10.11 Finding Taylor Polynomial Approximations of Functions

A Taylor polynomial approximates a function near a point $x=a$ by matching the function's value and derivatives at that point. The coefficient of the $n$th-degree term is $\frac{f^{(n)}(a)}{n!}$, and a Taylor polynomial centered at $x=0$ is called a Maclaurin polynomial. For AP Calculus BC, organize values in a derivative table before building the polynomial.

How Do Taylor Polynomial Approximations Work?

A Taylor polynomial approximates $f(x)$ near a center $x=a$ by using the values of $f$ and its derivatives at $a$. The $n$th-degree term has coefficient $\frac{f^{(n)}(a)}{n!}$ and power $(x-a)^n$, so each added term matches one more derivative at the center.

For AP Calculus BC, the core routine is: identify the center, compute the needed derivatives, evaluate them at the center, build the polynomial through the requested degree, and plug in a nearby $x$-value if the problem asks for an approximation.

Taylor polynomials lead directly into the Lagrange error bound and radius and interval of convergence later in Unit 10. Besides building and evaluating polynomials, you may be asked to recognize the pattern in the coefficients.

How Taylor Polynomial Terms Are Built

• The $n$th-degree term coefficient in a Taylor polynomial centered at $x=a$ is $\frac{f^{(n)}(a)}{n!}$.
• A Maclaurin polynomial is just a Taylor polynomial centered at $x=0$.
• The $n$th-degree Taylor polynomial is the $n$th partial sum of the full Taylor series.
• Higher-degree polynomials usually approximate the original function more accurately over an interval.
• Use a derivative table ($n$, $f^{(n)}(x)$, $f^{(n)}(a)$, $(x-a)^n$, full term) to stay organized.
• Taylor polynomials are most accurate near the center $x=a$ and tend to drift away from $f$ as $x$ moves farther from $a$.

The Taylor Polynomial Formula

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

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

Written out term by term, this is:

$$f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n$$

Here $f^{(n)}(a)$ is the $n$th derivative of the function evaluated at $a$, and $f^{(0)}(a)=f(a)$. The $n$th-degree Taylor polynomial is the $n$th partial sum of the infinite series, meaning you stop after the $(x-a)^n$ term.

Taylor series centered at $x=0$ are common enough to get their own name: Maclaurin series.

Breaking Down the Formula

Taylor series look intimidating at first. The trick is to organize the pieces in a table. Each row builds one term of the polynomial.

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & f(x) & f(a)&(x-a)^0 & \frac{f(a)}{1}\cdot(x-a)^0 \\ 1 & 1 & f'(x) & f'(a)&(x-a)^1 & \frac{f'(a)}{1}\cdot(x-a)^1 \\ 2 & 2 & f''(x) & f''(a)&(x-a)^2 & \frac{f''(a)}{2}\cdot(x-a)^2 \\ 3 & 6 & f'''(x) & f'''(a)&(x-a)^3 & \frac{f'''(a)}{6}\cdot(x-a)^3 \\ ... & ... & ... & ... & ... & ... \\ n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ \hline \end{array}$$

Fill in the table column by column, then add up the entries in the last column to get your polynomial.

Worked Example

Find the third-degree Maclaurin polynomial for $e^{5x}$.

Solution: Build the table. A Maclaurin polynomial is just a Taylor polynomial with $a=0$, so every derivative is evaluated at $0$.

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & e^{5x} & 1&1 & 1 \\ 1 & 1 & 5e^{5x} & 5&x & 5x \\ 2 & 2 & 25e^{5x} & 25&x^2 & 25x^2/2 \\ 3 & 6 & 125e^{5x} & 125&x^3 & 125x^3/6 \\ \hline \end{array}$$

Add the terms in the last column. The third-degree Maclaurin polynomial for $e^{5x}$ is:

$$1+5x+\frac{25}{2}x^2+\frac{125}{6}x^3$$

Building and Using Taylor Polynomials Step by Step

Problem Solving:

• Identify the center $a$ first. If the problem says "Maclaurin," use $a=0$.
• Take derivatives one degree at a time and evaluate each at $a$ before moving on. It is easy to lose track if you do all the derivatives first.
• Build the term using $\frac{f^{(n)}(a)}{n!}(x-a)^n$. Keep the factorial in the denominator separate from any constant that comes from the derivative.
• To approximate a function value near $a$, plug the given $x$ into your finished polynomial. The closer $x$ is to $a$, the more accurate the result.

Free Response:

• Show each derivative and its value at the center. Clear work makes your reasoning easy to follow.
• Leave coefficients as exact fractions rather than rounding, unless the problem asks for a decimal approximation.
• If a problem asks for a specific degree, stop at that degree. Do not add extra terms.

Common Trap:

• Watch the chain rule when taking repeated derivatives of functions like $e^{5x}$ or $\sin(2x)$. Each derivative pulls down another factor.
• For functions like $\ln(x)$ centered at $x=1$, keep the $(x-1)^n$ form. Do not switch to powers of $x$.

