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6.3 Taylor and Maclaurin Series

6.3 Taylor and Maclaurin Series

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Calculus II
Unit & Topic Study Guides

Taylor and Maclaurin Series

Taylor and Maclaurin series let you represent complex functions as infinite polynomial expressions centered at a specific point. This is incredibly useful because polynomials are easy to differentiate, integrate, and evaluate, while the original functions (like exe^x, sin(x)\sin(x), or ln(1+x)\ln(1+x)) can be much harder to work with directly.

The core idea: if you know all the derivatives of a function at a single point, you can reconstruct the entire function (within some radius) as a power series. The more terms you include, the better your approximation gets.

Construction of Taylor Polynomials

A Taylor polynomial approximates a function f(x)f(x) near a specific point aa by matching the function's value and its derivatives at that point. The nnth-degree Taylor polynomial Pn(x)P_n(x) is:

Pn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2++f(n)(a)n!(xa)nP_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n

Written in summation notation: Pn(x)=k=0nf(k)(a)k!(xa)kP_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k

A Maclaurin polynomial is just a Taylor polynomial centered at a=0a = 0:

Pn(x)=f(0)+f(0)x+f(0)2!x2++f(n)(0)n!xnP_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n

Notice each term has plain xkx^k instead of (xa)k(x-a)^k, since (x0)=x(x - 0) = x.

How to build a Taylor polynomial:

  1. Choose the center point aa and the degree nn of the polynomial.
  2. Compute the derivatives f(a),f(a),f(a),,f(n)(a)f(a), f'(a), f''(a), \ldots, f^{(n)}(a).
  3. Plug each derivative into the formula, dividing by the corresponding factorial.

Quick example: Find the 3rd-degree Maclaurin polynomial for exe^x.

Every derivative of exe^x is exe^x, and e0=1e^0 = 1, so:

P3(x)=1+x+x22!+x33!=1+x+x22+x36P_3(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}

Plugging in x=0.5x = 0.5: P3(0.5)=1+0.5+0.125+0.02081.6458P_3(0.5) = 1 + 0.5 + 0.125 + 0.0208 \approx 1.6458, while the true value of e0.51.6487e^{0.5} \approx 1.6487. That's pretty close with only four terms.

Construction of Taylor polynomials, Maclaurin and Taylor Series; Power Series - Wisewire

Interpretation of Taylor's Theorem

Taylor's theorem tells you exactly how the approximation error behaves. For a function f(x)f(x) that is (n+1)(n+1) times differentiable on an interval containing aa:

f(x)=Pn(x)+Rn(x)f(x) = P_n(x) + R_n(x)

where Rn(x)R_n(x) is the remainder term (also called the Lagrange remainder):

Rn(x)=f(n+1)(c)(n+1)!(xa)n+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

Here cc is some value between aa and xx. You typically don't know the exact value of cc, but the theorem guarantees such a cc exists.

The remainder Rn(x)R_n(x) represents the exact error between the true function value and the polynomial approximation. Two things shrink this error:

  • Increasing nn: More terms means a better fit, and the (n+1)!(n+1)! in the denominator grows fast.
  • Staying close to aa: The factor (xa)n+1(x-a)^{n+1} gets smaller when xx is near the center point.

Taylor polynomials are especially valuable when:

  • Direct evaluation of the function is difficult or impossible
  • You know the derivatives at a point but don't have a closed-form expression
  • You need a simpler expression for numerical computation (calculators use Taylor-type approximations internally for functions like sin(x)\sin(x))
Construction of Taylor polynomials, Datei:Taylor Approximation of sin(x).jpeg – Wikipedia

Error Analysis in Taylor Series

Bounding the error is where Taylor's theorem becomes a practical tool, not just a theoretical one. The goal is to figure out how many terms you need to achieve a desired accuracy.

Steps to bound the error using the Lagrange error bound:

  1. Identify the degree nn of your Taylor polynomial and the point xx where you're approximating.
  2. Find the (n+1)(n+1)th derivative f(n+1)(x)f^{(n+1)}(x).
  3. Determine MM, the maximum value of f(n+1)(t)|f^{(n+1)}(t)| for all tt between aa and xx.
  4. Apply the bound:

Rn(x)M(n+1)!xan+1|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}

Finding MM is often the trickiest step. For sin(x)\sin(x) and cos(x)\cos(x), all derivatives are bounded by 1, so M=1M = 1 always works. For exe^x on [0,x][0, x], the maximum of ete^t occurs at the right endpoint, so M=exM = e^x.

Three factors that control the error size:

  • Degree nn: Higher degree means the (n+1)!(n+1)! denominator grows rapidly, shrinking the bound.
  • Distance xa|x - a|: The closer xx is to the center, the smaller the error. Approximating e0.1e^{0.1} is far more accurate than e2e^{2} with the same number of terms.
  • Behavior of higher derivatives: If f(n+1)|f^{(n+1)}| is large, the error can be large too.

Alternating series error bound: When your Taylor series happens to be alternating (like those for sin(x)\sin(x), cos(x)\cos(x), or ln(1+x)\ln(1+x)), you can use a simpler bound:

Rn(x)an+1|R_n(x)| \leq |a_{n+1}|

where an+1a_{n+1} is the first omitted term. This bound is often tighter than the Lagrange bound and easier to compute.

Convergence and Applications of Taylor Series

When you let nn \to \infty, the Taylor polynomial becomes a Taylor series, an infinite power series:

f(x)=k=0f(k)(a)k!(xa)kf(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k

This series doesn't necessarily converge for all xx. You need to determine its radius of convergence RR, typically using the Ratio Test or Root Test. The series converges for xa<R|x - a| < R and diverges for xa>R|x - a| > R. At the endpoints x=a±Rx = a \pm R, you have to check convergence separately to find the full interval of convergence.

Within the interval of convergence, the Taylor series equals the original function (for the standard functions you'll encounter in this course). Some key Maclaurin series to know:

  • ex=k=0xkk!e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}, converges for all xx (R=R = \infty)
  • sin(x)=k=0(1)kx2k+1(2k+1)!\sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}, converges for all xx (R=R = \infty)
  • cos(x)=k=0(1)kx2k(2k)!\cos(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}, converges for all xx (R=R = \infty)
  • 11x=k=0xk\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k, converges for x<1|x| < 1
  • ln(1+x)=k=1(1)k+1xkk\ln(1+x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^k}{k}, converges for 1<x1-1 < x \leq 1

Common applications of Taylor series include:

  • Approximating function values when direct computation is impractical
  • Evaluating difficult integrals by integrating the series term-by-term (e.g., ex2dx\int e^{-x^2} dx has no elementary antiderivative, but you can integrate its Taylor series)
  • Solving differential equations by assuming a power series solution
  • Analyzing local behavior of functions near a point (for instance, determining limits of indeterminate forms)
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