Finding Taylor polynomial approximations of functions

AP Calculus BC· 35 min read

1. Definition of a Taylor Polynomial★★☆☆☆⏱ 10 min

📘 Definition

nth-degree Taylor polynomial centered at $a$

T_n(x)

The unique nth-degree polynomial that matches the value of $f (x)$ and its first $n$ derivatives at $x = a$ , given by the formula:

Example:

The first-degree Taylor polynomial of $f (x)$ at $a$ is exactly the tangent line approximation you learned earlier in calculus.

T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k

📐 Worked Example

Find the 0th-degree Taylor polynomial of $f (x) = e^{x}$ centered at $a = 1$ .

1
For $n = 0$ , the sum only includes the $k = 0$ term. Recall that the 0th derivative of $f$ is $f$ itself, and $0! = 1$ .
2
Evaluate $f^{(0)} (a) = f (1) = e^{1} = e$
3
Substitute into the Taylor formula:
4
$T_0(x) = \frac{f(1)}{0!} (x-1)^0 = \frac{e}{1} \cdot 1 = e$

2. Step-by-Step Calculation of Taylor Polynomials★★★☆☆⏱ 15 min

To avoid common mistakes, follow this consistent process to build any Taylor polynomial:

Compute all derivatives of $f (x)$ up to the $n$ th derivative
Evaluate each derivative at the center point $x = a$
Divide each evaluated derivative by the corresponding factorial $k!$ for the $k$ th term
Multiply each term by $(x - a)^{k}$ and add all terms together

📐 Worked Example

Find the 3rd-degree Taylor polynomial of $f (x) = ln (x)$ centered at $a = 1$ .

1
Compute derivatives up to 3rd order:
2
$f(x) = \ln x, \quad f'(x) = \frac{1}{x}, \quad f''(x) = -\frac{1}{x^2}, \quad f'''(x) = \frac{2}{x^3}$
3
Evaluate each derivative at $a = 1$ :
4
$f(1) = 0, \quad f'(1) = 1, \quad f''(1) = -1, \quad f'''(1) = 2$
5
Calculate coefficients for each term from $k = 0$ to $k = 3$ :
6
$c_0 = \frac{0}{0!} = 0, \quad c_1 = \frac{1}{1!} = 1, \quad c_2 = \frac{-1}{2!} = -\frac{1}{2}, \quad c_3 = \frac{2}{3!} = \frac{1}{3}$
7
Write the full polynomial:
8
$T_3(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}$

✓ Quick check

Test your understanding of coefficient calculation:

What is the coefficient of $(x - 2)^{4}$ in the 4th-degree Taylor polynomial of $f (x) = x^{4}$ centered at $a = 2$ ?
- 1
- 4
- 24
- \frac{1}{24}
Reveal answer
1 —
The 4th derivative of $f (x) = x^{4}$ is $4! = 24$ . The coefficient is $\frac{f ^{(4)} ( 2 )}{4 !} = \frac{24}{24} = 1$ .

3. Common Maclaurin Polynomials★★☆☆☆⏱ 10 min

Maclaurin polynomials (Taylor at $a = 0$ ) for basic functions appear constantly on the AP exam. You can use these standard forms to quickly construct new polynomials for related functions by substitution, instead of recalculating derivatives from scratch.

Function	General nth-degree Maclaurin Polynomial
$e^{x}$	$1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + ... + \frac{x ^{n}}{n !}$
$sin x$	$x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - ... + (- 1)^{k} \frac{x ^{2 k + 1}}{( 2 k + 1 )!}$
$cos x$	$1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - ... + (- 1)^{k} \frac{x ^{2 k}}{( 2 k )!}$
$\frac{1}{1 - x}$	$1 + x + x^{2} + ... + x^{n}$
$ln (1 + x)$	$x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - ... + (- 1)^{n + 1} \frac{x ^{n}}{n}$

📐 Worked Example

Find the 4th-degree Maclaurin polynomial for $f (x) = e^{- 2 x}$ using the standard form for $e^{u}$ .

