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[parent] Taylor series via division (Topic)

Let the real (or complex) functions $ f$ and $ g$ have the Taylor series

$\displaystyle f(x) = a_0+a_1(x-a)+a_2(x-a)^2+\ldots$
$\displaystyle g(x) = b_0+b_1(x-a)+b_2(x-a)^2+\ldots$
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 on $ I$. It is not hard to justify that if one divides 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 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

$\displaystyle a_0 = b_0c_0,\quad a_1 = b_0c_1+b_1c_0,\quad a_2 = b_0c_2+b_1c_1+b_2c_0,\quad \ldots$ (2)

Since for every index $ n$, the equation
$\displaystyle a_n = b_0c_n+b_1c_{n-1}+b_2c_{n-2}+\ldots+b_nc_0$
holds and $ b_0 \neq 0$, we get the recursion formula
$\displaystyle c_n = -\frac{b_1}{b_0}c_{n-1}-\frac{b_2}{b_0}c_{n-2}-\ldots-\frac{b_n}{b_0}c_0+\frac{a_n}{b_0}\quad\quad (n = 0,\,1,\,2,\,\ldots).$ (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 $ \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 right side of (4) as $ c_0+c_1x+c_2x^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

$\displaystyle c_0 = B_0 = 1,\quad c_1 = B_1 = -\frac{1}{2}.$
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
$\displaystyle \frac{B_0}{0!n!}+\frac{B_1}{1!(n-1)!}+\frac{B_2}{2!(n-2)!}+\ldots+\frac{B_{n-2}}{(n-2)!2!}+\frac{c_{n-1}}{(n-1)!} = 0,$
and multiplying this by $ n!$,
$\displaystyle {n\choose n}B_0+{n\choose n\!-\!1}{B_1}+{n\choose n\!-\!2}{B_2}+\ldots+{n\choose 2}B_{n-2}+{n\choose 1}c_{n-1} = 0.$
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
$\displaystyle \frac{1\!-\!2k}{2}+{2k\!+\!1\choose 2}{B_2}+{2k\!+\!1\choose 4}{B_4}+\ldots+ {2k\!+\!1\choose 2k\!-\!2}B_{2k\!-\!2}+{2k\!+\!1\choose 2k}B_{2k} = 0$
for the even Bernoulli numbers $ B_{2k}$ ( $ k = 1,\,2,\,\ldots$). It gives successively
$\displaystyle -\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)$.

Bibliography

1
ERNST LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset I. Second edition. WSOY, Helsinki (1950).



"Taylor series via division" is owned by pahio.
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See Also: binomial coefficient, Bernoulli number, Bernoulli polynomials and numbers, Ernst Lindelöf

Other names:  quotient of Taylor series, calculating Bernoulli numbers
Keywords:  Bernoulli numbers

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Cross-references: long division, expand, rational functions, bound, finite, terms, polynomial, power series, even, the odd Bernoulli numbers are zero, denominator, division, point, powers, expanded, numbers, Bernoulli numbers, calculate, equation, Cauchy multiplication rule, coefficients, series, derivatives, quotient, circle, interval, Taylor series, functions
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This is version 15 of Taylor series via division, born on 2007-12-20, modified 2008-04-07.
Object id is 10154, canonical name is TaylorSeriesViaDivision.
Accessed 848 times total.

Classification:
AMS MSC: 41A58 (Approximations and expansions :: Series expansions )
 26A24 (Real functions :: Functions of one variable :: Differentiation : general theory, generalized derivatives, mean-value theorems)
 30B10 (Functions of a complex variable :: Series expansions :: Power series )
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