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Taylor Series Construction · Tutorial 840 of 998

Quotient Taylor Series by Coefficient Recurrence

Use coefficient matching to construct quotient Taylor series, determine when the method applies, and check the resulting expansions.

Intermediate 10 min read

What You'll Learn

  • State the condition that allows a quotient series to be constructed near its center
  • Derive the coefficient recurrence by matching coefficients in a product
  • Calculate quotient-series coefficients exactly and verify them by multiplication
  • Use parity and the recurrence to compute the series for sine divided by cosine
  • Recognize why the recurrence gives a local expansion rather than an automatic global one

Why Divide Taylor Series by Matching Coefficients?

When two functions have Taylor series near a point, their product can be expanded by multiplying the series and collecting like powers. Division can be handled in a related way: represent the unknown quotient by a series, multiply it by the denominator series, and require the product to equal the numerator series.

This method is useful when the denominator has a nonzero value at the expansion center. The coefficients of the quotient are then determined one at a time. Each new coefficient depends only on the numerator coefficients and coefficients already found.

Key condition: For a quotient expanded at \(x=0\), the denominator must have a Taylor series there and its constant coefficient must be nonzero. For functions \(A\) and \(B\), this means \(B(0)\ne 0\). The method then constructs a quotient series near zero, where \(B(x)\) remains nonzero.

The Coefficient Recurrence

Write the numerator, denominator, and proposed quotient as power series:

$$ A(x)=\sum_{n=0}^{\infty}a_nx^n,\qquad B(x)=\sum_{n=0}^{\infty}b_nx^n,\qquad Q(x)=\sum_{n=0}^{\infty}c_nx^n. $$

We want \(B(x)Q(x)=A(x)\). By the coefficient rule for products of power series, the coefficient of \(x^n\) in \(B(x)Q(x)\) is \(\sum_{k=0}^{n}b_kc_{n-k}\). Therefore, matching coefficients requires

$$ \sum_{k=0}^{n}b_kc_{n-k}=a_n. $$

Separate the term containing the new coefficient \(c_n\). It is \(b_0c_n\), so when \(b_0\ne 0\), we can solve for \(c_n\).

Quotient coefficient recurrence: If \(b_0\ne 0\), then $$ c_0=\frac{a_0}{b_0}, \qquad c_n=\frac{1}{b_0}\left(a_n-\sum_{k=1}^{n}b_kc_{n-k}\right) \quad(n\ge 1). $$ In computing \(c_n\), only \(c_0,\ldots,c_{n-1}\) occur on the right.

The first few equations make the pattern explicit:

$$ \begin{aligned} b_0c_0&=a_0,\\ b_0c_1+b_1c_0&=a_1,\\ b_0c_2+b_1c_1+b_2c_0&=a_2,\\ b_0c_3+b_1c_2+b_2c_1+b_3c_0&=a_3. \end{aligned} $$

These equations are triangular: the \(n\)th equation introduces \(c_n\), but all its other quotient coefficients have already been determined. This is the central convenience of the recurrence.

Why the Recurrence Works

The derivation is coefficient matching, and it also shows uniqueness: if a quotient power series exists, its coefficients must satisfy these equations. Since \(b_0\ne0\), each equation has exactly one solution for the new coefficient. Thus the recurrence determines exactly one formal power series whose product with \(B\) has the coefficients of \(A\).

For functions, we also need the series to represent the quotient, not merely to satisfy a formal coefficient identity. Suppose \(A\) and \(B\) are represented by convergent power series near zero and \(B(0)=b_0\ne0\). Write \(B(x)=b_0(1+u(x))\), where \(u(0)=0\). By continuity, for sufficiently small \(|x|\), \(|u(x)|<1\). The geometric series for \(1/(1+u(x))\) then converges there. Multiplying this reciprocal by \(A(x)\) gives a convergent series for \(A(x)/B(x)\) in a sufficiently small neighborhood of zero. Its coefficients must obey the recurrence above.

Locality matters: The recurrence determines the coefficients, but it does not by itself give the interval of convergence. The quotient series represents the function on a neighborhood where the relevant series converge and the denominator is nonzero.

Worked Example: A Quotient with a Linear Denominator

Worked Example: Expanding \((1+x^2)/(2+x)\)

Find the first six coefficients \(c_0,\ldots,c_5\) in the series for \((1+x^2)/(2+x)\) about zero. The numerator coefficients are \(a_0=1\), \(a_1=0\), \(a_2=1\), and \(a_n=0\) for \(n\ge3\). The denominator has \(b_0=2\), \(b_1=1\), and \(b_n=0\) for \(n\ge2\).

