Any number>k+1, put in place of x, will render the sum of the

The greatest co-efficient of the equation plus unity, or any greater number, being substituted for x, will render the first term xTM greater than the arithmetical sum of all the others.

Ordinary limit of the Positive Roots.

315. The number obtained above may be considered a prime limit, since this number, or any greater number, rendering the first term superior to the sum of all the others, the results of the substitution of these numbers for x must be constantly positive; but this limit is commonly much too great, because, in general, the equation contains several positive terms. We will, therefore, seek for a limit suitable for all equations.

Let a denote the power of x, corresponding to the first nega. tive term which follows a", and we will consider the most unfavour. able case, viz. that in which all of the succeeding terms are nega. tive, and affected with the greatest of the negative co-efficients in the equation.

Let S be this co-efficient, and try to satisfy the condition

xm>Sxm-n+Sxm-n-1+ Sx+S;

or, dividing both members of this inequality by x",

Now by supposing a"-S or x="VS, the second member be

S

comes or 1, plus a series of positive fractions; but by making

S'

x="√S+1, or (supposing, for simplicity, VS=S', whence S=S'”), x=S'+1, the second member becomes

Moreover, every number >S'+1 or VS+1, will, when substi

smaller, since the numerators remaining the same, the denominator will increase. Hence VS+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 posi tive result for the first member.

n

Therefore VS+1, or 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. If the co-efficient of a term is 0, the term must still be counted.

Make n=1, in which case the first negative term is the second

1

term of the equation; the limit becomes √S+1, or S+1; that is, the greatest negative co-efficient plus unity.

Let n=2, then either the two first terms are positive, or the

2

term am-1 is wanting in the equation; the limit is then S+1. When n=3 the limit is 3√S+1...

Find the superior limits for the positive roots in the following examples:

24-5x+37x2-3x+39=0;

VS+1=√5+1=6;

x+7x-12x3-49x2+52x-13=0; VS+1= √49+1=8;

316. Let X=0, be the proposed equation; if in this equation we make x=x+u, x' being indeterminate, we shall obtain (Art. 296),

Conceive, that after successive trials we have determined a number

for a, which, substituted in X', Y',

Z
2

renders all these co-effi.

cients positive at the same time; this number will be greater than the greatest positive root of the equation X=0.

For, the co-efficients of the equation (1) being all positive, no positive number can verify it; therefore all of the real values of u must be negative; but from the equation x=x'+u, we have u=x—x'; and in order that the values of u corresponding to each of the values of x and x' (already determined) may be negative, it is absolutely necessary that the greatest positive value of a should be less than the value of x'.

EXAMPLE.

24-5x3-6x-19x+7=0.

As a' is indeterminate, the letter ≈ may be retained in the forma tion of the derived polynomials, and we have

The question is, as stated above, reduced to finding the smallest 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 the polynomial of the third degree, give a negative result; but 5 and any greater number, give a positive result.

Lastly, 5 substituted in X, gives a negative result, and so does 6; for the three first terms x4-5x3-6x2 are equivalent to the expres. sion x3(x— 5) — 6x2, which is reduced to 0 when x= :6; but x=7 evidently gives a positive result. Hence 7 is a superior limit of the positive roots of the proposed equation; and 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.

Applying this method to the equation

x3-3x2-8x3-25x2+4x-39=0,

the superior limit will be found to be 6.

We should find 7, for the superior limit of the positive roots of the equation

x3 — 5xa — 13x3+17x2—69=0.

This method is scarcely ever used, except in finding incommen surable roots.

317. It remains to determine the superior limit of the negative roots, and the inferior limits of the positive and negative roots. Hereafter we shall designate the superior limit of the positive roots of an equation by the letter L.

1st. If in the equation X=0, we make x=— -y, which gives 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 proposed equation; therefore, determining, by the known methods, the superior limit L' of the positive roots of the equation Y=0, we shall have L' for the superior limit (numerically) of the negative roots of the proposed equation.

2d. If in the equation X=0, we make x=- which gives the

9 y

1

equation Y=0, it follows from the relation x=— y

that the greatest

positive values of y correspond to the smallest of x; hence, desig. nating the superior limit of the positive roots of the equation Y=0

1 L"

by L", we shall have for the inferior limit of the positive roots

of the proposed equation.

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

1

y

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

will be the inferior limit (numerically) of the negative roots

of the proposed equation.

318. 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.

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 was of an odd degree. Hence the sum would not be equal to zero in either case.

This 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.