If it often the case that a power series may contain non-zero terms only for certain exponents following some pattern. In this section, I’m going to briefly introduce some common notation for particular cases of this phenomenon. Consider a series where all the odd terms are zero (centered at \(0\) for convenience).
\begin{equation*}
f(x) = c_0 + 0x + c_2x^2 + 0x^3 + c_4x^4 + 0x^5 + \ldots
\end{equation*}
I could similarly consider a series where all the even terms are zero.
\begin{equation*}
f(x) = 0 + c_1x + 0 + c_3x^3 + 0x^4 + c_45^5 + 0x^6 + \ldots
\end{equation*}
If I want to index all the even numbers, I can write \(k =
2n\) for \(n \in \NN\text{.}\) Similarly, I can index all the odd numbers by writing \(k = (2n+1)\) for \(n \in \NN\text{.}\) Using these tools, I could write a series with only odd or even non-zero terms. The series
\begin{equation*}
f(x) = \sum_{n=0}^\infty c_{2n} (x-\alpha)^{2n}
\end{equation*}
is a series with only even terms. The series.
\begin{equation*}
g(x) = \sum_{n=0}^\infty c_{2n-1} (x-\alpha)^{2n+1}
\end{equation*}
is a series with only odd terms.
Some extra care must be taken with calculating radii of convergence for these series. The formula
\begin{equation*}
R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}}
\right|
\end{equation*}
relies on the assumption that all \(c_n \neq 0\text{.}\) This is not true for these series with only odd term or even terms. For series with such patterns, I need a slightly different approach. Take the example series
\begin{equation*}
f(x) = \sum_{n=0}^\infty c_{2n} (x-\alpha)^{2n}
\end{equation*}
that only has even terms. Implicitly, even though I can assume \(c_{2n} \neq 0\text{,}\) all of the missing coefficient \(c_{2n+1}\) are zero. To use the calculation for radius of convergence, a series needs all non-zero coefficients. I can force this series into such a form by some clever manipulation of exponents. Using the laws of exponents, I adjust the power of \((x-\alpha)\text{.}\)
\begin{equation*}
f(x) = \sum_{n=0}^\infty c_{2n} ((x-\alpha)^2)^{n}
\end{equation*}
Then I also re-index the coefficients: what I called \(c_{2n}\) before, I’ll just call \(c_n\text{.}\)
\begin{equation*}
f(x) = \sum_{n=0}^\infty c_{n} ((x-\alpha)^2)^{n}
\end{equation*}
Then this is a series with all non-zero coefficients (with the assumption that the original \(c_{2n}\) were non-zero). However, it is now a power series in \(x^2\text{,}\) not \(x\text{.}\) Therefore, if I calculate a radius of convergence \(R\text{,}\) the resulting inequality is
\begin{equation*}
(x-\alpha)^2 \in \left( - R, R \right)
\end{equation*}
To get the actual bound on \(x\) itself requires manipulating the inequalities that bound \(x^2\text{.}\) The result is this.
\begin{equation*}
(x-\alpha) \in \left( - \sqrt{R}, \sqrt{R} \right) \implies
x \in \left( \alpha - \sqrt{R}, \alpha + \sqrt{R} \right)
\end{equation*}
In summary, there is often a way to get to a domain of convergence for this type of series, but it can be difficult and unwieldy.
The approach above doesn’t just work for even and odd terms. Using similar ideas, I could encode all sorts of patterns in the exponents of our power series. If a power series had non-zero terms only when the exponent was a power of \(3\text{,}\) I could write it as
\begin{equation*}
f(x) = \sum_{n=0}^\infty (x-\alpha)^{3n}\text{.}
\end{equation*}
If the power series had non-zero exponents only for every fifth number starting at \(7\text{,}\) I could write it as
\begin{equation*}
f(x) = \sum_{n=0}^\infty (x-\alpha)^{5n + 7}\text{.}
\end{equation*}
In either case, I’d have to use similar kinds of tricks to reduce it to a series in \(x^3\) or \(x^5\text{,}\) then calculate the radius, then manipulate the inequalities to figure out the actual bounds on \(x\text{.}\)
