AP Calculus BC: Differentiation and Integration of Power Series

Power series are more than just infinite polynomials—they are versatile tools for representing complex functions. The true power of calculus is revealed when you learn that you can differentiate and integrate these series just like finite polynomials, term by term. This technique allows you to derive new series from known ones, solve differential equations, and evaluate integrals that are otherwise impossible, forming a critical bridge to understanding Taylor and Maclaurin series in depth.

The Foundation: Term-by-Term Operations and Convergence

A power series centered at $x=a$ is an expression of the form ${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$. The most important property for calculus operations is its interval of convergence, which consists of all $x$-values for which the series converges. Crucially, within this interval, the series converges to a function $f(x)$.

The core theorem states: If a power series ${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$ has a radius of convergence $R>0$, then the function $f$ it defines is differentiable and integrable on the interval $(a-R,a+R)$. Moreover, the derivative and integral can be found by differentiating or integrating each term of the series individually.

Derivative: $$f^{\prime }(x)=\frac{d}{dx}\left[\sum _{n=0}^{\infty }{c}_n(x-a{)}^n\right]=\sum _{n=1}^{\infty }n{c}_n(x-a{)}^{n-1}$$ Integral: $$\int f(x)dx=\int \left[\sum _{n=0}^{\infty }{c}_n(x-a{)}^n\right]dx=C+\sum _{n=0}^{\infty }\frac{c_n}{n+1}(x-a{)}^{n+1}$$

The critical insight is that both the derived series and the integrated series have the same radius of convergence, $R$. However, convergence at the endpoints $x=a\pm R$ may differ and must be checked separately for each new series.

Term-by-Term Differentiation: Finding New Series

Differentiation is applied term-by-term to find power series for derivatives of known functions. This is often the simplest way to derive series for functions related by differentiation.

Example: You know the Maclaurin series for $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n$ for $|x|<1$. How do you find the series for $\frac{1}{(1-x{)}^2}$? Notice that $\frac{1}{(1-x{)}^2}$ is the derivative of $\frac{1}{1-x}$. Therefore, you can differentiate the geometric series term-by-term:

$$\frac{d}{dx}\left[\frac{1}{1-x}\right]=\frac{d}{dx}\left[\sum _{n=0}^{\infty }{x}^n\right]$$ $$\frac{1}{(1-x{)}^2}=\sum _{n=1}^{\infty }n{x}^{n-1}$$

You can re-index this series. Let $k=n-1$, so when $n=1$, $k=0$. Then: $$\frac{1}{(1-x{)}^2}=\sum _{k=0}^{\infty }(k+1)x^k=1+2x+3x^2+4x^3+...$$

Both series have the same radius of convergence, $R=1$. You must check the endpoints: the original series for $1/(1-x)$ diverges at $x=\pm 1$, so the differentiated series will also diverge at those endpoints.

Term-by-Term Integration: Deriving Series for New Functions

Integration term-by-term is exceptionally powerful for finding series representations of functions defined by integrals, most famously for the arctangent and natural logarithm functions.

Deriving the Arctangent Series: You start with the geometric series formula, substituting $-x^2$ for $x$: $$\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}=\sum _{n=0}^{\infty }(-x^2{)}^n=\sum _{n=0}^{\infty }(-1{)}^n{x}^{2n},\text{for }|x|<1.$$

You know that $\arctan (x)$ is an antiderivative of $\frac{1}{1+x^2}$. Therefore, you integrate the series term-by-term: $$\arctan (x)=\int \frac{1}{1+x^2}dx=\int \left[\sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}\right]dx$$ $$=C+\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1}$$

To find the constant of integration $C$, substitute a known value. Let $x=0$: $\arctan (0)=0$, and the series at $x=0$ is just $C$. Therefore, $C=0$. This gives you the famous Maclaurin series for arctangent: $$\arctan (x)=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$$

This series has a radius of convergence $R=1$. Remarkably, it converges at the endpoint $x=1$ (by the Alternating Series Test), giving you the beautiful formula: $\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$

Connecting to Taylor Series and Solving Differential Equations

These operations are not just tricks; they are foundational to building Taylor series. The standard method for finding a Taylor series involves calculating derivatives at a point. The term-by-term differentiation theorem guarantees that the series you build this way can, in fact, be differentiated to recover the derivative series.

Furthermore, power series methods are essential for solving certain differential equations, especially in engineering contexts. When faced with a differential equation that cannot be solved with elementary methods, you assume the solution can be written as a power series $y=\sum c_n{x}^n$. You then substitute the series, its derivative, and its second derivative into the equation. By matching coefficients for like powers of $x$ (a method called "equating coefficients"), you can derive a recurrence relation that defines all the coefficients $c_n$, thereby constructing the solution.

Example Scenario: You might use this method to solve Bessel's Equation, $x^2{y}^{\prime \prime }+x{y}^{\prime }+(x^2-{\nu }^2)y=0$, whose solutions (Bessel functions) are fundamental in fields like heat transfer and signal processing. The solution is not an elementary function but is defined by its power series, derived via term-by-term calculus.

Common Pitfalls

1. Forgetting the Radius of Convergence: The theorem guarantees the same radius of convergence, but not the same interval of convergence. You must test the endpoints $x=a\pm R$ separately for the new differentiated or integrated series. Convergence at an endpoint can be gained or lost through these operations.
2. Misapplying the Constant of Integration: When integrating term-by-term, you must include the constant $C$. This constant is an ordinary real number, not another series. You find it by evaluating both the integrated series (at a convenient $x$-value, often the center $a$) and the antiderivative function you know.
3. Incorrect Index Shifting During Differentiation: When you differentiate ${\sum }_{n=0}^{\infty }{c}_n{x}^n$, the derivative is ${\sum }_{n=1}^{\infty }n{c}_n{x}^{n-1}$. The new sum starts at $n=1$ because the constant term $c_0$ vanishes. Failing to adjust the starting index can lead to errors in subsequent coefficient matching.
4. Assuming Operations Work Outside the Interval: Term-by-term calculus is only valid within the open interval $(a-R,a+R)$. Attempting to use a series representation obtained this way at a point outside its radius of convergence is invalid and will yield incorrect results.

Summary

• Power series can be differentiated and integrated term by term within their open interval of convergence, and the resulting series share the same radius of convergence, $R$.
• This process is the primary method for deriving important series, such as the series for $\arctan (x)$ from the integral of the geometric series for $\frac{1}{1+x^2}$.
• You must always check the endpoints of the interval of convergence separately after differentiation or integration, as endpoint behavior can change.
• These operations provide the theoretical backbone for constructing Taylor series and are a practical tool for solving differential equations that arise in engineering and physics.
• Mastering term-by-term calculus transforms power series from static representations into dynamic, manipulable tools for analysis and problem-solving.