THE HIGHER POWERS OF COMPOUND QUANTITIES. 702. Powers of a greater number of degrees than squares and cubes are now to be considered. We have already explained the method in which they are represented by exponents. It will be found convenient to keep in mind that, in dealing with a compound root, it is inclosed in a parenthesis. For instance, (a + b)+ signifies that a + b is to be raised to the fourth power or degree, and that (a-b)7 expresses the seventh power of a-b; the subject of this section is to explain the nature of these powers, in which some peculiarities will be noticed. 703. If the root or first power be a + b, all the higher powers will be found by multiplying the last power found again by the root, as in the following example: a6+5a5b+10a4bb +10a3b3 + 5a2b4 + abs +ab+ 5a4bb + 10a3b3 +10a2b4 + 5ab5 + b6 (a+b)6 = a + 6a5b+15a4bb + 20a3b3 + 15a2b+ + 6ab5 + b3, &c. 704. In a similar manner are found the powers of the root a-b, the only difference being that the even or 2d, 4th, 6th, &c. terms will be found to be affected by the sign ab+ 5a4bb−10a3b3 +10aab1 — 5ab5 + b6 (a - b)6 = a - Ga3b + 15a1bb — 20a3b3 + 15aab1 – 6ab5 +bo, &c. 705. In this last example all the odd powers of b have the sign while the even powers retain the sign +. The reason is, that the powers of that letter ascend in the following series, b, bb, −b3, +b1, −b5, +b6, &c., which sufficiently indicates that the even powers must be affected by the sign, and the odd ones by the sign. The labour of the calculation being considerable, it is important to find a mode of performing the operation in an abridged manner. Now, if in the powers above determined we take away the numbers, or coefficients preceding each term, we shall observe the following order: first, in each succeeding term the powers of a decrease by unity, whereas the powers of b increase in the same proportion, so that the sum of the exponents of a and b is always the same, and always equal to the power of the exponent required; and, lastly, we find the term b by itself raised to the same power. Hence we know that if the tenth power of b were required, the terms without their coefficients would stand in the following order: a1o, aob, a§b2, a3⁄4b3, ab1, a5b5, a4b6, a3⁄4b7, a2b8, abo, b1o. To determine the coefficients or numbers by which these are to be multiplied, we may observe that, with regard to the first term, its coefficient is always unity; and that, in respect of the second, its coefficient is always the exponent of the power; but the order of the other coefficients is not so manifest, though there is a law by which they are governed, which the following table will show. From which we may see that the tenth power of a+b will be a10+10a%+ 45a8bb +120 a7b3 + 210a6b4 + 252a5b5 + 210a4b6 + 120a3b7 +45aab8 + 10ab9 + b10. 706. The sum of the coefficients in each power is equal to the number 2 raised to the same power, as will be seen by reference to the above table, and they increase from the beginning to the middle, and then decrease in the same order. In the even powers the greatest coefficients are exactly in the middle, but in the odd powers two coefficients equal and greater than the others are found in the middle belonging to the mean terms. shall hereafter touch upon the reason of the following rule for determining the coefficients We in all powers proposed. Let the power proposed be the seventh, then placing the exponent of the power as the numerator, and letting the denominator follow in the natural order of the numbers 1, 2, 3, 4, &c., we have the following fractions, 7, §, §, 4, 4 Now, as the first coefficient is always 1, the first fraction gives the second coefficient, the product of the two first fractions multiplied together gives the third coefficient, the product of the three first fractions represents the fourth coefficient, and so on; thus, for instance, the fifth coefficient will be the product of 7 × × × 1=35, &c. X This rule renders it unnecessary to find the preceding coefficients, and enables us to discover immediately the coefficients which belong to any power; and we can, by its aid, express any power of a+b however high; thus, the hundredth power of a + b, will be (a+b)100 = a100+ xab+ +a9862+ 1x2 a9769 + &c.; from which the law of the preceding terms is evident. 100 x 99 x 98 x 97 a96b4+, X 100 100 x 99 100 x 99 x 98 1x2x3 ON THE TRANSPOSITION OF LETTERS, WHEREON THE LAST RULE RESTS. 