Ex. 4. Given the series y=x+x3 +x2+x2+x2+, etc., to find the value of x in terms of y. Ans. x=y-y3+2y3—5y2+14y9—, etc. Ex. 5. Given the series y=x+3x2+5x3+7x2+9×3+, etc., to find the value of x in terms of y. Ans. x=y-3y2+13y3—67y*+381y5—, 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. Ex. 1. Given x+x2+z ̧x3+384x2+3840x5+, etc.=1, to find the value of x. If we calls the sum of the series, and proceed as in the last article, we shall have x=2s-s2+fs3-1s+3s-, etc. Substituting the value of s, we find x=1—18+38-312+2560−, etc., or x=0.446354 nearly. Ex. 2. Given 2x+3x3+4x3+5x+, etc., to find the value S 3s3 19s5 152s' + of x. or or Ans. x= 2 16 128 1024+, etc., 19 19 x=1-138+4086-1384+, etc.-0.2300 nearly. Ex. 3. Given x-3++, etc., to find the value of x. 5 7 x=3+31+3645+68+, etc.=.34625 nearly. Ex. 4. Given x++ + ++, etc., to find the value 2 3 4 or or 5 6 24 120 720+, etc., x=3—30+730-16000+375000-11260000+, 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 an-bn 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 ton; 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 a^-1, and, since there are n terms in the quotient, this quotient reduces to nan-1. n=2, Second. Suppose n to be a positive fraction, or n= p and q are positive whole numbers. Let Also, let bay, whence ba=yo, and b=y¶. Then, substituting, we have where 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 Substituting for x its value a, the fraction reduces to Third. Suppose n to be negative, and either integral or frac tional; or let n= -m. Then we shall have 1 1 an-bn a-m_b-m am &m 1 bm-am = a-b = Ɑ- b amfm Ɑ- b Now, when a=b, the first factor of the last expression re duces to 1 am or -a-2m, and the second factor (by one of the preceding cases) reduces to mam-1. Hence the expression becomes α ·×mam-1, or —ma-r 387. It is required to obtain a general formula expressing the value of (x+a)TM, whether m be positive or negative, integral or fractional. a=2(1+2); therefore (x+a)m=zm(1+")". Now x+a=x(1+· If then we obtain the development of (1+2)”, only to multiply it by xm to obtain that of (x+a)m. α Letz; then, to develop (1+2)m, assume х (1+2)=A+Bz+Cz2+Dz3+, etc., we have (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+z)m=1+Bz+Cz2+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)” — (1+n)m — B(z—n)-+C(z2 — n2)+D(z3—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+z)m—(1+n)m_B+C But when z=n, or 1+z=1+-n, the first member of equation ·=z+n, when z=n, becomes 2z. =z2+zn+n2, when z=n, becomes 3z2, etc. (6.) These values substituted in (5) give m(1+2)m-1=B+2Cz+3Dz2+4E≈3+, etc. Multiplying both members of Eq. (6) by 1+z, we have m(1+z)m=B+(2C+B)≈+(3D+2C)≈2+(4E+3D)z3+, 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)x+(3D+2C)22+(4E+3D)23+, 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 B=m. If in this equation we restore the value of z, which is we which is the general formula for the development of any binomial (x+a)m, whatever be the values of x and a, and whether m be positive or negative, integral or fractional; and this formula is known as the Binomial Theorem of Sir Isaac Newton. 388. When the Series is Finite.-The preceding development is a series of an infinite number of terms; but when m is a positive integer, the series will terminate at the (m+1)th term, and all the succeeding terms will become zero. For the second term of Eq. (11) contains the factor m, the third term the factor m-1, the fourth term the factor m-2, and the (m+2)d term contains the factor m-m, or 0, which reduces that term to 0; and since all the succeeding terms also contain the same factor, they also become 0. There will therefore remain only m+1 terms. When m is not a positive integer, it is evident that no one of the factors m, m−1, m—2, m-3, etc., can be equal to 0, so that in that case the development will be an infinite series. 389. Expansion of Binomials with negative integral Exponents. This is effected by substitution in formula (11). 1 Ex. 1. Expand or (a+b)-1 into an infinite series. a+b In (11) let m=-1, and we find the coefficient of the second term is -1, |