Taylor series via division
on an interval (or a circle in ) centered at . If , then also the quotient apparently has the derivatives of all orders (http://planetmath.org/HigherOrderDerivatives) on . It is not hard to justify that if one divides (http://planetmath.org/Division) the series of by the series of , the obtained series
is identically same as the Taylor series of on .
We consider the coefficients of (1) as undetermined . They can be determined by first multiplying, using Cauchy multiplication rule, the series (1) and the series of and then by comparing the gotten coefficients of powers (http://planetmath.org/GeneralAssociativity) of with the corresponding coefficients of the series of . Accordingly, we have the conditions
Since for every , the equation
holds and , we get the recurrence relation
Example. We will calculate the Bernoulli numbers, which are the numbers appearing in the Taylor series of expanded with the powers of :
This function has really all derivatives in the point , since in this point the inverse (http://planetmath.org/InverseNumber) naturally has the derivatives and the value 1 distinct from zero. Let us think the division of by the Taylor series of .
Corresponding to (1), we denote the of (4) as . When we now think this series and the series of the denominator of to be multiplied, the result must be , i.e. the coefficients of all powers of except the first power are 0. So the two first conditions corresponding to (2) are , ; thus
Setting the coefficient of equal to zero gives the formula
for . Putting here to (5) we obtain
and multiplying this by ,
This yields, by substituting the values of and and recalling that the odd Bernoulli numbers are zero (), the recursion formula
for the even Bernoulli numbers (). It gives successively
From here we obtain , , , 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 polynomial, because the number of the terms in the recursion formula (3) is, independently on , below a finite bound. Thus the method is applicable for expanding the rational functions to power series. For example, if we want to expand with the powers of , we write . The two first conditions corresponding to (2) are and , whence and . The coefficient of gives the condition , whence the simple recursion formula ; the use of this is much more comfortable than the long division .
- 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I. Second edition. WSOY, Helsinki (1950).
|Title||Taylor series via division|
|Date of creation||2014-12-02 17:46:48|
|Last modified on||2014-12-02 17:46:48|
|Last modified by||pahio (2872)|
|Synonym||quotient of Taylor series|
|Synonym||calculating Bernoulli numbers|