AP Calculus BC: Power Series and Representations

Power series are one of the most powerful ideas in calculus, transforming complex, non-polynomial functions into infinite sums that behave like polynomials. This concept is the bridge from algebra to advanced analysis and is indispensable in engineering for modeling oscillations, heat flow, and signal processing. Understanding how to find, manipulate, and use these series is essential for solving otherwise intractable problems.

What is a Power Series?

A power series is an infinite series of the form:

$$\sum _{n=0}^{\infty }{c}_n(x-a{)}^n=c_0+c_1(x-a)+c_2(x-a{)}^2+c_3(x-a{)}^3+\cdots$$

The numbers $c_n$ are the coefficients, and the fixed number $a$ is the center of the series. When $a=0$, the series simplifies to $\sum c_n{x}^n$, which is often easier to work with initially. Think of the center as the point around which the series is "expanded"; the series provides the best polynomial approximation of a function near this point $x=a$.

The central question for any power series is: for which values of $x$ does it converge to a finite number? A power series does not converge everywhere; its convergence depends entirely on the distance of $x$ from the center $a$. For some $x$, the terms grow too large, and the sum diverges to infinity. For others, the terms get vanishingly small, allowing the sum to settle on a specific value.

Determining Convergence: The Radius and Interval

To determine where a power series converges, we typically use the Ratio Test. For a given series $\sum c_n(x-a{)}^n$, we examine the limit:

$$L=\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}(x-a{)}^{n+1}}{c_n(x-a{)}^n}\right|=|x-a|\cdot \underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|$$

By the Ratio Test, the series converges absolutely if $L<1$, diverges if $L>1$, and the test is inconclusive if $L=1$. This inequality $L<1$ leads us directly to the concept of the radius of convergence, denoted by $R$.

The radius of convergence $R$ is the distance from the center $a$ within which the series converges absolutely. It is found by solving:

$$|x-a|\cdot \underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|<1\text{which simplifies to}|x-a|<R$$

where $R=1/{\lim }_{n\to \infty }|c_{n+1}/c_n|$, provided this limit exists. The set of all $x$ values satisfying $|x-a|<R$ is called the interval of convergence, but we must always check the endpoints $x=a\pm R$ separately using other convergence tests, as the series may converge or diverge at these points.

For example, consider ${\sum }_{n=1}^{\infty }\frac{x^n}{n}$. Applying the Ratio Test: $$L=\underset{n\to \infty }{\lim }\left|\frac{x^{n+1}}{n+1}\cdot \frac{n}{x^n}\right|=|x|\underset{n\to \infty }{\lim }\frac{n}{n+1}=|x|$$ The series converges when $|x|<1$, so $R=1$. We then test the endpoints: at $x=1$, the series is the divergent harmonic series $\sum 1/n$; at $x=-1$, it is the convergent alternating harmonic series. Thus, the interval of convergence is $[-1,1)$.

Representing Functions as Power Series

A major application is representing known functions as power series. This is often done by starting with a fundamental geometric series representation:

$$\frac{1}{1-u}=\sum _{n=0}^{\infty }{u}^n=1+u+u^2+u^3+\cdots ,\text{for }|u|<1$$

By cleverly substituting for $u$, we can find series for many related functions. This process is called manipulation of known series.

Example: Find a power series for $f(x)=\frac{1}{1+3x}$ centered at $a=0$. We match the form $\frac{1}{1-u}$. Here, $u=-3x$. Therefore, for $|-3x|<1$ (i.e., $|x|<1/3$): $$\frac{1}{1+3x}=\frac{1}{1-(-3x)}=\sum _{n=0}^{\infty }(-3x{)}^n=\sum _{n=0}^{\infty }(-3{)}^n{x}^n$$ The radius of convergence is $R=1/3$.

Other techniques include:

• Differentiation and Integration: If you have a series for $f(x)$, you can differentiate or integrate it term-by-term to get a series for $f^{\prime }(x)$ or $\int f(x)dx$.
• Multiplication and Division: Multiplying a known series by a polynomial or another series.
• Composition: Substituting a polynomial or another series into a known series, carefully respecting the radius of convergence.

Differentiation and Integration Term-by-Term

This is a powerful theorem that gives power series their utility. If a power series $\sum c_n(x-a{)}^n$ has a radius of convergence $R>0$, then the function $f$ it defines is differentiable (and integrable) on the interval $(a-R,a+R)$, and its derivative and integral can be found by term-by-term operations:

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

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

Crucially, both the differentiated and integrated series have the same radius of convergence $R$ as the original series. (Convergence at the endpoints may change and must be rechecked).

Example: Find a series for $g(x)=\frac{1}{(1-x{)}^2}$ from the geometric series. We know $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n$ for $|x|<1$. Notice that $\frac{1}{(1-x{)}^2}$ is the derivative of $\frac{1}{1-x}$. Differentiating term-by-term: $$\frac{d}{dx}\left[\frac{1}{1-x}\right]=\frac{d}{dx}\left[\sum _{n=0}^{\infty }{x}^n\right]$$ $$\frac{1}{(1-x{)}^2}=\sum _{n=1}^{\infty }n{x}^{n-1}=1+2x+3x^2+4x^3+\cdots$$ The new series also converges for $|x|<1$.

Common Pitfalls

1. Forgetting to Check Endpoints: The biggest error is stating the interval of convergence as simply $|x-a|<R$. You must always test the two endpoints $x=a\pm R$ individually. A series may converge at neither, one, or both. The radius $R$ only tells you the distance; it doesn't specify the status of the boundary.

1. Misapplying Term-by-Term Operations Outside the Interval: The theorem guarantees term-by-term differentiation and integration work only on the open interval $(a-R,a+R)$. You cannot assume the resulting series behaves the same at the endpoints. For example, integrating a series that converges at an endpoint may produce a series that diverges there. Always treat endpoints as new, separate cases.

1. Incorrect Manipulation of Known Series: When performing substitution (like letting $u=x^2$), you must update the condition for convergence. If $\frac{1}{1-u}$ converges for $|u|<1$, then $\frac{1}{1-x^2}$ converges for $|x^2|<1$, which means $|x|<1$. Also, ensure your algebraic manipulation is correct. A common mistake is to misidentify the "$u$" in the form $\frac{1}{1-u}$.

1. Confusing the Index of Summation After Operations: After differentiation, the sum starts at $n=1$. After integration, the "$n$" in the denominator becomes $n+1$. It's easy to make off-by-one errors. A good practice is to write out the first three terms of the original series, perform the operation on those terms, and then deduce the new general pattern.

Summary

• A power series $\sum c_n(x-a{)}^n$ is an infinite polynomial centered at $a$. Its convergence is not guaranteed for all $x$.
• The radius of convergence $R$ is found using the Ratio Test on the absolute value of the terms. The series converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$.
• The interval of convergence is determined by testing the endpoints $x=a\pm R$ separately, as convergence or divergence there depends on the specific series.
• Functions can be represented as power series by manipulating known series, most commonly starting from the geometric series $\frac{1}{1-u}=\sum u^n$ for $|u|<1$.
• Within the open interval of convergence, power series can be differentiated and integrated term-by-term. The resulting series have the same radius of convergence $R$ as the original, though endpoint behavior may change.