3. Let 4 / 50 be divided by 25. 407. Ans. 2 v 10. Ans. 23/20. Ans. For 5. Ans. 330. Ans. çab. I PROBLEM VIII. To Involve or Raise Surd Quantities to any Power. RAISE both the rational part and the surd part. Or mul. tiply the index of the quantity by the index of the power to which it is to be raised, and to the result annex the power of the rational parts, which will give the power required. EXAMPLES. 1. Required to find the square of zat. First, (*)*=**='s, and (az)==q1x2=a}=a. Therefore (zat)' = iga, is the square required. 2. Required to find the square of ļa). First, = $, and (a})2 = a* = ažja; Therefore (a})} = mia is the square required. 3. Required to find the cube of 6 or X 61, First, (3)} = {x} x { =?, and (61) = 6* = 6/6; Theref. (76)3 = 3 X 6/6 = V 6, the cube required. 4. Required the square of 2/2. Ans. 373. 6. Required the 3d power of į v 3. Ans. V3. 7. Required to find the 4th power of įv2. . 8. Required Ans. 434. Ans. 4 8. Required to find the mth power" of a". PROBLEM IX. To Evolve or. Extract the Roots of Surd Quantities*. EXTRACT both the rational part and the surd part. Or divide the index of the given quantity by the index of the root to be extracted; then to the result annex the root of .. the rational part, which will give the root required. EXAMPLES. I 1. Required to find the square root of 16V6. First, V 16 = 4, and (63)} =62=2 = 67; theref. (1616)7 = 4.67;= 496, is the sq. root required. 2. Required to find the cube root of 24V3. First, Vas = }, and (73)} = 31=3 = 36; theref. (45 V3}}=.38 = 1V3, is the cube root required. 3. Required the square root of 63. Ans. 66. 4. Required the cube root of a'b. Ans. 1/6. 5. Required the 4th root of 16a". Ans. 2Na. 6. Required to find the mth root of x 7. Required the square root of a’ - 6aVb+96. 62 = 0; * The square root of a binomial or residual surd, a + b, or d-b, may be found thus : Take var then va+b="to +7 i 2 2 ат с c and wa-b = V 2 2 Thus, the square root of 4 + 2V3 =1+V3; and the square root of 6-2V5 =V5-1.. But for the cube, or any higher root, do general rule is known. INFINITE INFINITE SERIES. An Infinite Series is formed either from division, dividing by a compound divisor, or by extracting the root of a compound surd quantity; and is such as, being continued, would run on infinitely, in the manner of a continued decimal fraction. But, by obtaining a few of the first terms, the law of the progression will be manifest; so that the series may thence be continued, without actually performing the whole operation. PROBLEM L To Reduce Fractional Quantities into Infinite Series by Division. Divide the numerator by the denominator, as in common division; then the operation, continued as far as may be thought necessary, will give the infinite series required. EXAMPLES. 2ab a + b 263 264 in + b)2ab .. ( 26 + + &c. a? 2ab + 262 a a 1 1 1 b 6 + a 5. Expand 1 7 into an infinite series. 1 + X Ans. 1-2x + 2x2 – 2x3 + 2x^, &c) a into an infinite series. (a + b) 26 362 Ans. 1 + a? 6. Expand 463 &c. a 1 7. Expand = }, into an infinite series. 1+1 PROBLEM II. To Reduce a Compound Surd into an Infinite Series. EXTRACT the root as in common arithmetic; then the operation, continued as far as may be thought necessary, will give the series required. But this method is chiefly of use in extracting the square root, the operation being too tedious for the higher powers. EXAMPLES. EXAMPLES. 1. Extract the root of a? _2 in an infinite series. x2 r4 76 5.78 a’ – 22 (a &c. 2a 8a3 16a: 128a? a? 2. Expand v1+1=2, into an infinite series. Ans. 1+-++-&c. 3. Expand VI-I into an infinite series. Ans. 1-1-1-1-T &c. 4. Expand va? + x into an infinite series. 5. Expand va? - 26x - xi? to an infinite series. PROBLEM III. To Extract any Root of a Binomial : or to Reduce @ Binomial Surd inta an Infinite Series. This will be done by substituting the particular letters of the binomial,' with their proper signs, in the following general theorem or formula, viz. - 2n (P+PQ) =p + AQ+ BQ+ CO+ &c. 2n and m m m m n n |