the independent number under consideration. Annex now to this independent number the next period on the right of it, and consider what is thus obtained as a new independent number; the two figures of the root already found will be the tens of the root of the new number; bringing down one figure of the right-hand period of it to the remainder after subtracting the nth power of the two figures of the root just found from the first independent number, and dividing by n times the n-1 power of the tens, now composed of two figures, a third figure of the root is obtained; proceeding in this manner, the entire root of the given number will at length be extracted.* 113. By employing the binomial theorem, we can raise any polynomial to any power, without the process of actual multiplication. For example, let it be required to raise x+a+b to the power 4. =x+4x3y+6x2y2+4xy3+y1, or putting for y its value, =x+4x3(a+b)+6x2(a+b)2+4x(a+b)3+(a+b)*. Expanding (a+b)2, (a+b)3, (a+b), by the binomial theorem, and performing the multiplications indicated, we shall arrive at the expansion of (x+a+b). It is manifest that we may apply a similar process to any polynomial. In the expression POWER OF A POLYNOMIAL. (a+b+c+d....)TM make x=b+c+d... the above power will be equal to (a+x)TM, and by the binomial theorem the term which contains a" in the development of this may be written m-n [a] Making y=c+d....... we have xTM-"=(b+y)"-", and developing this last power, the term containing b" may be put under the form *If there be decimals in the given number, ciphers must be annexed, if necessary, to make exact periods of decimals, on a principle similar to that explained in (Art. 90). If the index of the root to be extracted be composed of factors, it can be extracted by means of the successive roots, the degrees of which are expressed by these factors. For if the amp be required, we have Vap=a1p, √a"P=ap, and Var=a. mnp m The best way to extract roots of numbers of a degree higher than the square is by means of logarithms. This may be obtained from the ordinary form of the general term of the binomial formula by multiplying both numerator and denominator by 1.2.3 ... (m—n). It is evident that if this quantity be put in the place of 2- in the expression [a], the result will represent the assemblage of the terms which contain a"b" in the power of the given polynomial. This result, after canceling common factors, will be 1.2.3.4 1.2.3...nX1.2.3...n'x1.2.3... · (m—n — n')' ... [b] Making z=d+. we shall have ym▬▬n'=(c+z)m-n-"', and the term containing ca" will be substituting this expression for ym-a-n' in [b], we have 1.2.3...mab'cn''m-n-n-n/ 1.2.3. ...n×1.2.3...n' × 1.2.3...n" ×1.2.3...(m-n—n' —n'')' It is evident now, without carrying the reasoning farther, that if V be the general term of the development of (a+b+c+d...)TM, this term may be represented thus, V: 1.2.3.4....m× a"b"'c1'' ... = 1.2.3.. n, n', n"... being any positive whole numbers at pleasure, subjected only to the condition that their sum shall be equal to m. So that all the terms of the required development may be obtained by giving in this formula to n, n', n'....... all the entire positive values which satisfy the condition n+n'+n"... When one of these numbers is made zero, V takes an illusory form. If, for example, n=0, the series 1.2.3...n placed in the denominator is nonsensical, because factors increasing from one will never present us with a factor zero. To relieve this difficulty, let us recur to the general term [a] in the development of (a+x), and observe that the hypothesis n=0 reduces it to 1.2.3...0° But the hypothesis n=0 ought to give in this development the term which does not contain a, and this term is rm. Then, in order that this term shall be deduced from the formula [a], it is sufficient to consider the series 1.2.3...n as equivalent to 1 in this particular case of n=0. The same observation should be extended to the other series of factors contained in the denominator of V, and then V will give, without any exception, all the terms of the power of the polynomial a+b+c+, &c. TO EXTRACT THE mth ROOT OF A POLYNOMIAL. The problem is, having given a polynomial, P, which is the m another polynomial, p, to find the latter. .... power of Let us consider the two polynomials as arranged according to the decreasing exponents of some letter, x, and call a, b, c, the unknown terms of the root p. They must be such that, in raising a+b+c... to the power m, we obtain all the terms which compose P. But if we imagine that we have formed this power by successive multiplications, it is clear that, in the result, the term in which x has the highest exponent is the mth power of a; then we shall know the first term of the root sought, p, by extracting the mth root of the first term of the given polynomial, P. The first term of the root being found, it will be easy