10.11 Finding Taylor Polynomial Approximations of Functions

BC only. A Taylor polynomial matches a function's value and its first few derivatives at one point x = a, so it closely follows the function near a. The coefficient of (x − a)ⁿ is f⁽ⁿ⁾(a)/n!, and a Taylor polynomial centered at 0 is called a Maclaurin polynomial.

Key terms

Taylor polynomial
Maclaurin polynomial
center
factorial
nth derivative

The idea

This whole unit is BC only. The tangent line at x = a matches f's value and slope there. It's a degree-1 polynomial. A degree-2 polynomial can also match the second derivative, so it bends the way f bends. Keep going and you get polynomials that hug f more and more closely near a.

The degree-n Taylor polynomial for f centered at x = a is

Pₙ(x) = f(a) + f′(a)(x − a) + [f″(a)/2!](x − a)² + [f‴(a)/3!](x − a)³ + … + [f⁽ⁿ⁾(a)/n!](x − a)ⁿ.

Pₙ has the same value and the same first n derivatives as f at x = a. Centered at a = 0, it's called a Maclaurin polynomial.

Why the factorials

Differentiate c(x − a)ᵏ k times and you get c·k!. For the kth derivative of Pₙ at a to equal f⁽ᵏ⁾(a), the coefficient c must be f⁽ᵏ⁾(a)/k!. Remember 0! = 1 and 1! = 1, so the first two terms have no visible factorial.

This also works backward: if you know the coefficient cₖ of (x − a)ᵏ, then f⁽ᵏ⁾(a) = k!·cₖ. Exam questions often ask for a derivative value read off a given polynomial.

Building one

Find f(a), f′(a), f″(a), … up to the degree you need. Make a table.
Divide each derivative value by the matching factorial.
Attach the matching power of (x − a).
To approximate f at a point, plug that x into the polynomial.

How good is the approximation?

Close to the center, Taylor polynomials are very accurate, and they usually get better as the degree increases. Far from the center, they can be poor. You'll measure the error with the alternating series error bound (10.10) or the Lagrange error bound (10.12).

Because P₂ matches f″(a), the degree-2 polynomial has the same concavity as f at the center. That's why its graph bends along with f's graph near a, while the tangent line can't.

If a question asks for the “third-degree Taylor polynomial,” include terms up to (x − a)³, even if some coefficients are 0. If it asks for “the first three nonzero terms,” skip any zero terms and keep going until you have three.

Worked examples

Try each one yourself first, then open the solution.

Example 1
A Maclaurin polynomial
Find the third-degree Maclaurin polynomial for f(x) = e²ˣ.
Show the solutionHide the solution
1. Step 1: Derivatives: f = e²ˣ, f′ = 2e²ˣ, f″ = 4e²ˣ, f‴ = 8e²ˣ.
2. Step 2: At 0: 1, 2, 4, 8.
3. Step 3: Divide by factorials: 1/0! = 1, 2/1! = 2, 4/2! = 2, 8/3! = 4/3.
4. Step 4: P₃(x) = 1 + 2x + 2x² + (4/3)x³.
Answer: P₃(x) = 1 + 2x + 2x² + (4/3)x³
Example 2
From given derivative values
f(2) = 3, f′(2) = −1, f″(2) = 4 and f‴(2) = 12. Write the third-degree Taylor polynomial for f about x = 2, and use it to approximate f(2.1).
Show the solutionHide the solution
1. Step 1: Coefficients: 3, −1, 4/2! = 2, 12/3! = 2.
2. Step 2: P₃(x) = 3 − (x − 2) + 2(x − 2)² + 2(x − 2)³.
3. Step 3: At x = 2.1, x − 2 = 0.1: 3 − 0.1 + 2(0.01) + 2(0.001) = 3 − 0.1 + 0.02 + 0.002 = 2.922.
Answer: P₃(x) = 3 − (x − 2) + 2(x − 2)² + 2(x − 2)³; f(2.1) ≈ 2.922
Example 3
Trap: forgetting to divide by n!
Find the second-degree Taylor polynomial for f(x) = √x about x = 4, and use it to estimate √4.2.
Show the solutionHide the solution
1. Step 1: f(4) = 2. f′(x) = ½x^(−1/2), so f′(4) = 1/4. f″(x) = −¼x^(−3/2), so f″(4) = −¼ · (1/8) = −1/32.
2. Step 2: The x² coefficient is f″(4)/2! = −1/64. Using −1/32 (forgetting the 2!) is the trap.
3. Step 3: P₂(x) = 2 + (1/4)(x − 4) − (1/64)(x − 4)².
4. Step 4: √4.2 ≈ 2 + 0.05 − 0.04/64 = 2.049375. (A calculator gives 2.049390…)
Answer: P₂(x) = 2 + (x − 4)/4 − (x − 4)²/64; √4.2 ≈ 2.049375

Common mistakes

Forgetting the factorials, especially 2! in the x² term.
Writing powers of x instead of powers of (x − a) when the center isn't 0.
Plugging x into the derivatives instead of a.
Confusing “degree n” with “n nonzero terms.”

On the exam

Taylor questions are a regular part of BC free response: write a polynomial from given derivatives, approximate a value, then bound the error.
If you're given a Taylor polynomial and asked for f⁽ᵏ⁾(a), multiply the coefficient of (x − a)ᵏ by k!.

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