Now for a first approximation neglect the second and the following terms, and take simply whence we derive z= This value of z, substituted in the second term of (A), gives _m(m—1) am 22X b mam-1 2 Nm =az b=mam-1z +m= b=mam-1z b 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) 2maTM+(m—1)(N—am) ̄(m—1)N+(m+1)am° = =3x I 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 3N+a3 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 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 N+2a3 3N+2a5 2N+3a5' N=ax N3: 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. x= 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 x=.00000,71247 43 39 a+x=2.82842,71247 (true in the last place.) 4 44 EXPONENTIAL EQUATIONS. 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 = 26 1 If a=16, b=2, then 16x=2; from which we find x= 16+=2, or $16-2. 64' (Art. 20.) 1 If a= 1000' 1 x=-13, for (1100)=(1000)}_1 3' 1 then 2x=. from which we find x=-6, for 64' b=10, then (1000) * 4 =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 ; that is, let equal to 2+ some fraction. Represent this fractional part of x by 1 a' then our equation becomes EXPONENTIAL EQUATIONS. or (Arts. 5 and 21,) whence we 500 102 Involving to the power denoted by x', this becomes 10=5', then x=2+ > in which we find 2+ 10** x'=500, 1 103× 10x'=500, which is another exponential equation, from which we are to find the value of x'. Now 51=5<10, and 5o=25>10. Therefore x' is between 1 and 2. Let then we have in the same manner as above 1 51+=== =10, or 5×5"=10, or 5"=-2, or 5=2x", from which we are to find x". therefore x is between 2 and 3. We have 22-4<5, 23=8>5; Let 1 =2, 5 22 125 128 5 4 512 7256 or 2; or |