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a fortiori, if x is greater than pk +1, a" is so much greater than the sum of all the terms after a": and thus by increasing x, f(x) becomes greater and greater, and in the superior limit when x = ∞, f(x) also becomes infinite. Thus in

x3+6x2+7x + 4 = 0,

if we substitute 8 for x, x3 is greater than the sum of all the other terms. Hence it appears that for values of a greater than a certain assignable quantity, as a increases, ƒ(x) increases; and ultimately becomes infinite, when x = ∞.

If x = 0, then f(x) = p,; and if n is odd, f(x) = −∞, when a = ∞; hence if an equation is of uneven dimensions, f(x) changes sign from -∞ to + as a increases from - to +; and as such change of sign of f(x) can take place only when f(x) = 0, it follows that an equation of odd dimensions has at least one real root: and that root will be positive or negative according as the constant term of the equation is negative or positive.

Hence also it follows that an equation of even dimensions has two real roots if the last term is negative, of which one is positive and the other is negative; because f(x) will change its sign from to, and again from

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to, as a increases

As a continuously varies, so f(x) continuously varies; for let h be an increment of a, and let us suppose that f'(x) does not vanish; then by (21), Art. 116,

f(x+h)-f(x) = hf'(x+0h);

the right-hand member of which is infinitesimal, if h is infinitesimal; therefore f(x+h) − f(x) is infinitesimal; that is, f(x) varies continuously as its subject-variable continuously varies.

If f(x) is of even dimensions, f(x) has at least one minimum value; because f'(x) will be of odd dimensions, and will change sign from to +, as a increases from - ∞ to; and corresponding to the value of x at which the change of sign takes place will f(x) attain a minimum value. Thus much as to the values of f(x) when a is replaced by a real quantity.

172.] Having thus shewn that f(x) continuously increases up too, as a increases upwards from a certain real value, I will now shew that if for a we substitute in f(x) the most general

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form of an imaginary expression, viz. y + z√ −1, f(y + z √√−1) does not admit of a maximum or a minimum, but continuously either increases or decreases.

Let us replace a in f(x) by y+z√-1; then expanding by Taylor's Series, we have

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of which right-hand member the possible and impossible parts may be separated, and the whole expression may be equated to P+Q-1; if

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and where p and q are functions of two independent variables y

and z.

Р

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dy

d2p

;

(dy dz) = - (d22);

dz2

hence, (172) and (2) being necessarily of different signs, the

dy2

conditions of Art. 158 and 159 cannot be satisfied, and p does not admit of a maximum or minimum value; similarly also it is plain that a does not admit of a maximum or minimum; hence f(y+z√1) is not a function admitting of a maximum or minimum; and therefore goes on continuously increasing or decreasing, as the variables continuously vary.

-

173.] Hence we can shew that if f(x) is a function of a of the form (1), one or more values of x of the form y + z√√−1 exist, which when substituted for a render f(x) = 0; in other words, we can hence prove that every equation has a root. Let y+1 be replaced by r (cos +1 sin ), and f(y+z√−1) by R (COST+ √1 sin T), r and R being the moduli respectively of y+z√−1 and of ƒ (y+z√√−1), and 0 and T being real circular arcs; then, if it can be shewn that

some value of 0 is such as to render R = У and z are such as to make ƒ(y+z√−1) = 0, and it is proved that the equation f(x) = 0 has a root.

Since

f(y+1)
f(y + z √ −1) = R (COS T + √−1 sin r),

.. f(y-z√1) = R(COS T-√1 sin r); whence R2 = ƒ (y + z √√ −1) × ƒ (y—z √−1).

(7)

(8)

(9)

Now to take the most general case, let us suppose that all the derived-functions of f(x) up to the mth exclusively vanish when x = y + z√−1; viz. that

f'(y + z√ −1) = 0, ƒ” (y + z√√ −1) = 0,... ƒm−1(y + z√√ −1)=0; whence also,

f(y−x−1)=0,f”(y−xy−1)=0,...fm-1(y-~V-1)=0.

Then, by means of Leibnitz's Theorem, see Art. 55, taking the mth differential of R2 in (9), and neglecting terms which vanish, we have

dm.R2 = fm (y + z√ −1) (dy+dz √−1)TM ƒ(y—z√—1)

+fm(y−z√− 1) (dy — dz√−1)TM ƒ (y + z √−1). Let R, and p be the moduli respectively of fm (y+z√—1) and dy + √1 dz; and let TM and be the corresponding arcs,

T

fm(y+z√−1) = Rm (COS TMm + √√−1 sin TMm),

so that

(10)

fm (y-z√1) = Rm (COS T√1 sin TM);

Tm

(11)

dy ± dz√√−1 = p (cos 7+ √−1 sin 7) ;

(12)

.. (dy+dz√−1)m = pm (cos mr + √1 sin mT);

(13)

making which substitutions,

dm.R2 = 2 R Rm pm COS (Tm-T+MT).

