Multiplication of Power Series

Multiplying Two Power Series

Adding power series combines their coefficients term by term. Multiplication is different: a term of degree $n$ in the product can come from any pair of terms whose degrees add to $n$. Thus each product coefficient is a finite sum of cross-products. The main point is to prove that these coefficient sums really give a convergent power series for the pointwise product, not merely a formal expansion.

We use the absolute convergence of power series inside their radii, established in the Convergence Inside and Outside the Radius Theorem. The proof below makes explicit how that absolute convergence controls the infinitely many terms that arise when the two series are multiplied.

The Coefficients of the Product

Let two power series have the same center $a$:

A(x)=\sum_{n=0}^{\infty}a_n(x-a)^n, \qquad B(x)=\sum_{n=0}^{\infty}b_n(x-a)^n.

To obtain a term of degree $n$ in their product, choose a term of degree $k$ from $A$ and a term of degree $n-k$ from $B$. There are exactly $n+1$ such choices, with $k$ ranging from $0$ to $n$. This leads to the following finite coefficient formula.

Definition (Product Coefficients): Given coefficient sequences $(a_n)_{n\ge0}$ and $(b_n)_{n\ge0}$, define $$ d_n=\sum_{k=0}^{n}a_kb_{n-k}\qquad(n\ge0). $$ Each $d_n$ is a finite sum. These are the coefficients assigned to the product of the two power series.

For example, the first few coefficients are $d_0=a_0b_0$, $d_1=a_0b_1+a_1b_0$, and $d_2=a_0b_2+a_1b_1+a_2b_0$. The order of the terms in each finite sum does not affect its value. What still needs proof is that the series with coefficients $d_n$ converges to $A(x)B(x)$.

Theorem (Multiplication of Power Series): Suppose $A(x)=\sum_{n=0}^{\infty}a_n(x-a)^n$ and $B(x)=\sum_{n=0}^{\infty}b_n(x-a)^n$ have positive radii of convergence $R_A$ and $R_B$. Define $d_n=\sum_{k=0}^{n}a_kb_{n-k}$. For every $x$ such that $|x-a|<\min\{R_A,R_B\}$, the series $\sum_{n=0}^{\infty}d_n(x-a)^n$ converges absolutely and $$ A(x)B(x)=\sum_{n=0}^{\infty}d_n(x-a)^n. $$ In particular, the product series has radius of convergence at least $\min\{R_A,R_B\}$.

Proof. Fix $x$ with $|x-a|<\min\{R_A,R_B\}$, and write $h=x-a$. Since each original power series converges absolutely at $x$, the two nonnegative series

\sum_{i=0}^{\infty}\alpha_i =\sum_{i=0}^{\infty}|a_i h^i|, \qquad \sum_{j=0}^{\infty}\beta_j =\sum_{j=0}^{\infty}|b_j h^j|

have finite sums, say $U$ and $V$. For each $n$, the definition of $d_n$ and the triangle inequality give

|d_nh^n| =\left|\sum_{k=0}^{n}(a_kh^k)(b_{n-k}h^{n-k})\right| \leq \sum_{k=0}^{n}\alpha_k\beta_{n-k}.

Summing this inequality over $0\leq n\leq N$ bounds the partial sum of $\sum |d_nh^n|$ by the sum of $\alpha_i\beta_j$ over pairs with $i+j\leq N$. This is a finite subset of all pairs of nonnegative integers, so

\sum_{n=0}^{N}|d_nh^n| \leq \sum_{\substack{i,j\geq0\\i+j\leq N}}\alpha_i\beta_j \leq \left(\sum_{i=0}^{\infty}\alpha_i\right) \left(\sum_{j=0}^{\infty}\beta_j\right)=UV.

The partial sums of the nonnegative series $\sum |d_nh^n|$ are increasing and bounded above by $UV$. They therefore converge, proving absolute convergence of the product series at $x$.

It remains to prove the identity of sums. Define the triangular partial sums

T_N=\sum_{\substack{i,j\geq0\\i+j\leq N}}a_ib_jh^{i+j} =\sum_{n=0}^{N}d_nh^n

and the square partial products

P_N=\left(\sum_{i=0}^{N}a_ih^i\right) \left(\sum_{j=0}^{N}b_jh^j\right) =\sum_{0\leq i,j\leq N}a_ib_jh^{i+j}.

The partial sums of the original series converge to $A(x)$ and $B(x)$, so $P_N\to A(x)B(x)$. The difference between $P_N$ and $T_N$ consists exactly of the terms with $i,j\leq N$ and $i+j>N$. For such a pair, either $i>N/2$ or $j>N/2$. Consequently,

|P_N-T_N| \leq \left(\sum_{i>N/2}\alpha_i\right)V +U\left(\sum_{j>N/2}\beta_j\right).

Both tails on the right tend to zero because $\sum\alpha_i$ and $\sum\beta_j$ converge. Thus $P_N-T_N\to0$, and $T_N\to A(x)B(x)$. Since $T_N$ is the $N$th partial sum of $\sum d_nh^n$, this proves the product identity. The argument applies at every $x$ with $|x-a|<\min\{R_A,R_B\}$, which also proves the asserted lower bound for the radius. $\square$

Worked Examples

Worked Example: Multiplying a Polynomial by a Geometric Series

Consider, for $|x|<1/3$,

(1+2x-x^2)\sum_{n=0}^{\infty}3^nx^n.

