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# Maclaurin Series – Definition, Expansion Form, and Examples

The **Maclaurin series** is another important power series that you’ll learn and understand in calculus. This series allows us to find an approximation for a given function, $f(x)$. With its extensive applications in numerical methods and applied mathematics, it essential that we understand the Maclaurin series.

*The Maclaurin series is a power series that uses successive derivatives of the function and the values of these derivatives when the input is equal to zero. In fact, the Maclaurin series is a special type of the Taylor series. *

Our discussion focuses on what makes this power series unique. We’ll also cover the conditions we need to find the Maclaurin series representing different functions. Since this series is closely related to the Taylor series, keep your notes on this topic handy as well. By the end of the article, you’ll learn how to rewrite familiar functions in their Maclaurin series form!

**What is a Maclaurin series?**

**The Maclaurin series** is another polynomial approximation of a function. In fact, it is a special case of a Taylor series where each of the successive derivatives is evaluated at $x = 0$. Simply put, the Maclaurin series is the Taylor series of the function at $x = 0$.

\begin{aligned}\boldsymbol{e^x }&= \boldsymbol{\sum_{n = 0}^{\infty} \dfrac{x^n}{n!}} \\&= \boldsymbol{1+ \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}+ …}\\\boldsymbol{\sin x }&= \boldsymbol{(-1)^n\sum_{n = 0}^{\infty} \dfrac{x^{2n + 1}}{(2n)!}} \\&= \boldsymbol{x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^{7}}{7!}+ …}\\\boldsymbol{\dfrac{1}{1 – x}}&= \boldsymbol{\sum_{n = 0}^{\infty} x^n} \\&= \boldsymbol{1+ x + x^2 + x^3 + …} \end{aligned} |

These are just three examples of functions along with their Maclaurin series. The first equation shows the Maclaurin series of each of the functions in sigma notation while the second highlights the first three terms of each of the series.

**Understanding the Maclaurin series formula**

As we have mentioned, the Maclaurin series is a special case of the Taylor series. Let’s begin by recalling the general form of the function’s Taylor series.

\begin{aligned} f(x) &= \sum_{n = 0}^{\infty} \dfrac{f^{(n)}(c)}{n!} (x – c)^n \\&= f(c) + \dfrac{f^{\prime}(c)}{1!}( x – c) + \dfrac{f^{\prime\prime}(c)}{2!}( x – c)^2 + \dfrac{f^{\prime\prime\prime}(c)}{3!}( x – c)^3 + …\end{aligned}

The **Maclaurin series formula is simply the resulting expression when** $\boldsymbol{c = 0}$. Hence, we have the Maclaurin series formula as shown below.

\begin{aligned} f(x) &= \sum_{n = 0}^{\infty} \dfrac{f^{(n)}(0)}{n!} x^n \\&= f(0) + \dfrac{f^{\prime}(0)}{1!}x + \dfrac{f^{\prime\prime}(0)}{2!} x^2 + \dfrac{f^{\prime\prime\prime}(0)}{3!} x^3 + …\end{aligned} |

This means that we can find the polynomial approximation for the function, $f(x)$, using the Maclaurin series and the succeeding derivatives of $f(x)$ evaluated at $0$.

**How to find a Maclaurin series?**

Now that we know the general form of the Maclaurin series, we can try writing the Maclaurin series of different functions. Before we do so, check out the following pointers that may help you:

- Take the three succeeding derivatives of $f(x)$.
- Feel free to find more terms by differentiating the succeeding expressions as well.
- Evaluate $f(x)$, $f^{\prime}(x)$, $f^{\prime \prime}(x)$, $f^{\prime \prime \prime}(x)$, and more at $x = 0$.
- Write down the functions’ Maclaurin series by adding the resulting terms.

To check our current understanding, why don’t we confirm that $f(x) = e^x$ is equal to $1+ \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}+ …$? We can also confirm that the Maclaurin series’ sigma notation is $\sum_{n = 0}^{\infty} \dfrac{x^n}{n!}$.

Let’s begin by differentiating $e^x$ three times in a row using the derivative rule, $\dfrac{d}{dx} e^x$. Afterwards, evaluate $f(0)$, $f^{\prime}(0)$, $f^{\prime \prime}(0)$, and $f^{\prime \prime \prime}(0)$.

