The first line contains the powers of 1; the second line contains the coefficients, each of which is derived from the preceding one by multiplying by an appropriate factor; the third line contains the powers of (-2). The three parts of each term 4 lying in the same vertical column are multiplied together. Note especially the simple method of getting the coefficients; the first is n; multiplied by 2-1, it gives the second; this prod -2 uct multiplied in turn by 22 gives the third; this product 3 gives the fourth, and so on; that is, the first coeffi 9 cient is -;-(—,5) = 15; 15 (—2?) — — 18; —18 (-4) 2) (. 2 7 2 2 When such an expansion as this is applied to numerical computation, this use of multipliers can be carried a step further to advantage. We note that any complete term can be got by multiplying the preceding term by a simple factor; these multipliers are, for the expansion of (1+x)": To get the second term from the first: To get the fourth term from the third: and so on. n-1 2 nx, x, n = 2x, 3 1 Example: Find to eight decimals the value of √(3.9)* 2 Using the notation T, for first term, T2 for second term, T3 for third term, etc., we have T1=1, T2=nx=(-1) (−26) = 88%, T3= 2 1 whole computation is conveniently arranged as follows: 3 =80, T1= 32 T2, etc. The The work has been carried to nine decimals, that the eighth place may be correct. In an example of this sort it becomes evident after a few multipliers have been computed that the others will follow according to some simple rule; here the coefficients of. in the multipliers have successive odd numbers for numerators and successive even numbers for denominators. Find the value of each of the following to the number of decimal places indicated, carrying the work in each case one place further than is required in the result. 8. 1 T√ (130) 3 Ans. 1.7320508. Ans. 4.09818507. Ans. 1.2599210499. =(1+4)*(seven places). Ans. 0.1241722. 56. The Expansion of (a + bi)".—In the expansion of (a+bi)", where a and b are real and i=V-1, we have only to note that i=1, i3——i, i = 1, i, etc., the powers repeating in cycles of four. As the first term contains io, the third i2 and so on, every odd-numbered term is real, with signs alternately + and -; the even-numbered terms in the same way contain odd powers of i, equal to i, -i, i, -i alternately. Hence, collecting separately the real and imaginary parts of the expansion: Thus (a+bi)"=P+Qi, where P and Q are real. In the expansion of (a-bi)", the only difference is that b of (a+bi)" is replaced by -b. This change does not affect P, as only even powers of b occur in P, and merely changes the sign of Q, as only odd powers occur in Q; hence From this it follows that if we have any polynomial, f(x), with real coefficients, and substitute in it (a+bi) and (a−bi) in turn, after collecting the real and pure imaginary parts we shall have, if f(a+bi)=p +qi, f(a−bi) =p—qi, the two results being conjugate. If f(x) is a polynomial with real coefficients, f(a+bi)=p+qi, f(a−bi)=p—qi, where Ρ and q are real. 57. The Expansion of (a+b√c)”.—If a and b are rational, and Ve is irrational, odd powers of Vc are irrational, and even powers rational, so that a proof similar to that of the preceding article shows that (a+b√c)"=P+QVc, where P and Q are rational. The effect of changing b to b is the same as in the preceding case, so that (a-b√c)n=P-QVC. Also, if f(x) is a polynomial with rational coefficients, and the conjugate quadratic surds (a+b√c) and (a−b√c) are put in place of x, the results, f(a+bVc) and f(a-bVc), will be conjugate quadratic surds. 58. The Remainder Theorem.-Consider the cubic function f(x) = 2x3+x2-x+10. If we divide it by (x-2), the quotient will be 2x2+5x+9 and the remainder 28. Again, if we substitute 2 for x, we find f(2)=28. Dividing by (x-1), we get 12 for a remainder, and substituting 1 for x we get f(1) =12. We may try this for any number, and find always that dividing f(x) by (x-a) will leave f(a) for a remainder. This happens as follows: The first division shows that f(x) = 2x3+x2 −x+10= (x−2) (2x2+5x+9)+28. If we put 2 for x throughout, the identity gives f(2) = 28. In the general case, let the division of f(x) by (x-a) be carried out, as it always can be, until a remainder occurs not involving ; i. e., a constant number. Call this remainder R; then f(x) = (x− a)Q(x)+R, Q(x) being the quotient. This relation, true for any value of x, gives for the value a: f(a)=(a− a) Q (a)+R=R. It is evident that f(x) in this proof may stand for any polynomial, so that we have the theorem: If any polynomial, f(x), be divided by (x−a), the remainder found will be f(a). It therefore follows that: If for any polynomial, f(x), f(a) =0, then (x−a) is a factor of f(x); or if a is a root of any algebraic equation, f(x)=0, then (x-a) is a factor of its left member; and conversely: If (x-a) is a factor of any polynomial, f(x), then f(a)=0; or if (x-a) is a factor of the left member of any algebraic equa tion, then a is a root of the equation. |