Cauchy Products and Convolution Coefficients

Why Products Require a New Coefficient Rule

When two polynomials are multiplied, each coefficient in the result collects several products. For example, the coefficient of $x^2$ in $(1+2x)(1+3x)$ comes from both $1\cdot 3x$ and $2x\cdot 1$. The same collection rule applies to power series: contributions to a given power come from every pair of terms whose exponents add to that power.

For infinite series, however, it is not enough to use the finite-polynomial rule informally. We need a condition that makes the multiplication and regrouping legitimate. Absolute convergence provides that condition. The resulting coefficient rule is called the Cauchy product; its finite sums are also called convolution coefficients.

Definition of the Cauchy Product

Consider two power series with the same center $c$, with coefficients $a_n$ and $b_n$. For each nonnegative integer $n$, define $d_n$ by adding all products $a_kb_j$ for which $k+j=n$. Equivalently, $j=n-k$, so the sum is finite.

d_n=\sum_{k=0}^{n}a_kb_{n-k} =a_0b_n+a_1b_{n-1}+\cdots+a_nb_0.

Definition: The power series $\sum_{n=0}^{\infty}d_n(x-c)^n$, with $d_n=\sum_{k=0}^{n}a_kb_{n-k}$, is the Cauchy product of $\sum_{n=0}^{\infty}a_n(x-c)^n$ and $\sum_{n=0}^{\infty}b_n(x-c)^n$. The coefficient $d_n$ is the convolution of the two coefficient sequences at index $n$.

The definition itself involves only a finite sum for each coefficient. The important question is when the infinite series built from those coefficients actually represents the product of the two original functions. The following result gives a sufficient condition that is both natural and widely useful.

The Product Theorem and Its Justification

Theorem: Suppose both power series $\sum_{n=0}^{\infty}a_n(x-c)^n$ and $\sum_{n=0}^{\infty}b_n(x-c)^n$ converge absolutely at a point $x$. Then their product is represented at that point by the Cauchy product: $$ \left(\sum_{n=0}^{\infty}a_n(x-c)^n\right) \left(\sum_{n=0}^{\infty}b_n(x-c)^n\right) =\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_kb_{n-k}\right)(x-c)^n. $$

Here, absolute convergence means that the sums of the absolute values of the terms are finite at the chosen point. If the two power series have radii of convergence $R_1$ and $R_2$, respectively, both are absolutely convergent whenever $|x-c|<\min(R_1,R_2)$. At a boundary point, the theorem applies if both series converge absolutely there; being on a boundary alone does not guarantee that condition.

To see why the formula works, set $t=x-c$. Write the terms of the first series as $a_kt^k$ and those of the second as $b_jt^j$. Multiplying a term from each series produces $a_kb_jt^{k+j}$. Thus, all terms contributing to $t^n$ are exactly those with $k+j=n$, which gives the finite convolution sum. Absolute convergence is what allows all these products to be regrouped by the exponent $k+j$.

Check absolute convergence.
At the point in question, both $\sum |a_kt^k|$ and $\sum |b_jt^j|$ must be finite.

Form all pairwise products.
Each pair of terms contributes $a_kb_jt^{k+j}$.

Group by total exponent.
For exponent $n$, include exactly the pairs with $k+j=n$, producing $d_n$.

For a complete justification, the sum of the absolute values of all pairwise products is finite. Indeed, the finite partial sums of these nonnegative terms are bounded above by the product of the two absolute sums, so their total sum is finite:

\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}|a_kt^k b_jt^j| = \left(\sum_{k=0}^{\infty}|a_kt^k|\right) \left(\sum_{j=0}^{\infty}|b_jt^j|\right) <\infty.

An absolutely convergent sum can be regrouped without changing its value. Grouping the pairwise products according to $n=k+j$ therefore gives the product as the sum of its diagonal groups. Each group is finite, and its contribution is $d_nt^n$. This proves the asserted identity. It also shows that the product series itself converges absolutely, since

\sum_{n=0}^{\infty}|d_nt^n| \leq \sum_{n=0}^{\infty}\sum_{k=0}^{n}|a_kt^k b_{n-k}t^{n-k}| \leq \left(\sum_{k=0}^{\infty}|a_kt^k|\right) \left(\sum_{j=0}^{\infty}|b_jt^j|\right) <\infty.

Worked Products

Worked Example: A Linear Polynomial Times a Geometric Series

Find the Cauchy product of $1+2x$ and $\sum_{n=0}^{\infty}x^n$, and state where the representation is valid.

The polynomial has coefficients $a_0=1$, $a_1=2$, and $a_n=0$ for $n\geq 2$. The geometric series has $b_n=1$ for every $n\geq 0$, and converges absolutely when $|x|<1$. For $n=0$, the convolution gives $d_0=a_0b_0=1$. For $n\geq1$, only the terms $a_0b_n$ and $a_1b_{n-1}$ can be nonzero, so

d_n=1\cdot1+2\cdot1=3 \qquad(n\geq1).

The product series is therefore

(1+2x)\sum_{n=0}^{\infty}x^n =1+3\sum_{n=1}^{\infty}x^n =1+3x+3x^2+3x^3+\cdots,\qquad |x|<1.

