Sidebilder
PDF

Suppose that the quotient of S by S, is not exactly pi+913; if the series is recurring, it will be of an order superior to the second. Let us examine if

a' to'rtc'r? we can have S=;

Tatbr+cx + d.x3° We derive from this equation

a" + "'r

T=P+q? ta' +b'r+cʻz24; that is to say, after having obtained the first two terms of the quotient of 1, divided by the series Si, we shall find for a remainder a series, all of whose terms will contain x; and if we designate this remainder by Six?, we shall have

S. a"'+b'r

g=a'+b'x+c'22" This equality gives

Sa''

g=P.+917+" "ite ; hence, designating by Szt? the series which we find for a remainder after having carried the division of the series S by the series S, to the terms of the quotient pit9., we should have

S, a'"

S-a" 16" ** From this last equality we derive

[ocr errors]
[ocr errors]
[ocr errors]

Here the operations stop; for, returning to the generating fraction, we shall have the equations Iuni ,S , S

S, S,
g=p+
6X, =p+918+zra, a=p. +

itsit, =P:+92.7 ; and from these equations we derive

1 S 1

S, S. patex p+qx+5r Pat913+*** We have, then, only a few substitutions to make in order to obtain a fraction equal to S.

Without proceeding further, the reader will doubtless perceive that the successive operations for finding the quotients p+97, p.+91, &c., and for returning to the generating fraction, bear a striking analogy to those which are necessary in reducing an ordinary fraction to a contirued fraction, and in returning to the ordinary fraction. This observation will take the place of a general rule. If we arrive at a division which gives an exact quotient of the form ptqx, we know that the series is recurring. (See Contin. Fractions.)

EXAMPLE. Suppose we wish to determine whether the series of numbers 1, 2, 3, &c., be recurring. It is not this numerical series which we must consider, but the equation

S=1+2rt3.r? +42°+ ...
We perceive that the operations will be performed as follows :

[ocr errors][merged small][ocr errors][ocr errors][merged small][merged small][ocr errors]

= (1–2):

REMARK.-In finding a rule to determine whether a series is recurring, we have considered the series as derived from a fraction whose numerator is of a degree inferior to the denominator. But even if this last condition does not have place, it is easy to perceive that the same explications, and, consequently, the same rule, will always subsist.

329. PROBLEM.To find the general term of a recurring series. Suppose that the series has for a generating fraction

a'+b'rt... thrm-1

Sa+bur +...+kx". We can write this fraction thus :

F=(a'+b'z... thorn-1)(a+bx+...+krm)-1. It is evident, then, that by developing the power – 1, obtaining the product of the two factors in this equation, and taking in this product the part wbich contains x to any power whatsoever, we shall have the general term of the recurring series. But the problem is resolved ordinarily by another process, which I proceed to exhibit.

We divide first all the terms of the fraction F by k, and write it under the form

U a'rn-1 + B'rm-+...

V=" +ar"--+BX +... The fraction is supposed in all cases to be reduced to its most simple form, so that U has no common factor with V.

We decompose, then, the denominator into binomial factors, such as rta, whether it be by equating this denominator to zero, or by some other method, and then the fraction is regarded as resulting from the addition of many others, which have for denominators these different factors. We determine all these partial fractions, and then form the general term of the development of each; then, taking the sum of these general terms, we shall have the general term of the recurring series.

In this decomposition into partial fractions it is necessary carefully to dis

tinguish in V the simple factors from those which are raised to powers for each simple factor xta we shall take a fraction of the form

M

rta For each factor, such as (.x + b)", we might take one of the form

Arh-' + Birno +...

(x+b)" but it is more convenient to have only fractions with monomial numerators ; instead, therefore, of a fraction like the preceding, we take n, like the following:

N
N,

N,
(x+ba+(+bu-ite+6)--2*** tx+6'
M, N, N,... representing quantities independent of x.
Consequently, if we suppose that V=(r+a)(r+b)...., we can place
U M N

N

Now v=x+a+(+ba+ (x+6)-1 •••+x+6+.... and the question will be reduced, for the present, to the determination of the numerators M, N, N,, &c. But these have been determined in Art. 209, (3).

The preceding decomposition being effected, the determination of the general term of the recurring series does not offer any difficulty.

