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Maclaurin Polynomial — Definition, Formula & Examples

A Maclaurin polynomial is a polynomial that approximates a function near x=0x = 0, built entirely from the function's value and derivatives at zero. Adding more terms usually makes the approximation better, especially for xx close to 0.

If ff is nn times differentiable at 00, its nnth-degree Maclaurin polynomial PnP_n is the unique polynomial of degree at most nn satisfying Pn(k)(0)=f(k)(0)P_n^{(k)}(0) = f^{(k)}(0) for k=0,1,…,nk = 0, 1, \dots, n. It is the Taylor polynomial of ff centered at a=0a = 0.

Key Formula

Pn(x)=∑k=0nf(k)(0)k! xk,∣f(x)−Pn(x)∣≤M ∣x∣n+1(n+1)!P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}\,x^k, \qquad |f(x)-P_n(x)|\le \frac{M\,|x|^{n+1}}{(n+1)!}
Where:
  • Pn(x)P_n(x) = the Maclaurin polynomial of degree n
  • f(k)(0)f^{(k)}(0) = the kth derivative of f evaluated at zero (the 0th derivative is f itself)
  • k!k! = k factorial, the product of the whole numbers from 1 to k, with 0! equal to 1
  • nn = the degree, meaning the highest power of x kept
  • MM = any upper bound for the absolute value of the (n+1)th derivative of f on the interval between 0 and x

Worked Example

Problem: Find the 3rd-degree Maclaurin polynomial of f(x)=exf(x) = e^x. Use it to estimate ee, and bound the error.
Step 1: Every derivative of exe^x is exe^x, so f(k)(0)=e0=1f^{(k)}(0) = e^0 = 1 for all kk. Substitute into the formula with n=3n = 3.
P3(x)=1+x+x22+x36P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}
Step 2: Since e=e1e = e^1, evaluate at x=1x = 1.
P3(1)=1+1+12+16=83≈2.667P_3(1) = 1 + 1 + \frac{1}{2} + \frac{1}{6} = \frac{8}{3} \approx 2.667
Step 3: The 4th derivative is exe^x, which on [0,1][0, 1] is at most e<3e < 3. Take M=3M = 3, n=3n = 3, x=1x = 1.
∣e−P3(1)∣≤3⋅144!=324=18|e - P_3(1)| \le \frac{3 \cdot 1^4}{4!} = \frac{3}{24} = \frac{1}{8}
Answer: P3(x)=1+x+x22+x36P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}, giving e≈83≈2.667e \approx \frac{8}{3} \approx 2.667 with error at most 18=0.125\frac{1}{8} = 0.125. The actual error is about 0.0520.052, well within the bound.

Why It Matters

Lagrange error bound questions on Maclaurin polynomials appear regularly on the AP Calculus BC exam and in Calculus II courses. Physicists use the degree-1 polynomial sin⁡θ≈θ\sin\theta \approx \theta to simplify the pendulum equation. Polynomial approximations of this kind are a standard tool in numerical methods for computing functions like exe^x and sin⁡x\sin x.

Common Mistakes

Mistake: Writing the coefficient of xkx^k as f(k)(0)f^{(k)}(0) and forgetting to divide by k!k!. For exe^x this gives the wrong polynomial 1+x+x2+x31 + x + x^2 + x^3.
Correction: Each coefficient is f(k)(0)k!\frac{f^{(k)}(0)}{k!}. The factorial is what makes the kkth derivative of PnP_n at 0 match f(k)(0)f^{(k)}(0), because differentiating xkx^k exactly kk times produces k!k!.

Related Terms