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Power Series – Definition, General Form, and Examples
The power series is one of the most useful types of series in mathematical analysis. We can use power series to define different transcendental functions including the exponential and trigonometric functions. Understanding the power series will also make us appreciate how we have approximated functions’ values in our calculators and computers.
The power series allows us to approximate functions as the sum of the powers of the variable. We can think of power series as an infinite polynomial that leads to the approximation of common and new functions.
In this article, we’ll explore the definition of the power series and learn how to define common and new functions through this expansion. We’ll also show you how to confirm whether the given power series is convergent or divergent.
In order for us to do so, make sure to refresh your knowledge on the following:
- Review how we apply the ratio and root tests to check the series for convergence.
- Identify when the power series is absolutely convergent.
- Take a refresher on what makes a series convergent and divergent as well.
For now, let’s begin by understanding the unique components of a power series.
What is a power series?
The power series, centered at $c$, is a series represented by the general form shown below.
\begin{aligned}\sum_{n = 0}^{\infty} a_n(x -c)^n &= a_0 + a_1(x -c) + a_2( x- c)^2 + …\end{aligned}
The constants $\boldsymbol{a_n}$, where $n \geq 0$, are called the series’ coefficients and $\boldsymbol{c}$ represents the center. Notice something unique about this series as opposed to the ones we’ve learned in the past? The terms of the power series are now all in terms of $x$ when in the past, we’ve only worked with series that contains numbers. This will be the start of our understanding of series and expansions of common and new functions.
Examples of Power Series Expansion \begin{aligned}\sin x &= x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^7}{7!}+ …\\\cos x &= 1 – \dfrac{x^2}{2!} + \dfrac{x^4}{4!} – \dfrac{x^6}{6!}+ …\end{aligned} |
Here are two great examples of a power series- the power series of $\sin x$ and $\cos x$. Through this amazing series, we can now express transcendental functions such as sine and cosine functions as a series of polynomials. In the next sections, we’ll learn how to apply the power series formula and understand the process of expressing functions as a power series.
What is the power series formula?
We’ll show you two variants of the power series formula: 1) expression, when $\boldsymbol{x}$ is centered at zero and 2,) when $\boldsymbol{x}$ is centered at $\boldsymbol{c}$.
Suppose that $\{a_n\}$ is a sequence, $x$ is the variable, and $c$ represents a real number. We have the following power series formula:
Power series centered at the origin:
\begin{aligned}\boxed{\boldsymbol{\sum_{n = 0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3+…}}\end{aligned}
Power series centered at variable $\boldsymbol{c}$:
\begin{aligned}\boxed{\boldsymbol{\sum_{n = 0}^{\infty} a_n(x – c)^n = a_0 + a_1(x – c) + a_2(x – c)^2 + a_3(x – c)^3+…}}\end{aligned}
For a more precise definition, we establish that $x^0 = 1$ and $(x – c)^0 = 1$ even when $x = 0$ and $x = c$, respectively.
The simplest example of a power series and this occurs when $x$ is centered at the origin and when $a_n = 1$.
\begin{aligned}f(x) &= \sum_{n = 0}^{\infty} x^n\\&= 1 + x + x^2 + x^3 + …,\end{aligned}
This series is in fact geometric (with $x$ as the common ratio) and this converges when $-1 geometric series formula. This means that
\begin{aligned}\sum_{n = 0}^{\infty} x^n&= 1 + x + x^2 + x^3 + …\\&= \dfrac{1}{1 – x}\end{aligned},
when $|x| <1$ holds. This particular power series’ convergence was easy to determine since the resulting series is geometric. For more complex power series, let’s establish important rules when it comes to the convergence of the power series.
How to solve power series?
There are different types of problems involving the power series. When finding the radius and interval of convergence, use the conditions for the convergence of the power series. We can continue to apply the differential and integral rules we’ve learned in the past to find the derivative and integral of $f(x) = \sum_{n = 0}^{\infty} a_n(x – c)^n$.
