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Power Series Operations · Tutorial 755 of 998

Product Radii and Coefficient Cancellation

Use the Cauchy product to establish a guaranteed convergence region, then track coefficient cancellation to determine when the product has a larger radius.

Intermediate 10 min read

What You'll Learn

  • State the radius guarantee for the product of two power series with a common center.
  • Explain why absolute convergence permits regrouping terms into Cauchy product coefficients.
  • Compute product coefficients by convolution.
  • Distinguish a guaranteed lower bound on the product radius from its actual radius.
  • Identify cancellations that make the product radius larger than either factor radius.
  • Check the radius of a product using its coefficients and convergence behavior.

Why Product Radii Need Care

A power series has a radius that describes a disk around its center where the series converges. When two power series are multiplied, their product is again represented by a power series, but its radius cannot be determined merely by looking at the two original radii.

There is a reliable lower bound: the product series converges at least inside the smaller of the two factor radii. This does not say that the product radius must equal that smaller radius. The convolution of the factor coefficients can create cancellations, and those cancellations can remove enough terms to give the product a larger radius.

The distinction between a guaranteed region and the actual radius is useful whenever power series are multiplied. It prevents an unjustified conclusion about the product while giving a method for determining what happens in particular examples.

Power Series and Their Product Coefficients

Consider two power series with the same center \(c\):

$$ \sum_{n=0}^{\infty} a_n(x-c)^n \qquad\text{and}\qquad \sum_{n=0}^{\infty} b_n(x-c)^n. $$

Their radii of convergence will be denoted by \(R_a\) and \(R_b\). A series converges absolutely whenever \(|x-c|\) is strictly less than its radius. At points on the boundary, convergence must be checked separately; the radius alone does not determine the boundary behavior.

To find the coefficient of \((x-c)^n\) in the product, pair terms whose exponents add to \(n\). The resulting coefficient is

$$ d_n=\sum_{k=0}^{n} a_k b_{n-k}. $$

This finite sum is called the convolution of the two coefficient sequences. For instance, the first three coefficients are

$$ d_0=a_0b_0,\qquad d_1=a_0b_1+a_1b_0,\qquad d_2=a_0b_2+a_1b_1+a_2b_0. $$

Each coefficient combines all pairs of factor terms whose powers add to the required exponent. Different summands can have opposite signs, so a coefficient can be much smaller than the individual terms used to form it, or even equal to zero.

Definition: The Cauchy product of the two power series is the series \(\sum_{n=0}^{\infty}d_n(x-c)^n\), where \(d_n=\sum_{k=0}^{n}a_kb_{n-k}\). The formula defines the product coefficients; the convergence argument below establishes where this series represents the product of the original series.

The Guaranteed Radius of the Product

Theorem: If the two power series have radii \(R_a\) and \(R_b\), their Cauchy product converges absolutely whenever $$ |x-c|<\min(R_a,R_b). $$ On this region, its sum equals the product of the sums of the two original series. Consequently, the product series has radius at least \(\min(R_a,R_b)\).

The radius guarantee is an interior statement. It does not promise convergence at either factor's boundary, and it does not rule out convergence of the product outside the guaranteed region.

Why the Guarantee Holds

Let \(r=|x-c|\) and suppose \(r<\min(R_a,R_b)\). Both factor series then converge absolutely, so the two sums of absolute values below are finite:

$$ \sum_{k=0}^{\infty}|a_k|r^k<\infty, \qquad \sum_{j=0}^{\infty}|b_j|r^j<\infty. $$

For each product coefficient, the triangle inequality gives \(|d_n|\leq\sum_{k=0}^{n}|a_k||b_{n-k}|\). Summing these bounds over \(n\) yields a double sum whose terms are all nonnegative. It can be regrouped by the two indices, and its value factors into the product of the two finite sums above.

