n 46. Expansion of binomials with negative fractional exponents. р To obtain a formula for this case we have only to make negative in formula (18) obtained in the last article. This will be effected by making either p or n negative; therefore for n substitute -n, and we have In these formulæ ve may express the denominators as in (21). 3 24 3.5 26 3.5.7 XS 1 X2 =c1+ 4. )= ربه) 20+2.4***+2.4.6. c*+2.4.6.6. 2+ &c.) 47. Involution of polynomials. For example, to expand (x+y+z). In formula (15) make a=x+y, b=z, m=3, (w+y+z)=(x+y): +3(3+y)<z +3(x+y)zo +28. Then expanding (x+y) &c. (x+y)s=2 + 3x4y + 3xy? +yo, 28=23; The sum of these equations gives (x+y+z)=x+3x2y+3xy®+y8 + 3xʻz +6xyz +3y2z+3x22+ 3yz+ 23. To expand (a+b+c-d)*, we have by our formula (x+y) *** +4x®y +62'y + 4xy+y* in which make x=u+b, y=Cd; we shall find (a+b+c-d)= q* + 4a'b + 6a'b% + 4abs + 6+ + 4ac +12a2bc +12abc + 4b3c-4a’d—12a2bd-12ab'd—463d + ha?c + 12abc + 66%c% - 12a2cd 24abcd -125%cd +6a’da +12abd? +66%da +4ac3_12aced +12acd-4ad +4bc8-12bcd + 12bcd-4bd8 +64-4c9d+6cd24cd8 +d4. EXAMPLES. 1. (a+by+cya)3=a2 + 3a%by+ (3a%c + 3ab?)y® +(6abc + 63)ys +(3ac+36°C)y* +3bc%y$ +cRye. 2. (1-x+ 20%) 6 =1-62 +21.x2–50x3 +90x+ —126x5 +14126 -126.x"+90x9-5029+21210-6x11+x19. 3. (a® +ab+abs +68)4 = 212 +4a116 +10a1063 + 20a9b3 +31a864 + 40a7b5 + 442676 + 40a5b7 + 31a+b8 + 20a369 +10a%b10 + 4ab11 + 613. EXTRACTION OF ANY ROOT OF A NUMBER. 48. Any root of a number may be found by the binomial theorem by dividing the number into two parts and considering it as a binomial. Thus, to find the 3d root of 28, divide it into 27 and 1; then 1 3:28 =(27+18=27=(1+3)*=3(1+3)*: Now to develop (1+3+)*, in (20) Art. 45, make <= (77) + (3) - &c. + n=3, 27 + 1 27 1 1 =l+ 3 27 2 2.3 2.5 2.3.33 . 1 1 5 =1+ + &c. 81 6561 1594323 6641 The first three terms of this series are equal to or 1.0122 6561" nearly. The fourth term is less than .00001 ; so that if we wish to find the root to four places of decimals only, this term, as well as all the succeeding terms, may be neglected. Therefore we have We divided 28 into two such parts that one of them was an ex. act power of the third degree, and also the greatest contained in 28. The general method of proceeding is expressed in the following rule: To find the nth root of any number, first find by trial the nearest integral root (a); divide the given number into two parts, one of which is the nih power of (a); and consider these two parts as the terms of a binomial which may be developed by the binomial theorem. 49. Sometimes it is expedient to use the power next greater than the given number. Thus, to find the fifth root of 30 we may put 5 =(32—2)3=32+(1–2)=2(1–13)* (taking three terms of the series,) 3159 =2X =1.9744 to within .0001. 3200 The greatest integral power of the 5th degree less than 30 is 1, which would lead to a divergent series. 50. If the root is desired to a greater number of places, a greater number of terms of the series must be taken. By the following method, however, we may approximate to any degree of accuracy with the use of only the first two terms. Thus, to find the square root of 8, we have 1 1 1 &c. 2.22 (taking the first two terms,) 17 17 18 6 Taking 2.83 as a first approximation, we now divide 8 into two parts, one of which is the square of 2.83. We then have 89 V8=[(2.83):— 0.0089]}=2.83(1 1 89 =2.83(1 &c.) 2 ° (283):) =2.83-0.001572=2.828428, which is correct with. 1 in 0.000001 or 1069 4951184 Taking 2.828428 as a second approximation, we shall find V8=[(2.828428) 2—0.000004951184]1. (2828428) 4951184 =2.828428 (1– 2 (2828428) 4951184 =2.8284284 2 X2828428X1000000 1 =2.82842840.000000875254 =2.82842,71247,46, which is correct in the last 1 place ; that is, within 10139 2.828428x2x.22* It is readily shown that the third term will not affect the 12th place of decimals. This term will be 1 (4951184) (2628428)* which may be expressed approximately thus, 3 (5,000,000) 3x(5x106) 75 1 X 10^(8,000,000,000,000) 210X(8x1012)264x1013 1012° 51. The approximation may be effected also in the following manner.* Suppose the mth root of any number N is sought. Having found an approximate root a, let b be the difference between am and N, or N=am Eb 1 Then Nm will be equal to a, plus or minus some number which call z; that is, let Nm=a+z, or N=(a+z)". m(m—1) 2 am-22+ &c. or (A) * Francæur, Algèbre, no. 488. |