1
Start with the 4th-degree Maclaurin for $e^{u}$ :
2
$e^u \approx 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \frac{u^4}{4!}$
3
Substitute $u = - 2 x$ into the expression:
4
$1 + (-2x) + \frac{(-2x)^2}{2} + \frac{(-2x)^3}{6} + \frac{(-2x)^4}{24}$
5
Simplify each term to get the final polynomial:
6
$T_4(x) = 1 - 2x + 2x^2 - \frac{4}{3}x^3 + \frac{2}{3}x^4$

4. Common Pitfalls

Wrong move:

Forgetting to divide the kth derivative by $k!$ when calculating the coefficient.

Why:

The Taylor formula requires division by factorial to match the derivative of the polynomial. Skipping this step gives a coefficient that is off by a factor of $k!$ .

Correct move:

Always write the $k!$ denominator explicitly for each term, and double-check factorial values (e.g. $3! = 6$ , not 3).

Wrong move:

Writing $x^{k}$ instead of $(x - a)^{k}$ for a Taylor polynomial centered at $a \neq = 0$ .

Why:

Confusing Taylor and Maclaurin polynomials leads to an approximation that is only accurate near 0, not near the required center.

Correct move:

Always confirm the center first, and write $(x - a)$ for every non-constant term before simplifying.

Wrong move:

Making sign errors when taking repeated derivatives of trigonometric or logarithmic functions.

Why:

Signs flip every time you take a derivative of $sin x$ , $cos x$ , or negative powers of $x$ , so it is easy to end up with the wrong sign.

Correct move:

Write out each derivative one at a time, checking the sign and exponent at each step before evaluating at $a$ .

Wrong move:

Stopping at the $(n - 1)$ th term for an nth-degree polynomial, giving you only $n$ total terms.

Why:

Since counting starts at $k = 0$ , you need $n + 1$ total terms to get an nth-degree polynomial.

Correct move:

Count your terms after constructing the polynomial to confirm you have $n + 1$ terms.

Wrong move:

When substituting into a known Maclaurin, forgetting to raise the entire substituted expression to the correct power.

Why:

For example, substituting $3 x$ into $sin x$ often leads to the incorrect term $\frac{3 x ^{3}}{3 !}$ instead of $\frac{( 3 x ) ^{3}}{3 !}$ .

Correct move:

Always wrap the substituted expression in parentheses before raising it to the term power.

5. Quick Reference Cheatsheet

Concept	Key Formula	Exam Note
nth-degree Taylor (center $a$ )	$T_{n} (x) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$	Matches $f$ and first $n$ derivatives at $a$
nth-degree Maclaurin	$T_{n} (x) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( 0 )}{k !} x^{k}$	Taylor centered at $a = 0$
$e^{x}$ Maclaurin	$1 + x + \frac{x ^{2}}{2 !} + ... + \frac{x ^{n}}{n !}$	No alternating signs
$sin x$ Maclaurin	$x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - ...$	Only odd powers, alternating signs
$cos x$ Maclaurin	$1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - ...$	Only even powers, alternating signs

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · BC
6(a) Find 3rd degree Taylor
2022 · BC
2(b) Construct Maclaurin polynomial
2021 · BC
4(a) Find Taylor centered at $a = 1$

What's Next

Taylor polynomials are the foundation for all power series approximations in AP Calculus BC. Once you master how to construct them, you can move on to studying the error of these approximations using Lagrange error bounds, and extend finite Taylor polynomials to infinite Taylor series that converge to the original function on an interval of convergence. Taylor polynomials also have wide applications beyond the exam, used in physics, engineering, and computer science to approximate complicated functions that are hard to compute directly. Mastering this core skill will make every subsequent topic in infinite sequences and series much easier to understand.