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Taylor Series Construction · Tutorial 833 of 998

Termwise Differentiation and Integration as Construction Tools

Use differentiation and integration of power series to construct new expansions, justify the operations, and track where the resulting series represent their functions.

Intermediate 10 min read

What You'll Learn

  • State when a power series may be differentiated or integrated term by term.
  • Explain why these operations preserve the radius of convergence.
  • Construct a series for a derivative from the geometric series.
  • Integrate a series for the exponential function and verify the result.
  • Check endpoint convergence separately from convergence inside the radius.

Building New Series from Known Ones

A known power series is more than a formula for one function. Inside its interval of convergence, differentiating or integrating it term by term produces another power series. This is a practical way to construct expansions for derivatives and antiderivatives without recalculating all their Taylor coefficients from scratch.

The interval matters. These operations preserve the radius of convergence, but that statement concerns the open interval around the center. At an endpoint, the original series and the transformed series can behave differently. A valid construction therefore has two parts: perform the operation where the power series theorem permits it, then check any endpoints separately.

Key principle: A power series can be differentiated or integrated term by term at points strictly inside its radius of convergence. The resulting power series has the same radius, but endpoint behavior must be investigated separately.

The Termwise Operations Theorem

Let a power series centered at \(c\) be

$$ \sum_{n=0}^{\infty} a_n(x-c)^n $$

and suppose its radius of convergence is \(R>0\). It converges for \(|x-c|<R\). On this open interval it represents a function \(f\), and the following formulas hold:

$$ f'(x)=\sum_{n=1}^{\infty} n a_n(x-c)^{n-1} $$
$$ \int \! f(x)\,dx = C+\sum_{n=0}^{\infty}\frac{a_n}{n+1}(x-c)^{n+1}, \qquad |x-c|<R. $$

Here \(C\) is an arbitrary constant. The integrated series is an antiderivative of \(f\) on the open interval. Both transformed series have radius of convergence \(R\). At \(x=c-R\) or \(x=c+R\), however, these formulas do not by themselves say whether the transformed series converges or equals the corresponding derivative or antiderivative.

Why Termwise Differentiation Works

The justification uses uniform convergence on intervals strictly inside the radius. Fix any \(r\) with \(0<r<R\). Choose \(s\) so that \(r<s<R\). Since the original power series converges absolutely at distance \(s\), the sum \(\sum_{n=0}^{\infty}|a_n|s^n\) is finite.

For \(|x-c|\le r\) and \(n\ge1\), the absolute value of a differentiated term is bounded by

$$ n|a_n|r^{n-1} = \frac{|a_n|s^n}{s}\,n\left(\frac{r}{s}\right)^{n-1}. $$

The sequence \(n(r/s)^{n-1}\) is bounded because \(r/s<1\). Thus the differentiated terms are bounded by a constant times the summable terms \(|a_n|s^n\). The differentiated series therefore converges uniformly on \(|x-c|\le r\). The original series also converges uniformly there.

Let \(P_N(x)=\sum_{n=0}^{N}a_n(x-c)^n\) be the \(N\)th partial sum. Each \(P_N\) is a polynomial, so the fundamental theorem of calculus gives

$$ P_N(x)-P_N(c)=\int_c^x P_N'(t)\,dt. $$

As \(N\) increases, \(P_N(x)\) converges to \(f(x)\), and \(P_N'\) converges uniformly on the interval under consideration to the differentiated series \(g\). Uniform convergence allows the limit to pass through this finite-interval integral. Since \(P_N(c)=a_0\), we obtain

$$ f(x)-a_0=\int_c^x g(t)\,dt. $$

The uniform limit \(g\) is continuous. The fundamental theorem of calculus now implies \(f'(x)=g(x)\), proving the differentiation formula at every interior point.

For integration, uniform convergence of the original series on each such interval allows integration of its partial sums and passage to the limit. Integrating each power gives the stated coefficients. Finally, multiplying or dividing the coefficients by factors involving \(n\) does not change the radius: the corresponding \(n\)th-root factors tend to \(1\). Thus differentiation and integration preserve the radius of convergence.

Worked Example: Differentiate the Geometric Series

Worked Example: Differentiate the Geometric Series

For \(|x|<1\), the geometric series gives \(\sum_{n=0}^{\infty}x^n=1/(1-x)\). Differentiate both sides term by term inside this interval.

The \(n=0\) term is the constant \(1\), whose derivative is \(0\). For each \(n\ge1\), the derivative of \(x^n\) is \(nx^{n-1}\). Therefore

$$ \sum_{n=1}^{\infty}nx^{n-1} = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}, \qquad |x|<1. $$

The check is direct: differentiating \(1/(1-x)\) gives \(1/(1-x)^2\), exactly the function represented by the new series. For example, at \(x=0\), the series is \(1+0+0+\cdots=1\), and the function value is \(1/(1-0)^2=1\).

The endpoint checks are different from the interior argument. At \(x=1\), the terms \(nx^{n-1}\) equal \(n\), so the series diverges, and the function \(1/(1-x)^2\) is undefined. At \(x=-1\), the terms are \(n(-1)^{n-1}\), which do not tend to zero; the series diverges even though \(1/(1-(-1))^2=1/4\) is defined. The power series representation is therefore established on \((-1,1)\), not at either endpoint.

