Reciprocals of Power Series by Coefficient Recursion

Why Reciprocal Coefficients Can Be Found Recursively

A power series is often used to represent a function near a chosen center. If that function has a nonzero value at the center, its reciprocal is also defined nearby. The question is how to find the reciprocal's power-series coefficients without guessing them in advance.

The key is to multiply the original series by an unknown series and require the product to equal $1$. The constant coefficient determines the reciprocal's constant term. Then, in each successive degree, the product condition determines exactly one new coefficient. This gives a recursive method.

The Reciprocal-Series Setup

Let $c$ be the center of expansion, and write $t=x-c$. Suppose a power series converges for $|t|<R$, where $R>0$, and has the form

A(x)=\sum_{n=0}^{\infty} a_n t^n =a_0+a_1t+a_2t^2+\cdots, \qquad a_0\ne 0.

Because $A(c)=a_0\ne 0$, the function $A$ is nonzero in some neighborhood of $c$. The reciprocal function $1/A(x)$ is therefore defined there. A standard result for power series states that the reciprocal of a power-series function is itself represented by a power series in some neighborhood of any point where the function is nonzero. Write that reciprocal as

\frac{1}{A(x)}=\sum_{n=0}^{\infty} b_n t^n.

The coefficients $b_n$ are the quantities we want to determine. The condition $a_0\ne0$ is essential: if $a_0=0$, then $A(c)=0$, so a power series for $1/A(x)$ with a finite value at $c$ cannot satisfy $A(x)(1/A(x))=1$ there.

Definition: A reciprocal power series for $A$ at $c$ is a power series $\sum_{n=0}^{\infty}b_n(x-c)^n$ whose product with $A(x)$ equals $1$ wherever both series represent the functions near $c$.

Deriving the Coefficient Recursion

Multiply the two series. The Cauchy product rule says that the coefficient of $t^n$ in a product is the sum of all products of coefficients whose indices add to $n$. Since the product here must be the constant series $1$, its constant coefficient must be $1$, and each positive-degree coefficient must be $0$. In other words, for every $n\ge 0$,

\sum_{k=0}^{n}a_kb_{n-k} = \begin{cases} 1,&n=0,\\ 0,&n\ge 1. \end{cases}

For $n=0$, this gives $a_0b_0=1$, so $b_0=1/a_0$. For $n\ge1$, separate the term containing $b_n$, which occurs when $k=0$:

a_0b_n+\sum_{k=1}^{n}a_kb_{n-k}=0.

Solving for $b_n$ gives the recursion. The index $k$ in the sum ranges from $1$ to $n$, so every coefficient $b_{n-k}$ on the right has index less than $n$. Thus, once $b_0,\ldots,b_{n-1}$ are known, the recursion determines $b_n$.

Theorem: If $A(x)=\sum_{n=0}^{\infty}a_n(x-c)^n$ and $a_0\ne0$, then the coefficients of its reciprocal series satisfy $$ b_0=\frac{1}{a_0}, \qquad b_n=-\frac{1}{a_0}\sum_{k=1}^{n}a_kb_{n-k} \quad(n\ge1). $$ These equations determine the coefficients uniquely, one at a time.

Why the Recursion Works

The derivation also explains why the method is reliable. At degree $n$, the product coefficient contains $a_0b_n$, together with terms involving only earlier $b$-coefficients. Since $a_0\ne0$, the equation for the degree-$n$ coefficient can always be solved for $b_n$. No later coefficient is needed.

Set the constant product coefficient.
Require $a_0b_0=1$, which gives $b_0=1/a_0$.

Set each higher product coefficient to zero.
For degree $n\ge1$, require $\sum_{k=0}^{n}a_kb_{n-k}=0$.

Solve in increasing order.
Isolate $a_0b_n$ and use the coefficients already calculated.

