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Put a, b and obtain

f(a)(aa) (aa)

ƒ(b) = (b — a ̧) (b − a) .

(a -- a) F(a),

(b − a) F(b).

F(a) and F(b) have the same signs; for if they had opposite signs then there would be one root at least of F(x) = 0, between a and b,· which is excluded by hypothesis. By hypothesis f(a) and (b) have opposite signs. Therefore the two products

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have opposite signs; but b is greater than any of the a's; hence the second product is positive and consequently the first product is negative. Since a is less than a1, a, a,, the first product

can be negative only when r is odd. Q. E. D.

We have a geometrical illustration of this theorem in the graph From the table we have

in 823.

for x1,

f(x) =ƒ(— 1)=—22)(-1) and (2) have opposite

signs, and three roots, - 1, 1, 20 1.2, lie between −1 and 2;

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f(x) = ƒ( 2 )

=

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828. Algebraic solutions of the equations of the third and fourth degrees have been found. A method will now be given which enables one to obtain approximately the value of the real roots of an equation of any degree. This part of the subject is begun by proving Sturm's theorem, which has for its object, the determination of the situation and number of the real roots of an equation.

829. Sturm's Functions.-Let f(x) 0 be an equation whose equal roots have been removed, and let f(x) be the first derived function of x). Apply the process of finding the G. C. D. of f(x) and f(c), with the exception that signs of the remainders are changed, until the last remainder does not contain x.

Let f(x), f(x), . . . ƒ„(.) be the series of modified remainders thus derived. They are called Sturm's Functions.

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In-1 be the successive quotients which arise in performing the indicated operations; hence, we have the following relations:

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ƒn−2(x) = In-1£n−1 (X) — ƒ„(x).

Three inferences may be drawn from these relations:

I. The function f(x) is not zero. For, if f(x) is zero then f(x) and f(x) have a common divisor, 111, and therefore the equation f(x)=0 has a pair of equal roots, which is contrary to the hypothesis. II. Two consecutive functions can not vanish for the same value of x. For, if f(x) = f(x) = 0, then f(x) = 0, and so on until ƒ„(x) is zero, which is impossible by I.

III. If any function excepting f(c) vanishes for any value of x, the two adjacent functions have opposite signs. Thus if f(x) = 0, then we have

f(x) = f(x).

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830. Sturm's Theorem.-I. If, as x increases, f(x) passes through the value zero, Sturm's functions lose one change in sign. For let x, as it increases, pass through e, a root of f(x) = 0, then we have for x = c + h,

h

ƒ(c + h) = ƒ(c) +ƒ'(c) ;; +ƒ" (c) 'm; + · · ·

h2
2!

[2819]

Suppose that h is infinitesimal (2637); then, since f(c) = 0, the sign of f(ch) depends upon the sign of f'(c)·h (8668); thus we have

(1)

f(ch) = f(c) h;

hence f(ch) and f'(c) have the same sign where his positive. Therefore the function f(x), just after x passes through a root, c, has the same sign as ƒ(r) at a root.

Changing hinto - h we have from (1)

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hence the function f(x), just before x passes through a root, c, has the opposite sign of f(x) at a root. Q. E. D.

=

II. Sturm's functions neither lose nor gain a change of sign when x passes through a value which makes one of them except f(x) vanish. Let f(c) 0, then f(c) and fr+1(c) have contrary signs, and hence just before xc and also immediately after xc, the three functions ƒ-1(x), ƒ,(x), fr+1(x) have one permanence of sign and one variation of sign; for, when fr-1(x) and ƒ„(x) have contrary signs, ƒ,(x) and ƒr+1(x) have the same signs, and reciprocally. Q.E.D.

Theorem. The number of roots of f(x) between a and b is equal to the difference between the number of variations of signs in Sturm's functions when x = a and x = b.

COROLLARY. The total number of roots of f(x) = 0 is found by taking a = +∞ and b ∞, since the sign of each function is the

same as that of its first term (2825).

831. When the number of functions is greater by unity than the degree of the equation, the following theorem can be proved:

I. All the roots of f(x) = 0 are real, if the first terms of all the functions are positive.

II. There will be a pair of imaginary roots of f(x)=0 for every variation of sign in the first term of the function, if they are not all positive. Proof.-Use Descartes's rule (816) after putting x = -, examining the number of changes of signs in each case. EXAMPLE.-Locate the roots of the equation

x3-10x35x500.

+∞ and

Sturm's functions, calculated according to the preceding rule, are here tabulated.

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in the adjoining table. Theref(x) = +++++

is one variation of sign lost while x passes from -2 to -1; and no other.

1 ++

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+ +

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CALCULATION OF INCOMMENSURABLE ROOTS BY HORNER'S METHOD

832. Solve the equation (1) x3- 7x+7= 0. This example was selected by Lagrange on account of its difficulty.

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According to Sturm's theorem this equation has three real roots; one lies between 4 and 3, and two between 1 and 2, that is, between 1.3 and 1.4, and 1.6 and 1.7; of these we shall calculate the second.

Diminishing the roots of the equation (1) x3-7x+7= 0 by 1.3, we put x=+1.3, and have (2810, Ex.),

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1.3

3.38

2.6.12-1.93

1.3

4,3.9

Hence, the transformed equation is

(2) z3 +3.9 z2 — 1.93z+.097 = 0.

Since a root of this equation is so many hundredths we may neglect the powers of z above the first in determining the next figure in the root; thus -1.93z+.097 = 0 or z=.05. Hence 5 is the next figure of the root. Diminishing the root of (2) by .05 we have

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Continuing this process in a more compact form we have

.010375

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=.006.

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833. Second Illustration of Horner's Method. Find the numerical values of the real roots of the equation

x-4x+x2 + 6x + 2 = 0.

This equation has two roots between 2 and 3 (752). The method of calculating the least of these roots is given below:

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