Power Series for MATH 141

What is a Power Series?

Up until now, polynomials and infinite series have lived in separate worlds. Polynomials are easy to work with — you can add, multiply, differentiate, and integrate them without breaking a sweat. But polynomials are limited: they can only describe certain shapes and behaviors.

Infinite series, on the other hand, can represent much more — but they're just numbers. The series $\sum \frac{1}{n^2}$ converges to $\frac{\pi^2}{6}$, but that's a single value, not a function.

Power series bridge these two worlds. A power series is an "infinite polynomial" — it looks like a polynomial, but with infinitely many terms. And here's the magic: power series can represent functions that polynomials never could, like $e^x$, $\sin x$, and $\ln(1+x)$.

The Definition

A power series centered at $a$ (also called "centered at $a$" or "about $a$") is:

$$\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots$$

where $c_0, c_1, c_2, \ldots$ are constants called coefficients.

Special case — centered at 0:

$$\sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots$$

This is like a polynomial $c_0 + c_1 x + c_2 x^2 + \cdots + c_N x^N$, except it never stops.

Why Do We Care?

1. Functions become "infinitely smooth polynomials"

Many important functions can be written as power series:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$

This means we can use polynomial techniques on non-polynomial functions!

2. Easy differentiation and integration

Differentiating $e^{-x^2}$ is easy. Integrating it? Impossible with elementary functions. But as a power series, you can integrate term by term:

$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$

$$\int e^{-x^2}\,dx = x - \frac{x^3}{3} + \frac{x^5}{5 \cdot 2!} - \frac{x^7}{7 \cdot 3!} + \cdots$$

3. Approximations and computation

How does your calculator compute $\sin(0.5)$? It uses a power series! Truncating after a few terms gives excellent approximations.

The Big Question: Where Does It Converge?

Here's the crucial thing to understand: a power series is both a function and a series at the same time.

• As a function, it takes an input $x$ and (hopefully) produces an output
• As a series, it's an infinite sum that might converge or diverge

The catch is that not every input $x$ produces a convergent series. For some values of $x$, the series adds up to a finite number. For others, it blows up to infinity.

Consider the geometric power series $\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$

If we plug in $x = 0.5$: $$1 + 0.5 + 0.25 + 0.125 + \cdots = 2$$ The terms shrink rapidly, the series converges, and we get a finite answer.

If we plug in $x = 2$: $$1 + 2 + 4 + 8 + 16 + \cdots = \infty$$ The terms grow, the series diverges, and there's no finite answer.

Same series, different $x$ values, completely different outcomes!

The radius of convergence $R$ tells us exactly where the boundary is:

• The series converges absolutely for $|x - a| < R$ (inside the radius)
• The series diverges for $|x - a| > R$ (outside the radius)
• At $|x - a| = R$ (the endpoints), you must check separately

This creates an interval of convergence centered at $a$.

Visualizing the Radius of Convergence

Think of it like a "zone of validity" around the center $a$:

    diverges     converges     diverges
  ←──────────|─────●─────|──────────→
           a - R    a    a + R

Inside the interval $(a - R, a + R)$, the power series converges and represents a function. Outside, it diverges.

Three possibilities for $R$:

• $R = 0$: Only converges at the center point $x = a$
• $R = \infty$: Converges for all real $x$ (like $e^x$)
• $0 < R < \infty$: Converges on a finite interval (most common)

An Important Subtlety: The Series vs. The Function

Here's something that often confuses students: the underlying function might still exist outside the interval, but the series can't "reach" it there.

Consider $\sum_{n=0}^{\infty} x^n$, which equals $\frac{1}{1-x}$ for $|x| < 1$.

• At $x = 0.5$: The function $\frac{1}{1-0.5} = 2$, and the series gives $1 + 0.5 + 0.25 + \cdots = 2$ ✓ They match!
• At $x = 2$: The function $\frac{1}{1-2} = -1$ is perfectly well-defined, but the series $1 + 2 + 4 + 8 + \cdots$ diverges!

The function $\frac{1}{1-x}$ exists at $x = 2$ (it equals $-1$), but the power series can't compute it there — the series just blows up.

Think of it like a map: A map might only be accurate within 10 miles of your city. The rest of the world still exists, but your map shows blank space there. Similarly, the power series is a "representation" of the function that only works within the radius of convergence.

Key takeaway: Inside the interval, the series converges and equals the function. Outside the interval, the series diverges and represents nothing — even though the function itself might still be defined there.

Finding the Radius of Convergence

Use the Ratio Test or Root Test on the series, treating $x$ as a parameter.

Method (Ratio Test):

For $\sum c_n (x-a)^n$, compute:

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

The series converges when $L < 1$, which means:

$$|x - a| < \frac{1}{\lim_{n \to \infty} |c_{n+1}/c_n|}$$

So:

$$R = \frac{1}{\lim_{n \to \infty} |c_{n+1}/c_n|} = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right|$$

The Interval of Convergence: A Checklist

1. Find $R$ using the Ratio Test or Root Test
2. Write the open interval $(a - R, a + R)$
3. Check the left endpoint $x = a - R$ — plug in and test convergence
4. Check the right endpoint $x = a + R$ — plug in and test convergence
5. State the final interval including or excluding endpoints as appropriate

Common endpoint behaviors:

• Often one endpoint converges (conditionally) and the other diverges
• Both endpoints might converge, or both might diverge
• Use p-series, alternating series test, or comparison tests at endpoints

Summary: The Big Picture

Common Mistakes and Misunderstandings

❌ Mistake: Forgetting to check endpoints

Wrong: "The Ratio Test gives $|x| < 3$, so the interval is $(-3, 3)$."

Why it's wrong: The Ratio Test only tells you about the open interval. Endpoints require separate testing — they might converge!

Correct: After finding $R$, always test $x = a - R$ and $x = a + R$ individually.

❌ Mistake: Confusing radius with interval

Wrong: "The radius of convergence is $(-2, 4)$."

Why it's wrong: Radius is a single number $R$. The interval is $(a - R, a + R)$ possibly including endpoints.

Correct: "The radius is $R = 3$ and the interval of convergence is $(-2, 4]$" (for a series centered at $a = 1$).

❌ Mistake: Using wrong test at endpoints

Wrong: Using the Ratio Test at $x = R$ to check convergence.

Why it's wrong: At endpoints, the Ratio Test gives $L = 1$ (inconclusive). You need other tests.

Correct: At endpoints, use the Divergence Test, p-series test, Alternating Series Test, or comparison tests.

❌ Mistake: Thinking $R = 0$ means "doesn't converge anywhere"

Wrong: "$R = 0$ means the series never converges."

Why it's wrong: Even when $R = 0$, the series still converges at the center point $x = a$.

Correct: $R = 0$ means the series only converges at $x = a$ and nowhere else.