(14) As an example of Laplace's Theorem, let us take 3 sin (log *) 482 = d dx {4, (*)}2 = 2 sin (log ≈) cos (log *). ५ । 3 sin(log *) [ 2 – 3 { sin(log *) } * – sin(log ≈) cos(log≈)] {8-9 sin (log) - 2 sin (log x2) +3 sin (log 3)}. Whence e" = ≈ + sin (log ≈) 3 sin (log *) 48 {8 – 9sin (log ≈) – 2 sin (log-2) + 3 sin (log >3)} 1.2.3 p, (x) = cos e*, F(x) = e2. Therefore y = e* + e* cos (e*) — + e* cos (e*) {cos (e*) – 2 sin e*. (6*)} + ** cos (e3) {(cos e*) − 9 e cos (e) sin (e2) + &c. 1.2.3 SECT. 4. Expansion of Functions by particular methods. The preceding Theorems sometimes fail from the function which is to be expanded becoming infinite or indeterminate for particular values of the variable, and, more frequently, they become inapplicable from the complication of the processes necessary for determining the successive differential coefficients. Recourse must then be had to particular artifices depending on the nature of the function which is given. One of the most useful methods is to assume a series with indeterminate coefficients, and then to compare the differential of the function with that of the assumed series; so that by equating the coefficients of like powers of the variables conditions are found for determining the assumed coefficients. This method has the advantage of furnishing the law of dependence of any coefficient on those which precede it. Ex. (1) Let_u = €“. du dx (1) Assume u = a + a1 x + ax2 + &c. + a1a” + &c. Differentiating (1) we have (2) = a + 2a2x + &c. + na„x”−1 + (n + 1) a„ + 1?” + &c. (4) and substituting in (3) for the assumed series, it be Comparing now the coefficients of a" in (1) and (5) we find whence any coefficient is determined by means of those which precede it, except the first or a, the value of which is easily found by putting a = 0 in the original equation, in which case a = €. Therefore, forming the successive coefficients from this first one, = Then by the same process as before we find the coefficient of the general term to be given by the equation There remains to be determined a,, which is easily seen Therefore, assuming a series as in the preceding examples, we find for determining the coefficient of the general term, Also, it is easily seen, that a = 1, therefore Assume this to be equal to Á ̧ +  ̧x + Â1⁄2 x2 + &c. + A„¿c + &c. + &c. and take the logarithmic differentials of both expressions : equating these we have m {a ̧ + 2a ̧x + 3a ̧x2 + &c. + (n + 1) an+ 1 x" + &c.} A1 +2 Ax + 3 A3x2 + &c. + (n + 1) An+1 No” + &c. A + А1x + А2x2 + &c. + А„x” + &c. 0 Whence, multiplying up the denominators and equating the coefficients of like powers of x, we have (n + 1) a An+1 = (m − n) a ̧ A„ + {2m − (n − 1)} a ̧ à ̧-1 0 + {3m - (n - 2)} az An-2 + {4m − (n − 3)} a ̧ An-3 + &c. Also since u is reduced to a" when x = 0, we have A = α". u = A + A1x + А2x2 + &c. + A„x” + &c. 0 By taking the logarithmic differentials we find (n + 1) An+1 = a, 4, + 2α2 A„-1 + &c. + (n + 1) an+1 Ap⋅ Also since u = 6% when x = 0, A = €%, so that we have 0 In some cases the law of the coefficients is best found by proceeding to two differentiations. (6) Let it be required to expand cos na in ascending powers of cos x. Assume cos n x = a+a ̧ cosx+&c...+a, (cos x)2 + Differentiating, ... +ap+2 (COS x)P+2 + &c. -1 n sin na = {a1 + 2 a2 cos x + ..... + pa2 (cos x)3−1 + Differentiating again, n2 x ... +(p + 2) ap+2 (cos x)P+1+ &c. } sin æ. n cos n = a, cosx+...+pa (cos®)+...+(p+2)p+z(cos®)P+2 ... + (p + 1) (p + 2) ap+2 (COS x)2 + ...} (sin x)2. Putting 1 (cos x) for (sin x), and taking the coefficient of (cos x) we find it to be pap + p (p − 1) a, − (p + 1) (p + 2) Ap+3 ; and this must be equal to the coefficient of (cos x)" in the original series multiplied by n2: equating these we have the condition by means of which any coefficient is given in terms of that two places below it. There remain to be determined by other means the first two coefficients a, and a,. For this purpose make x = integer. (2r+1) π in the original equation, r being any Every term on the second side vanishes except the first, and there remains a = cos n (2r + 1) — · To find a1, make x = (2r+ 1) — in the second equation, when we obtain 2 |