Professor Baker's Calc 2 - Section 11.8: Power Series

Welcome back to Calc 2! Today's focus is on power series, a fundamental concept with applications in various areas of mathematics and physics. We'll be exploring how to determine the convergence of these series and how to represent functions using power series.

What is a Power Series?

A power series is a series of the form:

$$ \sum_{n=0}^{\infty} c_n x^n = c_0 + c_1x + c_2x^2 + c_3x^3 + ... $$

More generally, a power series centered at $a$ is given by:

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

Where the $c_n$ are coefficients and $x$ is a variable.

A crucial example is the geometric series:

$$ \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + ... $$

This geometric series converges when $ -1 < x < 1 $. Remember this - it's a building block!

Determining Convergence

To determine the convergence of a power series, we often use the Ratio Test. Let's look at some examples:

Example 1:

For what values of $x$ does the series $\sum_{n=1}^{\infty} \frac{(x - 3)^n}{n}$ converge?

Applying the Ratio Test:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x - 3)^{n+1}}{n+1} \cdot \frac{n}{(x - 3)^n} \right| = \lim_{n \to \infty} \left| (x - 3) \cdot \frac{n}{n+1} \right| = |x - 3| $$

For convergence, we need $|x - 3| < 1$, which implies $2 < x < 4$. Checking the endpoints: when $x = 2$, the series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ which converges (alternating harmonic series). When $x = 4$, the series is $\sum_{n=1}^{\infty} \frac{1}{n}$ which diverges (harmonic series).

Therefore, the interval of convergence is $2 \le x < 4$.

Example 2:

For what values of $x$ does the series $\sum_{n=0}^{\infty} n!x^n$ converge?

Applying the Ratio Test:

$$ \lim_{n \to \infty} \left| \frac{(n+1)!x^{n+1}}{n!x^n} \right| = \lim_{n \to \infty} |(n+1)x| $$

This limit is infinite for any $x \ne 0$. Therefore, the series only converges at $x = 0$.

Example 3:

For what values of $x$ does the series $\sum_{n=0}^{\infty} \frac{x^n}{(2n)!}$ converge?

Applying the Ratio Test:

$$ \lim_{n \to \infty} \left| \frac{x^{n+1}}{(2(n+1))!} \cdot \frac{(2n)!}{x^n} \right| = \lim_{n \to \infty} \left| \frac{x}{(2n+2)(2n+1)} \right| = 0 $$

Since the limit is 0 for all $x$, the series converges for all real numbers, i.e., $x \in \mathbb{R}$.

Key Theorem

For a power series $\sum_{n=0}^{\infty} c_n (x - a)^n$, there are only three possibilities:

  • The series converges only when $x = a$.
  • The series converges for all $x$.
  • There is a positive number $R$ such that the series converges if $|x - a| < R$ and diverges if $|x - a| > R$. $R$ is called the radius of convergence.

Representations of Functions Using Geometric Series

We can often represent functions as power series by manipulating geometric series. Recall that

$$ \frac{1}{1 - x} = 1 + x + x^2 + x^3 + ... = \sum_{n=0}^{\infty} x^n, \quad |x| < 1 $$

This is a powerful tool, and we'll explore how to use it in future lectures! Keep practicing, and you'll master power series in no time!