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a

14. It is required to convert

1, or its equal (a+c)3

(a + x) *, into an infinite series.

X

1千 F

F, &c.

4x2 4.703 4.7.104 Ans. a31

于 F
+

&c. За 3.602

3.6.9a3 3.6.9.12a4

1 75. It is required to convert

or its equal (1 + 5

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(1 + x) 7, into an infinite series.

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ба2 ? 6.11 x3 6.11.16x4
Ans. 1
+
+

&c.
5
5.10 5.10.15 5.10.15.20

a + x 16. It is required to convert

or its equal

(+) (a + x) (ao — xo) , into an infinite series.

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22

x2

Ans. 1 tot

2a2

+

324 325 526
+

+
8a4 8a5 166

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+

a

2a3

OF THE INDETERMINATE ANALYSIS.

In the common rules of Algebra, such questions are usually proposed as require some certain or definite answer; in which case it is necessary that there should be as many independent equations, expressing their conditions, as there are unknown quantities to be determined; or otherwise the problem would not be limited.

But in other branches of the science, questions frequently arise that involve a greater number of unknown quantities than there are equations to express them; in which instances they are called indeterminate or unlimited problems, being such as usually admit of an indefinite number of solutions ; although, when the question is proposed in integers, and the answers are required only in whole positive numbers, they are, in some cases, confined within certain limits, and in others the problem may become impossible.

PROBLEM 1.-To find the integral values of the unknown quantities x and y in the equation

ax -- by

- by = £c, or ax + by Where a and b are supposed to be given whole numbers, which admit of no common divisor, except when it is also a divisor of c.

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RULE 1.-Let wh denote a whole, or integral number; and reduce the equation to the form *by Ic

c - by wh, or x =

wh. 2. Throw all whole numbers out of that of these two expressions, to which'the question belongs, so that the numbers d and e, in the remaining parts, may be each less than

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a

a; then

e

a

а

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dy Łe

-dy wh, or

wn. 3. Take such a multiple of one of these last formulæ, corresponding with that abovementioned, as will make the coefficient of y nearly equal to a, and throw the whole numbers out of it as before.

ay Or, find the sum or difference of á, and the expression

ay above used, or any multiple of it that comes near

and the result, in either of these cases, will still be = wh, a whole number.

4. Proceed in the same manner with this last result; and so on, till the coefficient of y becomes 1, and the remainder = some number r ; then ytr

wh, =p, and y = ap Fr. Where

p may be 0, or any integral number whatever that makes y positive; and, as the value of y is now known, that of x may be found from the given equation, when the equation is possible.* NOTE. Any indeterminate equation of the form

by in which a and b are prime to each other, is always possible, and will admit of an infinite number of answers in whole numbers. But if the proposed equation be of the form

ax + by = c, the number of answers will always be limited; and, in some cases, the question is impossible ; both of which circum

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te,

* This rule is founded on the obvious principle, that the surn, difference, or product of any two whole numbers, is a whole number; and that if a number divides the whole of any other number and a part of it, it will also divide the remaining part.

stances may be readily discovered from the mode of solution above given.*

EXAMPLES.

1. Given 190 147 11, to find x and y in whole numbers. 14y +11

197 Here x =

wh., and also

wh. 19

19

19y 14y +11 5y - 11 Whence, by subtraction,

wh. 19 19

19 5y – 11

204

44 Also, x 4

y - 2+

wh. 19 19

19 And, by rejecting y - 2, which is a whole number,

y-6

Y 6

= wh.

P.
19
Whence we have

y = 19p + 6. 14y +11 14 (19p + 6) +11 266p + 95 And 2 = 19 19

19 14p + 5. Whence, if p be taken = -0, we shall have x =

5 and y=6, for their least values; the number of solutions being obviously indefinite.

