it is evident that the same quantity will be produced if in the product arising from the multiplication of the series for X by the infinite series we reject all terms containing negative powers of x and write a for x. If by A。, A1, A2, &c., we shall have for the approximate value of the integral, Hence we have the means of computing the coefficients A, A1, A2, &c., when the values of x are given equal to a, a1, a2, &c. (9). We have now to determine the values of x so that the polynomial which expresses the value of y may rise to the 2nth degree and the error of quadrature be zero at the same time. As f(x) is a polynomial of the 2nth degree, or less, if we denote by Q the quotient of this polynomial divided by 4(x) and by R the remainder we have f(x) = Q.4(x) + R. We now assume R for f(x), since for the values a, a1, a2, &c., y(x) is zero, and the error in the quadrature will be 2.4(x)dx. Since Q does not exceed the degree n -1 in order that the error may be zero we determine 4(x) by the conditions, By means of a known formula of reduction, or by integrating by parts, the integral x.q(x)dx can be reduced to multiple integrals of y(x). Thus S 4(x)dx = S4(x)dx Sx4(x)dx = xS 4(x)dx — S^4(x)dx2 3 = x2 S4(x)dx 1ƒ¢(x)dx — (m—1)æTM−2 ƒˆ4(x)dx2+(m—1)(m—2) m ·3 Ño4(x)da2+(−1)TM-1 (m−1)(m−2)... ....1. § ̈ 4(x)dxTM. .... Hence in order that sxy(x)dx, fx3y(x)dx, &c., become zero between the limits 0 and 1 the integrals sy(x)dx, ƒ3y(x)dx3. f(x)dx must vanish for the same limits. We have therefore to find a function such that the function itself and its 1st, 2nd, 3rd, and nth differentials vanish for x=0, and x=1. If Пl be the function sought II must have the factors 2+1 and (x—1)"+1, and inversely every function which has the factor +1.(x-1)+1, and besides only constant factors, fulfils the condition. Since y(x) is of the (n+1)th order, Пlx = +1(x)dx, is of the (2n+2)th order. If we put Пlx = x”+ 1.(x — 1)"+1.M, where M is a constant, we have Applying the rule for the differentiation of the product of two functions of Xx . If now we assume for n the values 0, 1, 2, 3, 4, &c., and reject terms containing negative powers of x the roots of this equation will give the values of the abscissas for the Gaussian method. These roots are all real and lie between 0 and 1, and when there is an odd number the middle one is. The approximate value of the integral is therefore Sydz = Aoy(a) + A1(a1) + A‚ç(α2) + ... + Any(an). The following are the values of the roots and coefficients for n = 0, n = 1, n = 2, and n = 3. Gauss has computed these roots and coefficients for all values of n, from n=0, to n=6, to sixteen places of decimals; and a calculation of the errors shows that the degree of approximation as compared with the Cotesian method is nearly doubled. Computing the area of article (6) by the Gaussian method, and putting n = 3, or using four ordinates, I find log 4(a) = log 4(a) = 9.9554751, log (a1) = log (a,) = 9.9936327, log4, 9.2403681, = log A, 9.5133143. A 34(a3) = 0.1569796 Area = 0.9566128, and error= +0.0000013. The Gaussian method has not, so far as I know, been brought much into use, and there are reasons for this in astronomical calculations, since in computing anomalies, radii vectores, &c., the intervals of time being equal, the method of differences gives us a valuable check on the numerical work. But it seems to me that this method may be advantageously applied to series of observations. Thus if one wishes to determine the barometric or thermometric curve for any place, and takes the time the sun is above the horizon for the interval of observation, and decides to make three observations a day, for which n=2, he will observe his instruments about one ninth of the interval after sunrise and before sunset, and at noon. Hence he will get as good a result as from five observations equally distributed over the interval. In the course of a year even the saving of labor in observing and reducing would be very great. While the need of a judicious arrangement of the observations is more apparent in meteorology, there can be no doubt that this question will at length be considered in astronomical observations; since in astronomy the objects to be observed are constantly increasing in number, and this increase will necessitate a more careful expenditure of the labor of the observer. EXTRACTION OF ROOTS. = BY ALEXANDER EVANS, ELKTON, MARYLAND. LET N the number whose root is to be extracted, r an approximate 1 root, ==the degree of the root and R=the true root. Then, approximately, n This formula is easily remembered, is applicable to roots of any degree, always converges, and rapidly; and to practical men, who often forget how to calculate a cube, and sometimes a square root, will be of value in the absence of a table of logarithms. Let it be required to find the square root of 2: supposer = 1.4: Then true to four places. By substituting 2 in the formula, we have .4142135623730950499; the true final figures being 488: see De Morgan, Budget of paradoxes, page 293; ANALYST, Vol. I, page 119. In this particular example the successive fractions are easily formed by taking the denominator of the first for the numerator of the second fraction, and double the numerator of the first for the denominator of the second. By proceeding in this manner, the value of 1/2 may, in a short time and with comparative ease, be found to one hundred places. This method therefore, will give in a single trial a value of the root sufficient for most purposes, and in a few trials a very accurate value. If v3 be taken the first trial gives √3 = = +1.7321: the true value = 1.73205. A second substitution of will give the root with great accuracy, the resulting fraction being 1881 figure. 086 = 1.7320508 true to last 18+ 25.1, and a second Required 26. Suppose 5; then /26 trial gives 5.0990, which is the root to four places. Required 121631/24. Suppose 5 for 1/24; then 4+ 24 = √/24 =4.9, and 14.7 = 216: the root is 14.6969385. 216: the root is 14.6969385. A second trial, using, 14.7 gives 14.696938 true to the last figure. Required 30. By the formula, supposing the root = 2, 30 5 + X 2 5X24 4 3 8 Required 37. Since 337 = 999 + 23×10 3 × (10)2 9.996666 x 3.3322222 = = 37: the true root is 3.3322218, being, in a single trial only 4 units in error in the ten millionth place. Required 1/500000. Since 219524288, suppose 1/500000=2, then 500000 19X 218 18 500000 36 The true root seems to be 1.99501. The advantage of this method appears to be, the ready remembrance of the formula; which for the square root is so that, at least for the square and cube roots, the extraction may be mentally performed. For example, the cube root of 37 has been obtained to six places in a single line of work, which could have been done "in the head;" while to extract it to six figures in the ordinary way requires a good deal of calculation. It may be noticed that if r be assumed too great, the first fraction of the formula will be too small, and the second too great; and the contrary if r be taken too small; so that there is a kind of compensation. If r be accurately taken,* R will be the exact root; thus, let it be required to find the 10th root of 1024. Suppose the root to be 2, then which we know to be the correct root. The fractions resulting for R, when r is not the exact root, are all in excess; and by assuming different values for r, different series of fractions may be found converging to R. Even if a very erroneous value be assumed for r, in a few trials R will result with some approach to accuracy; thus if 10 be assumed for 1/2, four substitutions will give 1.42, omitting the first fractional part until the last substitution. In the two examples following a little device is exhibited that is often useful. Required 91. Since 23 X 91 728, and 93729, *That the two members of equation (1) approach equality as the value of r approximates that of R may be shown as follows: Multiply the equation by nr-1 and we have N + (n − 1)TM" = nRrn−1 or NnRrn-1 — nrn + gums. Assume now r=R and we have NnR". or N=R... R= R — NA.—Ed. |