Finding Taylor or Maclaurin Series for a Function - AP Calculus BC Study Guide

The Core Idea: Finding Taylor or Maclaurin Series for a Function

This topic introduces a powerful method for representing a function as an infinite polynomial, known as a power series. The fundamental idea is that if a function has derivatives of all orders at a specific point, $x=c$, we can construct a unique power series centered at that point that is built from the values of these derivatives. This series, called a Taylor series, provides a polynomial approximation of the function near the center. The more terms we include in the polynomial, the better the approximation generally becomes.

The construction of this series relies on a specific formula where each term's coefficient is determined by the function's corresponding derivative at the center. A Maclaurin series is simply a special, more common case of a Taylor series where the center of expansion is $x=0$. This process allows us to work with complicated functions by using their more manageable polynomial representations.

Key Formulas

The construction of Taylor and Maclaurin series is based on two fundamental definitions.

Taylor Series for a Function $f$ about $x=c$

If a function $f$ has derivatives of all orders at $x=c$, its Taylor series is defined as:

$$\sum _{n=0}^{\infty }\frac{f^{(n)}(c)}{n!}(x-c{)}^n$$

When expanded, this series takes the form:

$$f(c)+f^{\prime }(c)(x-c)+\frac{f^{\prime \prime }(c)}{2!}(x-c{)}^2+\frac{f^{\prime \prime \prime }(c)}{3!}(x-c{)}^3+\cdots +\frac{f^{(n)}(c)}{n!}(x-c{)}^n+\dots$$

• $f^{(n)}(c)$ is the $n$-th derivative of the function $f$ evaluated at the center $x=c$. Note that $f^{(0)}(c)$ is simply $f(c)$.

• $n!$ is the factorial of $n$.

• $(x-c{)}^n$ is the power term, which centers the series at $x=c$.

Maclaurin Series for a Function $f$

A Maclaurin series is a Taylor series centered at $x=0$. To find the Maclaurin series, we set $c=0$ in the Taylor series formula:

$$\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$$

When expanded, this series takes the form:

$$f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\cdots +\frac{f^{(n)}(0)}{n!}{x}^n+\dots$$

Understanding the Prerequisite Condition

The ability to construct a Taylor series for a function $f$ about a center $x=c$ is entirely dependent on a single, critical condition: the function $f$ must have derivatives of all orders at $x=c$.

This means that you must be able to compute $f^{\prime }(c)$, $f^{\prime \prime }(c)$, $f^{\prime \prime \prime }(c)$, and so on, for all positive integers $n$. If any derivative fails to exist at $x=c$, a Taylor series for $f$ cannot be constructed at that center. For most functions encountered in AP Calculus, such as $e^x$, $\sin (x)$, $\cos (x)$, and polynomials, derivatives of all orders exist everywhere. However, it is the essential theoretical requirement that underpins the entire process. The formulas provided define the series that represents the function, but this representation is only possible if the derivatives needed to build the series coefficients exist.

Core Concepts & Rules

• Taylor Series Definition: A Taylor series for a function $f$ about $x=c$ is an infinite series given by ${\sum }_{n=0}^{\infty }\frac{f^{(n)}(c)}{n!}(x-c{)}^n$.

• Maclaurin Series Definition: A Maclaurin series is a specific type of Taylor series that is always centered at $x=0$. Its formula is ${\sum }_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$.

• Required Condition: To generate a Taylor series for a function $f$ about $x=c$, the function must have derivatives of all orders at $x=c$.

• Coefficient Calculation: The coefficient of the $(x-c{)}^n$ term in a Taylor series is always $\frac{f^{(n)}(c)}{n!}$. This value is calculated by finding the $n$-th derivative of $f$ and evaluating it at the center $c$, then dividing by $n!$.

• Structure: The series is a sum of terms, where the $n$-th term consists of three parts: the coefficient based on the $n$-th derivative ($\frac{f^{(n)}(c)}{n!}$), the power term ($(x-c{)}^n$), and the factorial in the denominator ($n!$).

Step-by-Step Example 1: Finding a Maclaurin Series

Problem: Find the Maclaurin series for the function $f(x)=\sin (x)$. Write out the first four non-zero terms and the general term.

Step 1: Find the first several derivatives of the function.

• $f(x)=\sin (x)$

• $f^{\prime }(x)=\cos (x)$

• $f^{\prime \prime }(x)=-\sin (x)$

• $f^{\prime \prime \prime }(x)=-\cos (x)$

• $f^{(4)}(x)=\sin (x)$

• $f^{(5)}(x)=\cos (x)$

Step 2: Evaluate these derivatives at the center, $c=0$.

• $f(0)=\sin (0)=0$

• $f^{\prime }(0)=\cos (0)=1$

• $f^{\prime \prime }(0)=-\sin (0)=0$

• $f^{\prime \prime \prime }(0)=-\cos (0)=-1$

• $f^{(4)}(0)=\sin (0)=0$

• $f^{(5)}(0)=\cos (0)=1$

Step 3: Substitute these values into the Maclaurin series formula.

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

$$\sin (x)=0+\frac{1}{1!}x+\frac{0}{2!}{x}^2+\frac{-1}{3!}{x}^3+\frac{0}{4!}{x}^4+\frac{1}{5!}{x}^5+\dots$$

Step 4: Write out the first four non-zero terms.