(14)

Now as f(y + z√1) does not admit of a maximum or minimum value, so R2, being the square of R, and thus necessarily a positive quantity, does not admit of a maximum value; but it must have a minimum value; that derived-function of it therefore, which is the first not to vanish, must be consistent with such a singular value. But on examining (14) is arbitrary, and dm.R2 will evidently change sign, when 7 and therefore when dy and dz receive certain values; as, for instance,

2k+1

m

change to T + , and the sign of d.R2 will be changed; and such a change is inconsistent with R2 being a minimum. The apparent difficulty however is obviated if one of the other factors of dm. R2 vanishes; but R, and p" do not vanish, and therefore it follows that the only condition which is consistent with R2 being a minimum value is, R= 0; this therefore is the minimum value; whence, by means of (7),

f(y + z√1)

=

0;

(15)

and therefore there always are some values of y and z such that equation (15) is satisfied; and therefore every equation of the form (1) has a root.

Also from (8) it follows that if R = 0, ƒ (y − z √ −1) = 0 ; and therefore if one of a conjugate pair is a root of an equation, the other conjugate is also a root.

z

If in the general form of the root = 0, the root is possible; but if z has a finite value, the root is imaginary.

174.] Taking a to be the general symbol for the root of an equation, we may thus prove that the equation is divisible by x-a without a remainder.

Let f(x)=0 be the equation; then, since a is a root, f(a) = 0. Observing now that Taylor's Series does not fail for a function of x, such as we have assumed f(x) to be, and that the (n+1)th derived-function vanishes, because f(x) is algebraical and of n dimensions, by means of (84), Art. 74, we have

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and because f(a) = 0, the second member is divisible by (x− a). The converse proposition is evident; viz. that if x-a is a factor of f(x), ƒ(a) = 0, or a is a root of the equation.

Hence we conclude that if any function of the algebraical form assumed in equation (1), Art. 170, vanishes for a particular value a of the variable, the function has a factor of the form x-a; and it is owing to its vanishing that the function vanishes.

175.] Hence also it follows, that every equation has as many roots as it has dimensions, and no more.

For dividing f(x) in equation (16) by (x-a), the highest power of a that remains is "-1, being that which is involved

in the last term of it, viz. in (x-a)"-1; whence an expression of n-1 dimensions results; which again, by virtue of Article 173, has a root, and therefore is again divisible by a factor of the form x-a; whereby the expression is depressed to one of n-2 dimensions; and if a similar process is repeated n-1 times, we shall finally have an expression of one dimension, which will give the last root; and thereby the equation will have been resolved into n factors.

Thus suppose the n roots to be a1, a2,

...

an

then

f(x) = (x−a1) (x −αş) ... (x — a1).

(17)

Also it is manifest that no other value than one of the n roots can, when substituted for x, make any simple factor, and thereby the whole expression, to vanish; and therefore f(x) has only n roots; some of which however may be equal; and therefore although all the n roots may not be different, yet there can never be fewer than n simple factors.

Again, if the coefficients of the several powers of a in f(x) are real, and f(x) has impossible roots, they must, as observed above, enter in pairs; so that, if a; and a, are two imaginary roots which are conjugate to each other,

a; = a + ß √ −1 = p {cos @ + √ −1 sin 0},

-

a; = a-ẞ√1 = p {cos 0 —

√1 sin 0};

.. (x-a¡) (x-a;) = (x− a)2 +ẞ2 = x2-2px cos 0+ p2, which quadratic expression is essentially positive; and by a similar composition of other conjugate factors we have, f(x)= (x — α1) (x —α2)..... {(x —α1)2 + ẞ12} {(x−a2)2 + ẞ22} ..... ; (18) and therefore f(x) is the product of factors, simple or quadratic.

176.] On the algebraical relation of f(x) to its derivedfunction.

Let f(x) be a function of the form (1), which has all its coefficients real quantities and all its roots real and unequal; and let the roots of f(x) be a1, a2,

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an; so that

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(19)

(20)

the upper or lower sign being taken in (19) according as n is odd or even, and the roots being arranged in order of magnitude, viz.

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