\begin{aligned}\boldsymbol{f^{(n)} (x)}\end{aligned} | \begin{aligned}\boldsymbol{f^{(n)}(0)}\end{aligned} |

\begin{aligned}f(x) &= e^x\end{aligned} | \begin{aligned}f(0) &= 1\end{aligned} |

\begin{aligned}f^{\prime}(x) &= e^x\end{aligned} | \begin{aligned}f^{\prime}(0) &= 1\end{aligned} |

\begin{aligned}f^{\prime \prime}(x) &= e^x\end{aligned} | \begin{aligned}f^{\prime \prime}(0) &= 1\end{aligned} |

\begin{aligned}f^{\prime \prime \prime}(x) &= e^x\end{aligned} | \begin{aligned}f^{\prime \prime \prime}(0) &= 1\end{aligned} |

\begin{aligned}. \\. \\.\end{aligned} | \begin{aligned}. \\. \\.\end{aligned} |

\begin{aligned}f^{(n)} (x) &= e^x\end{aligned} | \begin{aligned}f^{(n)} (0) &= 1\end{aligned} |

Substitute these expressions into the Maclaurin series formula to find the approximation for $y = e^x$.

\begin{aligned} e^x &= f(0) + \dfrac{f^{\prime}(0)}{1!}x + \dfrac{f^{\prime\prime}(0)}{2!} x^2 + \dfrac{f^{\prime\prime\prime}(0)}{3!} x^3 + …\dfrac{f^{(n)}(0)}{n!}+ …\\&=1 + \dfrac{1}{1!}x + \dfrac{1}{2!}x^2 + \dfrac{1}{3!}x^3 + …+ \dfrac{1}{n!}x^n+ …\\&= 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + … +\dfrac{x^n}{n!}+…\end{aligned}

We can write this series in sigma notation using the $n$th term of the Maclaurin series. Hence, we have $f(x) = e^x = \sum_{n = 0}^{\infty} \dfrac{x^n}{n!}$. Apply a similar process when writing the Maclaurin series of different functions. When you’re ready, head over to the sample problems below to master this topic!

*Example 1*

Find the Maclaurin series of $f(x) = \sin x$ up to its fourth-order then write the Maclaurin series in sigma notation.

__Solution__

Differentiate $f(x) = \sin x$ four times in a row to find the expressions for $f^{\prime}(x)$, f^{\prime\prime }(x)$, $f^{\prime \prime \prime }(x)$, and $f^{(4)}(x)$. Use the derivative rules for sine and cosine as shown below.

\begin{aligned}\dfrac{d}{dx} \sin x &= \cos x\\ \dfrac{d}{dx} \cos x &= -\sin x \end{aligned}

Evaluate each resulting expressions at $x = 0$. The table below summarizes our calculation.

\begin{aligned}\boldsymbol{f^{(n)} (x)}\end{aligned} | \begin{aligned}\boldsymbol{f^{(n)}(0)}\end{aligned} |

\begin{aligned}f(x) &= \sin x\end{aligned} | \begin{aligned}f(0) &= 0\end{aligned} |

\begin{aligned}f^{\prime}(x) &= \cos x\end{aligned} | \begin{aligned}f^{\prime}(0) &= 1\end{aligned} |

\begin{aligned}f^{\prime \prime}(x) &= -\sin x\end{aligned} | \begin{aligned}f^{\prime \prime}(0) &= 0\end{aligned} |

\begin{aligned}f^{\prime \prime \prime}(x) &= -\cos x\end{aligned} | \begin{aligned}f^{\prime \prime \prime}(0) &= -1\end{aligned} |

\begin{aligned}f^{(4)}(x) &= \sin x\end{aligned} | \begin{aligned}f^{(4)}(0) &= 0\end{aligned} |

The derivatives and their values when $x = 0$ will repeat its cycle for each four consecutive terms. Let’s show you the next four terms of the series to show you what we mean:

\begin{aligned}\boldsymbol{f^{(n)} (x)}\end{aligned} | \begin{aligned}\boldsymbol{f^{(n)}(0)}\end{aligned} |

\begin{aligned}f^{(5)}(x) &= \cos x\end{aligned} | \begin{aligned} f^{(5)}(0) &= 1\end{aligned} |

\begin{aligned}f^{(6)}(x) &= -\sin x\end{aligned} | \begin{aligned} f^{(6)}(0) &= 0\end{aligned} |

\begin{aligned}f^{(7)}(x) &= -\cos x\end{aligned} | \begin{aligned} f^{(7)}(0) &= -1\end{aligned} |

\begin{aligned}f^{(8)}(x) &= \sin x\end{aligned} | \begin{aligned}f^{(8)}(0) &= 0\end{aligned} |

Notice that when $n$ is even, $f^{(n)}(0)$ is zero? This means that when we use the Maclaurin series formula, we’ll be skipping the even powers. Let’s go ahead and confirm this by using the expressions shown in the two tables.