707. In the coefficients we have just been considering it will be found that each term is presented as many times as the letters whereof the term consists can be transposed; or, in other words, the coefficient of each term is equal to the number of transpositions that its letters admit. Thus, in the second power the term ba is taken twice, that is to say, its coefficient is 2, for the order of its letters ab or ba can be changed only twice. The term aa, whose letters can undergo no change, is hence only found once. In the third power of a + b the term aab can be written in three different ways, aab, aha, baa, and here the coefficient is 3. In the fourth power the coefficient of a3b must be 4, because aaab admits of four different arrangements, aaab, aaba, abaa, baaa, and so on. It thence becomes desirable to know in how many different ways a given number of different letters may be arranged. Now, beginning with the simplest case, namely, a and b, we see at a glance that only two transpositions, namely ab and ba, can take place. If we have three letters, we see that each of the three may take the first place, while the two others admit of two transpositions. Thus, making a the first letter, we have abc, acb; if b is the first, we have bac, bea; but if e is made the first, we have cab, cba. Hence the whole number of arrangements is 3 x 2 = 6. If four letters, abcd, occur, each may be placed first, and we know the three others are capable of six different arrangements; hence the whole number of transpositions is 4 x 6 = 24, or 4 x 3 x 2 x 1. If the number of letters be five, we have 5 × 24=120, or 5 x 4 × 3 × 2 × 1. Whatever, then, the number of letters, provided they be different, the number of transpositions is easily determined, and, up to the number ten, are subjoined in the following table: : Number of letters. Number of transpositions. 1-1 2x1=2 3×2×1=6 4 × 3 × 2 × 1 = 24 5 x 4 × 3 × 2 × 1 =120 6 x 5 x 4 x 3 x 2x1=720 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 8 x 7 x 6 x 5 x 4 x 3x2x1 = 40320 9 x 8 x 7 x 6 x 5 x 4 × 3 × 2 × 1 = 362880 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800 The numbers, however, in this table can only be used when the letters are all different; and if two or more of them are alike, the number of transpositions becomes much less; as, if the letters were all alike, there could be but one arrangement. Our next object, then, will be to find how the numbers in the table diminish from similarity of letters. We have seen that when two similar letters occur, only one arrangement can be made, consequently the number above found is reduced one half, or must be divided by 2. If these letters are alike, the six transpositions are reduced to one, whence the number in the table must be divided by 6 = 3 × 2 × 1. And, in the same way, when four letters are alike, we must divide the number in the table by 24-4 x 3 x 2 x 1, &c. 708. Thus there is no difficulty in ascertaining the number of transpositions the letters aaabbc may undergo; for, if they were all different, they would admit of 6 × 5 × 4 × 3 × 2 × 1 transpositions. But as a occurs three times, we must divide the number of transpositions by 3 × 2 × 1; and as b occurs twice, we must again divide by 2 × 1; the numbers required, therefore, will be 6x5x4×3×2×1 3x2x1x2x1 =5x4x360. 709. We shall now apply the rule in the example of the seventh power of a + b, or (a+b). The first term is a', which only occurs once; and, as all the other terms have seven letters, the number of transpositions for each term would be 7 × 6 × 5 × 4 × 3 × 2 × 1 if the letters were all different. But the second term ab contains six letters alike, hence x the product last mentioned must be divided by 6 × 5 × 4 × 3 × 2 × 1, whence the coefficient will be 7x6x5x4x3x2x1 5040 7 i or 7. 6x5x4x3x2x1 710. In the third term a5bb, the same letter a occurs five times and the same letter b twice, the total number of letters being seven all through the power. We have here, then, to divide the number which seven transpositions give by 5 x 4 × 3 × 2 × 1, and then by 2 x 1, 7x6x5x4x3x2x1 5040 21 whence we have the coefficient or 21. It will be unneces5x4x3x2x1 x2x1 240 1 sary to proceed with the remaining terms, the mode of finding the coefficient must be From what has been already said we shall find that the above rule enables us to find all the powers of roots consisting of more than two terms. Let us, for instance, apply them to the third power of a+b+c, the terms whereof must be formed by all the possible combinations of the three letters, each term having for its coefficient the number of its transpositions as above. The third power of a+b+c will be found by multiplication to be a3+ Saab + Saac + Sabb + 6abc + Sacc + b3 + 3bb + 3bcc + c3. obvious. 711. Now, suppose a=1, b-1, c-1, the cube of 1+1+1 or of 3 will be I +3 +3 +3 +6+3 +1+3 +3 +1=27, and the rule is thereby confirmed. THE EXPRESSION OF IRRATIONAL POWERS BY INFINITE SERIES. -1, we should have found for the cube of 712. If we had supposed a=1, b=1, and c=— 1 + 1-1, that is, of 1, 1+3−3+3 −6 + 3 + 1 · -3+3-1=1. 