Power series convergence
We’ll first understand the radius of convergence in every power series. We’ll focus on the series,
\begin{aligned}\sum_{n = 0}^{\infty} a_n(x – c)^n,\end{aligned}
where the domain of the function represents the values of $x$ in which the power series is convergent. We call that domain as the interval of convergence. When the interval, $(c -r, c + r)$, for some values $r >0$ then we call $\boldsymbol{r}$ as the radius of convergence. We can use the root test to find the radius of convergence as shown below.
\begin{aligned}r &= \lim_{n \rightarrow \infty} \dfrac{1}{\sqrt[n]{a_n}}\end{aligned}
Another way is through the ratio test as shown below.
\begin{aligned}r &= \lim_{n \rightarrow \infty} \left|\dfrac{a_n}{a_{n + 1}}\right|\end{aligned}
We can summarize the radius of convergence’s rule for the power series as shown below.
\begin{aligned}\boldsymbol{a -r | The power series converges. |
\begin{aligned}\boldsymbol{a -r a+ r}\end{aligned} | The power series diverges. |
Differentiating and integrating power series
When the power series has a radius of convergence at $\boldsymbol{r >0}$ then the differentiation of the power series (which will result in another power series) will also have a radius of convergence. We can differentiate the power series representation of $f(x)$, $f(x) = \sum_{n =0}^{\infty} a_n(x –c)^n$, by applying the power rule of differentiate.
\begin{aligned}f^{\prime}(x) &= \dfrac{d}{dx} \sum_{n = 0}^{\infty}a_n(x –c)^n\\&= \dfrac{d}{dx}\left(a_0 + a_1(x – c) + a_2(x – c)^2 + a_3(x – c)^3+…\right)\\&= a_1 + 2a_2(x –c) + 3a_3(x –c)^2+ …\\&= \sum_{n = 1}^{\infty} na_n(x – c)^{n -1}\end{aligned}
Notice how we changed $n = 0$ to $n=1$? That’s to account for the fact that $\dfrac{d}{dx}a_0 = 0$. Below are the second and third derivatives of $f(x)$.
\begin{aligned}f^{\prime\prime}(x) &=\sum_{n = 2}^{\infty} n(n -1)a_n(x – c)^{n -2} \\ f^{\prime\prime\prime }(x) &=\sum_{n = 3}^{\infty} n(n -1)(n -2)a_n(x – c)^{n -3} \end{aligned}
From this, we can see that the $n$ changes each time we differentiate the resulting function once more. We can integrate the function, $f(x)$, as well using the power rule for integration as shown below.
\begin{aligned}\int f(x)\phantom{x}dx &=\int \sum_{n = 0}^{\infty} a_n(x – c)^n \phantom{x}dx\\ &= \sum_{n = 0}^{\infty}\int a_n(x – c)^n \phantom{x}dx\\&=C +\sum_{n = 0}^{\infty}a_n\dfrac{(x- c)^{n +1}}{n+1} \end{aligned}
Both derivative and integral of $f(x)$ will have $r$ as the radius of convergence.
The best way to check our understanding is by working on actual problems, so when you’re ready, head over to the sample problems we’ve prepared for you!
Example 1
Determine the radius of convergence and interval of convergence of the power series, $\sum_{n = 0}^{\infty} n!x^n$.
Solution
Apply the ratio test to check the power series for its convergence. If we let $a_n$ represent the $n$th term of the power series, we have $a_n = n!x^n$. Take the limit of the ratio of $a_{n + 1}$ and $a_n$ as $n \rightarrow infty$ and assume that $x \neq 0$.
\begin{aligned}\lim_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right| &= \lim_{n \rightarrow \infty} \left|\dfrac{(n + 1)!x^{n +1}}{n!x^n}\right|\\&= \lim_{n \rightarrow \infty} \left|\dfrac{(n + 1)n!x^{n +1}}{n!x^n}\right|\\&= \lim_{n \rightarrow \infty} (n + 1)\left|x\right|\\ &=\infty\end{aligned}
From this, we can see that the series is divergent when $x \neq 0$. Since the power series is centered at $x = 0$, the power series must at least converge there. This means that the power series is convergent only when $x = 0$. In fact, this power series has an interval of convergence at $x =0$ and a radius of convergence at $r = 0$.
Example 2
What are the values of $x$ that will make the power series, $\sum_{n = 1}^{\infty} \dfrac{(x – 4)^n}{n}$, convergent?
