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Power Series Operations · Tutorial 749 of 998

Adding and Scaling Power Series at a Common Center

Add or scale coefficients of power series with a common center, and identify the points where the resulting series represents the corresponding function.

Intermediate 10 min read

What You'll Learn

  • Form the coefficient sequence for the sum of two power series with a common center
  • Scale every coefficient by a constant and interpret the resulting series
  • Prove the pointwise sum rule using partial sums
  • Determine where both original series converge, including endpoints
  • Distinguish a guaranteed representation domain from the full convergence set of the new series
  • Recognize when cancellation or a zero scalar changes what can be concluded

Why Adding and Scaling Are Coefficient Operations

A power series represents a function wherever it converges. If two power series use the same center, their terms involve matching powers of the same centered variable. This lets us add the coefficients of like powers, or multiply every coefficient by one constant.

The coefficient calculation is straightforward; the important question is where the new series actually represents the sum or scalar multiple of the original functions. That requires checking convergence of the original series at the point in question. Points strictly inside both intervals of convergence are covered by the standard absolute-convergence property of power series. Endpoints need separate attention because a power series may converge at one endpoint and diverge at the other.

Definition: Two power series have a common center \(c\) if they can be written as \(\sum_{n=0}^{\infty}a_n(x-c)^n\) and \(\sum_{n=0}^{\infty}b_n(x-c)^n\). Their coefficientwise sum is the power series \(\sum_{n=0}^{\infty}(a_n+b_n)(x-c)^n\). For a scalar \(\lambda\), the coefficientwise scalar multiple of the first series is \(\sum_{n=0}^{\infty}\lambda a_n(x-c)^n\).

The index \(n\) begins at \(0\), so the constant terms are added or scaled too. These operations act on coefficients of the same power. They are not the multiplication of two power series, which is a different operation.

The Pointwise Addition and Scaling Rule

Theorem: Let \(A(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\) and \(B(x)=\sum_{n=0}^{\infty}b_n(x-c)^n\). At every real number \(x\) where both series converge, the coefficientwise sum converges and equals \(A(x)+B(x)\). For any real scalar \(\lambda\), wherever the series for \(A\) converges, the coefficientwise scalar multiple converges and equals \(\lambda A(x)\). If \(\lambda\ne 0\), the scaled series converges at exactly the same points as the original series.

The condition on addition is pointwise: both original series must converge at the \(x\) being considered. Within the interior of both convergence intervals, this condition holds, and both series converge absolutely. At an endpoint, convergence must be checked for each series separately. When both converge there, their sum converges there as well, whether or not either convergence is absolute.

For scaling, multiplying the terms by a fixed nonzero number preserves convergence in both directions. If the scalar is zero, every coefficient becomes zero, so the resulting series converges everywhere. In that case, the function expression \(0A(x)\) refers to the original function only where \(A(x)\) is defined; the zero series itself is defined for all real \(x\).

Why the Addition Rule Works

The proof uses partial sums. Fix a real number \(x\) at which both original series converge, and write \(t=x-c\). Let \(A_N(x)\) and \(B_N(x)\) be their partial sums through index \(N\). The partial sum of the coefficientwise sum is exactly \(A_N(x)+B_N(x)\), because finite sums can be regrouped in this way:

$$ \sum_{n=0}^{N}(a_n+b_n)t^n = \sum_{n=0}^{N}a_nt^n+\sum_{n=0}^{N}b_nt^n = A_N(x)+B_N(x). $$

By the assumption that the original series converge, \(A_N(x)\to A(x)\) and \(B_N(x)\to B(x)\) as \(N\to\infty\). The sum rule for limits therefore gives \(A_N(x)+B_N(x)\to A(x)+B(x)\). Since these are precisely the partial sums of the coefficientwise sum, that series converges to \(A(x)+B(x)\).