The polynomial has coefficients $a_0=1$, $a_1=2$, $a_2=-1$, and $a_k=0$ for $k\ge3$. The second series has coefficients $b_n=3^n$ and radius $1/3$. The product coefficient formula gives $d_0=1$ and $d_1=3+2=5$. For $n\ge2$, the three potentially nonzero contributions are

d_n=3^n+2\cdot3^{n-1}-3^{n-2} =14\cdot3^{n-2}.

In particular, for $n=2$, this gives $d_2=9+6-1=14$, as the formula $14\cdot3^{n-2}$ also gives. Therefore the product expansion is

1+5x+14x^2+42x^3+126x^4+\cdots.

The calculation for $n\ge2$ includes the $a_2b_{n-2}$ contribution, while the separate calculation for $d_1$ does not: there is no term with index $n-2=-1$. This illustrates why the first coefficients should be checked rather than assuming one formula applies unchanged at every index.

Worked Example: A Product with Coefficients Counting Even Powers

For $|x|<1$, multiply the two geometric series

\left(\sum_{k=0}^{\infty}x^{2k}\right) \left(\sum_{j=0}^{\infty}x^j\right).

A contribution to the coefficient of $x^n$ occurs when $2k+j=n$, where both $k$ and $j$ are nonnegative integers. Thus $k$ can be any integer from $0$ through $\lfloor n/2\rfloor$, and for each such $k$, $j=n-2k$ is determined. There are $\lfloor n/2\rfloor+1$ choices, so

d_n=\lfloor n/2\rfloor+1.

The beginning of the product is therefore

1+x+2x^2+2x^3+3x^4+3x^5+\cdots.

For instance, the coefficient of $x^4$ comes from $k=0,1,2$, giving the three terms $x^4$, $x^2x^2$, and $x^4\cdot1$, each with coefficient $1$. The coefficient is $3=\lfloor4/2\rfloor+1$. The product theorem ensures that this coefficient calculation represents the pointwise product throughout the common interval $|x|<1$.

Worked Example: Convolution Produces Harmonic Sums

Consider the two power series

A(x)=\sum_{k=0}^{\infty}\frac{x^k}{k+1}, \qquad B(x)=\sum_{j=0}^{\infty}x^j.

Both have radius of convergence $1$. For $A$, this follows, for example, from the Root Criterion for a Power Series since $(1/(n+1))^{1/n}\to1$; $B$ is geometric. For $|x|<1$, the coefficient of $x^n$ in their product is

d_n=\sum_{k=0}^{n}\frac{1}{k+1} =1+\frac12+\cdots+\frac{1}{n+1}.

The first three coefficients are $d_0=1$, $d_1=1+1/2=3/2$, and $d_2=1+1/2+1/3=11/6$. Hence

A(x)B(x)=1+\frac32x+\frac{11}{6}x^2+\frac{25}{12}x^3+\cdots \qquad(|x|<1).

For $n=3$, the coefficient is $1+1/2+1/3+1/4=25/12$. The product coefficients need not be as simple as either original coefficient sequence: each one gathers all the cross-products that contribute to its degree.

Radius of Convergence and a Common Pitfall

The multiplication theorem guarantees convergence of the product series wherever both original series converge absolutely. It follows that the product radius is at least the smaller of the two original radii. It may be larger: the theorem gives a region in which the formula is valid, not necessarily the largest possible region. For example, if one factor is the zero series, its product is zero everywhere, regardless of the radius of the other factor.

A common error is to multiply corresponding coefficients and write $\sum a_nb_n(x-a)^n$. That operation does not, in general, represent the product of the functions. The coefficient of degree $n$ must include every pair of degrees adding to $n$, as in $\sum_{k=0}^{n}a_kb_{n-k}$. Even for degree $1$, the product coefficient is $a_0b_1+a_1b_0$, not just $a_1b_1$.

There is also an important difference between the finite formula for each coefficient and the infinite-series identity. The coefficient formula alone is a finite algebraic calculation; it does not establish that the resulting infinite series converges or that its sum equals the product. Absolute convergence inside the common radius supplies the control needed for those claims. The next step in studying products of series is to examine this same convolution operation for general convergent series, where there may be no power of $x-a$ to provide the extra structure.

Takeaway: The coefficient of degree $n$ in a product of power series is the finite convolution sum $\sum_{k=0}^{n}a_kb_{n-k}$. Inside the smaller original radius, absolute convergence makes the resulting series converge to the pointwise product.

Check Your Understanding

Use the coefficient formula and the multiplication theorem to answer the following questions.

Write the coefficients $d_0$, $d_1$, and $d_2$ of a product in terms of $a_n$ and $b_n$.
Why does the coefficient of degree $n$ involve only finitely many terms?
In the product theorem’s proof, what does absolute convergence allow us to bound?
What radius-of-convergence guarantee does the multiplication theorem provide?
For the product of $\sum_{k=0}^{\infty}x^{2k}$ and $\sum_{j=0}^{\infty}x^j$, why is the coefficient of $x^n$ equal to $\lfloor n/2\rfloor+1$?

What You'll Learn