Taylor series via division


Let the real (http://planetmath.org/RealFunction) (or complex (http://planetmath.org/ComplexFunction)) functions f and g have the Taylor seriesMathworldPlanetmath

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 ) centered at  x=a.  If  b00, then also the quotient f(x)g(x) apparently has the derivatives of all orders (http://planetmath.org/HigherOrderDerivatives) on I.  It is not hard to justify that if one divides (http://planetmath.org/Division) the series of f by the series of g, the obtained series

f(x)g(x)=c0+c1(x-a)+c2(x-a)2+ (1)

is identically same as the Taylor series of f(x)/g(x) on I.

We consider the coefficients cn of (1) as undetermined .  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 (http://planetmath.org/GeneralAssociativity) of x-a with the corresponding coefficients of the series of f.  Accordingly, we have the conditions

a0=b0c0,a1=b0c1+b1c0,a2=b0c2+b1c1+b2c0, (2)

Since for every n, the equation

an=b0cn+b1cn-1+b2cn-2++bnc0

holds and  b00,  we get the recurrence relation

cn=-b1b0cn-1-b2b0cn-2--bnb0c0+anb0  (n=0, 1, 2,). (3)

Example. We will calculate the Bernoulli numbersMathworldPlanetmathPlanetmath, which are the numbers Bn appearing in the Taylor series of xex-1 expanded with the powers of x:

xex-1=n=1Bnn!xn (4)

This function has really all derivatives in the point  x=0,  since in this point the inverse (http://planetmath.org/InverseNumber)  ex-1x=1+x2!+x23!+  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 of (4) as c0+c1x+c2x2+.  When we now think this series and the series x+x22!++xnn!+ of the denominator of xex-1 to be multiplied, the result must be x, i.e. the coefficients of all powers of x except the first power are 0.  So the two first conditions corresponding to (2) are  c0=1,  c1+12c0=0;  thus

c0=B0= 1,c1=B1=-12.

Setting the coefficient of xn equal to zero gives the formula

c0n!+c1(n-1)!++cn-22!+cn-1= 0 (5)

for  n2.  Putting here  ci=Bii!  to (5) we obtain

B00!n!+B11!(n-1)!+B22!(n-2)!++Bn-2(n-2)!2!+cn-1(n-1)!= 0,

and multiplying this by n!,

(nn)B0+(nn-1)B1+(nn-2)B2++(n2)Bn-2+(n1)cn-1= 0.

This yields, by substituting the values of B0 and B1 and recalling that the odd Bernoulli numbers are zero (n>2), the recursion formula

1-2k2+(2k+12)B2+(2k+14)B4++(2k+12k-2)B2k-2+(2k+12k)B2k= 0

for the even Bernoulli numbers B2k (k=1, 2,).  It gives successively

-12+3B2= 0,-32+10B2+5B4=0,-52+21B2+35B4+7B6= 0,

From here we obtain  B2=16,  B4=-130,  B6=142,  and so on.

Remark.  The method of using undetermined coefficients in division of power seriesMathworldPlanetmath 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 n, below a finite bound. Thus the method is applicable for expanding the rational functions to power series. For example, if we want to expand 11+x2 with the powers of x-1, we write  1+x2=2+2(x-1)+(x-1)2.  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-1-12cn-2; the use of this is much more comfortable than the long division1:(2+2(x-1)+(x-1)2).

References

  • 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I. Second edition.  WSOY, Helsinki (1950).
Title Taylor series via division
Canonical name TaylorSeriesViaDivision
Date of creation 2014-12-02 17:46:48
Last modified on 2014-12-02 17:46:48
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 20
Author pahio (2872)
Entry type Topic
Classification msc 30B10
Classification msc 26A24
Classification msc 41A58
Synonym quotient of Taylor series
Synonym calculating Bernoulli numbers
Related topic BinomialCoefficient
Related topic BernoulliNumber
Related topic BernoulliPolynomialsAndNumbers
Related topic ErnstLindelof