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Dividing both members of this inequality by am, we have

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=

x = √√S + 1, or for simplicity, making VSS
x= S+1,
S= S'n, and x = S' + 1,

which gives,

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the expression for the sum of all the terms is

is the

(S' + 1)m

the last term. Hence,

(Art. 192),

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Moreover, every number >S' + 1 or S+1, will, when substituted for x, render the sum of the fractions

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still smaller, since the numerators remaining the same, the de nominators will increase.

n

Hence, S + 1, and any greater number, will render the first term a greater than the arithmetical sum of all the negative terms of the equation, and will consequently give a positive result for the first member. Therefore,

Unity increased by that root of the greatest negative co-efficient whose index is the number of terms which precede the first negative term, is a superior limit of the positive roots of the equation. co-efficient of a term is 0, the term must still be counted.

Make n

=

If the

1, in which case the first negative term is the sec

ond term of the equation; the limit becomes

√√S+1=S+ 1;

that is, the greatest negative co-efficient plus unity.

2

Let n = 2; then, the limit is S+1. When n = 3, the limit is

3

+1.

EXAMPLES.

1. What is the superior limit of the positive roots in the equation

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Ans. VS+1=√√5+1 = 6.

2 What is the superior limit of the positive roots in the equation

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3. What is the superior limit of the positive roots in the equation

x2 + 11x2 - 25x — 67 = 0.

In this example, we see that the second term is wanting, that is, its co-efficient is zero; but the term must still be counted in fixing the value of n. We also see, that the largest negative co-efficient of x is found in the last term where the exponent of x is Hence,

zero.

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and therefore, 6 is the least whole number that will certainly fulfil the conditions.

Smallest Limit in Entire Numbers.

318. In Art. 316, it was shown that the greatest co-efficient of x plus unity, is a superior limit of the positive roots. In the last article we found a limit still less; and we now propose to find the smallest limit in whole numbers.

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be the proposed equation. If in this equation we make x=x+u, being indeterminate, we shall obtain (Art. 297),

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Let us suppose, that after successive trials we have determined a number for r', which substituted in

Z'

X', Y',

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renders all these co-efficients positive at the same time; this number will be greater than the greatest positive root of the equation X = 0.

For, if the co-efficients of equation (1) are all positive, no positive number can verify it; therefore, all of the real values of u must be negative. But from the equation

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and in order that every value of u, corresponding to each of the values of x and x', may be negative, it is necessary that the greatest positive value of x should be less than the value of x'.

Let

x45x3

EXAMPLES

6x2

19x+7=0.

As is indeterminate, we may, to avoid the inconvenience of writing the primes, retain the letter x in the formation of the derived polynomials; and we have

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The question is now reduced to finding the smallest entire number which, substituted in place of x, will render all of these polynomials positive.

It is plain that 2 and every number >2, will render the polynomial of the first degree positive.

But 2, substituted in the polynomial of the second degree, gives a negative result; and 3, or any number > 3, gives a positive result.

Now 3 and 4, substituted in succession in the polynomial of the third degree, give negative results; but 5, and any greater number, gives a positive result.

Lastly, 5 substituted in X, gives a negative result, and so does 6; for the first three terms, x4 5x3 6x2, are equivalent to the expression x3 (x — 5) — 6x2, which reduces to 0 when x = 6; but x = 7 evidently gives a positive result. Hence 7, which here

stands for x', is a superior limit of the positive roots of the given equation. Since it has been shown that 6 gives a negative result, it follows that there is at least one real root between 6 and 7. 2. Applying this method to the equation

ენ 3x4 8x3 25x2 + 4x 39 = 0,

--

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the superior limit is found to be 6.

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3. We find 7 to be the superior limit of the positive roots of the equation

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This method is seldom used, except in finding incommensurable

roots.

Superior Limit of negative Roots.-Inferior Limit of positive and negative Roots.

319. Having found the superior limit of the positive roots, it only remains to find the inferior limit, and the superior and inferior limits of the negative roots.

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superior limit of positive roots.

inferior limit of positive roots.

superior limit (that is, numerically) of negative roots.

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inferior limit of negative roots.

1

1st. If in any equation X = 0, we make x =

we have a

y

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derived equation Y=0. We know from the relation a

y

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smallest

the greatest positive value of y will correspond to the of x; hence, designating the superior limit of the positive roots

1

of the equation Y=0 by L, we shall have =L', the inferior

limit of the positive roots of the given equation.

L

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2d. If in the equation X=0, we make x = the transformed equation Y = 0, it is clear that the positive roots of this new equation, taken with the sign, will give the negative roots of the given equation; therefore, determining, by the known methods, the superior limit L of the positive roots of the equation Y= 0, we shall have L= L", the superior limit (numerically) of the negative roots of the proposed equation.

3d. Finally, if we replace x, in the given equation, by

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and find the superior limit L of the transformed equation Y = 0,

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will be the inferior limit (numerically) of the

L

negative roots of the given equation.

Consequences deduced from the preceding Principles.

First.

320. Every equation in which there are no variations in the signs, that is, in which all the terms are positive, must have all of its real roots negative; for, every positive number substituted for x, will render the first member essentially positive.

Second.

321. Every complete equation, having its terms alternately positive and negative, must have its real roots all positive; for, every negative number substituted for x in the proposed equation, would render all the terms positive, if the equation was of an even degree, and all of them negative, if it were of an odd degree. Hence, their sum could not be equal to zero in either case.

This principle is also true for every incomplete equation, in which there results, by substituting -y for x, an equation having all of its terms affected with the same sign.

Third.

322. Every equation of an odd degree, the co-efficients of which are real, has at least one real root affected with a sign contrary to that of its last term.

For, let

xm + Pxm-1 + Tx± U = 0,

...

be the proposed equation; and first consider the case in which the last term is negative.

By making a = 0, the first member becomes U. But by giving a value to x equal to the greatest co-efficient plus unity, or (K+1), the first term am will become greater than the arithmetical sum of all the others (Art. 315), the result of this substitution will therefore be positive; hence, there is at least one

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