$$ \begin{aligned} \sum_{n=0}^{\infty}|d_n|r^n &\leq \sum_{n=0}^{\infty}\sum_{k=0}^{n} |a_k||b_{n-k}|r^n\\ &=\sum_{k=0}^{\infty}\sum_{j=0}^{\infty} |a_k|r^k\,|b_j|r^j\\ &=\left(\sum_{k=0}^{\infty}|a_k|r^k\right) \left(\sum_{j=0}^{\infty}|b_j|r^j\right)<\infty. \end{aligned} $$

Here the change of indices is \(j=n-k\), so \(n=k+j\). The finite total absolute sum justifies regrouping the terms of the product: terms can be grouped by their total exponent \(n\) without changing the sum. Grouping that way gives exactly the coefficients \(d_n\). Thus the Cauchy product converges absolutely and has sum equal to the product of the factor sums. Since this holds for every \(r<\min(R_a,R_b)\), the product radius is at least the smaller factor radius.

Worked Example: A Product with Unequal Factor Radii

Worked Example: A Product with Unequal Factor Radii

Find the product coefficients and radius for

$$ \left(\sum_{n=0}^{\infty}x^n\right) \left(\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n\right). $$

The geometric series show that the first factor has radius \(1\) and the second has radius \(2\). For the coefficient \(d_n\), convolve \(a_k=1\) with \(b_{n-k}=2^{-(n-k)}\):

$$ \begin{aligned} d_n &=\sum_{k=0}^{n}1\cdot 2^{-(n-k)} =\sum_{j=0}^{n}2^{-j}\\ &=\frac{1-2^{-(n+1)}}{1-\frac12} =2-2^{-n}. \end{aligned} $$

In particular, \(d_0=1\), \(d_1=\frac32\), and \(d_2=\frac74\). The product series begins \(1+\frac32x+\frac74x^2+\cdots\). The coefficient formula also gives the product of the geometric sums whenever \(|x|<1\):

$$ \left(\sum_{n=0}^{\infty}x^n\right) \left(\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n\right) =\frac{1}{(1-x)(1-x/2)}. $$

The theorem guarantees convergence for \(|x|<1\). In fact, \(d_n=2-2^{-n}\) tends to \(2\). If \(|x|>1\), the terms \(d_nx^n\) do not tend to zero, so the series diverges. At \(|x|=1\), their absolute values tend to \(2\), so the terms again do not tend to zero. The product radius is therefore exactly \(1\), the smaller factor radius.

Check: The first convolution coefficients are \(d_0=1\), \(d_1=1+\frac12=\frac32\), and \(d_2=1+\frac12+\frac14=\frac74\), agreeing with the formula.

Cancellation Can Increase the Radius

The theorem provides a region where the product is guaranteed to converge, not a claim that the product radius equals the smaller factor radius. To determine the actual radius, inspect the product coefficients. If convolution makes many coefficients zero, or makes the remaining coefficients small enough, the product series may converge much farther than either factor series.

Worked Example: Two Radius-One Factors Whose Product Is Constant

Consider the two series centered at zero

$$ \sum_{n=0}^{\infty}a_nx^n=\frac{1+x}{1-x}, \qquad \sum_{n=0}^{\infty}b_nx^n=\frac{1-x}{1+x}, \qquad |x|<1. $$

Using geometric series, the first coefficients are \(a_0=1\) and \(a_n=2\) for \(n\geq1\). For the second, \(b_0=1\) and \(b_n=2(-1)^n\) for \(n\geq1\). Both coefficient sequences have terms whose absolute values do not tend to zero, so each series diverges when \(|x|=1\); each has radius \(1\).

Now calculate the product coefficient \(d_n\). For \(n\geq1\), the two end terms in the convolution are \(a_0b_n=2(-1)^n\) and \(a_nb_0=2\). The interior terms, when present, are \(a_kb_{n-k}=4(-1)^{n-k}\). Thus

$$ d_n=2(-1)^n+2+4\sum_{j=1}^{n-1}(-1)^j. $$

If \(n\) is even, the alternating sum has an odd number of terms and equals \(-1\), giving \(d_n=2+2-4=0\). If \(n\) is odd, that sum has an even number of terms and equals \(0\), giving \(d_n=-2+2=0\). Also \(d_0=a_0b_0=1\). Therefore the Cauchy product is the constant series \(1\), with infinite radius of convergence.

