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3

get a 1.5 ± nearly,
= nearly, so that we have a =

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3

Putting for a in a3- 7a + 7 it becomes

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2

8 2

27 84

56

1

+

=

8 8

8

8'

which being small, the given equation

has roots which are nearly equal to each other, and which

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Hence, putting for a in the expression for k, it will be

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in which we must assume h such that the numerical value of k shall be small in comparison to that of h.

To find the nearly equal roots; we observe that since h must be small, and as the coefficient of the simple power of h is small in comparison to that of h, it will clearly answer our purpose to assume − 1 + 36h2 = 0, which gives h2

1

36'

1

6

, or, extracting the square root, we get h consequently, the equation is reduced to k =

=

= or h =

1

6

1

Substituting

1

6.

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for h in this equation, we get = ; also, substituting

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used for the new values of h in the above expression for k..

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for the new value of k, and 75276

= 0.1920 +, which is correct in

all its figures, is the new value of h; substituting this new

value of h in the expression for k, we get .00002147 for the new value of k, and adding this to 0.1920 we get 0.19202147, which is correct in all its figures, for the new value of h;

consequently, if we add this to

3

2

=

1.5, we shall have

1.69202147 for one of the nearly equal roots of the equation.

To get the remaining one of the nearly equal roots, we

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= 0.144 nearly, in the expression for k, and

-

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we shall get 0.000893 nearly for the new value of k, and of course we shall have – 0.144 + 0.000893 = 0.143107 for the new value of h very nearly. Substituting this value of h in the expression for k, we get = 0.00000286 for the new value of k, and thence 0.143107 +0.00000286 = -0.14310414 is the new value of h; and adding this to a = 1.5 we get 1.5 0.14310414 = 1.35689586, which is correct in all its figures, for the sought root.

We will now proceed to find the remaining root of the given equation; thus, because the coefficients of h2 and h3 in the numerator of k are large in comparison to the coefficient of h and 1, if we regard h as being somewhat great, we may clearly assume Sh3 + 36h2 = 0 or h

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=

- 1-2h
2+72h+24h2

9

; conse

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-0.05, which is numerically very small in comparison to h, as it ought to be.

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Hence, we shall have a +h+k 1.5-4.5 — 0.05 = — 3.05 nearly, for an approximate value of the sought root. We will now proceed to find the remaining figures of the root, after the manner of Horner, as exhibited in the scheme (E), Ex. 4, p. 115.

Thus, writing the coefficients in their order, and supplying that of the deficient term by 0, we have the following work, which we hope will be easily understood from the examples given at the page cited (above), etc.

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If one or more of the first figures of a root of an equation are known, it is clear from this example that the elegant method of Horner enables us to find the remaining figures with great facility.

The reason for our method of extracting the root is evident from the consideration that it consists in a succession of transformations of the given equation, united in one scheme, by putting down only the coefficients with their proper signs in the successive transformations.

Thus, having found 3 for the first part of the root, we put y 3 for x in the given equation, and it will become y3-9y2+20y+10, where the coefficients agree with those in the scheme. To get the next transformation, we

observe that, on account of the minuteness of y, we may take 20y+ 1 = 0 for the preceding equation nearly, which gives y = - = 0.05 nearly; but as this will be found

1

20

to be numerically too great, we put y=0.04.

To get the second transformation, we put y' -0.04 for y in the preceding transformed equation, and thence get y's 9.12y' + 20.7248y' 0.814464 + 1 = y3 — 9.12y' + 20.7248y' +0.185536 = 0, as required. Dividing -0.185536 by 20.7248, we get — 0.008 for the quotient; then, putting y' — 0.008 for y' in the preceding transformed equation, we proceed to get the third transformed equation, and so on, until a sufficient number of the figures of the root have been found. It may be noticed, that after four transformations, we have obtained the last four figures of the root, by divid ing -0.000153305180029 by 20.8875565643 the coefficient of y'"; and it is clear that we may proceed in a similar way in all cases, after more than half of the number of decimal places in the root have been found.

4th. If a stands for a root of the equation X = 0, and a+h for any value of the independent variable x, it is clear from (3) (given at p. 229), that the value of X which corresponds to a + h, will be expressed by X = Ah + A1h2 + A etc. Because the right member of this equation is exactly divisible by haa, it follows that X must be X exactly divisible by x-a, or that the quotient

=

a

must.

be exact. Hence, because a may represent any root of the equation X 0, it follows that, if we subtract the roots of the equation separately from the independent variable x, the remainders will be factors of X.

Since the quotient

X

x a

is of less dimensions in x by a

unit than in X, and that it contains all the roots of the equation X = 0 except a, it follows that, having found the root a, of the equation X = 0, we may find the remaining roots from

the more simple equation

X

x a
a

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Thus, because the equation - 3x2 — x + 3 = 0 is satis

fied by putting 1 for x, we get

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3x2

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x

x + 3 1

= x2 + 4x

+3; and since the equation a2 + 4x + 3 = 0 is satisfied by x2+4x+3 c + 1

putting 1 for x, we get

= 3; and by putting -3 for x, the equation +30 is also satisfied; consequently, 1, -1, and 3 are the roots of the equation; and we shall have a3-3x2-x+3=(x-1)x(x+1) × (x+3). Again, if a is a root of the equations X = 0, and A = 0, then the above general value of X will become X = A1h2+ Ah etc.; and because the right member of this equation is exactly divisible by h2 = (x - a), it follows that the quoX tient must be exact. (x-a)3

Hence, because X is exactly divisible by (x − a)2=(x—a). (x − a), the equation X = 0 has two roots, whose common value is a.

=

If a is a root of the equations X 0, A = 0, A1 = 0, A, 0, etc., it may be shown, in like manner, that the equation X = 0 has as many equal roots whose common value is a, as there are successive equations which have a for a common root.

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Thus, if X is represented by — x3-3x2 + 5x — 2, and 1+h is put for x, the function is reduced to 3h3 + h1, which shows that the equation — x3-3x2 + 5x-2=0 has three equal roots, whose common value is 1; and since — 2 is the remaining root of the equation, we get a1 — x3-3x2 +5 − 2 = ( − 1)3 x ( + 2).

Similarly, the equation + 2x + x3 = (x + 1)2 × x3 = 0 has three equal roots, whose common value is 0; and two other equal roots, whose common value is - 1.

5th. If we represent an equation of the nth degree in which the coefficients and the absolute term are real positive or negative (numbers or) quantities, by Ax" + A∞2¬1 + A ̧ï¤ñ−з + Azan−3 + . . . . + An-1o + A=0; then, if the roots of the equation are all real, there will be n roots, or the number of roots equals the degree of the equation.

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For if a, b, c, d, etc., represent the roots, then, since x-α, x — b, x— c, x-d, etc. (as has been shown), must be factors of the first member of the equation, it is clear that there must be n such factors, or that our proposition is true; and we shall have Ax2 + A1-1+ An- + Agen-3 + .... +

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