the coefficients B, D, F, &c., of the odd powers of x becoming zero. We might, therefore, have assumed ART. 317. EVOLUTION BY INDETERMINATE COEFFICIENTS. Ex. Extract the square root of a2+x2. Assume (a2+x2)=A+Bx+Cx2+Dx3+Ex1+, &c. Squaring both sides so as to obtain quantities of the same form, we find a2+x2=A2+2ABx+2AC x2+2AD|x3+2AEx1+, &c. From which, by equating the corresponding coefficients, we A2=a2, 2AB=0, 2AC+B2= 2AD+2BC=0, &c. get ART. 318. DECOMPOSITION OF RATIONAL FRACTIONS.- Fractions whose denominators can be separated into certain factors, may often be decomposed into other fractions whose denominators shall consist of one or more of these factors. We shall illustrate the method of operation by an example. Decompose 5x14 x2-6x+8 into two other fractions whose denom inators shall be the factors of x2-6x+8. Since x2-6x+8=(x−2)(x—4), (Art. 234, Prop. 2nd), or 5x14 x2-6x+8 __A(x-4)+B(x-2); 5x-14=A(x-4)+B(x−2)=(A+B)x−4A-2B. Now since this equation is true for any value whatever of x, we may equate the coefficients of the corresponding terms, 275 Lame A+B-5; -4A-2B=-14; whence, A=2, and B-3.4 EXAMPLES FOR PRACTICE. By the method of Indeterminate Coefficients show that, 1. • uc. 1+ 24 = A + (13-34) x + (b ~ 3 B) x2 + -8. 1+2x 1+2x 2. =1+3x+4x2+7x+114+18x+, &c. ཉ་ 1-3x+2x2 B-A - 2 AW 214 [x1+, &c. 54 4-31+ &c. 7. √(1+x+x2)=1+2+ i+, &c. like 6/h. 8. √(1+x+x2+x3+, &c.) =1+2+, 4 4.6.2 se 6 8 1 2 Art. 318. 8x4 5 3 10. = + x2-4x+2x-2° 2—1)(x−2) ̄ x1—aa 4a3(x—a) 4a3(x+a) 2a2(x2+a2)* - X-2 x+21 1 1 管 BINOMIAL THEOREM, WHEN THE EXPONENT IS FRACTIONAL OR NEGATIVE. ART. 319. We shall now proceed to prove the truth of the Binomial Theorem generally; that is, to show that (a+b)"=a"+na"-1b+n(n-1) am-262+ an 1.2 n(n-1)(n-2) an-3b3+, &c. 1.2.3 whether n be integral or fractional, positive or negative. .. (a+b)"a" Hence, if we can find the law of the expansion of (1+x)”, we may obtain that of (a+b)", by writing for x, and multiplying b a by an. We shall, therefore, first prove that (1+r)”=1++*(n−1)+ 1.2 n(n−1)(n−2)x3+, &c. 1.2.3 The proof may be divided into two parts: 1st. To show that (1+x)"=1+nx+, &c. 2nd. To find the general law of the coefficients. FIRST. To prove that the coëfficient of the second term of the expansion of (1+x)" is n, whether n be integral or fractional, positive or negative. Let the index be positive and integral; then, since by multiplication we know that (1+x)=1+2x+, &c., (1+x)3=1+3x+, &c. ; let us assume that (1+x)"1=1+(n−1)x+, &c. or, Multiply both sides of this equality by 1+x, then (1+x)”-1(1+x)={1+(n−1)x+, &c.)}(1+x); Hence, if the proposition is true for any one index n—1, it it will be true for the next higher index n. Now, by multiplication, it is true for the index 3, it is therefore true for the index 3+1=4; and therefore true for the index 4+1=5, and so on. D Hence, by continued induction, it is always true for n when it is integral and positive. Also, let (1+x)=1+ax+, &c., =1+Ax, where Ax is put to represent all the terms after the first. By equating the coëfficients of the like powers of x (Art. 314), .. (1+x)"=1+nx+, &c., whatever be the value of n. .. (a+b)"=a" (1+o ) "=a"(1+no+, &c.), a =a"+na"-'b+, &c., and the first two terms of the series are determined. SECOND. To find the general law of the coefficients. Let (1+x)=1+nx+Вx2+Сx3+Dx1+, &c., where B, C, D, &c., depend upon n. For x, put xz, and consider (x+2) as one term, then {1+(x+2)}" =1+n(x+2)+B(x+2)2+C(x+2)3+, &c... But (a+b)"=an+na”-1b+, &c. ; ••• (x+2)2=x2+2xz+, &c.; (x+2)=x+4x3z+, &c. ; .. {1+(x+2)}"=1+nx+Bx2+Cx3+D+, &c., +(n+2Bx+3Cx2+4Dx3+, &c.)z+; &c., =(1+x)"+(n+2Bx+3Cx2+4Dx3+, &c.)z+, &c., (A). But, considering (1+x) as one term, (1+x+2)"={(1+x)+z}"; and {(1+x)+2}”=(1+x)"+n(1+x)n ̄1z+, &c. Equating the coefficients of z in (A) and (B), n+2Bx+3Cx2+4Dx3+, &c., =n(1+x)n-1 ; multiplying both sides by 1+x, we have n+2Bx+3Cx2+4Dx3+, &c.) =n(1+x)" (B). +nx+2Bx2+3 Cx3+, &c.) =n(1+nx+Bx2+Сx3+, &c). By equating the coëfficients of the same powers of x, we have 2B+n=n2.. 2B-n2-n=n(n-1). Hence, generally, if N is any coëfficient, M the one which next precedes it, and r-1 the largest factor in the denominator of M, _M{n—(r—1)}__M(n+1—r) N= we have g •.(1+)"=1+n + (n−1), b 1.2 1.2.3 and .•. putting - for x, (a+b)"=a" (1+0)", α _b__n(n−1)b2___n(n−1)(n—2)b3+ &c.,}, =a^{1+n+: α 1.2 a2 1.2.3 a3 =a"+na21b+n(n−1) an-2b2+n(n—1) (n—2) an-3⁄4b3+, &c. 1.2 1.2.3 Ifb be put for b, then since the odd powers of -b are negative (Art. 193) and the even powers positive, (a—b)”—a”—na”-1b+ _n(n—1) an—2p2__n(n−1)(n—2) an-sz3+,&c., 1.2 1.2.3 |