AP Calculus BC: Maclaurin Series for Common Functions

Maclaurin series are your gateway to approximating complex functions with simple polynomials, predicting system behavior in engineering, or evaluating integrals that are otherwise impossible. These powerful tools transform transcendental functions like $e^x$ and $\sin (x)$ into infinite polynomials, providing the foundational language for much of applied calculus, physics, and engineering. Mastering a short list of key series and the techniques to manipulate them is a core skill for the AP Calculus BC exam and future technical coursework.

What is a Maclaurin Series?

A Maclaurin series is a specific type of Taylor series—an infinite sum of polynomial terms—that is centered at $x=0$. The general form for a function $f(x)$ is:

$$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\cdots =\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$$

Here, $f^{(n)}(0)$ denotes the $n$th derivative of $f$ evaluated at $x=0$, and $n!$ is $n$ factorial. The series represents $f(x)$ exactly within its interval of convergence. The magic lies in the fact that by knowing all the function's derivatives at a single point (zero), you can reconstruct the entire function around that point. For the AP exam, you are not typically required to derive every series from this formula repeatedly; instead, you memorize a few essential forms and learn to build others from them.

The Core Series to Memorize

Your first task is to commit four fundamental Maclaurin series to memory. These are the building blocks for almost all related problems.

1. Exponential Function: $e^x$

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots =\sum _{n=0}^{\infty }\frac{x^n}{n!}$$ Its interval of convergence is $(-\infty ,\infty )$. Notice the simple pattern: the coefficient of $x^n$ is simply $1/n!$.

1. Sine Function: $\sin (x)$

$$\sin (x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$$ Only odd powers of $x$ appear, and the signs alternate starting with a positive $x$.

1. Cosine Function: $\cos (x)$

$$\cos (x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}$$ This series contains only even powers of $x$, with alternating signs starting with a positive $1$.

1. Geometric Series: $\frac{1}{1-x}$

$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots =\sum _{n=0}^{\infty }{x}^n$$ This is the simplest form, where the coefficients are all $1$. Its interval of convergence is $|x|<1$.

Think of these series as your mathematical toolkit. Just as a carpenter uses a few basic tools to create complex structures, you will use these four series to construct representations for many other functions.

Deriving New Series Through Manipulation

You will rarely be asked to generate a series from the derivative definition on the exam. Instead, you manipulate the known series using substitution, algebraic operations, differentiation, and integration.

1. Substitution

This is the most straightforward technique. You replace the $x$ in a known series with a more complicated expression.

• Example: Find the Maclaurin series for $e^{-x^2}$.

Start with the series for $e^x$: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ Substitute "$-x^2$" for every "$x$": $$e^{-x^2}=1+(-x^2)+\frac{(-x^2{)}^2}{2!}+\frac{(-x^2{)}^3}{3!}+\cdots$$ $$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{n!}$$

2. Algebraic Manipulation (Multiplication & Division)

Use algebra to transform a given function into a form that matches a known series.

• Example: Find the first four non-zero terms for $x\cos (3x)$.

Start with the series for $\cos (u)$: $\cos (u)=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots$ Substitute $u=3x$: $\cos (3x)=1-\frac{(3x{)}^2}{2!}+\frac{(3x{)}^4}{4!}-\cdots =1-\frac{9x^2}{2}+\frac{81x^4}{24}-\cdots$ Now multiply every term by $x$: $$x\cos (3x)=x-\frac{9x^3}{2}+\frac{81x^5}{24}-\cdots$$

3. Differentiation and Integration

If the derivative or integral of a function is a known form, you can find the series term-by-term.

• Example (Differentiation): The derivative of $\sin (x)$ is $\cos (x)$. Therefore, you can differentiate the series for $\sin (x)$ term-by-term to get the series for $\cos (x)$.

$$\frac{d}{dx}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \right)=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\cdots =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$

• Example (Integration): Find the Maclaurin series for $\ln (1+x)$.

