Binomial Theorem and Logarithms

Now for a first approximation neglect the second and the following terms, and take simply

whence we derive

This value of z, substituted in the second term of (A), gives

_m(m—1)

am 22X

b mam-1

Nm =az

b=mam-1z

+m=

b=mam-1z

mam

In this value substitute the value of b=±(N—am) and it becomes

±2a(N-aTM)

+2a(N-am)

(m—1)b

Finally, substituting this value of z, we have

2a(N-aTM)
(m−1)N+(m+1)am

2maTM+(m—1)(N—am) ̄(m—1)N+(m+1)am°

=3x

N2 ax

24+9

8+27

=ax

With this formula we may, by successive substitutions, approximate

to any degree of accuracy.

For example, find the square root of 8. In the formula make m=2, we have

(m+1)N+(m-1)am
(m−1)N+(m+1)am°

3N+a3
N+3a2

In this let N=8, and substitute as a first approximation, a=3. Then we have

;=3x

33 99

35 35

(24)

=2.829.

Then substituting a=2.829, we have

82.829×

24+8.003241
8+24.009723

which agrees with the root found in Art. 50.

-2.82842,71247,5

In the same way we may approximate very rapidly to any root of a number. The general formula will become for the particular

cases

N-ax 3N+a3 Nax2N+a3

= N+3a2'

5N+3a1
3N+5a4'

N+2a3

3N+2a5

2N+3a5'

N=ax

N3:
Ns=ax

This very convenient method of approximation was first investigated in a general manner by HUTTON. See his "Tracts," Vol. I. p. 45, et seq. A similar method, however, had been before employed in the extraction of the cube root by Mr. J. DODSON, and also by Dr. Halley.

&c.*

52. An important facilitation in the ordinary mode of extracting the square root is also suggested by considering the binomial square. Let N represent the number whose square root is sought, and a an approximate root correct to n places of decimals; and let x be the correction to be added to a, or the difference between a and the true root; then the first significant figure of a must be in the (n+1)th place, and a+x represents the true root of N; that is,

N=(a+x)2=a3+2ax+x2.

Now since the first significant figure of x is in the (n+1)th place, the first significant figure of x is at least in the (2n+1)th place. Hence, if we desire the root to 2n places only we may neglect x3, and the above expression becomes

N=a2+2ax,

* A first approximation may be found to five places of figures with the aid of logarithms, (for which see Chapter VI.,) and by the above methods we may then extend it to any required number of decimals.

whence we have

Here N=8

a=2.82842

That is, having found an approximate root a, subtract its square from the given number, and divide the remainder by 2a for the remaining figures of the root, which will be accurate to twice as many places as there are in a.

To illustrate this, take the above example, viz: to find the square root of 8.

82.82842,71247,

2a=5.65684

N-a-.00004,03036

N-a9
2a

x=.00000,71247

a+x=2.82842,71247 (true in the last place.) 4

EXPONENTIAL EQUATIONS.

CHAPTER V.

53. An exponential equation is one in which the unknown quantity is an exponent. Thus,

ax=b

is an exponential equation from which, a and b being known, the value of x may be found.

It is evident that if b is some exact power or root of a, the value of x is the exponent of this power or root. For example, if a=2, b=8, the equation becomes

2x=8,

in which x is evidently =3, since 23=8.

If a=2, b=

1 1
64'

If a=16, b=2, then 16x=2; from which we find x=

16+=2, or $16-2.

64'

(Art. 20.)

If a= 1000'

x=-13, for (1100)=(1000)}_1

then 2x=. from which we find x=-6, for

64'

b=10, then (1000) *

=10.

since

=10; from which we find

54. When b is not an exact power or root of a, the solution of the equation is not so readily effected. There are methods, however, of approximating to the value of x as nearly as we please. The following is that of Lagrange. Take the equation

10=500.

Now 102-100 and 103=1000; that is, 102 is too small and 10s is too great. Therefore the value of x is between 2 and 3, and is