Worked Example: A Quadratic Polynomial

Worked Example: The Reciprocal of $1+x-x^2$

Find the first six coefficients, $b_0$ through $b_5$, of the reciprocal series about $x=0$. Here $a_0=1$, $a_1=1$, $a_2=-1$, and $a_n=0$ for $n\ge3$. The recursion simplifies to $b_n=-(b_{n-1}-b_{n-2})$ for $n\ge2$; for $n=1$, it gives $b_1=-b_0$.

b_0=1,\qquad b_1=-1.

For the next coefficients, apply the recursion in order:

\begin{aligned} b_2&=-(b_1-b_0)=-(-1-1)=2,\\ b_3&=-(b_2-b_1)=-(2-(-1))=-3,\\ b_4&=-(b_3-b_2)=-(-3-2)=5,\\ b_5&=-(b_4-b_3)=-(5-(-3))=-8. \end{aligned}

Thus the reciprocal series begins $1-x+2x^2-3x^3+5x^4-8x^5+\cdots$. Check by multiplying the truncated series by $1+x-x^2$: the coefficients of degrees $0$ through $5$ are, respectively,

1,\quad -1+1,\quad 2-1-1,\quad -3+2+1,\quad 5-3-2,\quad -8+5+3.

These values are $1,0,0,0,0,0$, as required through degree $5$.

Worked Example: A Nonunit Constant Term

Worked Example: The Reciprocal of $2-3x+x^2$

Find $b_0$ through $b_4$ for the reciprocal series about $x=0$. The coefficient equation for degree $n\ge2$ is $2b_n-3b_{n-1}+b_{n-2}=0$. For degree $1$, it is $2b_1-3b_0=0$. The constant equation gives $b_0=1/2$, and the degree-one equation gives $b_1=3b_0/2=3/4$.

\begin{aligned} b_2&=\frac{3b_1-b_0}{2} =\frac{9/4-1/2}{2}=\frac{7}{8},\\ b_3&=\frac{3b_2-b_1}{2} =\frac{21/8-3/4}{2}=\frac{15}{16},\\ b_4&=\frac{3b_3-b_2}{2} =\frac{45/16-7/8}{2}=\frac{31}{32}. \end{aligned}

For verification, the product coefficients from degree $0$ to degree $4$ are:

\begin{aligned} 2b_0&=1,\\ 2b_1-3b_0&=\frac32-\frac32=0,\\ 2b_2-3b_1+b_0&=\frac74-\frac94+\frac12=0,\\ 2b_3-3b_2+b_1&=\frac{15}{8}-\frac{21}{8}+\frac{6}{8}=0,\\ 2b_4-3b_3+b_2&=\frac{31}{16}-\frac{45}{16}+\frac{14}{16}=0. \end{aligned}

Hence the computed coefficients give product $1$ through degree $4$.

Worked Example: An Infinite Input Series

Worked Example: A Series with a Geometric Tail

Consider $A(x)=1+2x+x^2+x^3+x^4+\cdots$, which converges for $|x|<1$. Its coefficients are $a_0=1$, $a_1=2$, and $a_n=1$ for $n\ge2$. Calculate $b_0$ through $b_5$ for its reciprocal.

The constant term gives $b_0=1$. The next coefficient equation gives $b_1+2b_0=0$, so $b_1=-2$. For degree $n\ge2$, the coefficient equation is $b_n+2b_{n-1}+b_{n-2}+\cdots+b_0=0$. Therefore:

\begin{aligned} b_2&=-(2b_1+b_0)=-(-4+1)=3,\\ b_3&=-(2b_2+b_1+b_0)=-(6-2+1)=-5,\\ b_4&=-(2b_3+b_2+b_1+b_0)=-(-10+3-2+1)=8,\\ b_5&=-(2b_4+b_3+b_2+b_1+b_0)=-(16-5+3-2+1)=-13. \end{aligned}

Multiplying back checks the result: the degree-zero coefficient is $1$, and the coefficients from degree $1$ through degree $5$ are

\begin{aligned} b_1+2b_0&=-2+2=0,\\ b_2+2b_1+b_0&=3-4+1=0,\\ b_3+2b_2+b_1+b_0&=-5+6-2+1=0,\\ b_4+2b_3+b_2+b_1+b_0&=8-10+3-2+1=0,\\ b_5+2b_4+b_3+b_2+b_1+b_0&=-13+16-5+3-2+1=0. \end{aligned}

So the reciprocal begins $1-2x+3x^2-5x^3+8x^4-13x^5+\cdots$. The calculation uses only finitely many input coefficients to find any specified finite list of output coefficients.