2. Given 2x + 3y = 25, to determine x and y in whole positive numbers. 25 Зу

1 Here x =

– 2

2 Hence, since x must be a whole number, it follows that 1-Y

must also be a whole number. 2

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* That the coefficients a and b, when these two formulæ are possible, should have no cominon divisor, which is not at the same time a divisor of c, is evident; for if a = md, and b me, we shall have ax

t

by mdx + mey=c; and consequently, dlat tey But d, e, x, y, being

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m

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т

supposed to be whole numbers, must also be a whole number, which it cannot be except when m is a divisor of C.

Hence, if it were required to pay 1001. in guineas and moidores only, the question would be impossible; since, in the equation 21:+277 2000, which represents the conditions of the problem, the coefficients, 21 and 27, are each divisible by 3, whilst the absolute term 2000 is not divisible by it. See my Treatise on Atgebra, for the method of resolving questions of this kind by means of Continued Fractions.

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1-Y

-0, p

-1, P

2, p =

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12 -- y +

= 12 – (1 – 2p) +p=12 + 3p-1, ,

2
We shall have x = 11 + 3p, and

11 + 3p, and y=1- 2p; Where p may be any whole number whatever, that will render the values of x and y in these two equations positive. But it is evident, from the value of Y,

that

р

must be either O or negative; and consequently, from that of x, that it must be 0,-1, -2, or - 3. Whence, if p

3,
11, = 8, a 5, x= 2,
Then

y=l, y -- 3, y = 5, y Which are all the answers in whole positive numbers that the question admits of. 3. Given 3x = 8y — 16, to find the values of x' and

y

in whole numbers. 8y

24-1

2y-1 Here x = 2y -- 5+

=

-wh. 3

3

3
2y - 1

x2 =
3
3

3 Or, by rejecting y, which is a whole number, there will

5, y ==7.

wh.; or

4y - 2

Also,

Y-2 yt

= wh.

y - 2

wh.= p.

remain 3

Therefore, y=3p + 2, 8y - 16 8 (3p +2) - 16

24p And x =

8p. 3

3

3 Where, if p be put 1, we shall have x = 8 and y = 5 for their least values ; the number of answers being, as in the first question, indefinite.

4. Given 21x + 174 2000, to find all the possible values of x and y in whole numbers. 2000 – 177

5 - 174
Here x =
21

21

5 - 177
Or, omitting the 95,

.
21
214
5 - 174

4y + 5 Consequently, by addition, -

21
21

21
4y + 5

4+20y Also,

1+

= wh; 21 21

21

174 = 95.+

wh. ;

wh.;

20y + 25

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wh.;

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be put

х

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4 + 20y Or, by rejecting the whole number 1,

21
21% 4 + 20y
And, by subtraction,

wh
21
21

21 Whence y=21p + 4, 2000 – 179 2000 17 (21p + 4) And x =

92 -- 17p. 21

21 Where, 'if P

0, we shall have the least value of y= 4, and the corresponding or greatest value of x = 92.

And the rest of the answers will be found by adding 21 continually to the least value of y, and subtracting 17 from the greatest value of x; which being done, we shall obtain the six following results :

92 75 58 41

41 | 24 7 25 | 46

46 67 88 109 These being all the solutions, in whole numbers, that the question admits of.

Note 1.-When there are three or more unknown quantities, and only one equation by which they can be determined, as

ax + by + cz d, it will be proper first to find the limit of the quantity that has the greatest coefficient, and then to ascertain the different values of the former, from 1 up to that extent, as in the following question :

5. Given 3x + 5y +72 = 100, to find all the different values of x, y, and 2 in whole numbers.*

Here each of the least integer values of ic and y are 1, by the question ; whence it follows, that 100 – 5-3 100- 8 92

137. 7

7

7 Consequently, z cannot be greater than 13, which is also the limit of the number of answers; though they may be considerably less

* If any indeterminate equation, of the kind above given, has one or more of its coefficients, as c, negative, the equation may be put under the form

ax +-by=d+cz, in which case it is evident that an indefinite number of values may be given to the second side of the equation by means of the indefinite quantity z; and consequently, also, to x and y in the first. And if the coefficients a, b, c, in any such equation, have a common divisor, while d has not, the question, as in the first case, becomes impossible.

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