Note that $\frac{d}{dx}[\ln (1+x)]=\frac{1}{1+x}$. We know the series for $\frac{1}{1-u}$ is $1+u+u^2+\cdots$. Substitute $u=-x$: $\frac{1}{1+x}=\frac{1}{1-(-x)}=1+(-x)+(-x{)}^2+(-x{)}^3+\cdots =1-x+x^2-x^3+\cdots$ for $|x|<1$. Now integrate term-by-term from $0$ to $x$: $$\ln (1+x)={\int }_0^x(1-t+t^2-t^3+\cdots )dt={\left[t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots \right]}_0^x$$ $$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots =\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}{x}^n}{n}$$

Applications and Approximations

The primary application of these finite polynomial approximations is to estimate function values and solve otherwise intractable problems. By taking the first few terms of an infinite series, you get a polynomial that is very close to the original function near $x=0$.

• Engineering Scenario: An electrical engineer might need to compute $e^{-0.1}$ without a calculator. Using the series:

$$e^{-0.1}\approx 1+(-0.1)+\frac{(-0.1{)}^2}{2!}+\frac{(-0.1{)}^3}{3!}=1-0.1+0.005-\mathrm{0.000166...}\approx \mathrm{0.904833...}$$ The actual value is about $0.904837$, so our 4-term approximation is accurate to 5 decimal places.

• Calculus Application: Evaluate ${\lim }_{x\to 0}\frac{\sin (x)-x}{x^3}$.

Instead of L'Hôpital's Rule (which would require multiple derivatives), substitute the series: $$\sin (x)-x=(x-\frac{x^3}{3!}+\cdots )-x=-\frac{x^3}{6}+\cdots$$ Therefore, $\frac{\sin (x)-x}{x^3}=\frac{-\frac{x^3}{6}+\cdots }{x^3}=-\frac{1}{6}+\cdots$ As $x\to 0$, the higher-order terms vanish, so the limit is $-\frac{1}{6}$.

Common Pitfalls

1. Ignoring the Interval of Convergence: Every power series has a radius where it is valid. The series for $\frac{1}{1-x}$ only converges for $|x|<1$. If you use it to find a series for $\ln (1+x)$ via integration, the resulting series also only converges for $|x|<1$ (though it converges at $x=1$ as well). Always state or consider the interval of convergence.

• Correction: After finding a series through manipulation, determine its new radius of convergence. For substitution, if $|u|<R$ is the condition for the original series, then the new series converges where $|(\text{new expression})|<R$.

1. Incorrect Substitution: When substituting a complex expression like $5x^2$ for $x$, you must replace every $x$ in the series, including those in the exponents and factorials.

• Correction: Write the known series with a placeholder, e.g., $e^u=1+u+\frac{u^2}{2!}+\cdots$. Then explicitly set $u=5x^2$ and simplify: $1+(5x^2)+\frac{(5x^2{)}^2}{2!}+\cdots$.

1. Mishandling Signs in Trig Series: The alternating signs in the $\sin (x)$ and $\cos (x)$ series are crucial. A common error is to forget that $(-1{)}^n$ produces a positive term when $n$ is even and a negative term when $n$ is odd.

• Correction: Write out the first few terms explicitly from the summation notation to see the pattern. For $\cos (x)$, when $n=1$, the term is $\frac{(-1{)}^1{x}^2}{2!}=-\frac{x^2}{2!}$.

1. Integrating/Differentiating Without Adjusting the Interval: When you find a series by integrating, a constant of integration appears. For a Maclaurin series, this constant is chosen so the series equals the function at $x=0$.

• Correction: Use definite integration from $0$ to $x$, or find the constant by evaluating your integrated series at $x=0$ and setting it equal to the known value of $f(0)$.

Summary

• A Maclaurin series is a Taylor series centered at $x=0$, representing a function as an infinite sum of polynomial terms based on its derivatives at zero.
• Memorize the four cornerstone series: $e^x=\sum \frac{x^n}{n!}$, $\sin (x)=\sum \frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$, $\cos (x)=\sum \frac{(-1{)}^n{x}^{2n}}{(2n)!}$, and $\frac{1}{1-x}=\sum x^n$ for $|x|<1$.
• Generate new series through substitution (replacing $x$ with another expression), algebraic manipulation (multiplication, division), and term-by-term differentiation or integration of known series.
• These series are powerful for approximating function values, evaluating limits, and solving integrals that are difficult with standard methods.
• Always pay attention to the interval of convergence associated with any series you derive or use, as it dictates where your representation is valid.