A Common Mistake: Omitting the Factor $a_0$

When $a_0\ne1$, the recursion must divide by $a_0$. For example, take $A(x)=2+3x+x^2$. A mistaken calculation might set $b_0=1/2$ correctly but then write $b_1=-3b_0=-3/2$, forgetting the factor $a_0$ multiplying $b_1$. That proposed coefficient fails the degree-one product condition:

2\left(-\frac32\right)+3\left(\frac12\right) =-3+\frac32=-\frac32\ne0.

The correct equation is $2b_1+3b_0=0$, so $b_1=-3b_0/2=-3/4$. Substitution gives $2(-3/4)+3(1/2)=0$, as required. Always write the full coefficient equation before isolating the new coefficient.

Using the Recursion Carefully

The recursion is useful for symbolic work and for computing as many coefficients as needed. It does not, by itself, identify the exact radius of convergence of the reciprocal series. The original series may converge on a known interval or disk, but the reciprocal can encounter zeros of $A$ that restrict its domain. A few calculated coefficients cannot establish that radius.

For reliable calculations, keep the coefficients indexed by degree, include every pair of indices whose sum is the degree being considered, and verify by multiplication. In particular, the coefficient of degree $n$ in the product is not just $a_nb_0$ and $a_0b_n$; it includes all intermediate terms $a_kb_{n-k}$.

Key takeaway: A nonzero constant coefficient makes the reciprocal coefficients uniquely computable. Start with $b_0=1/a_0$, then use the product equation in each degree to find the next coefficient.

Check Your Understanding

Use the coefficient equations to answer these questions. Check each proposed sequence by multiplying through the degrees requested.

For $A(x)=1-2x$, find the reciprocal coefficients $b_0,b_1,b_2,b_3$.
For $A(x)=3+x$, find $b_0,b_1,b_2$. Why must the recursion divide by $3$?
Write the coefficient equation for degree $3$ in the product of $\sum_{n=0}^{\infty}a_nx^n$ and $\sum_{n=0}^{\infty}b_nx^n$, when their product is $1$.
Explain why $a_0\ne0$ is necessary for a reciprocal power series with finite constant coefficient.
For an infinite input series, which input coefficients are needed to calculate $b_4$?

Reciprocals of Power Series by Coefficient Recursion

What You'll Learn

Why Reciprocal Coefficients Can Be Found Recursively

The Reciprocal-Series Setup

Deriving the Coefficient Recursion

Why the Recursion Works

Worked Example: A Quadratic Polynomial

Worked Example: The Reciprocal of \(1+x-x^2\)

Worked Example: A Nonunit Constant Term

Worked Example: The Reciprocal of \(2-3x+x^2\)

Worked Example: An Infinite Input Series

Worked Example: A Series with a Geometric Tail

A Common Mistake: Omitting the Factor \(a_0\)

Using the Recursion Carefully

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Reciprocals of Power Series by Coefficient Recursion

What You'll Learn

Why Reciprocal Coefficients Can Be Found Recursively

The Reciprocal-Series Setup

Deriving the Coefficient Recursion

Why the Recursion Works

Worked Example: A Quadratic Polynomial

Worked Example: The Reciprocal of \(1+x-x^2\)

Worked Example: A Nonunit Constant Term

Worked Example: The Reciprocal of \(2-3x+x^2\)

Worked Example: An Infinite Input Series

Worked Example: A Series with a Geometric Tail

A Common Mistake: Omitting the Factor \(a_0\)

Using the Recursion Carefully

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Previous

Next