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Taylor Series Construction · Tutorial 839 of 998

Products of Taylor Series by Coefficient Convolution

Use coefficient convolution to multiply Taylor series and determine where the resulting expansion represents the product.

Intermediate 10 min read

What You'll Learn

  • Define the Cauchy product of two power series.
  • Compute product coefficients by finite convolution sums.
  • Prove why absolute convergence permits the factors to be multiplied and regrouped.
  • Find product expansions for an exponential-trigonometric product and geometric-series examples.
  • Distinguish the guaranteed interval of convergence from possible extension beyond it.

Why Products Require a New Coefficient Rule

When two polynomials are multiplied, each coefficient of the product collects all pairs of terms whose degrees add to the same total. The same pattern applies to power series, but an infinite sum requires care: we need conditions that justify multiplying and regrouping its terms.

The resulting rule is called the Cauchy product. It gives a systematic way to construct Taylor series for products without repeatedly differentiating the product function.

Key idea: If the coefficients of two power series are \(a_n\) and \(b_n\), then the coefficient of degree \(n\) in their product is the finite sum of all products \(a_kb_{n-k}\) for \(k=0,\ldots,n\).

Definition of the Cauchy Product

Consider two power series centered at the same point, written using \(t=x-a\):

$$ \sum_{n=0}^{\infty}a_nt^n \qquad\text{and}\qquad \sum_{n=0}^{\infty}b_nt^n. $$
Definition: The Cauchy product of these series is the power series \(\sum_{n=0}^{\infty}c_nt^n\) whose coefficients are $$ c_n=\sum_{k=0}^{n}a_kb_{n-k}. $$ The sum defining each \(c_n\) has only \(n+1\) terms, so each product coefficient is determined by a finite calculation.

For example, the first few coefficients are

$$ c_0=a_0b_0,\qquad c_1=a_0b_1+a_1b_0,\qquad c_2=a_0b_2+a_1b_1+a_2b_0. $$

The indices in each sum pair terms whose powers add to \(n\): \(t^k t^{n-k}=t^n\). This is exactly the rule obtained by multiplying polynomials and collecting like powers.

Why the Rule Works

For polynomials, multiplying and regrouping finitely many terms is always valid. For infinite series, the corresponding argument is valid when both series converge absolutely at the value of \(t\). Absolute convergence means that the series of absolute values converges.

Theorem (Cauchy product): If \(\sum_{n=0}^{\infty}a_nt^n\) and \(\sum_{n=0}^{\infty}b_nt^n\) both converge absolutely at a given \(t\), then their Cauchy product converges absolutely there, and $$ \left(\sum_{n=0}^{\infty}a_nt^n\right) \left(\sum_{n=0}^{\infty}b_nt^n\right) = \sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_kb_{n-k}\right)t^n. $$

Here is a justification of the theorem. The absolute values of all pairwise products have finite total sum, because

$$ \sum_{k=0}^{\infty}\sum_{m=0}^{\infty} |a_kb_m t^{k+m}| = \left(\sum_{k=0}^{\infty}|a_kt^k|\right) \left(\sum_{m=0}^{\infty}|b_mt^m|\right) <\infty. $$

Thus these products can be regrouped according to their total degree \(n=k+m\). The terms with total degree \(n\) are precisely \(a_kb_{n-k}t^n\), for \(k=0,\ldots,n\). Regrouping gives the stated coefficient formula and preserves the sum. More explicitly, the partial sum through degree \(N\) collects the pairs with \(k+m\leq N\). As \(N\) increases, these collections include every fixed pair, and absolute convergence ensures that the sum of the omitted terms tends to zero. Their limit is therefore the product of the two original sums.

Where the Product Expansion Is Guaranteed to Work

Suppose the first series has radius of convergence \(R_1\) and the second has radius \(R_2\). A power series converges absolutely whenever \(|t|\) is smaller than its radius. Therefore both series converge absolutely when \(|t|<\min(R_1,R_2)\), and the Cauchy product theorem applies throughout that interval.

Convergence guarantee: The Cauchy product represents the product of the two functions at every point satisfying $$ |x-a|<\min(R_1,R_2). $$ This is a guaranteed interval, not necessarily the full radius of convergence of the product series. Cancellation among the coefficients can make that radius larger.

The endpoints require separate attention. At an endpoint, one or both original series may fail to converge absolutely, so the theorem above does not automatically apply. If both converge absolutely there, the theorem still applies. If absolute convergence is absent, the convergence and value of the Cauchy product must be checked using an appropriate argument for that case; they cannot be inferred from the interior result alone.

Worked Example: The Product \(e^x\cos x\)

Worked Example: The Maclaurin Series for \(e^x\cos x\)

The Maclaurin series for the exponential and cosine are

$$ e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}, \qquad \cos x=\sum_{j=0}^{\infty}\frac{(-1)^j x^{2j}}{(2j)!}. $$

The exponential series has radius \(\infty\), and the cosine series also has radius \(\infty\). Both converge absolutely for every real \(x\), so their Cauchy product represents \(e^x\cos x\) for every real \(x\).

