The value of x is evidently between 3 and 4, since 33-27 and 4'=256; hence, taking the logarithms of both sides of the equation, we have x log. x= log. 100=2.* Then, as the difference of the results is to the difference of the assumed numbers, so is the least error to a correction of the assumed number corresponding to the least error; that is, 098451: 1 :: 002689: ·00273; hence x=3.6-·00273—3·59727, nearly. Again, by forming the value of x for x=3.5972, we find the error to be -0000841, and for x=3·5973, the error is +0000149; hence, as 000099: 0001 ::·0000149: 0000151; therefore, x=3.5973-0000151=3.5972849, the value nearly. 326. If, at Prop. V., Art. 245, we suppose the second terms a1, a2, A3, &c., of the binomials to be all positive instead of negative, and all equal to a, then the products two and two will all become a2; those three and three, a3, and so on; and, by recurring to Art. 203, we perceive that the number of combinations or products two and two, if we suppose that there are n binomials, will n(n-1) n(n-1)(n-2) be expressed by the number three and three by and so on. Hence, where n is a whole number, or 1.2 1.2.3 n(n-1) (x+a)"=x+nax"1+ 1.2 -a2x22+, &c., +a”, .... (1) (a+x)=a+na"¬1x+Aa"¬2x2+Ba"-323+, &c. . . by reversing the order of the terms, and disregarding the particular form of the coefficients after the second term. CASE II. If the exponent be fractional, we have m (a+x)=√(a+x)TM=√ a"+maTM¬1x+AaTM ̄222+, &c. * In equations of this kind the following method may be adopted: Let x=a; then x log. x= log. a; put log. x=y, and log. a=b; then xy=b, and log. x+ log. y= log. b; hence y+log. y= log. b. Now y may be found by double position, as above, and then s becomes known. When a is less than unity, put 1 1 .. y log. b= log. y, and if log. b=c, and log. y=z; then cy=z, and log. c+ log. y= log. ≈, or log. c+z= log. z. Hence z may be found by the preceding method, and then y and z become known. Applying the rule at Art. 113 for extracting the root of a polynomial, the m first term of the root will be a; the divisor of the second term of the given polynomial, n(a) Mm-L(m will be a n m m- =na m n ; and the quotient or second term of the root x. When the two terms of the root thus found are raised to the nth power, and subtracted from the given polynomial according to the rule, the first two terms of the latter will be canceled, and the next m- m m-3 highest power of a to be divided by the constant divisor na" will be a multiplied by x2, and the quotient, which is the third term of the root, will the same form, so far as regards the exponents, as when the exponent is a whole number. The coefficients will be examined for this and the next case together. CASE III. When the exponent is negative, either entire or fractional, as a consequence of what has just been demonstrated, we have But if the division be effected according to the ordinary rules, the quotient will be indefinite, and of the form then, whatever be the exponent of a binomial, its development, as to the coefficients of the first two terms and the exponents of all, is of the same form, viz., that indicated by equation (1). Now, to examine the coefficients of the other terms, for the sake of generality, I shall consider two consecutive terms of any rank whatever, and I shall write (a+x)=a+maTM ̄1x...+Ma¬"x"+NaTM-"-1x2+1+, &c. Let us change throughout r into x+y; as the unknown coefficients contain neither a nor r, the above expression becomes (a+x+y)=a+ma"-1(x+y).... By changing a into a+y, we should have found (a+y+r)=(a+y)+m(a+y)m-1x... ....+M(a+y)" "x"+N(a+y)"-"-1x"+1+, &c. In the two preceding equalities the first members are equal, therefore the second members must be equal also; and this is the case whatever values x and y may have. Then, if they be arranged according to the powers of y, they must be identical. It is true, they contain binomials, but we know the first two terms of each of these binomials, so that we can form the part which, in each second member, contains y to the first degree, and that will suffice for our purpose. Designating it by Yy in the one and by Y'y in the other, it is easy to find Ymam-1 Y' mam-1. .... ·+Mnam¬¬1+N(n+1)aTM¬n¬1Ãa.......... These two quantities must be equal, whatever be the value of r; the coefficients, therefore, of the same powers of r must be equal. Considering only those which pertain to am-n-1", we have We see by this according to what law, in the development (1), any coefficient whatever is formed from the preceding. It is the same that we have found for the case of a positive exponent (Art. 107, IV.); and as we have seen that the first two terms are composed in the same manner, whatever be the exponent m, it will be so also with all the other terms. An abbreviate notation, sometimes employed to express the coefficients of the binomial formula, is the initial letter B of the word binomial, with the exponent of the power of the binomial before it, and the order of the coefficient above. Thus, the coefficient of the 1° term, if the exponent be n, is ex pressed by B; of the 2°, "B; of the 3°, "B, &c.; of the th term n(n-1)...(n-k+1) k by B, or otherwise simply n 1.2.3...k and proceeding by the method of undetermined coefficients, explained at Art. 209, we find From which we perceive that each coefficient is obtained by multiplying the preceding by b Here the series is a simple geometrical progression. Proceeding in a similar manner with the fraction Here each coefficient from the 3° is the sum of the two preceding, multi each term will be composed of the three preceding, multiplied respectively by Finally, it becomes now evident that in general a fraction of the form produces a series, each term of which from the (m+1)th is composed of the Series of this form are called recurrent, and the assemblage of quantities by which it is necessary to multiply several consecutive terms to obtain the following term, is called the scale of relation of the terms. 328. PROBLEM.-A recurring series being given, to return to the generating fraction. In this enunciation it is supposed that the recurring series is arranged with respect to an indeterminate x. Let S=A+Br+Cx2+... be such a series, having for a scale of relation [pr3, qx2, rx]. Since this scale contains three terms, the generating fraction is of the form a'+b'x+c'r2 If this fraction had been given, we have seen that the scale of relation would and then we perceive that the three terms in r of the denominator can be at once obtained by taking those of the scale of relation with contrary signs. Thus, we can put the generating fraction under the form and since, after clearing it of fractions, the equation ought to be identical in form, we derive from it, having regard only to the first three terms, For example, let S=1-2x-x2 — 5x3 + 4x1 — ... be a recurring series, whose scale of relation is [+23, +4x2, −2x]. Taking the above formula, we shall have A=1, B=-2, c=−1, p=1, q=4, r=—2, 329. PROBLEM.-A series being given, to determine whether it be recurring, and, in this case, to return to the generating fraction. Let the given series be S=A+Br+Cx2+Dr3+.... Let us determine first whether it be equal to a fraction of the form a' a+br' the quotient, therefore, of (1), divided by the series, ought to be exact, and of the form p+qx. Then the generating fraction will be expressed thus: If the division does not stop at the second term this series will not be recurring, or else it will arise from a more complicated fraction. that is to say, dividing (1) by the series S, if we stop the division after we have obtained as a quotient terms of the form p+qx, the series S12, which is the remainder that we then have, and which is always divisible by x2, will be S1 a" Sa'+b'x' such that, after we have removed this factor, we must have = that is to say, the new division ought to terminate at the second term in the quotient; and then, to find the generating fraction, we shall have the two equations |