Check by direct multiplication: the constant term is $1$; the coefficient of $x$ is $1+2=3$; and for each $x^n$ with $n\geq2$, the contributions are $x^n$ from the $1$ and $2x^{n-1}\cdot x$, whose coefficient is $1+2=3$. The Cauchy product coefficients agree with the direct multiplication, and the absolute-convergence condition holds throughout $|x|<1$.

Worked Example: Two Geometric Series

Multiply $\sum_{n=0}^{\infty}(2x)^n$ by $\sum_{n=0}^{\infty}(-x)^n$ and find a formula for the coefficient of $x^n$.

The first series is absolutely convergent when $|2x|<1$, or $|x|<\tfrac12$. The second is absolutely convergent when $|x|<1$. Thus both converge absolutely on their shared interval $|x|<\tfrac12$. The coefficients are $a_k=2^k$ and $b_j=(-1)^j$, so the coefficient of $x^n$ is

d_n=\sum_{k=0}^{n}2^k(-1)^{n-k} =(-1)^n\sum_{k=0}^{n}(-2)^k =(-1)^n\frac{1-(-2)^{n+1}}{1-(-2)} =\frac{2^{n+1}+(-1)^n}{3}.

For instance, $d_0=(2+1)/3=1$, $d_1=(4-1)/3=1$, $d_2=(8+1)/3=3$, and $d_3=(16-1)/3=5$. Hence the product begins $1+x+3x^2+5x^3+\cdots$, with the general coefficient given above.

To check the result, use the geometric-series identities in the common interval of convergence:

\frac{1}{1-2x}\frac{1}{1+x} =\frac{1}{3}\left(\frac{2}{1-2x}+\frac{1}{1+x}\right).

Indeed, putting the right-hand side over a common denominator gives numerator $\bigl(2(1+x)+(1-2x)\bigr)/3=3/3=1$. Its coefficient of $x^n$ is $\frac{2}{3}2^n+\frac{1}{3}(-1)^n=(2^{n+1}+(-1)^n)/3$, matching the convolution calculation. Both series and their product representation are valid for $|x|<\tfrac12$.

Worked Example: A Quadratic Polynomial Times a Geometric Series

Find the product series for $(1-x+x^2)\sum_{n=0}^{\infty}x^n$.

Here $a_0=1$, $a_1=-1$, $a_2=1$, and all later $a_k$ are zero. Also $b_n=1$ for every $n\geq0$. The first three coefficients require attention to the finite starting point of the series:

d_0=a_0b_0=1, \qquad d_1=a_0b_1+a_1b_0=1-1=0,

$$ d_2=a_0b_2+a_1b_1+a_2b_0=1-1+1=1. $$

For every $n\geq2$, all three polynomial coefficients contribute, and each geometric coefficient is $1$. Thus $d_n=1-1+1=1$ for all $n\geq2$. The product series is

(1-x+x^2)\sum_{n=0}^{\infty}x^n =1+0x+\sum_{n=2}^{\infty}x^n =1+x^2+x^3+x^4+\cdots,\qquad |x|<1.

Check algebraically: $1-x+x^2=(1-x)+x^2$, so dividing by $1-x$ gives $1+x^2/(1-x)$. Expanding the remaining geometric series gives $1+x^2+x^3+x^4+\cdots$, exactly as calculated by convolution. The geometric series converges absolutely when $|x|<1$, so the product formula is justified there.

A Common Mistake: Matching Only Equal Indices

A tempting but incorrect rule is to multiply corresponding coefficients and use $a_nb_n$ as the coefficient of $(x-c)^n$. This overlooks the cross-products. For example, with $1+x$ and $1+x$, that rule would assign the coefficient of $x$ as $1\cdot1=1$. But direct multiplication gives

$$ (1+x)(1+x)=1+x+x+x^2=1+2x+x^2. $$

The correct coefficient of $x$ is $a_0b_1+a_1b_0=1+1=2$, not $a_1b_1=1$. In general, the coefficient at degree $n$ must include every pair of indices that add to $n$, not just the pair where both indices equal $n$. A second mistake is to use the formula without checking convergence: the convolution coefficients can always be defined by finite sums, but the theorem identifying their series with the product requires absolute convergence of both original series at the point considered.

How to Use the Cauchy Product Reliably

For a product calculation, first identify the shared center and establish where both series converge absolutely. Then calculate one coefficient at a time using the finite convolution sum. If one factor is a polynomial, only its finitely many nonzero coefficients contribute, which often makes the calculation short. Finally, check the first few coefficients by direct multiplication or verify the resulting expression by an algebraic identity where one is available.

The convolution rule matters because it turns multiplication of represented functions into a precise operation on their coefficients. It is the infinite-series counterpart of polynomial multiplication, with absolute convergence supplying the justification for collecting infinitely many terms. The essential bookkeeping is finite at each degree; the essential analytic condition is that the two series can be multiplied and regrouped without changing their sum.

Check Your Understanding

Use the convolution formula and state any convergence condition needed.

For $1+3x$ times $\sum_{n=0}^{\infty}x^n$, find the coefficients of $x^0$, $x^1$, and $x^2$.
Write the convolution sum for the coefficient of $(x-c)^4$ in the product of two power series centered at $c$.
Why does absolute convergence justify grouping pairwise products by their total exponent?
For $\sum_{n=0}^{\infty}(3x)^n$ and $\sum_{n=0}^{\infty}(-x)^n$, what is the interval on which both are absolutely convergent?
In multiplying two series, what contributions make up the coefficient of $x^3$?