In the cosine series, odd-degree coefficients are zero. Thus the degree-\(n\) product coefficient is

$$ c_n=\sum_{j=0}^{\lfloor n/2\rfloor} \frac{1}{(n-2j)!}\frac{(-1)^j}{(2j)!}. $$

For \(n=0\) through \(n=5\), the coefficients are computed as follows:

$$ \begin{aligned} c_0&=1,\\ c_1&=\frac{1}{1!}=1,\\ c_2&=\frac{1}{2!}-\frac{1}{0!2!} =\frac12-\frac12=0,\\ c_3&=\frac{1}{3!}-\frac{1}{1!2!} =\frac16-\frac12=-\frac13,\\ c_4&=\frac{1}{4!}-\frac{1}{2!2!}+\frac{1}{0!4!} =\frac{1}{24}-\frac14+\frac{1}{24}=-\frac16,\\ c_5&=\frac{1}{5!}-\frac{1}{3!2!}+\frac{1}{1!4!} =\frac{1}{120}-\frac{1}{12}+\frac{1}{24}=-\frac{1}{30}. \end{aligned} $$

Consequently, the beginning of the product series is

$$ e^x\cos x = 1+x-\frac{x^3}{3}-\frac{x^4}{6}-\frac{x^5}{30}+\cdots. $$

To check the coefficient calculation, multiply the two series through degree five. The exponential coefficients through that degree are \(1,1,\frac12,\frac16,\frac1{24},\frac1{120}\); the cosine coefficients are \(1,0,-\frac12,0,\frac1{24},0\). Collecting each degree gives, in order, \(1,1,0,-\frac13,-\frac16,-\frac1{30}\), exactly the coefficients above. Since both original series converge absolutely for every real \(x\), the expansion is valid on all of \(\mathbb R\).

Worked Example: Multiplying Two Geometric Series

Worked Example: The Product \(1/(1-x^2)\)

For \(|x|<1\), the geometric-series formula gives

$$ \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k, \qquad \frac{1}{1+x}=\sum_{m=0}^{\infty}(-1)^m x^m. $$

Both series converge absolutely when \(|x|<1\). The coefficient of \(x^n\) in their Cauchy product is

$$ c_n=\sum_{k=0}^{n}1\cdot(-1)^{n-k}. $$

If \(n\) is even, this is a sum of alternating signs beginning and ending with \(1\), so \(c_n=1\). If \(n\) is odd, each \(1\) cancels with a \(-1\), so \(c_n=0\). Therefore

$$ \left(\sum_{k=0}^{\infty}x^k\right) \left(\sum_{m=0}^{\infty}(-1)^m x^m\right) = 1+x^2+x^4+x^6+\cdots = \sum_{j=0}^{\infty}x^{2j}. $$

Check the result by using the geometric-series formula on the final series and simplifying the original product:

$$ \sum_{j=0}^{\infty}x^{2j} =\frac{1}{1-x^2}, \qquad \frac{1}{1-x}\frac{1}{1+x} =\frac{1}{(1-x)(1+x)} =\frac{1}{1-x^2}. $$

The two expressions agree for \(|x|<1\), the interval where both starting series converge absolutely.

Worked Example: Cancellation Can Increase the Radius

Worked Example: Multiplying \(1-x\) by a Geometric Series

For \(|x|<1\), write

$$ 1-x=1-x+0x^2+0x^3+\cdots, \qquad \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n. $$

The first series has coefficients \(a_0=1\), \(a_1=-1\), and \(a_k=0\) for \(k\geq2\). The second has \(b_n=1\) for every \(n\geq0\). The convolution gives \(c_0=a_0b_0=1\). For \(n\geq1\),

$$ c_n=a_0b_n+a_1b_{n-1}=1-1=0. $$

Thus the product series is just \(1\). Checking against the functions on the interval where the geometric expansion is valid gives

$$ (1-x)\sum_{n=0}^{\infty}x^n =(1-x)\frac{1}{1-x}=1, \qquad |x|<1. $$

The product series \(1+0x+0x^2+\cdots\) converges for every real \(x\), even though the geometric series used to construct it converges only for \(|x|<1\). The Cauchy product theorem guarantees the identity on the shared interval; the resulting constant series then has a larger interval of convergence. In particular, the original geometric series does not converge at \(x=1\), so its value there cannot be used in the multiplication argument.

Common Pitfall: Matching the Wrong Indices

A frequent error is to multiply only coefficients with the same index, writing \(c_n=a_nb_n\). That omits the cross terms. For instance, the degree-two coefficient is not just \(a_2b_2\): it receives contributions from the degree pairs \(0+2\), \(1+1\), and \(2+0\).

Correction: To obtain the coefficient of degree \(n\), include every pair of degrees that sum to \(n\): $$ c_n=a_0b_n+a_1b_{n-1}+\cdots+a_nb_0. $$

Another important caution is to distinguish the interval where the theorem guarantees the product identity from the full convergence interval of the resulting series. The product series may converge farther because of cancellation, as the preceding example shows. Conversely, convergence at an endpoint cannot be concluded merely from the fact that both series work in the interior.

Using Convolution to Construct Taylor Series

If two functions are represented near the same center \(a\) by Taylor series, the Cauchy product gives the coefficients of their product. For \(f(a+t)=\sum a_nt^n\) and \(g(a+t)=\sum b_nt^n\), the product coefficients are \(c_n=\sum_{k=0}^n a_kb_{n-k}\). On the shared open interval of convergence, the theorem shows that this series equals \(f(a+t)g(a+t)\); it is therefore the Taylor expansion of that product there.

The practical method is finite at each degree: list the coefficient pairs that add to the target degree, multiply each pair, and add the results. Then establish the interval where both input series converge absolutely. Only after that should any claim about endpoints or a larger interval be considered.

Check Your Understanding

Use the definition and convergence conditions to answer the following questions.

  1. Write the convolution formula for the coefficient of \(t^4\) in the product of \(\sum a_nt^n\) and \(\sum b_nt^n\).
  2. What condition on the two series at a particular value of \(t\) guarantees that the Cauchy product equals their product?
  3. If the two series have radii \(3\) and \(5\), on what open interval around their common center is the product identity guaranteed?
  4. Why can the product series have a larger radius of convergence than either input series?
  5. Does convergence of both series at an endpoint by itself guarantee that their Cauchy product converges to the product of their sums? Explain.

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