Each partial fraction can be put under the form P(p+r)-, designating by à an entire positive number, which can be equal to 1. If we develop this power, we readily find that the term affected with 2* is -21-2-1)(-2-2)...(-2-n+1),

Pp-1-1x1. 1.2.3...n It is the sum of like expressions, all containing 2", and resulting from the different partial fractions which compose the general term required.

When the denominator of the generating fraction contains imaginary factors, these factors introduce imaginary quantities into the general term. If we suppose, however, that the coefficients of the numerator and denominator of the proposed fraction are all real (and they are always taken so), it is evident, a priori, that, as we find the development of this fraction by division, the general term can not embrace any imaginary factors ; consequently, we are sure that all the imaginary quantities which arise from the factors of the denominator will disappear.

SUMMATION OF SERIES. The summation of series is the finding of a finite expression equal to the proposed series, even when the series is infinite, and in many cases this finite expression is found by subtraction.

EXAMPLES.

1 1 1 (1) Required the sum of the series Fotostat.... to infinity.

1 1 1 1 1 1 Let S=;totita tätät ............ ad infinitum.

1 1 1 1 1 1 .. S-1= totitätätit ............ ad infinitum.

Hence, by subtracting the latter from the former, we have the required sum:

1 1 1 1 1
1.2+23+3.4+4.5+5.6+ ......=l.

1 1 1 (2) Required the sum of the series toitzat..... to n terms.

[ocr errors]
[ocr errors]

When n is infinitely great, then we have

1 1 1 1 13+at +hot ... ad infinitun

1 1 1 1 (3) Sum the series 3- 4+3.5-4.6+ ....

[ocr errors]
[ocr errors][merged small][merged small]
[merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][merged small]

DIFFERENCE SERIES. 330. Let there be the arithmetical progression

a, atd, a + 28, a +38.... If we begin with a new term, b, and add to it successively each term of the above, we obtain

6, 6+a, b+ 2a +0, 5+3a +38, 6+4a+68..., which is called a difference series of the 2° order, and so on, as in the following scheme :

Order of a

1° term. De term. Series.

[ocr errors][merged small]
[ocr errors]

I. a ato, a+28, ...a+in—1)8.
II. b, bta, 6+2a+0...6+(n-1)a+!"

1.2
C, ctb, c+26+a...ct(n-1)6+!"
-L, AL(n-2)(n-1), (x-3)(x-2)(n-1),

1.2.3 &c. &c.

EXAMPLE.
I. order, 2, 5, 8, 11, 14 ..
II. order, 4, 6, 11, 19, 30..

III. order, 5, 9, 15, 26, 45 .. 331. From the manner in which these difference series are formed, it is evident that if we subtract from one another the successive terms of any order, we obtain the terms of the preceding, and continuing in this way till we subtract the successive terms of the first from one another, we obtain between them the constant difference d.

332. If the order of a series be unknown, its order may be found from what has been said above. Thus the series

5, 9, 15, 26, 45; taking the difference of the consecutive terms,

4, 6, 11, 19

2, 5, 8

3, 3, 3, after three subtractions of consecutive terms presents a constant difference, and is, therefore, a series of the 3° order.

333. To separate the roots of an equation by means of difference series.
The rila term of a series of the order m would be expressed by
k+(x-1)/+

L. (2)(x-1), (r—m)... (x-2)(x-1),

1 8 +....t- 1.2...mi which, arranged according to the powers of z, would be of the form

Mrm+Axm-1+B.zm-2.... +Gr+K; that is, of the form of the first member of an equation of the mth degree, X=0.

If, now, we give to x the values ...-4, -3, -2, -1, -0, 1, 2, 3, 4, .... representing the values which the polynomial X assumes by

X__, X_3, X_2, X_1, X., X1, X2, &c. ....... (1) these quantities will form a difference series, since x denotes the order of the . term in a series of which X is the general term. There is no objection to x being negative, as a series may be continued below as well as above the first term, observing the same law in a contrary sense.

Taking a sufficient number of terms of the series (1) to obtain, by subtraction of its successive terms, the series of next lower order, and from this, in the same manner, that of the next lower order still, till we arrive at constant differences, the terms of the series (1) may be extended indefinitely to the right and left by forming them according to (Art. 330), without the trouble of substituting numerical values for x, and calculating the corresponding values of X. Those values of X which have contrary signs will (Art. 252, Cor. 1). have one or an odd number of roots between them. Take, for example, the equation

9x+ - 3.x3 – 130.x2—177+260=0.

DD

« ForrigeFortsett »