For scaling, the partial sum through \(N\) is \(\sum_{n=0}^{N}\lambda a_nt^n=\lambda A_N(x)\). If \(A_N(x)\to A(x)\), then \(\lambda A_N(x)\to\lambda A(x)\), which proves the claimed value. If \(\lambda\ne0\), convergence of the scaled partial sums also implies convergence of the original ones, because \(A_N(x)=\frac{1}{\lambda}\bigl(\lambda A_N(x)\bigr)\). This proves the converse as well.

Worked Example: Adding Two Geometric Series

Worked Example: Adding Two Geometric Series

Consider two series with center \(c=1\):

$$ P(x)=\sum_{n=0}^{\infty}\frac{(x-1)^n}{2^n}, \qquad Q(x)=\sum_{n=0}^{\infty}\frac{(x-1)^n}{3^n}. $$

The first is geometric with ratio \((x-1)/2\), so it converges when \(|x-1|<2\). The second has ratio \((x-1)/3\), so it converges when \(|x-1|<3\). Both converge whenever \(|x-1|<2\). In that interval, adding the coefficients gives:

$$ P(x)+Q(x) = \sum_{n=0}^{\infty} \left(\frac{1}{2^n}+\frac{1}{3^n}\right)(x-1)^n. $$

For example, take \(x=\frac{3}{2}\), so \(x-1=\frac12\). The first series has ratio \(\frac14\), and the second has ratio \(\frac16\). Thus:

$$ P\left(\frac32\right) = \sum_{n=0}^{\infty}\left(\frac14\right)^n = \frac{1}{1-\frac14} = \frac43, \qquad Q\left(\frac32\right) = \sum_{n=0}^{\infty}\left(\frac16\right)^n = \frac{1}{1-\frac16} = \frac65. $$

The coefficientwise sum gives the same value:

$$ \sum_{n=0}^{\infty} \left(\frac{1}{2^n}+\frac{1}{3^n}\right)\left(\frac12\right)^n = \sum_{n=0}^{\infty}\left(\frac14\right)^n + \sum_{n=0}^{\infty}\left(\frac16\right)^n = \frac43+\frac65 = \frac{20}{15}+\frac{18}{15} = \frac{38}{15}. $$

Checking by adding the function values gives \(\frac43+\frac65=\frac{38}{15}\), as required. The common convergence interval used here is \((-1,3)\), since \(|x-1|<2\) is equivalent to \(-1<x<3\). At \(x=3\), the terms of \(P\) are all \(1\), so they do not tend to zero. At \(x=-1\), the terms of \(P\) are \((-1)^n\), which also do not tend to zero. Thus \(P\) diverges at both endpoints; these conclusions use the actual terms at each endpoint, not a mistaken assumption that the two series have identical endpoint behavior.

Endpoints Must Be Checked Series by Series

A shared center does not imply shared endpoint behavior. One series might converge at an endpoint while the other diverges there. The addition rule applies at that point only if both converge. A useful method is to write the terms at each endpoint explicitly, then determine whether each original series converges before invoking the theorem.

Worked Example: Adding and Scaling with Convergent Endpoints

Define two series centered at \(0\):

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

Form \(G(x)=P(x)-2Q(x)\). The coefficient of \(x^n\) is \(\frac{1}{(n+1)^2}-\frac{2(-1)^n}{(n+1)^2}\), so:

$$ G(x) = \sum_{n=0}^{\infty} \frac{1-2(-1)^n}{(n+1)^2}x^n. $$

At \(x=1\), the absolute values of the terms of \(P\) and \(Q\) are both \(1/(n+1)^2\). The series \(\sum_{n=0}^{\infty}1/(n+1)^2\) converges, so both original series converge absolutely there. The combined coefficient is \(-1/(n+1)^2\) when \(n\) is even and \(3/(n+1)^2\) when \(n\) is odd. Its absolute value is at most \(3/(n+1)^2\), so \(G(1)\) converges absolutely too, and the theorem gives \(G(1)=P(1)-2Q(1)\).