\begin{aligned} \sin x &= f(0) + \dfrac{f^{\prime}(0)}{1!}x + \dfrac{f^{\prime\prime}(0)}{2!} x^2 + \dfrac{f^{\prime\prime\prime}(0)}{3!} x^3 +\dfrac{f^{(4)}(0)}{4!} x^4\\&+\dfrac{f^{(5)}(0)}{5!} x^5 +\dfrac{f^{(6)}(0)}{6!} x^6+ +\dfrac{f^{(7)}(0)}{7!} x^7+ \dfrac{f^{(8)}(0)}{8!} x^8+…\\&= 0 + \dfrac{1}{1!}x + \dfrac{0}{2!} x^2 + \dfrac{-1}{3!} x^3 +\dfrac{0}{4!} x^4 +\dfrac{1}{5!} x^5 +\dfrac{0}{6!} x^6+ +\dfrac{-1}{7!} x^7+ \dfrac{0}{8!} x^8+…\\&= x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^7}{7!}+ … \end{aligned}

This means that we’re only dealing with odd powers, so we’re only concerned with powers that can be expressed as $(2n + 1)$. Since the operation alternates from negative to positive, the sigma notation will have a factor of $(-1)^n$.

\begin{aligned} \sin x &= x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^7}{7!}+ …\\&= \sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2n + 1}}{(2n + 1)!} \end{aligned}

Hence, we’ve shown the Maclaurin series of $\sin x$ as well as its sigma notation.

*Example 2*

Use the result from the previous example to find the Maclaurin series of $g(x) = x\sin x$. Finalize your answer by writing the Maclaurin series in sigma notation.

__Solution__

From the previous example, we have $\sin x = x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^5}{7!}+ …$. To find the Maclaurin expansion of $g(x) = x \sin x$, simply multiply the expansion by $x$.

\begin{aligned}g(x) &= x \sin x \\&= x \left(x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^7}{7!}+ …\right)\\&= x^2 – \dfrac{x^4}{3!} + \dfrac{x^6}{5!}- \dfrac{x^8}{7!} + …\end{aligned}

Let’s now work on $g(x)$’s sigma notation given that $\sin x = \sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2n + 1}}{(2n + 1)!}$.

\begin{aligned}g(x) &= x \sin x \\&= x \left(\sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2n + 1}}{(2n + 1)!}\right)\\&=\sum_{n =0}^{\infty} (-1)^n \dfrac{x \cdot x^{2n + 1}}{(2n + 1)!} \\&= \sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2(n + 1)}}{(2n + 1)!}\end{aligned}

This example highlights the fact that we can use common functions’ Maclaurin series expansions to find the Maclaurin series of more complex functions.

*Example 3*

Use the fourth Maclaurin polynomial for $f(x) = \ln (3 + 4x)$ to approximate $\ln 7$.

__Solution__

The $n$th Maclaurin polynomial is simply the terms of the series from $f(0) $ to $\dfrac{f^{(n)}(0)}{n!}$. Meaning, for the fourth Maclaurin polynomial, we only need $\ln(4 + 3x)$’s expansion until the term containing $\dfrac{f^{(4)}(0)}{4!}x^4$.

As always, we begin by finding the succeeding derivatives of $\ln (3 + 4x)$ and evaluate the resulting expressions at $x = 0$. The table below summarizes the calculation for you.

\begin{aligned}\boldsymbol{f^{(n)} (x)}\end{aligned} | \begin{aligned}\boldsymbol{f^{(n)}(0)}\end{aligned} |

\begin{aligned}f(x) &= \ln(2 + 4x)\end{aligned} | \begin{aligned}f(0) &= \ln 3\end{aligned} |

\begin{aligned}f^{\prime}(x) &= 4(3 + 4x)^{-1}\end{aligned} | \begin{aligned}f^{\prime}(0) &= \dfrac{4}{3}\end{aligned} |

\begin{aligned}f^{\prime \prime}(x) &= -4^2(3 + 4x)^{-2}\end{aligned} | \begin{aligned}f^{\prime \prime}(0) &= -\dfrac{4^2}{3^2}\end{aligned} |

\begin{aligned}f^{\prime \prime \prime}(x) &= -4^3(2)(3 + 4x)^{-3}\end{aligned} | \begin{aligned}f^{\prime \prime \prime}(0) &= \dfrac{4^3(2)}{3^3}\end{aligned} |

\begin{aligned}f^{(4)}(x) &= -4^4(2)(3)(3 + 4x)^{-3} \end{aligned} | \begin{aligned}f^{(4)}(0) &= \dfrac{-4^4(2)(3) }{3^4}\end{aligned} |

Now that we have the first four succeeding derivatives of $\ln(4 + 3x)$, write down the fourth Maclaurin polynomial of the function.

