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

$$f(x)=a_0+a_1(x-a)+a_2{(x-a)}^2+\mathrm{\dots }$$

$$g(x)=b_0+b_1(x-a)+b_2{(x-a)}^2+\mathrm{\dots }$$

on an interval $I$ (or a circle in $ℂ$) centered at  $x=a$.  If  $b_0\ne 0$, 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

${\displaystyle \frac{f(x)}{g(x)}}=c_0+c_1(x-a)+c_2{(x-a)}^2+\mathrm{\dots }$(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

$a_0=b_0{c}_0,a_1=b_0{c}_1+b_1{c}_0,a_2=b_0{c}_2+b_1{c}_1+b_2{c}_0,\mathrm{\dots }$(2)

Since for every $n$, the equation

$$a_n=b_0{c}_n+b_1{c}_{n-1}+b_2{c}_{n-2}+\mathrm{\dots }+b_n{c}_0$$

holds and  $b_0\ne 0$,  we get the recurrence relation

$c_n=-{\displaystyle \frac{b_1}{b_0}}{c}_{n-1}-{\displaystyle \frac{b_2}{b_0}}{c}_{n-2}-\mathrm{\dots }-{\displaystyle \frac{b_n}{b_0}}{c}_0+{\displaystyle \frac{a_n}{b_0}}\mathit{\hspace{1em}\hspace{1em}}(n=0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$(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 $x$:

${\displaystyle \frac{x}{e^x-1}}={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{B_n}{n!}}{x}^n$(4)

This function has really all derivatives in the point  $x=0$,  since in this point the inverse (http://planetmath.org/InverseNumber)  $\frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\mathrm{\dots }$  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 $c_0+c_{1}x+c_2{x}^2+\mathrm{\dots }$.  When we now think this series and the series $x+\frac{x^2}{2!}+\mathrm{\dots }+\frac{x^n}{n!}+\mathrm{\dots }$ 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

$$c_0=B_0=\mathrm{\hspace{0.33em}1},c_1=B_1=-\frac{1}{2}.$$

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

${\displaystyle \frac{c_0}{n!}}+{\displaystyle \frac{c_1}{(n-1)!}}+\mathrm{\dots }+{\displaystyle \frac{c_{n-2}}{2!}}+c_{n-1}=\mathrm{\hspace{0.33em}0}$(5)

for  $n\ge 2$.  Putting here  $c_i=\frac{B_i}{i!}$  to (5) we obtain

$$\frac{B_0}{0!n!}+\frac{B_1}{1!(n-1)!}+\frac{B_2}{2!(n-2)!}+\mathrm{\dots }+\frac{B_{n-2}}{(n-2)!2!}+\frac{c_{n-1}}{(n-1)!}=\mathrm{\hspace{0.33em}0},$$

and multiplying this by $n!$,

$$\left(\genfrac{}{}{0pt}{}{n}{n}\right)B_0+\left(\genfrac{}{}{0pt}{}{n}{n-1}\right)B_1+\left(\genfrac{}{}{0pt}{}{n}{n-2}\right)B_2+\mathrm{\dots }+\left(\genfrac{}{}{0pt}{}{n}{2}\right)B_{n-2}+\left(\genfrac{}{}{0pt}{}{n}{1}\right)c_{n-1}=\mathrm{\hspace{0.33em}0}.$$

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

$$\frac{1-2k}{2}+\left(\genfrac{}{}{0pt}{}{2k+1}{2}\right)B_2+\left(\genfrac{}{}{0pt}{}{2k+1}{4}\right)B_4+\mathrm{\dots }+\left(\genfrac{}{}{0pt}{}{2k+1}{2k-2}\right)B_{2k-2}+\left(\genfrac{}{}{0pt}{}{2k+1}{2k}\right)B_{2k}=\mathrm{\hspace{0.33em}0}$$

for the even Bernoulli numbers $B_{2k}$ ($k=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }$).  It gives successively

$$-\frac{1}{2}+3B_2=\mathrm{\hspace{0.33em}0},-\frac{3}{2}+10B_2+5B_4=0,-\frac{5}{2}+21B_2+35B_4+7B_6=\mathrm{\hspace{0.33em}0},\mathrm{\dots }$$

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  $1:(2+2(x-1)+{(x-1)}^2)$.