At \(x=-1\), the absolute values of the terms of both original series are again \(1/(n+1)^2\), so both converge absolutely. The combined series also converges absolutely by the same bound. Therefore \(G(-1)=P(-1)-2Q(-1)\). For \(|x|<1\), all three series converge absolutely by comparison with a convergent geometric series or by the standard interior convergence property of power series. Consequently the resulting series represents \(P(x)-2Q(x)\) throughout \([-1,1]\).

This is the exact convergence interval of the resulting series. For \(|x|>1\), consider its terms at odd indices \(n=2k+1\). Their absolute values are \(3|x|^{2k+1}/(2k+2)^2\). The ratio of successive such magnitudes tends to \(|x|^2>1\), so they do not tend to zero. Since terms of a convergent series must tend to zero, the series diverges whenever \(|x|>1\).

The checks at \(1\) and \(-1\) are separate: both original series happen to converge at both endpoints in this example, but that conclusion comes from the \(1/(n+1)^2\) bound, not just from knowing their center.

Worked Example: Scaling a Geometric Series

Worked Example: Scaling a Geometric Series

Consider the series centered at \(-2\):

$$ H(x)=\sum_{n=0}^{\infty}\frac{(x+2)^n}{4^n}. $$

It is geometric with ratio \((x+2)/4\), so it converges when \(|x+2|<4\), or \(-6<x<2\). Scale its coefficients by \(-3\):

$$ -3H(x) = \sum_{n=0}^{\infty}\frac{-3(x+2)^n}{4^n}. $$

At \(x=0\), the ratio in the original series is \(2/4=1/2\). The original value is:

$$ H(0) = \sum_{n=0}^{\infty}\left(\frac12\right)^n = \frac{1}{1-\frac12} = 2. $$

The scaled series at \(x=0\) is \(\sum_{n=0}^{\infty}-3(1/2)^n\), whose value is:

$$ \sum_{n=0}^{\infty}-3\left(\frac12\right)^n = -3\sum_{n=0}^{\infty}\left(\frac12\right)^n = -3(2) = -6. $$

This equals \(-3H(0)=-3(2)=-6\), confirming the scaling rule. At the endpoints \(x=-6\) and \(x=2\), the scaled series has terms \(-3(-1)^n\) and \(-3\), respectively. Neither sequence tends to zero, so the scaled series diverges at both endpoints, just as the original does.

Common Pitfall: Confusing a Guaranteed Domain with the Only Possible Domain

The addition theorem guarantees that the coefficientwise sum represents \(A(x)+B(x)\) wherever both original series converge. It does not say that this is the only place where the coefficientwise sum can converge. Coefficients can cancel, so the combined series may converge where the original series do not.

For a concrete example, take \(\sum_{n=0}^{\infty}x^n\) and \(\sum_{n=0}^{\infty}(-1)x^n\). At \(x=1\), their terms are \(1\) and \(-1\), respectively. Neither series converges, because its terms do not tend to zero. But adding coefficients gives \(1+(-1)=0\) for every \(n\), so the resulting series is \(\sum_{n=0}^{\infty}0\cdot x^n\), which converges to \(0\) at every real \(x\).

Takeaway: The coefficientwise sum represents the sum of the two original functions wherever both original series converge. The combined series may have additional convergence points, but at such points the original functions need not both be defined.

A related scaling error is to assume that multiplying by zero preserves the original convergence set. It does not: every coefficient becomes zero, and the zero series converges everywhere. For a nonzero scalar, convergence is preserved exactly. For addition, check both original series at the point of interest before asserting equality with the sum of their function values.

Check Your Understanding

Use the common-center condition and the pointwise convergence requirements to answer these questions.

  1. For two power series centered at \(c\), what is the coefficient of \((x-c)^n\) in their coefficientwise sum?
  2. State the pointwise condition under which the coefficientwise sum represents the sum of the two power-series functions.
  3. Why does multiplying a convergent series by a nonzero scalar preserve convergence, while multiplying by zero can change its convergence set?
  4. If both original series converge at an endpoint, what can be concluded about their coefficientwise sum there?
  5. Can the coefficientwise sum converge at a point where one or both original series diverge? Explain why this does not contradict the addition theorem.

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