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Power Series Operations · Tutorial 773 of 998

Mixed Power-Series Operations Workshop

Combine power-series operations by calculating coefficients carefully and checking that each step is valid inside the relevant interval of convergence.

Intermediate 11 min read

What You'll Learn

  • Add power series by combining coefficients of matching powers.
  • Compute Cauchy-product coefficients as finite convolutions.
  • Differentiate and integrate power series term by term on interior intervals.
  • Track the common interval where combined operations are justified.
  • Check calculated coefficients by reversing or differentiating an operation.
  • Recognize and correct common errors in product coefficients.

Combining Operations Without Losing Track

Power-series calculations often involve several operations in succession: adding two series, multiplying them, and then differentiating or integrating the result. The central task is to keep the coefficients aligned with the correct powers of \(x\). A second task is to make sure the values of \(x\) under consideration lie in a region where every operation is valid.

For example, if two power series converge for different ranges, their sum and product can at least be handled where both original series converge. Termwise differentiation and integration are justified at interior points of a series’ interval of convergence; endpoints require separate analysis. This tutorial combines these established rules in complete calculations, rather than treating each operation in isolation.

Working principle: Calculate coefficients using the operation’s rule, and track convergence using the intersection of the intervals where the input series and the operation are valid. Unless endpoint convergence is checked separately, make conclusions only for interior points.

Rules for a Mixed Calculation

Write two power series centered at zero as \(A(x)=\sum_{n=0}^{\infty}a_nx^n\) and \(B(x)=\sum_{n=0}^{\infty}b_nx^n\). Here \(a_n\) and \(b_n\) are the coefficients of \(x^n\). The same reasoning applies to series centered at another point after expressing each series in powers of the corresponding centered variable.

For addition, the coefficient of \(x^n\) in \(A(x)+B(x)\) is \(a_n+b_n\). For a product, the coefficient of \(x^n\) is the finite sum \(\sum_{k=0}^{n}a_kb_{n-k}\), called the Cauchy-product coefficient. The index limits matter: only pairs of powers whose exponents add to \(n\) contribute.

$$ A(x)+B(x)=\sum_{n=0}^{\infty}(a_n+b_n)x^n, \qquad A(x)B(x)=\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}a_kb_{n-k}\right)x^n. $$

For a power series with radius of convergence \(R\), termwise differentiation and integration are valid for \(|x|<R\). In particular, differentiating \(\sum_{n=0}^{\infty}a_nx^n\) gives coefficients \((n+1)a_{n+1}\) at \(x^n\); integrating from \(0\) to \(x\) gives coefficient \(a_n/(n+1)\) at \(x^{n+1}\). Both resulting series have the same radius of convergence as the original series. These rules do not by themselves determine convergence at \(x=R\) or \(x=-R\).

$$ \frac{d}{dx}\sum_{n=0}^{\infty}a_nx^n = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^n, \qquad |x|<R, $$
$$ \int_0^x\sum_{n=0}^{\infty}a_nt^n\,dt = \sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}, \qquad |x|<R. $$

When both original series converge for \(|x|<r\), their sum and Cauchy product are valid there. The combined series can sometimes have a larger radius because of cancellation or other special coefficient patterns, but that possibility is not needed to justify the calculation on the shared interval. If differentiation or integration is then applied, the resulting operation is valid throughout that same open interval.

Why the Product Coefficients Are Convolutions

The coefficient rule for a product follows by multiplying terms and grouping together powers with the same exponent. Suppose \(|x|<r\), where both input series converge absolutely. Absolute convergence ensures that the terms in their product may be grouped by total exponent. Each product \(a_kx^k b_jx^j\) contributes to the power \(x^{k+j}\). To obtain \(x^n\), the nonnegative indices must satisfy \(k+j=n\), so \(j=n-k\) and \(k\) ranges from \(0\) to \(n\).

1
Multiply terms.
Each pair of terms contributes \(a_kb_jx^{k+j}\).
2
Fix a target power.
For the coefficient of \(x^n\), require \(k+j=n\).
3
Sum all contributing pairs.
Taking \(j=n-k\) gives \(\sum_{k=0}^{n}a_kb_{n-k}\).

More formally, the double series of absolute values is bounded by the product of the two convergent absolute-value sums:

$$ \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} |a_kb_jx^{k+j}| = \left(\sum_{k=0}^{\infty}|a_kx^k|\right) \left(\sum_{j=0}^{\infty}|b_jx^j|\right) <\infty. $$

Thus rearranging and grouping its terms by \(k+j=n\) is valid. For each \(n\), there are only \(n+1\) pairs, so the coefficient of \(x^n\) is exactly the finite convolution sum stated above. The absolute-convergence condition is important: it is what licenses this regrouping for the infinite product.

