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Section 6.1 Power Series and Functions

References.

We have already seen one important type of a power series: the geometric series

\begin{equation*} \sum_{n=0}^\infty a x^n = a + ax + ax^2 + \cdots + ax^n + \cdots \end{equation*}

and determined that

  • it converges for some values of \(x\) (\(|x| < 1\text{,}\) so \(-1 < x < 1\)) but not others, and

  • for \(x\) values giving convergence, the value is \(\ds\frac{a}{1-x}\text{.}\)

These two questions will arise for other power series:

  1. for which \(x\) values does the series converge, and

  2. when it does converge, what is the value of the sum? That is, what function does the power series give?

Definition 6.1.2. Power Series.

A Power Series is a series of the form

\begin{equation*} \sum_{n=0}^\infty c_n x^n = c_0 + c_1 x + c_2 x^2 + \cdots + c_n x^n + \cdots \end{equation*}

where \(x\) is some number, and the \(c_n\) are constants: that is, they do not depending on \(x\text{.}\)

More generally, powers of \(x-a\) can be used for some constant \(a\text{,}\) so the most general power series is of the form

\begin{equation*} \sum_{n=0}^\infty c_n (x-a)^n = c_0 + c_1 (x-a) + c_2 (x-a)^2 + \cdots + c_n (x-a)^n + \cdots \end{equation*}

The constant \(a\) is called the center of the series.

For which values of \(x\) does one get convergence of the series

\begin{equation*} \sum_{n=1}^\infty \frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n} + \cdots \end{equation*}
Answer.
The sum converges for \(-1 \leq x < 1\) and diverges otherwise.

For which values of \(x\) does one get convergence of the series

\begin{equation*} \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots \end{equation*}
Answer.
The sum converges for all \(x\text{;}\) that is \(|x| < \infty\text{.}\)

For which values of \(x\) does one get convergence of the series

\begin{equation*} \sum_{n=0}^\infty n! x^n = 1 + x + 2x^2 + 6x^3 + \cdots \end{equation*}
Answer.
The sum converges only for \(x = 0\text{;}\) that is “\(|x| \leq 0\)”.

Two patterns are worth noting in the above examples:

  1. One primarily gets convergence for \(|x|\) “small enough”, divergence for sufficiently large \(|x|\text{.}\)

  2. Convergence and divergence is shown primarily by the Ratio Test (or the Root Test).

  3. Exceptions to the previous two observations occur at the two borderline points where \(|x|\) has the largest value for which convergence might occur.

  4. These two borderline \(x\) values give \(\rho = 1\) in the Ratio Test (or the Root Test), so those tests give no answer; thus to determine convergence we must use some other method, like the Alternating Series Test or one of the comparison tests.

The above patterns are in fact universal:

The first case is silent on two values of \(x\text{:}\) \(a+R\) and \(a-R\text{.}\) At each of these, one can have either convergence or divergence: see for example the “50-50” case of Example 6.1.3 above.

The number \(R\) in case (i) is called the Radius of Convergence. In fact, we can make sense of a radius of convergence in every case:

  • in case (ii), we say the radius of convergence is \(R=\infty\text{;}\)

  • in the boring case (iii), we say the radius of convergence is \(R=0\text{.}\)

Also, in every case the \(x\) values giving convergence form an interval, which we call the Interval of Convergence:

  • In case (i), the interval of convergence can be \((a-R,a+R)\text{,}\) \([a-R,a+R)\text{,}\) \((a-R,a+R]\text{,}\) or \([a-R,a+R]\text{.}\)

  • In case (ii) it is \((-\infty,\infty)\)

  • In case (iii) it is just \([a,a]\text{.}\)

Find the radius of convergence and interval of convergence of the series

\begin{equation*} \sum_{n=0}^\infty (-2)^n \sqrt{n} (x-2)^n. \end{equation*}
Answer.
The root test shows that the sum converges for \(|x-2| < 1/2\) and diverges for \(|x-2| > 1/2\text{,}\) so the radius of convergence is \(R = 1/2\text{.}\) Then testing the end cases \(x = 3/2,5/3\text{,}\) where \(|x-2| = 1/2\text{,}\) gives divergence in each case, so the interval of convergence is \((3/2, 5/2)\text{.}\)

Find the radius of convergence and interval of convergence for the series

\begin{equation*} \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2} \end{equation*}

Answer.
The series for \(J_0(x)\) converges for all \(x\text{,}\) so the radius of convergence is \(R=\infty\text{,}\) and so with no extra work needed, the interval of convergence is \((-\infty,\infty)\text{.}\)

Study Guide.

Study Calculus Volume 2, Section 6.1 2 ; in particular

  • The defintions of a Power Series, its Center and its Radius of Convergence.

  • Theorem 1 about the possibilites for which \(x\) values give convergence.

  • Examples 1, 2 and 3 (focus on “radius” more than “interval”)

  • Checkpoints 1 and 3

  • and one or several exercises from each of the following ranges: 1–4, 5 and 6, 13–16, 23–26, 29–32.

https://openstax.org/books/calculus-volume-2/pages/6-1-power-series-and-functions
https://openstax.org/books/calculus-volume-2/pages/6-1-power-series-and-functions