383. The reversion of a series is the finding the value of the unknown quantity contained in an infinite series by means of another series involving the powers of some other quantity. This may be accomplished by the method of undetermined coefficients in a mode similar to that employed in Art. 379. Ex. 1. Given the series y=x+x2+x3+, etc., to find the value of x in terms of y. Assume x=Ay+By2+Cy3+Dy1+, etc. Find, by involution, the values of x2, x3, x2, and x3, carrying each result only to the term containing y. Then, substituting these values for x, x2, x3, etc., in the given equation, we shall have y=Ay+B |y2+C y3+D y3+, etc. + A3 Since this is an identical equation, we place the coefficients of the like powers of y in the two members equal to each other, and we obtain A=+1, B=−1, C=+1, D=−1, E=+1, etc. Hence we have x=y—y2+y3—y2+y3—, etc., Ans. Ex. 4. Given the series y=x+x3 +x2+x2 +x9+, etc., to find the value of x in terms of y. Ans. x=y—y3+2y3 — 5y1 +14y3—, etc. Ex. 5. Given the series y=x+3x2+5x3+7x+9x5+, etc., to find the value of x in terms of y. Ans. x=y—3y2+13y3—67y2+381y3—, etc. 384. When the sum of a series is known, we may sometimes obtain the approximate value of the unknown quantity by reverting the series. 48 38 1 Ex. 1. Given x+3x2+1 ̧x2+33 4x2 + 3840x5+, etc.=1, to find the value of x. If we call s the sum of the series, and proceed as in the last article, we shall have x=2s-s2+3s3-1s+255-, etc. Substituting the value of s, we find 96 x=1-16+36-12+2660-, etc., or x=0.446354 nearly. Ex. 2. Given 2x+3x3+4x3+5x+, etc. =1, to find the value of x. S 383 1985 152s7 2 16 128 1024 +, etc., x=1-128 +4096-16384+, etc.=0.2300 nearly. Ans. x= 3 or or 2 3 15 351 17 x = 3 + 3 1 + 36 45+767637+, etc.=.34625 nearly. +36 Ex. 4. Given x+2+3+4++, etc., to find the value of x. or or Ans. x=s 2+624120 720 +, etc., 50 x=3-30+730-13000 + 376000-1120000+, etc., x=0.1812692 nearly. Binomial Theorem. 385. In Art. 353 the binomial theorem was demonstrated for the case in which m is a positive whole number. By means of the method of undetermined coefficients we can prove that this formula is true, whether m is positive or negative, entire or fractional. The demonstration of this theorem depends upon the following proposition: 386. The value of when a=b, is in all cases nan—1 whether n is positive or negative, integral or fractional. First. It was shown in Art. 83 that when n is a positive whole number, a"-b" is exactly divisible by a-b, and the quotient is an-1+an-2b+an-3b2+... +bn-1. The number of terms in this quotient is equal to n; for b is contained in all the terms except the first, and the exponents of b are 1, 2, 3, etc., to n-1, so that the number of terms containing b is n-1, and the whole number of terms is equal to n. Now, when a=b, each term of the above quotient becomes an-1, and, since there are n terms in the quotient, this quotient reduces to nan−1. , Second. Suppose n to be a positive fraction, or n=2, where p and q are positive whole numbers. Let Also, let ba=y, whence ba=yo, and b=y!. Then, substituting, we have XP —YP an bn But p and q are positive integers; therefore, when a=b, and, consequently, x=y, according to case first, the numerator of the last fraction becomes px2-1, and the denominator becomes qx9-1; that is, the fraction reduces to 1 Substituting for x its value a, the fraction reduces to Third. Suppose n to be negative, and either integral or fractional; or let n= m. Then we shall have Now, when a=b, the first factor of the last expression re 1 a2m, duces to or α 2m, and the second factor (by one of the preceding cases) reduces to mam-1. Hence the expression becomes a-2m ×mam-1, or —ma¬m▬1, or nan-1. 387. It is required to obtain a general formula expressing the value of (x+a)m, whether m be positive or negative, integral or fractional. Now x+ɑ=x(1+ а X therefore (x+a)m=xm(1+ α we have If then we obtain the development of (1+2)", only to multiply it by xm to obtain that of (x+a)m. а Let=2; then, to develop (1+2), assume (1+2)=A+Bz+Cz2+Dz3+, etc., (1.) in which A, B, C, D, etc., are coefficients independent of z, and we are to determine their values. Now this equation must be true for any value of z; it must therefore be true when z=0, in which case A=1. Substituting this value of A in Eq. (1), it becomes (1+2)=1+B+C+Dz2+, etc. (2.) Since Eq. (2) is to be true for all values of z, let z=n; then (2) becomes (1+n)m=1+Bn+Cn2+Dn3+, etc. (3.) Subtracting (3) from (2), member from member, we have (1+z)m—(1+n)m—B(z—n)+C(z2 —n2)+D(23—n3)+, etc. (4.) Dividing the first member of (4) by (1+z)−(1+n), and the second by its equal z-n, we have (1+2)m−(1+n)m=B+ But when z=n, or 1+2=1+n, the first member of equation 3 =2+zn+n2, when z=n, becomes 322, etc. These values substituted in (5) give m(1+z)m−1=B+2C≈+3D≈2+4E23+, etc. (6.) Multiplying both members of Eq. (6) by 1+2, we have m(1+2)m=B+(2C+B)≈+(3D+2C)≈2+(4E+3D)1⁄23+, etc. (7.) If we multiply Eq. (2) by m, we have m(1+z)m=m+mBz+mCz2+mDz3+, etc. (8.) The first members of Eq. (7) and (8) are equal; hence their second members are also equal, and we have m+mBz+mCz2+mDz3+, etc.= B+(2C+B)≈+(3D+2C)22+(4E+3D)≈3+, etc. (9.) This equation is an identical equation; that is, it is true for all values of z. Therefore the coefficients of the like powers of z in the two members are equal each to each, and we have Substituting these values in (2), we have (1+z)m=1+mz+ m(m—1), m (m—1) (m—2) 23+, etc. (10.) 2 + 2.3 а If in this equation we restore the value of %, which is a we 3+, etc.; |