Worked Example: Add Two Series and Differentiate

Worked Example: Add Two Series and Differentiate

Let \(F(x)=\sum_{n=0}^{\infty}\frac{n+1}{3^n}x^n\) and \(G(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^n}x^n\). Find the first four coefficients of the derivative of \(H(x)=F(x)+G(x)\), and state where the calculation is valid.

For \(F\), the ratio of successive coefficient magnitudes tends to \(1/3\), so its radius is \(3\). For \(G\), that ratio is \(1/2\), so its radius is \(2\). Both series converge absolutely for \(|x|<2\). Addition and termwise differentiation are therefore justified on that common open interval.

The coefficients of \(F\), for powers \(x^0\) through \(x^3\), are \(1,\frac{2}{3},\frac{1}{3},\frac{4}{27}\). The corresponding coefficients of \(G\) are \(1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8}\). Thus the first four coefficients of \(H\) are calculated by matching powers:

$$ h_0=1+1=2,\quad h_1=\frac{2}{3}-\frac{1}{2}=\frac{1}{6},\quad h_2=\frac{1}{3}+\frac{1}{4}=\frac{7}{12},\quad h_3=\frac{4}{27}-\frac{1}{8}=\frac{32-27}{216}=\frac{5}{216}. $$

When \(H\) is differentiated, the coefficient of \(x^n\) is \((n+1)h_{n+1}\). Therefore the coefficients through \(x^3\) are:

$$ \begin{aligned} [x^0]H'(x)&=1\cdot h_1=\frac{1}{6},\\ [x^1]H'(x)&=2\cdot h_2=2\cdot\frac{7}{12}=\frac{7}{6},\\ [x^2]H'(x)&=3\cdot h_3=3\cdot\frac{5}{216}=\frac{5}{72},\\ [x^3]H'(x)&=4h_4 =4\left(\frac{5}{81}+\frac{1}{16}\right) =4\left(\frac{80+81}{1296}\right) =\frac{161}{324}. \end{aligned} $$

Here \(h_4\) uses the next coefficients: the \(x^4\) coefficient of \(F\) is \(5/81\), and that of \(G\) is \(1/16\). The result begins \(\frac{1}{6}+\frac{7}{6}x+\frac{5}{72}x^2+\frac{161}{324}x^3+\cdots\), valid for \(|x|<2\). As a check, differentiating the terms \(h_1x+h_2x^2+h_3x^3+h_4x^4\) gives \(h_1+2h_2x+3h_3x^2+4h_4x^3\), which reproduces each displayed coefficient.

Worked Example: Form a Product, Then Integrate

Worked Example: Form a Product, Then Integrate

Using the same \(F\) and \(G\), calculate the product coefficients through \(x^3\), then integrate the product from \(0\) to \(x\). The radii of the two input series are \(3\) and \(2\), respectively. Their product series is justified for \(|x|<2\) by absolute convergence, and termwise integration is valid there.

Write \(a_0=1,a_1=\frac{2}{3},a_2=\frac{1}{3},a_3=\frac{4}{27}\) for \(F\), and \(b_0=1,b_1=-\frac{1}{2},b_2=\frac{1}{4},b_3=-\frac{1}{8}\) for \(G\). For \(P(x)=F(x)G(x)=\sum_{n=0}^{\infty}p_nx^n\), use \(p_n=\sum_{k=0}^{n}a_kb_{n-k}\):

$$ \begin{aligned} p_0&=a_0b_0=1,\\ p_1&=a_0b_1+a_1b_0=-\frac{1}{2}+\frac{2}{3}=\frac{1}{6},\\ p_2&=a_0b_2+a_1b_1+a_2b_0 =\frac{1}{4}-\frac{1}{3}+\frac{1}{3} =\frac{1}{4},\\ p_3&=a_0b_3+a_1b_2+a_2b_1+a_3b_0\\ &=-\frac{1}{8}+\frac{1}{6}-\frac{1}{6}+\frac{4}{27} =\frac{-27+32}{216} =\frac{5}{216}. \end{aligned} $$

Integrating each term \(p_nx^n\) from \(0\) to \(x\) gives \(p_nx^{n+1}/(n+1)\). The first four terms of the integral are consequently:

$$ \int_0^x P(t)\,dt = x+\frac{1}{12}x^2+\frac{1}{12}x^3+\frac{5}{864}x^4+\cdots, \qquad |x|<2. $$

For example, the \(x^3\) coefficient is \(p_2/3=(1/4)/3=1/12\), and the \(x^4\) coefficient is \(p_3/4=(5/216)/4=5/864\). To check, differentiate the displayed integrated series: its first terms become \(1+\frac{1}{6}x+\frac{1}{4}x^2+\frac{5}{216}x^3+\cdots\), exactly the product coefficients just calculated.

