10.14 Finding Taylor or Maclaurin Series for a Function

TLDR

A Taylor series writes a function as an infinite power series built from its derivatives at a center point: $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$. A Maclaurin series is just a Taylor series centered at $x=0$. The fastest way to find these on the AP Calculus BC exam is to memorize a few common Maclaurin series and then substitute, multiply, differentiate, or integrate to match the function you are given.

How Do You Find a Taylor or Maclaurin Series?

To find a Taylor or Maclaurin series, start from the derivative formula $\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$ or transform a known series. A Maclaurin series is the special case where the center is $a=0$.

For AP Calculus BC, known series are usually faster than repeated differentiation. Use the geometric series for $\frac{1}{1-x}$ and memorize the Maclaurin series for $e^x$, $\sin x$, and $\cos x$, then substitute, multiply, differentiate, or integrate to match the function.

Finding a Taylor or Maclaurin series is a BC-only skill, so it only appears if you take AP Calculus BC. It builds on two earlier ideas: approximating functions with Taylor polynomials and representing functions as power series.

The Taylor Formula and Maclaurin Series to Memorize

• The Taylor series of $f$ centered at $x=a$ is $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$, where $f^{(0)}(a)=f(a)$.
• A Maclaurin series is a Taylor series with center $a=0$.
• A Taylor polynomial is a partial sum (finite piece) of the full Taylor series.
• Memorize the Maclaurin series for $\frac{1}{1-x}$, $e^x$, $\sin x$, and $\cos x$; most other series come from these by substitution or algebra.
• The series for $\frac{1}{1-x}$ is a geometric series, and the series for $\sin x$, $\cos x$, and $e^x$ are the building blocks for many other functions.
• Pattern recognition in derivatives is the key skill when you have to build a series from scratch.

What's a Taylor Series?

A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. It lets you approximate functions and estimate their values near that point.

Taylor Series: For a function $f(x)$, its Taylor series centered at $x = a$ is:

$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}\cdot(x-a)^n=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n$$

Here $f^{(n)}(a)$ is the $n^{\text{th}}$ derivative of the function evaluated at $a$, and $f^{(0)}(a)=f(a)$.

When the center is $x = 0$, this special case is called a Maclaurin series. Taylor series centered at $0$ come up so often that they get their own name.

A Taylor polynomial for $f(x)$ is a partial sum of the Taylor series for $f(x)$. In other words, a Taylor polynomial is a finite polynomial with a limited number of terms, while a Taylor series is an infinite summation of terms. To build a Taylor polynomial of degree $n$ for $f(x)$ at $x = c$, evaluate $f$ and its first $n$ derivatives at $x = c$.

To form the full Taylor series, it helps to find a pattern that describes the $n$th derivative of $f$ at $x = c$. Pattern recognition is the most useful skill here.

Important Maclaurin Series to Remember

Focus on Maclaurin series (center $x = 0$). If a problem gives you a different center and you already know the Maclaurin form, you can replace $x$ with $(x - c)$ in each term and simplify.

$$\begin{array}{ |c|c|c| } \hline f(x)& Series \enspace Representation & Expanded \enspace Form \\\\ \frac{1}{1-x} & \sum_{n=0}^\infty{x^n} & 1+x+x^2+x^3+... \\\\ \frac{1}{1+x} & \sum_{n=0}^\infty{(-x)^n} & 1-x+x^2-x^3+... \\\\ \frac{1}{1+{x}^2} & \sum_{n=0}^\infty{(-x)^{2n}} & 1-x^2+x^4-x^6+... \\\\ \frac{1}{1-{x}^2} & \sum_{n=0}^\infty{x^{2n}} & 1+x^2+x^4+x^6+... \\\\ e^x & \sum_{n=0}^\infty\frac{x^n}{n!} & 1+x+\frac{x^2}{2}+\frac{x^3}{3!}+... \\\\ sin(x) & \sum_{n=0}^\infty{(-1)^n\frac{x^{2n+1}}{(2n+1)!}} & x-\frac{x^3}{3!}+\frac{x^5}{5!}+... \\\\ cos(x) & \sum_{n=0}^\infty{(-1)^n\frac{x^{2n}}{(2n)!}} & 1-\frac{x^2}{2}+\frac{x^4}{4!}+... \\\\ ln(1+x) & \sum_{n=1}^\infty{(-1)^{n+1}\frac{x^{n}}{n}} & x-\frac{x^2}{2}+\frac{x^3}{3}+... \\\\ (1+x)^a & \sum_{n=0}^\infty{a \choose n}x^n & 1+ax+\frac{a(a-1)}{2}x^2+\frac{a(a-1)(a-2)}{3!}x^3+... \\\\ \hline \end{array}$$

Two things to notice:

1. The first four are variations of the geometric series, with the multiplying factor being various powers of $x$. The series $(1+x)^a$ is the binomial series for any value of $a$.
2. The Maclaurin series for $\sin(x)$, $\cos(x)$, and $e^x$ are the foundation for building the Maclaurin series of many other functions.