For this denominator the recurrence becomes \(2c_0=a_0\) and, for \(n\ge1\), \(2c_n+c_{n-1}=a_n\). Compute in order:

$$ \begin{aligned} 2c_0&=1 &&\Longrightarrow c_0=\frac12,\\ 2c_1+c_0&=0 &&\Longrightarrow c_1=-\frac14,\\ 2c_2+c_1&=1 &&\Longrightarrow c_2=\frac{1+\frac14}{2}=\frac58,\\ 2c_3+c_2&=0 &&\Longrightarrow c_3=-\frac{5}{16},\\ 2c_4+c_3&=0 &&\Longrightarrow c_4=\frac{5}{32},\\ 2c_5+c_4&=0 &&\Longrightarrow c_5=-\frac{5}{64}. \end{aligned} $$

Thus the series begins

$$ \frac{1+x^2}{2+x} = \frac12-\frac14x+\frac58x^2-\frac{5}{16}x^3 +\frac{5}{32}x^4-\frac{5}{64}x^5+\cdots. $$

Check by multiplying the displayed terms by \(2+x\). The coefficients of \(x^0\) through \(x^5\) are, respectively,

$$ \begin{aligned} x^0 &: 2\left(\frac12\right)=1,\\ x^1 &: 2\left(-\frac14\right)+\frac12=0,\\ x^2 &: 2\left(\frac58\right)-\frac14=1,\\ x^3 &: 2\left(-\frac{5}{16}\right)+\frac58=0,\\ x^4 &: 2\left(\frac{5}{32}\right)-\frac{5}{16}=0,\\ x^5 &: 2\left(-\frac{5}{64}\right)+\frac{5}{32}=0. \end{aligned} $$

These agree with \(1+x^2\) through the computed order. The denominator is nonzero for \(|x|<2\), and in fact its reciprocal has a geometric-series expansion there, so this quotient series converges for \(|x|<2\).

Worked Example: Dividing an Exponential Series

Worked Example: Expanding \(e^x/(1+x)\)

Use the Maclaurin series for \(e^x\), whose coefficient is \(a_n=1/n!\), and the denominator \(1+x\), for which \(b_0=b_1=1\) and all later \(b_n\) are zero. The recurrence gives \(c_0=1\), and for \(n\ge1\), \(c_n=1/n!-c_{n-1}\).

Apply it successively:

$$ \begin{aligned} c_0&=1,\\ c_1&=1-1=0,\\ c_2&=\frac{1}{2}-0=\frac12,\\ c_3&=\frac{1}{6}-\frac12=-\frac13,\\ c_4&=\frac{1}{24}-\left(-\frac13\right)=\frac{1}{24}+\frac{8}{24}=\frac38. \end{aligned} $$

So the expansion begins

$$ \frac{e^x}{1+x}=1+\frac12x^2-\frac13x^3+\frac38x^4+\cdots. $$

To check, multiply by \(1+x\). The resulting coefficients through \(x^4\) are \(1\) at degree zero, \(1\) at degree one, \(1/2\) at degree two, \(-1/3+1/2=1/6\) at degree three, and \(3/8-1/3=9/24-8/24=1/24\) at degree four. These are exactly the coefficients of \(e^x\) through that order. Since \(1+x\ne0\) near zero, the expansion represents the quotient locally.

Worked Example: Constructing the Series for Sine Divided by Cosine

Worked Example: Finding the Maclaurin Coefficients of \(\sin x/\cos x\)

Use the known Maclaurin coefficients of sine and cosine. For the numerator, \(a_1=1\), \(a_3=-1/6\), \(a_5=1/120\), \(a_7=-1/5040\), \(a_9=1/362880\), and the other \(a_n\) through degree nine are zero. For the denominator, \(b_0=1\), \(b_2=-1/2\), \(b_4=1/24\), \(b_6=-1/720\), \(b_8=1/40320\); the odd coefficients are zero.

Let \(c_n\) be the quotient coefficients. The recurrence is \(c_n=a_n-\sum_{k=1}^n b_kc_{n-k}\), since \(b_0=1\). First, \(c_0=0\), and the first odd coefficients are found in order:

$$ \begin{aligned} c_1&=1,\\ c_2&=0,\\ c_3&=-\frac16-\left(-\frac12c_1\right) =-\frac16+\frac12=\frac13,\\ c_5&=\frac1{120}-\left(-\frac12c_3+\frac1{24}c_1\right) =\frac1{120}-\left(-\frac16+\frac1{24}\right) =\frac1{120}+\frac18=\frac2{15},\\ c_7&=-\frac1{5040}-\left(-\frac12c_5+\frac1{24}c_3-\frac1{720}c_1\right)\\ &=-\frac1{5040}-\left(-\frac1{15}+\frac1{72}-\frac1{720}\right) =-\frac1{5040}+\frac{13}{240} =\frac{17}{315}. \end{aligned} $$

The even coefficients through this order are zero. Indeed, the numerator and denominator have zero odd coefficients, and the denominator has only even powers. Starting with \(c_0=0\), the recurrence for each even index uses only earlier even-index quotient coefficients, so induction gives \(c_2=c_4=c_6=c_8=0\).