Check: On \(|x|<1\), multiplying the displayed rational expressions gives \(\frac{1+x}{1-x}\cdot\frac{1-x}{1+x}=1\), consistent with the computed coefficients. Each factor has radius \(1\), while the product has infinite radius because every coefficient after the constant term cancels.

The Smaller-Radius Bound Can Be Exact

Cancellation need not increase the radius. In the next example, the product coefficients grow at a rate that makes the smaller factor radius the exact product radius. This also shows why the theorem gives a lower bound rather than an equality.

Worked Example: A Product Whose Radius Equals the Smaller Factor Radius

Take

$$ \left(\sum_{n=0}^{\infty}x^n\right) \left(\sum_{n=0}^{\infty}(2x)^n\right). $$

The first radius is \(1\), and the second is \(\frac12\). The convolution coefficient is

$$ \begin{aligned} d_n &=\sum_{k=0}^{n}1^k2^{n-k} =\sum_{j=0}^{n}2^j\\ &=2^{n+1}-1. \end{aligned} $$

The first values are \(d_0=1\), \(d_1=3\), and \(d_2=7\). On \(|x|<\frac12\), both factors converge and their product is \(\frac{1}{(1-x)(1-2x)}\). For \(|x|>\frac12\), \(d_n|x|^n=(2^{n+1}-1)|x|^n\) does not tend to zero. At \(|x|=\frac12\), the absolute values of the terms are \(2-2^{-n}\), which tend to \(2\), so the product series diverges there as well. Its radius is exactly \(\frac12\).

Check: Convolution gives \(d_1=1\cdot2+1\cdot1=3\) and \(d_2=1\cdot4+1\cdot2+1\cdot1=7\), matching \(2^{n+1}-1\).

A Common Mistake: Multiplying Matching Coefficients

A frequent error is to assume that the coefficient of \(x^n\) in a product is simply \(a_nb_n\). That pairs only one contribution and misses all other pairs of powers that add to \(n\).

In the first worked example, this incorrect shortcut would give the coefficient of \(x\) as \(a_1b_1=1\cdot\frac12=\frac12\). But \(x\) can arise either from the constant term of the first factor times the \(x\)-term of the second, or from the \(x\)-term of the first times the constant term of the second. The correct computation is

$$ d_1=a_0b_1+a_1b_0 =1\cdot\frac12+1\cdot1 =\frac32. $$

Another mistake is to conclude that the product radius always equals the smaller factor radius. The example with reciprocal factors shows why that fails: the smaller radius is a guaranteed minimum, while coefficient cancellation can make the actual radius larger.

What to Remember

Key takeaway: For power series with a common center, the Cauchy product converges and represents the product inside the smaller factor radius. Its actual radius can be greater, because convolution may cancel coefficients; determine that radius from the product coefficients rather than assuming equality.

This distinction makes product-radius arguments both useful and careful. The guaranteed interior region follows from absolute convergence. Beyond it, the coefficient sequence decides the answer: persistent growth can keep the radius at the lower bound, while extensive cancellation can enlarge it.

Check Your Understanding

Answer each question using the coefficient convolution and the radius guarantee.

  1. Two power series centered at the same point have radii \(3\) and \(5\). What region is guaranteed to contain an absolutely convergent Cauchy product?
  2. Write the convolution formula for the coefficient of \((x-c)^3\).
  3. Why does absolute convergence of both factor series allow the product terms to be regrouped by total exponent?
  4. In the example with reciprocal factors, what happens to each product coefficient after the constant coefficient, and what does that imply about the product radius?
  5. For the product \(\left(\sum x^n\right)\left(\sum (2x)^n\right)\), compute the coefficient of \(x^2\) by convolution and state the product radius.

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