Worked Example: Integrate the Exponential Series

Worked Example: Integrate the Exponential Series

The Maclaurin series for the exponential function is \(\sum_{n=0}^{\infty}x^n/n!=e^x\), and it converges for every real \(x\). Integrate from \(0\) to \(x\), which fixes the constant of integration:

$$ \int_0^x e^t\,dt = \int_0^x\left(\sum_{n=0}^{\infty}\frac{t^n}{n!}\right)dt = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)n!} = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)!}. $$

By the fundamental theorem of calculus, the integral is \(e^x-e^0=e^x-1\). Hence the construction gives

$$ e^x-1=\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)!}, \qquad x\in\mathbb R. $$

To check the result, differentiate the series. Its derivative is \(\sum_{n=0}^{\infty}x^n/n!=e^x\), which is the derivative of \(e^x-1\). Both sides are \(0\) at \(x=0\), so the constant also agrees. The radius remains infinite, so there are no finite endpoints to check.

Worked Example: Integrate the Geometric Series Twice

Worked Example: Integrate the Geometric Series Twice

Start with \(\sum_{n=0}^{\infty}t^n=1/(1-t)\) for \(|t|<1\). Integrate from \(0\) to \(x\), then integrate the resulting series once more from \(0\) to \(x\). Termwise integration is valid for every \(|x|<1\):

$$ F(x) = \sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+1)(n+2)}, \qquad |x|<1. $$

The construction gives \(F(0)=0\), \(F'(0)=0\), and \(F''(x)=1/(1-x)\). To identify the function, first integrate \(1/(1-t)\) from \(0\) to \(x\): this gives \(F'(x)=-\ln(1-x)\) for \(|x|<1\). Integrating once more, with \(F(0)=0\), gives

$$ F(x)=(1-x)\ln(1-x)+x, \qquad |x|<1. $$

Differentiate this expression to check it. The product rule gives

$$ F'(x)=-\ln(1-x)-1+1=-\ln(1-x), \qquad F''(x)=\frac{1}{1-x}. $$

It also satisfies \(F(0)=\ln(1)+0=0\) and \(F'(0)=-\ln(1)=0\), matching the construction.

The radius is \(1\), but the endpoint \(x=1\) can be checked separately. Substitution into the series gives positive terms, and the partial sums telescope:

$$ \sum_{n=0}^{N}\frac{1}{(n+1)(n+2)} = \sum_{n=0}^{N}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) = 1-\frac{1}{N+2} \longrightarrow 1. $$

The function also has limit \(1\) as \(x\) approaches \(1\) from below: \(x\to1\), and \((1-x)\ln(1-x)\to0\). Thus the endpoint series converges to the continuous limiting value \(F(1)=1\). This endpoint conclusion required its own check; it did not follow merely from termwise integration inside \((-1,1)\).

Endpoint Behavior and a Common Mistake

Inside the radius, the termwise formulas are reliable. At the boundary, convergence of the original series does not ensure convergence of its derivative series. For instance, the known logarithm expansion \(\sum_{n=1}^{\infty}(-1)^{n+1}x^n/n\) converges at \(x=1\). Differentiating it gives the geometric series \(\sum_{n=1}^{\infty}(-1)^{n+1}x^{n-1}\) for \(|x|<1\).

A tempting but incorrect argument is: “The original series converges at \(x=1\), so its differentiated series also converges there.” Substituting \(x=1\) into the differentiated series gives \(1-1+1-1+\cdots\). Its terms do not tend to zero, so it diverges, even though the derivative function \(1/(1+x)\) has the value \(1/2\) at \(x=1\). The correction is to use termwise differentiation only in the open interval and then test the endpoint series independently.

Endpoint rule: The radius of convergence identifies the open interval where termwise differentiation and integration are justified. It does not settle endpoint convergence or endpoint equality. Substitute each endpoint into the transformed series and analyze it separately.

When These Construction Tools Help

Termwise operations are useful when a function is difficult to expand directly but its derivative or antiderivative has a known series. Differentiation can produce expansions for derivatives and rational functions; integration can create series for accumulated quantities or antiderivatives. Repeating the operations can build further expansions, as in the twice-integrated example.

For any construction, keep three questions distinct:

  • What is the starting power series, and where does it converge?
  • What transformed coefficients result from differentiation or integration?
  • What happens at each endpoint of the original interval?

The first two questions are governed by the power series theorem. The last requires a separate convergence and function-value check whenever an endpoint is relevant.

Check Your Understanding

Answer each question, giving a reason for your conclusion.

  1. If a power series has radius \(R\), on what set is termwise differentiation guaranteed to represent the derivative?
  2. What coefficients result when \(\sum_{n=0}^{\infty}a_n(x-c)^n\) is integrated term by term?
  3. Why does termwise differentiation preserve the radius of convergence?
  4. Can convergence of a power series at an endpoint guarantee convergence of its differentiated series there? Explain.
  5. For the twice-integrated geometric series, how can the value at \(x=1\) be checked without relying on the open-interval theorem?

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