Solution
Similar with our previous example, the $n$th term of the power series will be $a_n = \dfrac{(x – 4)^n}{n}$. Apply the ratio test on the power series as shown below.
\begin{aligned}\lim_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right| &= \lim_{n \rightarrow \infty} \left|\dfrac{\dfrac{(x -4)^{n +1}}{n +1}}{\dfrac{(x -4)^{n}}{n}}\right|\\&= \lim_{n \rightarrow \infty} \left|\dfrac{(x -4)^{n +1}}{n +1} \cdot \dfrac{n}{(x – 4)^n}\right|\\&= \lim_{n \rightarrow \infty} \left|x – 4\right| \dfrac{n}{n +1}\\&= \lim_{n \rightarrow \infty}|x – 4|\dfrac{1}{1 + \dfrac{1}{n}}\\&= |x -4|\lim_{n \rightarrow \infty}\dfrac{1}{1 + \dfrac{1}{n}}\\&=|x – 4|\end{aligned}
From this, we can see that the series is absolutely convergent. This means that the power series is convergent when $|x+ 4| <1$ and divergent when $|x – 4| > 1$.
\begin{aligned} |x – 4| < 1 \phantom{xx}\\-1 < x- 4<1\\3 < x < 5\end{aligned}
Through the ratio test, we’ve shown that the power series is convergent when $3< x< 5$ and divergent when $x < 3$ or $x >5$. Now, what happens when $x = 3$ or when $x = 5$? The ratio test does not give us any information about this, so we’ll have to figure it out by substituting $x =3$ and $x =5$ into the power series.
\begin{aligned}\boldsymbol{x = 3}\end{aligned} | \begin{aligned}\boldsymbol{x = 5}\end{aligned} |
\begin{aligned}\sum_{n = 1}^{\infty} \dfrac{(3 – 4)^n}{n} &= \sum_{n =1}^{\infty} \dfrac{(-1)^n}{n}\\&\Rightarrow \text{converges by alternating series test}\end{aligned} | \begin{aligned}\sum_{n = 1}^{\infty} \dfrac{(5 – 4)^n}{n} &= \sum_{n =1}^{\infty} \dfrac{(1)^n}{n}\\&\Rightarrow \text{harmonic series & is divergent}\end{aligned} |
This means that the power series is convergent for $3 \leq x < 5$.
Example 3
Determine a power series representation for the following function.
\begin{aligned}f(x) = \dfrac{1}{1 + x^3}\end{aligned}
What is its interval of convergence?
Solution
We can rewrite $f(x) = \dfrac{1}{1 + x^3}$ as $f(x) = \dfrac{1}{1 – (-x^3)}$. Keep in mind that we can replace $-x^3$ with $u$ and have the expression shown below.
\begin{aligned}f(x) &= \dfrac{1}{1 – (-x^3)} \\&= \dfrac{1}{1 –u}\end{aligned}
In our earlier discussion, we’ve shown that $\sum_{n = 0}^{\infty} x^n = \dfrac{1}{1 – x}$, so we’ll use this to rewrite the previous expression.
\begin{aligned}\dfrac{1}{1 –u} &= \sum_{n=0}^{\infty} u^n\\&= \sum_{n=0}^{\infty} (-x^3)^n\end{aligned}
This is only true when $|-x^3| <1$. Let’s simplify this series further as shown below.
\begin{aligned}f(x) &= \sum_{n = 0}^{\infty} (-1)^n x^{3n}, \phantom{x}|x^3| <1 \Rightarrow |x|<1\end{aligned}
Hence, we have $f(x) = \sum_{n = 0}^{\infty} (-1)^nx^{3n}$ with an interval of convergence of $|x|<1$.
Practice Questions
1. What are the values of $x$ that will make the power series, $\sum_{n = 0}^{\infty} \dfrac{x^n}{n!}$,convergent?
2. What are the values of $x$ that will make the power series, $\sum_{n = 1}^{\infty} \dfrac{(x – 3)^n}{n}$, convergent?
3. Determine the radius of convergence and interval of convergence of the power series, $\sum_{n = 0}^{\infty} nx^n$.
4. Determine the radius of convergence and interval of convergence of the power series, $\sum_{n = 1}^{\infty} \dfrac{(n!)^2}{n^n}(x – 3)^n$.
Answer Key
1. The series converges for all real numbers meaning, $x \in (-\infty, \infty)$.
2. The series converges when $2 \leq x < 4$.
3. The radius of convergence is $r = 1$ and the interval of convergence is $x \in (-1, 1)$.
4. The radius of convergence is $r = 0$ and the series converges only when $x =3$.