For degree nine, include all the nonzero denominator coefficients that contribute:

$$ \begin{aligned} c_9 &=\frac1{362880} -\left(-\frac12c_7+\frac1{24}c_5-\frac1{720}c_3+\frac1{40320}c_1\right)\\ &=\frac1{362880} -\left(-\frac{17}{630}+\frac1{180}-\frac1{2160}+\frac1{40320}\right). \end{aligned} $$

The parenthesized sum, using denominator \(1814400\), is

$$ \frac{-48960+10080-840+45}{1814400} =-\frac{39675}{1814400} =-\frac{529}{24192}. $$

Therefore

$$ c_9=\frac1{362880}+\frac{529}{24192} =\frac{1+7935}{362880} =\frac{7936}{362880} =\frac{62}{2835}. $$

The resulting expansion is

$$ \frac{\sin x}{\cos x} =x+\frac{x^3}{3}+\frac{2x^5}{15} +\frac{17x^7}{315}+\frac{62x^9}{2835}+\cdots. $$

Check by multiplying the quotient series by \(\cos x\). For degrees one, three, five, seven, and nine, the product coefficients are, in turn,

$$ \begin{aligned} x^1 &: c_1=1,\\ x^3 &: c_3-\frac12c_1=\frac13-\frac12=-\frac16,\\ x^5 &: c_5-\frac12c_3+\frac1{24}c_1 =\frac2{15}-\frac16+\frac1{24}=\frac1{120},\\ x^7 &: c_7-\frac12c_5+\frac1{24}c_3-\frac1{720}c_1 =\frac{17}{315}-\frac1{15}+\frac1{72}-\frac1{720} =-\frac1{5040},\\ x^9 &: c_9-\frac12c_7+\frac1{24}c_5-\frac1{720}c_3+\frac1{40320}c_1 =\frac1{362880}. \end{aligned} $$

The even-degree coefficients vanish, as required. The product therefore agrees with the sine series through degree nine. Since \(\cos 0=1\), continuity ensures that \(\cos x\ne0\) on some neighborhood of zero, so the quotient and its series are defined there.

A Common Mistake: Dividing Coefficients Separately

A quotient of power series is not found by dividing each numerator coefficient by the corresponding denominator coefficient. Product coefficients combine several terms, so the quotient recurrence must account for the contributions from earlier quotient coefficients.

For example, in the sine-over-cosine calculation, a mistaken calculation might set \(c_3=a_3=-1/6\), on the assumption that the degree-three quotient coefficient comes directly from the degree-three numerator coefficient. This misses the contribution of the denominator’s \(x^2\) term multiplied by the quotient’s \(x\) term. The correct degree-three product equation is

$$ c_3-\frac12c_1=-\frac16. $$

Since \(c_1=1\), it gives \(c_3=-1/6+1/2=1/3\). Substitution checks the correction: \(1/3-(1/2)(1)=-1/6\), as required.

Practical rule: At each degree, write the product-coefficient equation first. Then solve for the new quotient coefficient. This prevents terms from lower degrees in the denominator from being overlooked.

When the Recurrence Cannot Be Used as Written

If \(b_0=0\), the displayed recurrence cannot be solved for \(c_n\) by dividing by \(b_0\). This is not a minor arithmetic problem: the denominator vanishes at the expansion center, so the quotient may be undefined or may require a different treatment. For instance, common powers of \(x\) might be factored from numerator and denominator before expanding, but the quotient must still be checked to be defined near the center.

When \(b_0\ne0\), the recurrence gives the coefficients in sequence, but a finite calculation only determines a finite part of the series. It does not establish that the series converges everywhere. Establishing a full interval or radius of convergence requires attention to where the original series converge and where the denominator is nonzero.

Check Your Understanding

Use coefficient matching and show enough work to make each recurrence step clear.

  1. For \(A(x)=1-x\) and \(B(x)=1+2x\), write the recurrence for the quotient coefficients \(c_n\).
  2. Why must the constant coefficient \(b_0\) of the denominator be nonzero to use this recurrence at zero?
  3. For \(e^x/(1+2x)\), calculate \(c_0,c_1,c_2\).
  4. In a quotient expansion, what equation determines the coefficient \(c_n\), and which quotient coefficients can appear on its right-hand side?
  5. Explain why matching coefficients alone does not establish that a quotient series converges on the entire real line.

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