\begin{aligned}f(x) &= f(0) + \dfrac{f^{\prime}(0)}{1!}x + \dfrac{f^{\prime\prime}(0)}{2!} x^2 + \dfrac{f^{\prime\prime\prime}(0)}{3!} x^3 +\dfrac{f^{(4)}(0)}{4!} x^4\\&= \ln 3 + \dfrac{4}{3\cdot 1!}x -\dfrac{4^2}{3^2\cdot 2!}x^2 + \dfrac{4^3(2)}{3^3\cdot 3!}x^3 – \dfrac{4^4(2)(3)}{3^4 \cdot 4!}x^4 \end{aligned}

Since we want to estimate $\ln (7)$ using the Maclaurin polynomial, we begin by writing $\ln 7$ as a function of $f(x)$.

\begin{aligned}\ln 7 &= \ln( 3 +4 \cdot 1) \\&= f(1)\end{aligned}

Evaluate the Maclaurin polynomial at $x = 1$ to estimate $\ln 7$.

\begin{aligned}\ln 7 &\approx \ln 3 + \dfrac{4}{3\cdot 1!}(1) -\dfrac{4^2}{3^2\cdot 2!}(1)^2 + \dfrac{4^3(2)}{3^3\cdot 3!}(1)^3 – \dfrac{4^4(2)(3 + 4x)^{-4}}{3^4 \cdot 4!}(1)^4\\&= \ln 3 + \dfrac{4}{3\cdot 1!} -\dfrac{4^2}{3^2\cdot 2!}+ \dfrac{4^3(2)}{3^3\cdot 3!}- \dfrac{4^4(2)}{3^4 \cdot 4!}\\&= \ln 3 + \dfrac{4}{3} – \dfrac{8}{9} + \dfrac{64}{81} – \dfrac{64}{81}\\ &\approx 1.543 \end{aligned}

This example shows how it’s possible for us to estimate values using Maclaurin polynomials. In fact, this also highlights how we numerically approximate values of transcendental functions. Of course, the higher the order of the Maclaurin polynomial, the more accurate the approximations are.

**Practice Questions**

1. Find the Maclaurin series of $f(x) = e^{-x}$ up to its fourth-order. Write the Maclaurin series in sigma notation as well.

2. Find the Maclaurin series of $f(x) = e^{2x}$ up to its fourth-order then write the Maclaurin series in sigma notation.

3. Find the Maclaurin series of $f(x) = \cos x$ up to its fourth-order then write the Maclaurin series in sigma notation.

4. Use the result from the previous problem to find the Maclaurin series of $g(x) = x\cos x$. Finalize your answer by writing the Maclaurin series in sigma notation.

5. What is the seventh Maclaurin polynomial of $f(x) = \sin (3x)$? Use the result to approximate $\sin\left(\dfrac{\pi}{2}\right)$.

6. Use the fifth Maclaurin polynomial for $f(x) = \sin x$ to approximate $\sin\left(\dfrac{\pi}{12}\right)$.

**Answer Key**

1. Find the Maclaurin series of $f(x) = e^{-x}$ up to its fourth-order. Write the Maclaurin series in sigma notation as well.

2. $\begin{aligned}f(x) &= 1 + 2x + 2x^2 + \dfrac{4x^3}{3} + \dfrac{2x^4}{3} + …\\&= \sum_{n = 0 }^{\infty} \dfrac{2^{n}x^n}{n!}\end{aligned}$

3.$ \begin{aligned} \cos x &= 1 – \dfrac{x^2}{2!} + \dfrac{x^4}{4!}- \dfrac{x^6}{6!} …\\&= \sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!} \end{aligned}$

4.$ \begin{aligned} x\cos x &= x – \dfrac{x^3}{2!} + \dfrac{x^5}{4!}- …\\&= \sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2n + 1}}{(2n)!} \\&=x \sum_{n =0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!} \end{aligned}$

5.$ \begin{aligned} \sin (3x) &= 3x – \dfrac{27}{6}x^3 + \dfrac{81}{40}x^5 – \dfrac{243}{560}x^7\\\\ \sin\left(\dfrac{\pi}{2}\right) &= \sin\left(3 \cdot \dfrac{\pi}{6}\right)\\ &= 3\left(\dfrac{\pi}{6} \right ) – \dfrac{27}{6}\left(\dfrac{\pi}{6} \right )^3 + \dfrac{81}{40}\left(\dfrac{\pi}{6} \right )^5 – \dfrac{243}{560}\left(\dfrac{\pi}{6} \right )^7 \\&\approx 0.998\end{aligned}$

6.$ \begin{aligned} \sin \left(\dfrac{\pi}{12}\right) &= \dfrac{\pi}{12} – \dfrac{1}{3!}\left(\dfrac{\pi}{12} \right)^3 + \dfrac{1}{5!} \left(\dfrac{\pi}{12}\right)^5 \\&\approx 0.2588\end{aligned}$