Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

283. Every number which exceeds the greatest of the positive roots of an equation, is called a superior limit of the positive roots.

From this definition, it follows, that this limit is susceptible of an infinite number of values. For, when a number is found to exceed the greatest positive root, every number greater than this, is also a superior limit. The term, however, is generally applied to that value nearest the value of the root.

Since the greatest of the positive roots will, when substituted for x, merely reduce the first member to zero, it follows, that we shall be sure of obtaining a superior limit of the positive roots by finding a number, which substituted in place of x, renders the first member positive, and which at the same time is such, that every greater number will also give a positive result; hence, The greatest co-efficient of x plus 1, is a superior limit of the positive roots.

Ordinary Limit of the Positive Roots.

284. The limit of the positive roots obtained in the last article, is commonly much too great, because, in general, the equation contains several positive terms. We will, therefore, seek for a limit suitable to all equations.

Let am-n denote that power of x that enters the first negative term which follows am, and let us consider the most unfavorable case, viz., that in which all the succeeding terms are negative, and the co-efficient of each is equal to the greatest of the nega tive co-efficients in the equation.

Let S denote this co-efficient. What conditions will render

[blocks in formation]

Dividing both members of this inequality by am, we have

[blocks in formation]

n

x=/S+1, or for simplicity, making /S= S.

which gives,

S S'", and S'+1,
S=S'",

x =

[blocks in formation]

Moreover, every number > S'+ 1 or 2 S+1, will, when substituted for x, render the sum of the fractions

[blocks in formation]

still smaller, since the numerators remain the same, while the denominators are increased. Hence, this sum will also be less.

n

Hence, S+1, and every greater number, being substituted for x, will render the first term a greater than the arithmetical sum of all the negative terms of the equation, and will conse quently give a positive result for the first member. Therefore,

That root of the numerical value of the greatest negative co-efficient whose index is equal to the number of terms which precede the first negative term, increased by 1, 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 terin of the equation; the limit becomes

√ S + 1 = S + 1;

that is, the greatest negative co-efficient plus 1.

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

[blocks in formation]

EXAMPLES.

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

24

5x337x2 3x+390?

Ans. "√S +1 = √√/ 5 + 1 = 6.

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

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

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

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

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 zero. Hence,

[blocks in formation]

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

Smallest Limit in Entire Numbers.

285. In Art. 282, it was shown that the greatest co-efficient of a plus 1, 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.

[blocks in formation]
[ocr errors]

be the proposed equation. If in this equation we make x + being arbitrary, we shall obtain (Art. 264),

[blocks in formation]

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

[blocks in formation]

renders, at the same time, all these co-efficients positive, this num ber will in general 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 value of u can satisfy it; therefore, all the real values of u must be negative. But from the equation

[blocks in formation]

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 a should be less than the value of x'. Hence, this value of x' is a superior limit of the positive If we now substitute in succession for x in X the values

roots.

[ocr errors]

1,

x'

2, x' - 3, &c., until a value is found which will make X negative, then the last number which rendered it positive will be the least superior limit of the positive roots in whole numbers.

Let

EXAMPLE.

x4 - 5x3 6x2 19x+7=0.

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

[blocks in formation]

have,
6x2

X = x2+ 5x3

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

19x+7,

[blocks in formation]

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.

Bat 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, is the least limit in entire numbers. We see that 7 is a superior limit, and that 6 is not; hence, 7 is the least limit, as above shown.

x=

2. Applying this method to the equation,

x5 3x4 Sx3 25x2+4x

the superior limit is found to be 6.

39 = 0,

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

ენ 5x4 13x3 + 17x2

[blocks in formation]

This method is seldom used, except in finding incommensurable roots.

Superior Limit of Negative Roots.—Inferior Limit of Posi tive and Negative Roots.

286. Having found the superior limit of the positive rcots, it remains to find the inferior limit, and the superior and inferior limits of the negative roots, numerically considered. First, If, in any equation,

[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small]
« ΠροηγούμενηΣυνέχεια »