and since all the succeeding terms also contain the same factor, they all become 0. There will, therefore, remain m+1 terms only. The literal part of the (m+1)th term will be am_mbm=aobm=bm, (Art. 26,) and the coefficient will be m(m-1)(m-2)......... (m—m—2))(m—m-1)) (m-2)(m-1)m or reversing the order of the factors in the denominator mm-1)(m-2).........2.1 -1. m(m-1)(m—2). 2 Hence the (m+1)th term, or last term, is simply bm, which is analogous to the first term am. It might be shown in the same manner that the term preceding the last is mabm—1, and in general, that the terms equally remote from the first and last have equal coefficients. But this appears more readily by taking the development of (b+a)", which is of course identical with that of (a+b)m. We have at once (b+a)"=m+mbm-a+ =?q 2 37. When m is not a positive integer it is evident that no one of the factors m, m-1, m-2, m-3, &c. can be equal to 0); so that the development will, in that case, be an infinite series.* * This celebrated theorem is generally ascribed to Newton, but Hutton, in his History of Logarithms, shows that it was used in a particular form by several mathematicians before the time of Newton. He mentions Lucas de Burgo, who, as early as 1470, extracted the cube root by the coefficients of the power of a binomial, but who did not go to any higher roots. Hutton proceeds ; “And this is the first mention I have seen made of this law of the coefficients of the powers of a binomial, commonly called Sir Isaac Newton's binomial theorem, though it is very evident that Sir Isaac was not the first inventor of it: the part of it properly belonging to him seems to be only the extending it to fractional indices, which was indeed an immediate effect of the general method of denoting all roots like powers with fractional exponents, the theorem being not at all altered. However, it appears that our author Briggs was the first who taught the rule for generating the coefficients of the terms successively one from CHAPTER IV. APPLICATION OF THE BINOMIAL THEOREM. 38. The most simple application of the binomial theorem is the expansion of a binomial with a positive integral exponent. To take the most simple case, let us find the development of (a+b). Making m=2 in (11), we have ) (a+b)=a? +2ab+ a+ 2 2.3 2.1 Here the coefficient of the 3d term is or 1, and ao=1; there. 2 fore, this term is simply bạ. The coefficient of the 4th term is 2.1.0 2 -X0=0; therefore, this term is equal to 0, and all the 2.3 6 succeeding terms are equal to 0, since they all contain the factor 2-2 or 0. Therefore, our development contains but three terms, or (a+b)=a+2ab+b?, which agrees with the result obtained by the actual multiplication of (a+b)(a+b). The 3d power of a +b will be found from the formula in the same manner; by making m=3, we shall find (a+b)=a+3aRb+3ab +6%. The 5th term, and all the succeeding terms vanish, and there remain only four terms, or m+1 terms, as was shown in Art. 36. another, of any power of a binomial, independent of those of any other power.” It seems, nevertheless, that Newton was not acquainted with what had been done by Briggs. In another place Hutton makes the following remark: “But I do not wonder that Briggs' remark was unknown to Newton, who owed almost every thing to genius and deep meditation, but very little to reading : and I have no doubt that he made the discovery himself, without any light from Briggs, and that he thought it was new for all powers in general, as it was indeed for roots and quanti. ties with fractional and irrational exponents.”—Hutton's Introduclion to Mathematical Tables. and since all the succeeding terms also contain the same factor, they all become 0. There will, therefore, remain m+1 terms only. The literal part of the (m+1)th term will be am-mbm=aobm=bm, (Art. 26,) and the coefficient will be m(m-1)(m—2).. (m-(m—2))(m—(m1)) 2 3 4 (m-2)(m-1)m or reversing the order of the factors in the denominator m(m-1)(m—2). 2 ==1. m(m—1)(m—2). 2 Hence the (m+1)th term, or last term, is simply bm, which is analogous to the first term am. It might be shown in the same manner that the term preceding the last is mabm-, and in general, that the terms equally remote from the first and last have equal coefficients. But this appears more readily by taking the development of (6+a), which is of course identical with that of (a+b)m. We have at once - 2 37. When m is not a positive integer it is evident that no one of the factors m, m—1, m-2, m-3, &c. can be equal to 0); so that the development will, in that case, be an infinite series.