Practice

Try these on your own, then check the worked solutions.

Problems

1. Find the fifth-degree Maclaurin polynomial for $f(x)=\cos(x)$.
2. Find the third-degree Taylor polynomial for $f(x)=\ln(x)$ about $x=1$.
3. Find the fourth-degree Taylor polynomial about $x=2$ for $f(x)=\sqrt{x}$.

Solution for Question 1

Build your table and fill in the values:

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \text{cos}(x) & 1&1 & 1 \\ 1 & 1 & -\text{sin}(x) & 0&x & 0 \\ 2 & 2 & -\text{cos}(x) & -1&x^2 & -x^2/2 \\ 3 & 6 & \text{sin}(x) & 0&x^3 & 0 \\ 4 & 24 & \text{cos}(x) & 1&x^4 & x^4/24 \\ 5 & 120 & -\text{sin}(x) & 0&x^5 & 0 \\ \hline \end{array}$$

Putting it together, the fifth-degree Maclaurin polynomial for $f(x)=\cos(x)$ is

$$1-\frac{x^2}{2}+\frac{x^4}{24}$$

Notice that the odd-degree terms all drop out because the sine values at $0$ are zero. This is why $\cos x$ has only even powers.

Solution for Question 2

Here the $(x-a)^n$ column uses $(x-1)$ since the center is $x=1$.

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \text{ln}(x) & 0&1 & 0 \\ 1 & 1 & 1/x & 1&(x-1) & (x-1) \\ 2 & 2 & -1/x^2 & -1&(x-1)^2 & -\frac{1}{2}(x-1)^2 \\ 3 & 6 & 2/x^3 & 2&(x-1)^3 & \frac{1}{3}(x-1)^3 \\ \hline \end{array}$$

The third-degree term is $\frac{f'''(1)}{3!}(x-1)^3 = \frac{2}{6}(x-1)^3 = \frac{1}{3}(x-1)^3$. The polynomial is:

$$(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3$$

Solution for Question 3

One more table, this time centered at $x=2$.

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \sqrt{x} & \sqrt{2}&1 & \sqrt{2} \\ 1 & 1 & \frac{1}{2\sqrt{x}} & \frac{1}{2\sqrt{2}}&(x-2) & \frac{1}{2\sqrt{2}}\cdot(x-2)\\ 2 & 2 & -\frac{1}{4\sqrt{x^3}} & -\frac{1}{4\sqrt{8}}&(x-2)^2 & -\frac{1}{8\sqrt{8}}\cdot(x-2)^2 \\ 3 & 6 & \frac{3}{8\sqrt{x^5}} & \frac{3}{8\sqrt{32}}&(x-2)^3 & \frac{3}{48\sqrt{32}}\cdot (x-2)^3 \\ 4 & 24 & -\frac{15}{16\sqrt{x^7}} & -\frac{15}{16\sqrt{128}}&(x-2)^4 & -\frac{15}{384\sqrt{128}}\cdot(x-2)^4 \\ \hline \end{array}$$

Putting it all together:

$$\sqrt{2}+\frac{1}{2\sqrt{2}}\cdot(x-2)-\frac{1}{8\sqrt{8}}\cdot(x-2)^2+\frac{3}{48\sqrt{32}}\cdot (x-2)^3 -\frac{15}{384\sqrt{128}}\cdot(x-2)^4$$

Simplify a few terms using exponent rules:

$$\sqrt{2}+\frac{x-2}{2\sqrt{2}}-\frac{(x-2)^2}{16\sqrt{2}}+\frac{(x-2)^3}{64\sqrt{2}} -\frac{5(x-2)^4}{1024\sqrt{2}}$$

This is the fourth-degree Taylor polynomial centered at $x=2$ for $\sqrt{x}$.

Common Misconceptions

• Misconception: A Taylor polynomial equals the function everywhere.
  Actually: It only approximates $f$, and it is most accurate near the center $x=a$. Accuracy usually gets worse as $x$ moves away from $a$.
• Misconception: The coefficient is just $f^{(n)}(a)$, without a factorial.
  Actually: The coefficient is $\frac{f^{(n)}(a)}{n!}$, not just $f^{(n)}(a)$. Skipping the $n!$ is one of the most common errors.
• Misconception: The degree tells you where the center is.
  Actually: A "third-degree" polynomial stops at the $(x-a)^3$ term. The degree tells you how far to go, not where the center is.
• Misconception: When the center is not zero, you can drop the $(x-a)$ form.
  Actually: For a Taylor polynomial about $x=1$, keep the powers as $(x-1)^n$. Only Maclaurin polynomials use plain powers of $x$.
• Misconception: More terms always help equally.
  Actually: Higher degree usually improves the fit, but only over the interval where the series converges. Far outside that range, adding terms may not help.
• Misconception: You should evaluate derivatives at $x$ instead of $a$.
  Actually: You must plug in the center $a$ for every derivative before building the term.

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