713. In subsection 705. we have shown the method of finding any power of a + b. Suppose the exponent undetermined, but expressed by n, we shall have the rule there laid down n-l x 77-2 If the same power of the root a-b were required, we should have only to change the signs of the even terms, and should have + 1an-21 62-1x1 n-2 n-1 n12 a α n-3 R-464, &c. X α 4 714. These formulæ are useful from the facility they afford in expressing all kinds of radicals. It has already been seen that all irrational quantities may assume the form of powers whose exponents are fractional, and that Va=a*; =a*; and Va-a. We have, also, then (a+b)=(a+b)3; Y(a+b)=(a+b)}; and √(a + b) = (a + b)‡, &c. Whence, if the square root of a + b is required, we have only in the general formula to substitute the fraction for the exponent n, and we shall have, first, for the coefficients The square root, then, of a + b may be expressed in the following manner, (a+b)=Va+}0 − x}b0a + }x}x}ơ3@ −}x}x}x}+% 715. Hence, if a be a square number, the value of a may be assigned, and the square root of a + b may also be expressed by an infinite series without any radical sign. Suppose, for 64 -TIS X &c. instance, a=cc, we shall have a=c; then ✓ (cc + b) = c + { × So that there is no number whose square root may not be extracted in the same way; for every number may be resolved into two parts, one whereof is a square, represented by cc. Thus, if we require the square root of 6, make 6=4+2, then cc-4, c=2, b=2, whence results √/6=2+1+-1, &c.; and, taking only the two leading terms of this series, we find 2=g, whose square is greater than 6; but if we take three terms, we shall have 278, whose square 15 is still too small. As in this example approaches very nearly to the true value of √6, we will take for 6 its equivalent quantity 2-1. Thus cc=3; c={; b=}; and using the two leading terms, we find √6={+}× -=18, and the square of this fraction being 240, exceeds the 6 only by 716. Now, taking 6=240-, so that c=18 and b, and still confining ourselves to the two leading terms, we have 6=1+1× Now 6, when reduced to the same denominator, is-33049600; the error, 1600 717. In a similar way may be expressed the cube root of a + b by an infinite series. For, as ✅(a+b)=(a+b)3, we shall have in the general formula n =}, and the coefficients will be 1= n-2 a" = Ya; a"-1_va; aa -}; "==−}}, &c., and for the powers of a we have = &c. Then (a+b) = But if a be a cube or a=c3, we have b сс a + } xb+} x bb a=c and the radi 63 cal signs will vanish, for we shall have (c3 +b)=c+} × 2-1 × 8+ × 25-243 C1+ &c. 718. Thus we arrive at a formula, enabling us by approximation, as it is called, to find the cube root of any number, because every number may be resolved into two parts, as c3 + b, whereof the first is a cube. If, for example, we are required to determine the cube root of 2, we represent 2 by 1+1, so that c=1 and b=1, consequently √2=1+}−}+f†, &c. The two leading terms of which series make 13, the cube of which, , is too great by 1. Let us then make 2=-; we shall now have c, and 6=-19, and therefore 72=1+1× 16' These two terms give -- the cube whereof is 7537. But 2= 14; so that the error is 7075 Thus we may approximate the root; and the faster, as a greater number of terms is taken. 719. It has already been seen that may be expressed by a-1. For the same reason, 1 + may be represented by (a+b)−1; hence the fraction may be considered as a power of a+b, namely, that whose exponent is -1; hence we conclude that the series already found as the value of (a+b)" will extend to this case. is the same as (a+b)−1, let us assume, in the general formula, n= -1; then 1; n-2 n-3 == 1 ; a2 ; a an-2 1 = an-3=11, &c. a3; a' &c., which is the same series before found 721. Now, being the same as (a+b)-2, let us reduce it to an infinite series; for which purpose we must suppose n=— -2, and we have for the coefficients Now =2; x=3; 7××1=4; 7×××1=5, &c. We have, therefore,+2=-23+301-43+56-667+78, &c. To proceed, let us take n=—3, · 1 ; " ~ 2 — — }; " — 3 = −, &c., and the powers of a become a” &c., which gives as, 1 1 b b2 (a+b)3 = 67 63 all 3+63-10+15-21+28-36 +45 &c. 63 = 1 or of (a+b)-3. The coefficients will aši an—1 723. From the cases considered, we are able to conclude that for any negative power of a + b, we shall have by means of which formula all such fractions may be transformed into infinite series, substituting also fractions or fractional exponents for m, in order to express irrational quantities. 724. In further illustration of this subject, we recal to mind that a+b -+, &c.; now this series, therefore, multiplied by a + b, ought to produce 1; which |