its terms, or when a limit to the sum can be found when the number of the terms is infinite; otherwise the series is not summable.

Theorem. The integral of the (x+1)th term of a series, will be equal to the sum of all the preceding x terms of that series. For let there be the series a,

b,

d, &c. Then the successive sums are a, a+b, a+b+c, a+b+c+d, &c.

Now since any term of the upper series is the difference of the corresponding term of the lower series, and the preceding term of the said lower series ; therefore, calling the upper series the series of differences, and the lower series the series of sums, any term, as x+1, in the series of differences, will be the difference between the ath term and the (x+1)th term in the series of sums; but what is called the difference of a function of x, is another function of 1+1, which is the difference between the first function of x, and the same function as the first of 71: therefore, reciprocally, since a certain function of x, is the integral of a difference which is a function of x+1, the integral of the (x + 1)th term in the series of differences will be the xth term in the series of sums; that is, the sum of a terms of the series of differences. Q.E.D.

Problem. To find the sum of a given number of terms of a series, the law of the series being given,

Let x represent the number which indicates the number of the terms. Find the general term of the series, and the integral of the (x+1)th term by such of the former rules as applies, then the integral corrected, if need be, is the sum of the series.

N. B. If the general term of the series is not expressed : factorials, it will be most convenient to reduce it to factorials, ard this may be done by Problem 1, when the genera! *erm is formed of binomial factors.

Notation.

Let rrepresent the xth term, y the (x+1)th term of the series ;

also let A denote the integral of y, and the integral corrected, or the sum required

Examples. 1. Find the sum of 100 ternis of the series, 1+2+3+&c

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2'(x+1) 100.101 Whence Σ =

-=5050. 2

2 2. Find the sum of 50 terms of the series 1.2.3+2.3.4+3.4.5+ 4 5.6+ &c.

Here t=2311,

y=(x+1),

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3. Find the sum of 20 terms of the series 1° +2° +39 +46 + &c.

Here t= x = 3,211 - 2117 by decomposition,

y=x+1)21 - (+1)",

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4. Find the sum of x terms of a series, whose general term is represented by ar' +6x®+cr+d. By decomposition ax:= 40.32311 – 3.4211 + x3lr),

bro=\(......x211),

and cx =cl.......... l), ..t = ax311+(6-3a).x211 + (a-b+c)x1+d,

y=alr+1)311+(1—32)(x+1)2l1 +(a-b+c)(x+1)+1+d(x+1), Ay=as it. -3a)

+ dix 4.1

2.1 whence,

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More Examples for Iractice. 5. Find the sum of 15 terms of the series 59+74+99+&c. 6. Find the sum of x terms of the series of products 1.23 +39.4 +59.69+7.89 +&c.

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