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Taylor Polynomial — Definition, Formula & Examples

Taylor Polynomial
nth Degree Taylor Polynomial

An approximation of a function using terms from the function's Taylor series. An nth degree Taylor polynomial uses all the Taylor series terms up to and including the term using the nth derivative.

 

nth degree Taylor polynomial formula: sum from k=0 to n of f^(k)(a)/k! (x-a)^k = f(a) + f'(a)(x-a) + f''(a)/2!(x-a)² +...

 

 

See also

Taylor series remainder, sigma notation, factorial, nth derivative

Key Formula

Pn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2++f(n)(a)n!(xa)n=k=0nf(k)(a)k!(xa)kP_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k
Where:
  • Pn(x)P_n(x) = The nth-degree Taylor polynomial evaluated at x
  • f(a)f(a) = The value of the original function at the center point a
  • f(k)(a)f^{(k)}(a) = The kth derivative of f evaluated at a
  • aa = The center point around which the polynomial is built
  • nn = The degree of the polynomial (highest power of (x − a) included)
  • k!k! = k factorial, the product of all positive integers from 1 to k

Worked Example

Problem: Find the 3rd-degree Taylor polynomial for f(x) = eˣ centered at a = 0.
Step 1: Compute the derivatives of f(x) = eˣ. Every derivative of eˣ is eˣ.
f(x)=ex,f(x)=ex,f(x)=ex,f(x)=exf(x) = e^x,\quad f'(x) = e^x,\quad f''(x) = e^x,\quad f'''(x) = e^x
Step 2: Evaluate each derivative at a = 0. Since e⁰ = 1, every derivative at 0 equals 1.
f(0)=1,f(0)=1,f(0)=1,f(0)=1f(0) = 1,\quad f'(0) = 1,\quad f''(0) = 1,\quad f'''(0) = 1
Step 3: Substitute into the Taylor polynomial formula with n = 3 and a = 0.
P3(x)=10!x0+11!x1+12!x2+13!x3P_3(x) = \frac{1}{0!}x^0 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3
Step 4: Simplify the factorials and write the final polynomial.
P3(x)=1+x+x22+x36P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}
Answer: The 3rd-degree Taylor polynomial for eˣ centered at 0 is P₃(x) = 1 + x + x²/2 + x³/6.

Another Example

This example differs because the center is not 0 (making it a non-Maclaurin Taylor polynomial), and some derivative values are 0, which eliminates terms from the polynomial.

Problem: Find the 2nd-degree Taylor polynomial for f(x) = sin(x) centered at a = π/2.
Step 1: Compute the needed derivatives of sin(x).
f(x)=sinx,f(x)=cosx,f(x)=sinxf(x) = \sin x,\quad f'(x) = \cos x,\quad f''(x) = -\sin x
Step 2: Evaluate each derivative at a = π/2.
f ⁣(π2)=1,f ⁣(π2)=0,f ⁣(π2)=1f\!\left(\frac{\pi}{2}\right) = 1,\quad f'\!\left(\frac{\pi}{2}\right) = 0,\quad f''\!\left(\frac{\pi}{2}\right) = -1
Step 3: Substitute into the formula with n = 2 and a = π/2.
P2(x)=1+0 ⁣(xπ2)+12! ⁣(xπ2)2P_2(x) = 1 + 0\cdot\!\left(x - \frac{\pi}{2}\right) + \frac{-1}{2!}\!\left(x - \frac{\pi}{2}\right)^2
Step 4: Simplify.
P2(x)=112 ⁣(xπ2)2P_2(x) = 1 - \frac{1}{2}\!\left(x - \frac{\pi}{2}\right)^2
Answer: The 2nd-degree Taylor polynomial for sin(x) centered at π/2 is P₂(x) = 1 − (1/2)(x − π/2)².

Frequently Asked Questions

What is the difference between a Taylor polynomial and a Taylor series?
A Taylor polynomial has a finite number of terms — it stops at the nth-degree term. A Taylor series is the infinite sum you get by letting n go to infinity. The Taylor polynomial is therefore a partial sum of the Taylor series. As you increase n, the polynomial generally becomes a better approximation of the function.
What is a Maclaurin polynomial?
A Maclaurin polynomial is simply a Taylor polynomial centered at a = 0. The formula is the same, but every (x − a) factor reduces to just x. For example, the Maclaurin polynomial for eˣ is 1 + x + x²/2! + ⋯ + xⁿ/n!.
How do you choose the degree n of a Taylor polynomial?
The choice depends on how accurate you need the approximation to be. Higher-degree polynomials match more derivatives of the original function, so they tend to be more accurate near the center. You can use the Taylor Series Remainder (Lagrange remainder) to bound the error and determine the smallest n that meets your accuracy requirement.

Taylor Polynomial vs. Taylor Series

Taylor PolynomialTaylor Series
Number of termsFinite (n + 1 terms)Infinite
Formulak=0nf(k)(a)k!(xa)k\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^kk=0f(k)(a)k!(xa)k\sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k
Result typeA polynomial of degree ≤ nAn infinite series (power series)
AccuracyApproximation; error can be bounded by the remainderExact (equals the function where the series converges)
When to useWhen you need a practical, computable approximationWhen you need an exact representation or convergence analysis

Why It Matters

Taylor polynomials appear throughout AP Calculus BC and college-level calculus as a core tool for approximating functions that are otherwise hard to compute by hand. Engineers and scientists use low-degree Taylor polynomials constantly — for instance, the approximation sin(x) ≈ x for small x comes directly from the 1st-degree Taylor polynomial. Understanding Taylor polynomials also lays the groundwork for error analysis using the Lagrange remainder, a topic frequently tested on exams.

Common Mistakes

Mistake: Forgetting the factorial in the denominator of each term.
Correction: Every term in a Taylor polynomial has the form f⁽ᵏ⁾(a) / k! · (x − a)ᵏ. The k! is essential. Without it, the coefficients are wrong and the polynomial will not match the function's derivatives at the center point.
Mistake: Confusing the center a with 0 and writing xᵏ instead of (x − a)ᵏ when a ≠ 0.
Correction: When the center is not zero, you must use (x − a) as the base in every term. Replacing (x − a) with x produces the Maclaurin polynomial for a different function, not the Taylor polynomial centered at a.

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