which the root may be carried to any length, as in the preceding rule, and it will be the series required. EXAMPLES.-1. Convert x2+z+ to an infinite series. a2-x2 be converted into an infinite series. Ans. a x6 &c. 2a 8a3 16as 3. Change a2+b into an infinite series. Ans. a+ 4. Express 1++ in an infinite series. 13. SIR ISAAC NEWTON'S BINOMIAL THEOREM ". For readily finding the powers and roots of binomial quantities. RULE I. Let P=the first term of any given binomial, Q= the quotient arising from the second term being divided This theorem was first discovered by Sir I. Newton in 1669, and sent (in the above form) in a letter dated June 13th, 1676, to Mr. Oldenburgh, at that time Secretary of the Royal Society, in order that it might be communicated to M. Leibnitz. As early as the beginning of the 16th century, Stifelius and others knew how to determine the integral powers of a binomial, not merely by continued multiplication of the root, but also by means of a table, which Stifelius had formed by addition, wherein were arranged the coefficients of the terms of any power within the limits of the table. Vieta seems also to have m by the first; then will PQ=the second term. Let the index of the power or root required to be found, viz. m=its nume understood the law of the coefficients, but the method of generating them successively one from another, was first taught by Mr. Henry Briggs, Savilian Professor of Geometry at Oxford, about the year 1600: thus the theorem, as far as it relates to powers, appears to have been complete, wanting only the algebraic form; this Newton gave it, and likewise extended its application and use to the extraction of roots of every description, by infinite series, which probably never was thought of before his time. The theorem was obtained at first by induction, and for some time no demonstration of it appears to have been attempted; several mathematicians have however since given demonstrations, of which the following is perhaps the most simple. Let 1+x" =1+px + qx2 + rx 3 + sx4 +, &c. 1+ y)" =1+py+qy2 + ry3 + sya +, &c. } each to n + 1 terms. Then by subtraction 1+x − 1 + y)2 = p.x−y+q.x 2 —y3 +r.x 3 —y3 +; p.x—y+q.x2 —y2 + r.x3 — y3 that is, (by actual division; see the preceding note,) x-y ] + x1 − 1 + 1+ y.l+x \n — 2+, &c. (to n terms) =p+q.x + y + r«x2 + xy + y2 + s.x3 + x2y + xy2+ y3 +, &c. to n terms. ..... Let x=y, then n.1+x^ − 1 =p+2qx+3rx2 + 4 sx3 +, &c. to n terms, whence n.1+x}" = • • • • · p + 2 qx+3rx2 + 4 sx3 +, &c. X 1+x =p+2qx+3rx2 + 4 sx3 +, &c. + px+2qx2 + 3 rx3 +, &c. S 2 } = p+2q+p.x+3r+ 2 q.x2 + 4 s +3 r„x3 +,&c. (A). But because 1+a^=1+ px+qx2+rx3 +, &c. by the above assumption, therefore n.1+x)'=n+ npx+nqx2 +nrx3 +, &c. (B) wherefore the two series A and B (being each equal to n.1+x)") are equal to one another, and consequently the coefficients of the same powers of x will be equal; that is, 1. p=n, 2. 2q+p=np, or 2q+n=n2, ·.· 2 q=n2 —n=n.n—1, and Now since a+b=ax1+ • a+b=a" x 1+ +"=(by substituting rator, nits denominator; then P+PQ will express the given binomial with the index of the required power or root placed over it. II. Let each of the letters A, B, C, D, &c. represent the value of the term in a series, which immediately precedes the term in which that letter stands. III. Then will the root or power of the binomial P+PQ be expressed by the following series, viz. P+ m n m-n AQ+ 2n BQ IV. If the terms and index of any binomial, with their proper signs, be substituted respectively for those in the above general form, then will the series which arises express the power or root required. EXAMPLES.-1. To extract the square root of a2—z2 in an infinite series. 2 Here P=a2, Q=——, and (since is the index of the square root) m=1, n=2; then P+PQa2-z, and P==(a+=) a=the first term .... = A. -.an —2 b2 +, &c. in which, if P be substituted for a", Q for and 2 m A, B, C, &c. for the preceding terms, the series will become P m If the index be a positive whole number, the series will terminate at the +1th term ; but if it be negative, or fractional, the series will not terminate all which is manifest from the above examples. Here y2 √y2+22 'y2+z2 in an infinite series. = y2 × y2 + z 2) − +; we first find y2+z2)—, and then multiply the resulting series by y; wherefore in the m present case P=y3, Q==", m=−1, n=2, and P+PQ " = 1 y'+z") —+, then P= =y")—+=(y−'=) —=—= y the first term.. A. &c. &c. This series multiplied by y3, according to what was pre 4. To involve 12, or its equal 11+1, to the cube. 1 Here P=11,Q== m=3, n=1; then, as before, P+PQ =P+AQ+ BQ+ m m-n n 2n 3 n 33× =)+1, where (since the coefficient of the next term will 11 be 0) the series must evidently terminate. Wherefore collecting the above terms, (1331+363+33+1=) 1728 is the cube of 12, as was required. |