Simplifying the expression from Step 3, we get:

$$\sin (x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$$

Step 5: Determine the general term and write the series in sigma notation.

Observe the pattern:

• The powers are odd: $1,3,5,7,\dots$, which can be written as $2n+1$.

• The signs alternate: $+,-,+,-,\dots$, which can be written as $(-1{)}^n$.

• The factorial in the denominator matches the power: $(2n+1)!$.

The series starts at $n=0$: for $n=0$, we get $\frac{(-1{)}^0{x}^{2(0)+1}}{(2(0)+1)!}=\frac{x^1}{1!}=x$. This matches our first term.

Therefore, the general term is $\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$.

The Maclaurin series is:

$$\sin (x)=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$$

Step-by-Step Example 2: Exam-Style Application from a Table

Problem: A function $g$ has derivatives of all orders. The values of $g$ and its first three derivatives at $x=-2$ are given in the table below.

Write the third-degree Taylor polynomial for $g$ about $x=-2$.

Step 1: Identify the center and the required degree.

The problem asks for the Taylor polynomial about $x=-2$, so the center is $c=-2$.

The required degree is 3.

Step 2: Recall the formula for a Taylor series.

The general formula is ${\sum }_{n=0}^{\infty }\frac{g^{(n)}(c)}{n!}(x-c{)}^n$.

For a third-degree polynomial, we need the terms from $n=0$ to $n=3$.

$$P_3(x)=g(c)+g^{\prime }(c)(x-c)+\frac{g^{\prime \prime }(c)}{2!}(x-c{)}^2+\frac{g^{\prime \prime \prime }(c)}{3!}(x-c{)}^3$$

Step 3: Substitute the given values from the table with $c=-2$.

From the table:

• $g(-2)=6$

• $g^{\prime }(-2)=4$

• $g^{\prime \prime }(-2)=-12$

• $g^{\prime \prime \prime }(-2)=18$

Substitute these into the formula:

$$P_3(x)=6+4(x-(-2))+\frac{-12}{2!}(x-(-2){)}^2+\frac{18}{3!}(x-(-2){)}^3$$

Step 4: Simplify the coefficients.

• $\frac{-12}{2!}=\frac{-12}{2}=-6$

• $\frac{18}{3!}=\frac{18}{6}=3$

Step 5: Write the final polynomial.

$$P_3(x)=6+4(x+2)-6(x+2{)}^2+3(x+2{)}^3$$

This is the third-degree Taylor polynomial for $g$ about $x=-2$.

Using Your Calculator

This topic is primarily analytical, and a calculator cannot be used to generate a Taylor or Maclaurin series directly from a function. The process of finding derivatives and evaluating them at the center must be done by hand.

However, a graphing calculator is an excellent tool for checking your work. After you have derived a Taylor polynomial, you can verify its accuracy visually.

To check the result of Example 1:

1. In Y1, enter the original function: Y1 = sin(x).

2. In Y2, enter the Taylor polynomial you found. For example, the first few terms: $Y2=X-X^3/3!+X^5/5!$. (Note: $3!$ can be entered as 3 MATH -> PRB -> 4:!`). 3. Graph both functions. A good viewing window would be centered around the series center, $x=0. For example, $Xmin=-pi$, $Xmax=pi$.

3. Observe the graphs. The graph of your polynomial Y2 should be almost indistinguishable from the graph of the original function Y1 very close to the center x=0. The more terms you add to your polynomial in Y2, the wider the interval on which the graphs will match. If the graphs do not match near the center, you have likely made a calculation error in your derivatives or coefficients.

AP Exam Quick Hit

Common Question Types

• Direct Series Construction: "Find the first four nonzero terms and the general term of the Taylor series for $f(x)=\ln (x)$ about $x=1$." This requires the full process of finding derivatives, evaluating at the center, and identifying the pattern.

• Polynomial from a Table: "The function $f$ has derivatives of all orders for all real numbers. Values of $f$ and its first three derivatives at $x=2$ are given in the table. Write the second-degree Taylor polynomial for $f$ about $x=2$." This tests direct application of the formula without the differentiation step.

• Finding a Specific Coefficient: "Find the coefficient of the $(x-\pi /2{)}^3$ term in the Taylor series for $f(x)=\cos (x)$ about $x=\pi /2$." This isolates one part of the process, requiring the student to calculate $\frac{f^{\prime \prime \prime }(\pi /2)}{3!}$.

Common Mistakes

• Forgetting the Factorial: A very frequent error is to write the coefficient as just $f^{(n)}(c)$ instead of $\frac{f^{(n)}(c)}{n!}$. Always remember to divide by the factorial.

• Forgetting the Power Term: Omitting the $(x-c{)}^n$ term and just writing a series of coefficients.

• Incorrect Center: When finding a Taylor series about a non-zero center $c$, students sometimes mistakenly evaluate the derivatives at $x=0$ as if it were a Maclaurin series.

• Derivative Errors: Simple arithmetic or sign errors when calculating higher-order derivatives can cascade through the entire problem. Double-check your differentiation.

• Ignoring the "n=0" Term: The series begins with $n=0$, which corresponds to the term $\frac{f^{(0)}(c)}{0!}(x-c{)}^0=f(c)$. Students sometimes start their series at $n=1$, forgetting the initial constant term.