Looking at point (2), the patterns connect: $\sin(x)$ is an odd function, so its series has only odd powers, while $\cos(x)$ is even, so its series has only even powers. The series for $e^x$ uses all powers. If you are short on time, $\sin(x)$, $\cos(x)$, and $e^x$ are the must-know series.

Building Series by Substitution and Shifting the Center

Problem Solving: Start by checking whether the function matches a known Maclaurin series. If it does, substitution is usually faster than computing derivatives one by one.

Question 1. Find the Taylor series for $f(x) = \cos(3x)$ centered at $x = 0$.

Centered at $x = 0$ means this is a Maclaurin series. Pull the series for $\cos(x)$:

$$\sum_{n=0}^\infty{(-1)^n\frac{x^{2n}}{(2n)!}}$$

Replace $x$ with $3x$ and simplify:

$$\sum_{n=0}^\infty{(-1)^n\frac{(3x)^{2n}}{(2n)!}}=\sum_{n=0}^\infty{(-1)^n\frac{3^{2n}x^{2n}}{(2n)!}}=\sum_{n=0}^\infty{(-1)^n\frac{9^{n}x^{2n}}{(2n)!}}$$

That is the full answer. The more comfortable you are recalling common Maclaurin series, the faster these go.

Question 2a. Find the Taylor series centered at $x = 5$ for $f(x)=e^{2x}$.

Compute a few derivatives and look for a pattern:

$$f(x) = e^{2x}$$ $$f'(x) = 2e^{2x}$$ $$f''(x) = 4e^{2x} = (2)^2e^{2x}$$ $$f^{(3)}(x) = 8e^{2x} = (2)^3e^{2x}$$ $$f^{(4)}(x) = 16e^{2x} = (2)^4e^{2x}$$

The $n$th derivative brings out a factor of $2^n$. Using the structure of the $e^x$ series:

$$\sum_{n=0}^\infty\frac{x^n}{n!} \Rightarrow \sum_{n=0}^\infty\frac{f^{(n)}(5)}{n!}(x-5)^n$$

The center is $x = 5$, so write $(x-5)^n$ instead of $x^n$ and evaluate the $e^{2x}$ factor at $x = 5$. Leave the $n$ exponents alone:

$$\sum_{n=0}^\infty\frac{f^{(n)}(5)}{n!}(x-5)^n=\sum_{n=0}^\infty\frac{2^ne^{10}(x-5)^n}{n!}$$

Question 2b. List the first four terms of that series.

Plug in $n = 0, 1, 2, 3$:

$$\begin{array}{ |c|c| } \hline n & Term \\\\ 0 & \frac{(2)^0e^{10}(x-5)^0}{0!}=e^{10} \\\\ 1 & \frac{(2)^1e^{10}(x-5)^1}{1!} =2e^{10}(x-5)\\\\ 2 & \frac{(2)^2e^{10}(x-5)^2}{2!} =\frac{4e^{10}(x-5)^2}{2}=2e^{10}(x-5)^2\\\\ 3 & \frac{(2)^3e^{10}(x-5)^3}{3!}=\frac{8e^{10}(x-5)^3}{3*2*1}=\frac{4e^{10}(x-5)^3}{3} \\\\ \hline \end{array}$$

Common Trap: When you shift a known Maclaurin series to a new center, change only the $(x-c)$ base and any constant factors that depend on $c$. Do not change the $n$ in the exponents or factorials.

Common Misconceptions

• Misconception: A Taylor polynomial and a Taylor series are the same.
  Actually: A Taylor polynomial and a Taylor series are not the same. The polynomial is a finite partial sum; the series is the full infinite sum.
• Misconception: A Maclaurin series is a separate type of series.
  Actually: A Maclaurin series is not a separate type of series. It is just a Taylor series with center $a = 0$.
• Misconception: When substituting something like $3x$ into a known series, you do not need to replace $x$ inside exponents.
  Actually: When you substitute something like $3x$ into a known series, replace $x$ everywhere it appears, including inside exponents, then simplify (for example $(3x)^{2n} = 9^n x^{2n}$).
• Misconception: Changing the center to $x = c$ means adding $c$ to the index $n$.
  Actually: Changing the center to $x = c$ means replacing $x$ with $(x-c)$ in the power terms, not adding $c$ to the index $n$.
• Misconception: Knowing the series for $\sin x$, $\cos x$, and $e^x$ means you do not need pattern recognition.
  Actually: Knowing the series for $\sin x$, $\cos x$, and $e^x$ does not replace pattern recognition. When a function does not match a known form, you still need to differentiate and find the pattern in $\frac{f^{(n)}(a)}{n!}$.

Related AP Calculus Guides

• Unit 10 Review: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.4 Integral Test for Convergence