and equating the coefficients of the like powers of x, we have 2 as = (A+B) a, and A B = 0; .'. A = B, and A + B = 2 a, or 2 B =2a; hence B a + x a -x a - X = Ans. + + + + etc. a2 a3 a1 BINOMIAL THEOREM. 136. By actual multiplication we have (1+x)2 = 1+2x + x2; (1+x)3 = 1 + 3x + 3x2 + x3 ; (1 + x)* = 1 + 4x + etc., and hence generally we see that the form of the expansion of (1 + x)”, when n is a positive integer, is 1 + n x + A x2 + Вx3 + Сx+ where the coefficient of the second term is n, and the coefficients A, B, C, etc., are entirely independent of x. But if the exponent be a positive fraction, as m (1 + x)" = (1 + x)TM = / (1 + mx + A'x2 + B'x3 + C'x+...). And since (1+y)" = 1+ny + Ay2+ By + Cy+..., y = ax + bx2 + cx2 + the nth root of 1+mx+Px2+Q x3 + is of the form ... Again, if the exponent be negative, as — n, where n may be either integer or fractional, then we have 1 (1 + x)" 1 + n x + A x2+Bx3 +Сx* + . = 1 nx + P' x2 + Q'203 +... ... by actual division; therefore generally the form of the expansion of 1 + n x + A x2 + B x3 +С x2 + . . . (1). (1 + x)" =3 Now if we substitute x + y for x in equation (1) we have (1 + x + y)” = 1 + n (x + y) + A (x + y)2 + B (x + y)3 + • Bx3 + Cx+...) = A x2 + 1 + nx + + But (1+x+y)" = {(1 + y) +x}" = (1 + y)" {1+ +nx+n(n-1) xy + + A x2 (1 + y)"−2+ Bx3 (1 + y)" — : +Cx*(1+ y)*~* + • • Cy+...) By3 + n A, xy+ n B, xy+... A x2+A (n-2) x y + A A x y +.....(3), where A, B, etc., A2, B2, etc., are what A, B, etc. become when n is changed into n 1, n 2, etc. Now the series (2) and (3) must be identical; hence equating the coefficients of the same powers, or combinations of powers, of x and y, in both, we get sequently the coefficients of x y and xy in (3) are identical, each being Finding now the values of B, and A,, we can find the coefficients of xy3, x'y', and x3y in (3) to be respectively n (n − 1) (n − 2) (n − 3) n (n − 1) (n − 2) (n − 3) and Comparing either of these with the coefficients 4c, 6 c, 4 c of the corresponding powers of x and y in C= (2), we get In this manner the law of the coefficients can be obtained, and we have n (n − 1) n (n − 1) (n − 2) 3 + ... (4) ; = 1 + n + a 1.2 n (n − 1) x2 1.2 a2 + n (n 1 1.2.3 =a" + na"-1x + n (n − 1) (n − 2) 2.3 1.2.3 2 By a little consideration, it will be found that the pth term of the 1. Expand (a — x) in a series by the binomial theorem. ( a − x ) = {a ( 1 − Z ) } 1 = a * (1 − 2), and : 1 x \ α a + .:.(a− x)*=a3 {1 1 2 - a with (1+x)", we have x = 1 a 2.4 a 1 x 1 1.3 etc.} 2. Let it be required to expand (a2 + x2)2 in a series. 4. Find the fifth term of the expansion of (a — x3) ̄‡. + etc. 1155 Ans. 2048 EXPONENTIAL THEOREM. 137. Let it be required to develop a in a series. 1 = (a− 1) - ( a* Now = = 1 1 ..... ... .... (a − 1)' + ¦ 3 1+Ax+Bx2 + Сx3 + Dx+ + z for x gives 1 + A (x+≈) + B (x + 2)2 + C (x + z)3 + etc. + Az +2B x z + 3 C x2 z+4 Dx3z+etc. = (1+Ax+B+Cx3+...)× (1+Az+B22+C**+...), or a*+: = 1+ Ax+B x2+С x3+D x + eto. + Az+A2 x z +ABx2z + AC+ etc. The theorem (3) will find its application presently in the calculation of logarithms. NATURE AND PROPERTIES OF LOGARITHMS. 138. We have already seen in the Arithmetic that logarithms are a series of numbers in arithmetical progression corresponding to another series in geometrical progression. Thus in the two series 0, 1, 1, 2, 3, 4, 5, 6, etc. 10, 100, 1000, 10000, 100000, 1000000, etc. the logarithm of 1 is 0, of 10 is 1, of 100 is 2, of 1000 is 3, and so on. But the best method of considering logarithms is derived from the following definition: A logarithm of a number is the index of the power to which a given quantity must be raised so as to be equal to that number. Thus in the equation a* = n, x is the logarithm of n to the base a, and is usually written x = logan. A system of logarithms is a series of values of x corresponding to different values of n, the base a remaining the same; but since a may have different values, it is obvious that there may be as many systems of logarithms as we please. In every system of logarithms the logarithm of its base is 1; for a' = a. In every system of logarithms the logarithm of 1 is 0; for a° 1. Properties of Logarithms. = 1. The logarithm of the product of any number of factors is equal to the sum of the logarithms of those factors. Let But a = P, a" = Q, a = R and a S; then we have = P Q R S; ..log. PQRS=x+y+z+ v = loge P+ loga Q + loga R + loga S. 2. The logarithm of a fractional quantity is equal to the excess of the logarithm of the numerator above the logarithm of the denominator. Let a* = N and a = N'; then x = log, N, y = loga N'. |