* to This celebrated theorem is generally ascribed to Newton, but Hutton, in his History of Logarithms, shows that it was used in a particular form by several mathematicians before the time of Newton. He mentions Lucas de Burgo, who, as early as 1470, extracted the cube root by the coefficients of the power of a binomial, but who did not go any higher roots. Hutton proceeds ; “And this is the first mention I have seen made of this law of the coefficients of the powers of a binomial, commonly called Sir Isaac Newton's binomial theorem, though it is very evident that Sir Isaac was not the first inventor of it: the part of it properly belonging to him seems to be only the extending it to fractional indices, which was indeed an immediate effect of the general method of denoting all roots like powers with fractional exponents, the theorem being not at all altered. However, it appears that our author Briggs was the first who taught the rule for generating the coefficients of the terms successively one from CHAPTER IV. A PPLICATION OF THE BINOMIAL THEOREM. 2 38. The most simple application of the binomial theorem is the expansion of a binomial with a positive integral exponent. To take the most simple case, let us find the development of (a+b). Making m=2 in (11), we have 2(2-1)(242) (a+b)2=a+2ab + a-138 + &c. 2 2 . 3 2.1 Here the coefficient of the 3d term is or 1, and ao=1; there 2 fore, this term is simply bo. The coefficient of the 4th term is 2.1.0 X0=0; therefore, this term is equal to 0, and all the 2.3 6 succeeding terms are equal to 0, since they all contain the factor 2-2 or 0. Therefore, our development contains but three terms, or (a+b)2=a2 + 2ab +63, which agrees with the result obtained by the actual multiplication of (a+b)(a+b). The 3d power of a +b will be found from the formula in the same manner; by making m=3, we shall find (a+b=a+3a'b +3ab+6%. The 5th term, and all the succeeding terms vanish, and there remain only four terms, or m+1 terms, as was shown in Art. 36. another, of any power of a binomial, independent of those of any other power.” It seems, nevertheless, that Newton was not acquainted with what had been done by Briggs. In another place Hutton makes the following remark: “But I do not wonder that Briggs' remark was unknown to Newton, who owed almost every thing to genius and deep meditation, but very little to reading: and I have no doubt that he made the discovery himself, without any light from Briggs, and that he thought it was new for all powers in general, as it was indeed for roots and quantities with fractional and irrational exponents.”—Hutton's Introduclion to Mathematical Tables. 39. To find the 6th power of a +b, in (11) make m=6; then (a+b)=a +6ab+15a+ba+20a83+ &c. Here we have found the first half of the coefficients, but it is unnecessary to find the remaining ones by the formula, since, as was shown in Art. 36, the coefficients of terms equally distant from the first and last terms are equal to each other. Therefore the remaining coefficients will be the same as those already found, but taken in the reverse order. Hence the complete expansion is (a+b)=a +6a+b+15a*b*+20a%b3+15a%b4+6ab5 +66. The literal part of the expression is easily put down, observing in all cases that the exponents of a commence at the highest in the first term, and decrease regularly by 1, and that those of b commence in the second term with 1, and increase regularly by 1 to the highest. The expansion of (a−b)is found in the same manner from (14) to be (a−b)=q8_6a5b+15a*b2—20a3b3 +150%b46ab5 +66. for 40. Any binomial may be involved by substituting in our formulæ, a, b and m. For example, required the 5th power of 2x-3y. In (14) substitute for a, b and m the values a=2x, b=3y, m=5, and we obtain (2x^3y)5=(2x)5—5(2x)*(3y)+10(2x)?(3y)—10(2x)?(3y)8+ 5(2x)(3y)-(3y), EXAMPLES. 1. Find the 7th power of 1–2y. In (13) put z=2y, m=7, then (1+y)=1+14y+84yo+280y + 560y4+672y +448y6128y. 2. Find the 4th power of 2+4xy. We have (2+4xy)=24(1+2xy)=16(1+2xy)". (1+2xy)=1+8xy + 24xʻy+32x'y8 +16x4y*, 512x*y3 + 256x*y*. |