Page images
PDF
EPUB

The first seven convergents formed as explained in Art. 472 are

4 9 13 48 61 170 1421

2

1' 2' 3 11' 14' 39' 326

1

The error in taking the last of these is less than
1
1

and is

(326)2'

therefore less than and still less than 00001. (320)2 102400' Thus the seventh convergent gives the value to at least four places of decimals.

and

482. Every periodic continued fraction is equal to one of the roots of a quadratic equation of which the coefficients are rational.

or

Let x denote the continued fraction, and y the periodic part, and suppose that

x=a+

...

1 1

b+c+

1

n+

y=m+

where a, b, c, h, k, m, n, u, v are positive integers.

1 1 1 h+k+ y 1 1 1

u+v+ y

...

Let P, P be the convergents to x corresponding to the quoq q

tients h, k respectively; then since y is the complete quotient,

we have x= py+p; whence y= p-qx

qy+q

q'x —p'

r ri Let -9 be the convergents to y corresponding to the quoS s' r'y+r tients u, v respectively; then y = s'y + s

Substituting for y in terms of x and simplifying we obtain a quadratic of which the coefficients are rational.

A+ √B

The equation s'y2+(s-r')y-r=0, which gives the value of y, has its roots real and of opposite signs; if the positive value of be substituted in x= py+P, on rationalizing the denomiy q'y + q' nator the value of x is of the form integers, B being positive since the value of y is real.

where A, B, C are

2 с

1 1 1 1

2+ 3+ 2+ 3+

Let x be the value of the continued fraction; then

1

x-1=

Example. Express 1 +

[blocks in formation]

1

2+3+(x−1)

whence 2x2+2x-7=0.

The continued fraction is equal to the positive root of this equation, and is therefore equal to

v15-1

2

EXAMPLES XLV. b.

Express the following surds as continued fractions, and find the sixth convergent to each:

2. √5.
6. 13.

10. 4√2.

1

14.

√33

17. Find limits of the error when

18. Find limits of the error when

19. Find the first convergent to places of decimals.

[blocks in formation]

...

3.

√6.

7. √14.

11. 3√5.

16

15.

V5

28. Find the value of 5+

as a surd.

6+ 6+ 6+

1 1 1

4. √8.

8. √22.

12. 4√10.

16.

268

65

916

191

101 that is correct to five

20. Find the first convergent to √15 that is correct to five places of decimals.

Express as a continued fraction the positive root of each of the following equations :

21. x2+2x-1=0. 22. x2-4x-3=0. 23. 7x2-8x-3=0. 24. Express each root of x2-5x+3=0 as a continued fraction. 25. Find the value of 3+

1 1 1

17 Vii

is taken for 17.

is taken for 23.

3+ 1+ 3+

1 1 1 1 1
1......
1+ 2+ 3+ 1+ 2+ 3+
1 1 1 1
1+1+1+ 10+

CHAPTER XLVI.

SUMMATION OF SERIES.

483. EXAMPLES of the summation of certain series (Arithmetic and Geometric) have occurred in previous chapters. We will now consider methods for summing other series.

484. A series

up + u 1 + U2+ Uz +

in which from and after a certain term each term is equal to the sum of a fixed number of the preceding terms multiplied respectively by certain constants is called a recurring series. A recurring series is of the 1st, 2nd, or rth order according as one, two, or r constants are required as multipliers.

485. In the series

or

[ocr errors]

1+2x+3x2+4x3 +5x1+......,

each term after the second is equal to the sum of the two preceding terms multiplied respectively by the constants 2x, and -x2; these quantities being called constants because they are the same for all values of n. Thus

5x=2x.4x3+(−x2). 3x2;
u4=2xUz-x22;

that is,

and generally when n is greater than 1, each term is connected with the two that immediately precede it by the equation

un=2xun-1-x2un-2, Un-2xun-1+x2un-2=0.

In this equation the coefficients of un, un-1, and un-2, taken with their proper signs, form what is called the scale of relation.

Thus the series

1+2x+3x2+4x3+5x1+

is a recurring series in which the scale of relation is

1-2x+x2.

486. If the scale of relation of a recurring series is given, any term can be found when a sufficient number of the preceding terms are known. As the method of procedure is the same however many terms the scale of relation may consist of, the following illustration will be sufficient.

If

we have

1-px-qx2-rxs

is the scale of relation of the series

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

ɑnx2=px . ɑn-1xn−1 +qx2. An-2x2−2+rx3. An-3x2−3,

or

an pan-1+qan-2+ran-3;

thus any coefficient can be found when the coefficients of the three preceding terms are known.

487. Conversely, if a sufficient number of the terms of a series be given, the scale of relation may be found.

Example. Find the scale of relation of the recurring series

2+5x+13x2+35x3+97x2+275x+

[ocr errors]

This is plainly not a series of the first order. If it be of the second order, to obtain p and q we have the equations

13=5p+2q, and 35=13p+5q;

whence p=5, and q=-6. By using these values of p and q, we can obtain the fifth and sixth coefficients; hence they are correct, and the scale of relation is

1-5x+6x2.

488. If the scale of relation consists of 3 terms it involves 2 constants, p and q; and we must have 2 equations to determine р and 1. To obtain the first of these we must know at least 3 terms of the series, and to obtain the second we must have one more term given. Thus to obtain a scale of relation involving two constants we must have at least 4 terms given.

If the scale of relation be 1-px-qx2-rx3, to find the 3 constants we must have 3 equations. To obtain the first of these

we must know at least 4 terms of the series, and to obtain the other two we must have two more terms given; hence to find a scale of relation involving 3 constants, at least 6 terms of the series must be given.

Generally, to find a scale of relation involving m constants, we must know at least 2m consecutive terms.

Conversely, if 2m consecutive terms are given, we may assume for the scale of relation

1-p1x-P2x2-P1x3-.

-Pmxm

489. To find the sum of n terms of a recurring series.

The method of finding the sum is the same whatever be the scale of relation; for simplicity we shall suppose it to contain only two constants.

Let the series be

......

ɑ+ɑ1x+ɑ2x2+ɑ3x3+

and let the sum be S; let the scale of relation be 1-px-qx2; so that for every value of n greater than 1, we have

an-pan-1-Jan-2=0.

Now S=α+ ax+ a‚x2+...+ ɑn-12”-1,

-px S= -pax—pаx2— ... —pan-2x′′-1 —pɑn-1x”,
-qx2S=

Hence

[ocr errors]

(1-px-qx2)S=ɑ+(a1−pа)x− (pan-1+qan-2)x” — Jan-12n+1,

for the coefficient of every other power of x is zero in of the relation

consequence

an-pan-1-qan-2=0.

... S=a+(a−pa)x _ (pan-1+Jan−2) x2+Jan-12”+1
1-px-qx2

1-px-qx2

Thus the sum of a recurring series is a fraction whose denominator is the scale of relation.

490. If the second fraction in the result of the last article decreases indefinitely as n increases indefinitely, the sum of an infinite number of terms of a recurring series of the second order reduces to ɑ+(a1—pɑ)x ̧ `

1-px-qx2

« PreviousContinue »