Worked Example: Differentiate and Integrate Before Adding

Worked Example: Differentiate and Integrate Before Adding

Find the first four coefficients of \(K(x)=F'(x)+\int_0^xG(t)\,dt\), with \(F\) and \(G\) as above. Both component operations are valid for \(|x|<2\): differentiation and integration preserve the respective radii, and the resulting series can be added on their common interval.

From the differentiation rule, the coefficient of \(x^n\) in \(F'\) is \((n+1)a_{n+1}\). The coefficients through \(x^3\) are computed from \(a_1=\frac23,a_2=\frac13,a_3=\frac4{27},a_4=\frac5{81}\):

$$ [x^0]F'=\frac{2}{3},\quad [x^1]F'=\frac{2}{3},\quad [x^2]F'=\frac{4}{9},\quad [x^3]F'=\frac{20}{81}. $$

For the integral, \(b_nx^n\) becomes \(b_nx^{n+1}/(n+1)\). Its coefficients at \(x^0,x^1,x^2,x^3\) are therefore \(0,1,-\frac14,\frac1{12}\). Adding matching coefficients gives:

$$ \begin{aligned} [x^0]K&=\frac{2}{3}+0=\frac{2}{3},\\ [x^1]K&=\frac{2}{3}+1=\frac{5}{3},\\ [x^2]K&=\frac{4}{9}-\frac{1}{4}=\frac{16-9}{36}=\frac{7}{36},\\ [x^3]K&=\frac{20}{81}+\frac{1}{12} =\frac{80+27}{324}=\frac{107}{324}. \end{aligned} $$

Thus \(K(x)=\frac23+\frac53x+\frac7{36}x^2+\frac{107}{324}x^3+\cdots\), for \(|x|<2\). As a check, differentiate this result: the first terms of \(K'\) are \(\frac53+\frac7{18}x+\frac{107}{108}x^2+\cdots\). Independently, differentiating the definition gives \(K'=F''+G\). Its first three coefficients are \(\frac23+1=\frac53\), \(\frac89-\frac12=\frac7{18}\), and \(\frac{20}{27}+\frac14=\frac{107}{108}\), agreeing with the differentiated result.

A Common Error: Omitting Product Terms

A frequent mistake is to calculate the coefficient of \(x^n\) in a product using only one pair of terms. For the product \(FG\) above, taking just \(a_2b_0\) would give \(1/3\) as the coefficient of \(x^2\). That is incomplete: \(a_0b_2\) and \(a_1b_1\) also produce \(x^2\). The correct calculation is:

$$ [x^2](FG) =a_0b_2+a_1b_1+a_2b_0 =\frac14-\frac13+\frac13 =\frac14. $$

Another source of error is shifting the index incorrectly after differentiation or integration. Differentiation sends \(a_{n+1}x^{n+1}\) to \((n+1)a_{n+1}x^n\); integration sends \(a_nx^n\) to \(a_nx^{n+1}/(n+1)\). Writing the transformed power before assigning its coefficient helps prevent off-by-one errors.

Convergence Checks in Practice

A coefficient calculation alone does not establish that a series identity is valid at every real number. For the examples above, the original series have radii \(3\) and \(2\), so the shared open interval is \((-2,2)\). All the additions, products, differentiations, and integrations used in those examples are justified there. The calculations make no claim about \(x=2\) or \(x=-2\).

For new problems, record the valid interval at the beginning and carry it through the calculation. If the inputs have radii \(R_A\) and \(R_B\), the shared interior interval is \(|x|<\min(R_A,R_B)\). Addition and multiplication are valid on that interval, and termwise differentiation or integration of the resulting power series is also valid at its interior points. A boundary point needs a separate convergence check; it should not be included simply because the formal coefficient calculation still makes sense.

Takeaway: In a mixed power-series problem, keep two records: a coefficient calculation that shows how each power is formed, and a convergence statement that identifies where the operations are justified.

Check Your Understanding

For each question, keep both the coefficient rule and its convergence condition in view.

  1. If \(A(x)=\sum_{n=0}^{\infty}a_nx^n\) and \(B(x)=\sum_{n=0}^{\infty}b_nx^n\), what finite sum gives the coefficient of \(x^3\) in \(A(x)B(x)\)?
  2. Suppose \(A\) and \(B\) have radii \(5\) and \(2\). On what open interval are addition and the Cauchy product guaranteed to be valid?
  3. If the coefficient of \(x^4\) in a series is \(c_4\), what term does it contribute to the derivative, and what term does it contribute to an integral from \(0\) to \(x\)?
  4. For the \(F\) and \(G\) in the examples, calculate the coefficient of \(x^2\) in \(FG\) using every contributing pair.
  5. Why does validity of a power-series operation for \(|x|<R\) not automatically establish validity at \(x=R\)?

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