# Taylor series via division

Let the real (http://planetmath.org/RealFunction) (or complex (http://planetmath.org/ComplexFunction)) functions $f$ and $g$ have the Taylor series  $f(x)\;=\;a_{0}+a_{1}(x-a)+a_{2}(x-a)^{2}+\ldots$
 $g(x)\;=\;b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}+\ldots$

on an interval $I$ (or a circle in $\mathbb{C}$) centered at  $x=a$.  If  $b_{0}\neq 0$, then also the quotient $\displaystyle\frac{f(x)}{g(x)}$ 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

 $\displaystyle\frac{f(x)}{g(x)}\;=\;c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\ldots$ (1)

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

We consider the coefficients $c_{n}$ 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

 $\displaystyle a_{0}\;=\;b_{0}c_{0},\quad a_{1}\;=\;b_{0}c_{1}+b_{1}c_{0},\quad a% _{2}\;=\;b_{0}c_{2}+b_{1}c_{1}+b_{2}c_{0},\quad\ldots$ (2)

Since for every $n$, the equation

 $a_{n}\;=\;b_{0}c_{n}+b_{1}c_{n-1}+b_{2}c_{n-2}+\ldots+b_{n}c_{0}$

holds and  $b_{0}\neq 0$,  we get the recurrence relation

 $\displaystyle c_{n}\;=\;-\frac{b_{1}}{b_{0}}c_{n-1}-\frac{b_{2}}{b_{0}}c_{n-2}% -\ldots-\frac{b_{n}}{b_{0}}c_{0}+\frac{a_{n}}{b_{0}}\quad\quad(n=0,\,1,\,2,\,% \ldots).$ (3)

Example. We will calculate the Bernoulli numbers   , which are the numbers $B_{n}$ appearing in the Taylor series of $\displaystyle\frac{x}{e^{x}-1}$ expanded with the powers of $x$:

 $\displaystyle\frac{x}{e^{x}-1}\;=\;\sum_{n=1}^{\infty}\frac{B_{n}}{n!}x^{n}$ (4)

This function has really all derivatives in the point  $x=0$,  since in this point the inverse (http://planetmath.org/InverseNumber)  $\frac{e^{x}-1}{x}=1+\frac{x}{2!}+\frac{x^{2}}{3!}+\ldots$  naturally has the derivatives and the value 1 distinct from zero.  Let us think the division of $x$ by the Taylor series of $e^{x}\!-\!1$.

Corresponding to (1), we denote the of (4) as $c_{0}+c_{1}x+c_{2}x^{2}+\ldots$.  When we now think this series and the series $x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}+\ldots$ of the denominator of $\displaystyle\frac{x}{e^{x}-1}$ 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  $c_{0}=1$,  $c_{1}+\frac{1}{2}c_{0}=0$;  thus

 $c_{0}\;=\;B_{0}\;=\;1,\qquad c_{1}\;=\;B_{1}\;=\;-\frac{1}{2}.$

Setting the coefficient of $x^{n}$ equal to zero gives the formula

 $\displaystyle\frac{c_{0}}{n!}+\frac{c_{1}}{(n-1)!}+\ldots+\frac{c_{n-2}}{2!}+c% _{n-1}\;=\;0$ (5)

for  $n\geq 2$.  Putting here  $c_{i}=\frac{B_{i}}{i!}$  to (5) we obtain

 $\frac{B_{0}}{0!n!}+\frac{B_{1}}{1!(n-1)!}+\frac{B_{2}}{2!(n-2)!}+\ldots+\frac{% B_{n-2}}{(n-2)!2!}+\frac{c_{n-1}}{(n-1)!}\;=\;0,$

and multiplying this by $n!$,

 ${n\choose n}B_{0}+{n\choose n\!-\!1}{B_{1}}+{n\choose n\!-\!2}{B_{2}}+\ldots+{% n\choose 2}B_{n-2}+{n\choose 1}c_{n-1}\;=\;0.$

This yields, by substituting the values of $B_{0}$ and $B_{1}$ and recalling that the odd Bernoulli numbers are zero ($n>2$), the recursion formula

 $\frac{1\!-\!2k}{2}+{2k\!+\!1\choose 2}{B_{2}}+{2k\!+\!1\choose 4}{B_{4}}+% \ldots+{2k\!+\!1\choose 2k\!-\!2}B_{2k\!-\!2}+{2k\!+\!1\choose 2k}B_{2k}\;=\;0$

for the even Bernoulli numbers $B_{2k}$ ($k=1,\,2,\,\ldots$).  It gives successively

 $-\frac{1}{2}+3B_{2}\;=\;0,\quad-\frac{3}{2}+10B_{2}+5B_{4}=0,\quad-\frac{5}{2}% +21B_{2}+35B_{4}+7B_{6}\;=\;0,\quad\ldots$

From here we obtain  $B_{2}=\frac{1}{6}$,  $B_{4}=-\frac{1}{30}$,  $B_{6}=\frac{1}{42}$,  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 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 $\frac{1}{1+x^{2}}$ with the powers of $x\!-\!1$, we write  $1+x^{2}=2\!+\!2(x\!-\!1)\!+\!(x\!-\!1)^{2}$.  The two first conditions corresponding to (2) are  $2c_{0}=1$  and  $2c_{1}+2c_{0}=0$,  whence  $c_{0}=\frac{1}{2}$  and  $c_{1}=-\frac{1}{2}$.  The coefficient of $(x\!-\!1)^{n}$ gives the condition $2c_{n}+2c_{n-1}+c_{n-2}=0$,  whence the simple recursion formula  $c_{n}=-c_{n-1}-\frac{1}{2}c_{n-2}$; 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).
 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