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Taylor series via division∗ pahio† 2013-03-22 0:10:22 Let the real (or complex) functions f and g have the Taylor series f (x) = a0 + a1 (x − a) + a2 (x − a)2 + . . . g(x) = b0 + b1 (x − a) + b2 (x − a)2 + . . . on an interval I (or a circle in C) centered at x = a. If b0 6= 0, then also the f (x) quotient apparently has the derivatives of all orders on I. It is not hard g(x) to justify that if one divides the series of f by the series of g, the obtained series f (x) = c0 + c1 (x − a) + c2 (x − a)2 + . . . g(x) (1) is identically same as the Taylor series of f (x)/g(x) on I. We consider the coefficients cn of (1) as undetermined constants. They can be determined by first multiplying, using Cauchy multiplication rule, the series (1) and the series of g and then by comparing the gotten coefficients of powers of x−a with the corresponding coefficients of the series of f . Accordingly, we have the conditions a0 = b0 c0 , a1 = b0 c1 + b1 c0 , a2 = b0 c2 + b1 c1 + b2 c0 , ... (2) (n = 0, 1, 2, . . .). (3) Since for every index n, the equation an = b0 cn + b1 cn−1 + b2 cn−2 + . . . + bn c0 holds and b0 6= 0, we get the recurrence relation cn = − b1 b2 bn an cn−1 − cn−2 − . . . − c0 + b0 b0 b0 b0 ∗ hTaylorSeriesViaDivisioni created: h2013-03-2i by: hpahioi version: h40154i Privacy setting: h1i hTopici h30B10i h26A24i h41A58i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Example. We will calculate the Bernoulli numbers, which are the numbers x Bn appearing in the Taylor series of x expanded with the powers of x: e −1 ∞ X x Bn n = x ex − 1 n! n=1 (4) This function has really all derivatives in the point x = 0, since in this point x 2 x the inverse e x−1 = 1 + 2! + x3! + . . . naturally has the derivatives and the value 1 distinct from zero. Let us think the division of x by the Taylor series of ex−1. Corresponding to (1), we denote the right side of (4) as c0 + c1 x + c2 x2 + . . .. n 2 When we now think this series and the series x + x2! + . . . + xn! + . . . of the x denominator of x to be multiplied, the result must be x, i.e. the coefficients e −1 of all powers of x except the first power are 0. So the two first conditions corresponding to (2) are c0 = 1, c1 + 21 c0 = 0; thus c0 = B0 = 1, 1 c1 = B1 = − . 2 Setting the coefficient of xn equal to zero gives the formula c0 c1 cn−2 + + ... + + cn−1 = 0 n! (n − 1)! 2! for n ≥ 2. Putting here ci = Bi i! (5) to (5) we obtain B1 B2 Bn−2 cn−1 B0 + + + ... + + = 0, 0!n! 1!(n − 1)! 2!(n − 2)! (n − 2)!2! (n − 1)! and multiplying this by n!, n n n n n B0 + B1 + B2 + . . . + Bn−2 + cn−1 = 0. n n−1 n−2 2 1 This yields, by substituting the values of B0 and B1 and recalling that the odd Bernoulli numbers are zero (n > 2), the recursion formula 1−2k 2k+1 2k+1 2k+1 2k+1 + B2 + B4 + . . . + B2k−2 + B2k = 0 2 2 4 2k−2 2k for the even Bernoulli numbers B2k (k = 1, 2, . . .). It gives successively 1 − + 3B2 = 0, 2 3 − + 10B2 + 5B4 = 0, 2 5 − + 21B2 + 35B4 + 7B6 = 0, 2 1 From here we obtain B2 = 61 , B4 = − 30 , B6 = 1 42 , ... and so on. Remark. The method of using undetermined coefficients in division of power series is especially simple in the case that the denominator in (1) is a 2 polynomial, because the number of the terms in the recursion formula (3) is, independently on n, below a finite bound. Thus the method is applicable for expanding the rational functions to power series. For example, if we want to 1 2 2 expand 1+x 2 with the powers of x−1, we write 1 + x = 2+2(x−1)+(x−1) . The two first conditions corresponding to (2) are 2c0 = 1 and 2c1 +2c0 = 0, whence c0 = 12 and c1 = − 12 . The coefficient of (x−1)n gives the condition 2cn +2cn−1 + cn−2 = 0, whence the simple recursion formula cn = −cn−1 − 12 cn−2 ; the use of this is much more comfortable than the long division 1 : (2+2(x−1)+(x−1)2 ). References [1] Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I. Second edition. WSOY, Helsinki (1950). 3