to obtain the second; but I prefer to show at once how, when we know several successive terms of the root setting out from the first, we can determine the term which comes immediately after. Let u represent the sum of the known terms, and v that of the unknown; then P (u+v)", or, developing, P=u+muTM1v+kuTM¬£v2+k'uTM¬3v3+, &c. I have not exhibited the composition of the coefficients k, k'.., this not being necessary, as will appear. From this equality, by subtracting uTM from both the equals, we obtain P-umu+kum-2y2+k'um¬3y3+, &c. The first of these equals, P-um, is a quantity which we can calculate by forming the mth power of the known quantity u, and subtracting it from the polynomial P. The second is a sum of products, by means of which we can easily assign the composition of the first term of the remainder P-u, and, consequently, discover the first term of the unknown part, v. .... First, if we develop um-1, it is clear, by the rules of multiplication alone, that the first term of the development, that is, the one which contains r, with the highest exponent, will be am-1; then, if we call ƒ the first term of r, the first term of the product mum-ly will be mam-lf. By a similar course of reasoning, we perceive that the first terms in the developments of the other products will be respectively kam-f2, k'aTM-3ƒ3,. These terms, abstraction being made of the coefficients which have no influence upon the degree of 1, can be deduced from the term mam--lf, by suppressing in it one or more factors equal to a, and replacing them by as many factors equal to f. But ƒ being of a degree inferior to a with respect to x, these changes can give only terms of a degree inferior to maTM-lf. Then, after having subtracted from the given polynomial P the mth power of the part u of the root already found, the first term of the remainder is equal to the product of m times the power m—1 of the first term a of the root by the first of those terms which remain still to be found. Therefore, dividing the first term of the remainder by m times the power m-1 of the first term of the root, the quotient will be a new term of this root. This conclusion furnishes the means of discovering successively all the terms of the root as soon as the first is known. To have the second term, b, subtract from the given polynomial P the m power of the first term of the root, then divide the first term of the remainder by mam-1; to have the third term, c, of the root, subtract from P the mth power of a+b, then divide the first term of this remainder by maTM-1, and so on. If in any part of the process, the remainder being arranged according to the powers of x, its first term is not divisible by m times the m-1 power of the first term of the root, the given polynomial will not have an exact root of the degree m. Observe, also, that the mth root of the last term of the given polynomial ought to be the last term of the root. Therefore, if the process leads to a term in the root of a less degree than this, the given polynomial is not an exact power of the order m. We may arrange according to the ascending powers of a letter, r, as was remarked at (Art. 80, III.), when treating of the square root, and the above observation will undergo the same modification as there stated. It would be superfluous to speak of the case where the letter of arrangement, z, enters, with the same exponent, into several terms. The method of proceeding in such a case has been already sufficiently indicated in previous articles. EXAMPLES. (1) Extract the 5th root of 32.5—80x80x3-40x2+10x—1. (2) Extract the 5th root of 729-2916.x2+4860.x-4320x6-576x1o+64x12. Ans. 3-2x2. (3) Extract the fifth root of 20+15x-16-5x-14+90.x-12—60.x-10. 10+282x-8 ·252x6505x-496x2495 +275x80x+15x-10. - 465.x2 Ans. 3-ra. 114. In the observations made upon the expansion of (x+a)”, we have supposed n to be a positive integer. The binomial theorem, however, is applicable, whatever may be the nature of the quantity n, whether it be positive or negative, integral or fractional. When n is a positive integer, the series consists of n+1 terms; in every other case the series never terminates, and the development of (x+a)" constitutes what is called an infinite series. Before proceeding to consider this extension of the theorem, we may remark, that in all our reasonings with regard to a quantity such as (x+a)", we may confine our attention to the more simple form (1+a)", to which the former may always be reduced. For, Suppose n, where r and s are any whole numbers whatever, 2 Then (x+a) becomes (x+a), and substituting for n in the series, S A perfectly rigorous demonstration of the binomial theorem for any exponent whatever, integral or fractional, positive or negative, will be found towards the close of this treatise. This expansion may be obtained by substituting, in the general form (Art. 110), 1 for |