Applications of Taylor Series

What You’ll Learn

In this lesson you’ll see Taylor series in action: approximating function values, estimating errors, and understanding why these approximations work so well in practice.

The Concept

Taylor Polynomials

The nth-degree Taylor polynomial centered at a is the first n + 1 terms of the Taylor series

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

This is your approximation. The more terms you keep, the better it matches f(x) near the center.

The Remainder (Error Bound)

The error when using the polynomial instead of the actual function is bounded by the Lagrange remainder

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

for some c between a and x. You don’t know exactly what c is, but you can bound the (n+1)th derivative on the interval to get a worst-case error estimate.

The key insight: the (n+1) factorial in the denominator grows extremely fast, which is why Taylor approximations get accurate so quickly with just a few terms.

Worked Examples

Example 1: Approximating e

Use the Maclaurin series for e to the x to approximate e (that is, evaluate at x = 1) using terms up to degree 4.

$$e \approx P_4(1) = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}$$ $$= 1 + 1 + 0.5 + 0.16667 + 0.04167 = 2.70833$$

The actual value is e = 2.71828…, so the error is about 0.01. Not bad for just 5 terms.

We can bound the error using the remainder. The 5th derivative of e to the x is still e to the x, and on [0, 1] the maximum is e itself (less than 3)

$$|R_4(1)| \leq \frac{3}{5!} \cdot 1^5 = \frac{3}{120} = 0.025$$

The actual error (0.00995) is well within this bound.

degree 1 error degree 2 degree 3 degree 4

Each curve shows how the error grows as you move away from x = 0. Higher-degree polynomials (green, blue) keep the error tiny over a much wider range. Near x = 0, all approximations are excellent.

Example 2: Approximating sin(0.1)

Use the Maclaurin series for sin x with the first two nonzero terms

$$\sin x \approx x - \frac{x^3}{3!} = x - \frac{x^3}{6}$$

At x = 0.1

$$\sin(0.1) \approx 0.1 - \frac{(0.1)^3}{6} = 0.1 - \frac{0.001}{6} = 0.1 - 0.000167 = 0.099833$$

The actual value is 0.0998334…, so the error is less than 0.0000004. For small x, even two terms give extraordinary accuracy because the next term involves x to the 5th, which is (0.1)⁵ = 0.00001, divided by 120.

Example 3: Approximating square root of 1.1

Use the Taylor polynomial for f(x) = sqrt(x) centered at x = 1, degree 2.

The derivatives at x = 1

$$f(1) = 1 \qquad f'(1) = \frac{1}{2} \qquad f''(1) = -\frac{1}{4}$$

The degree-2 polynomial is

$$P_2(x) = 1 + \frac{1}{2}(x - 1) + \frac{-1/4}{2!}(x - 1)^2 = 1 + \frac{1}{2}(x - 1) - \frac{1}{8}(x - 1)^2$$

At x = 1.1

$$P_2(1.1) = 1 + \frac{1}{2}(0.1) - \frac{1}{8}(0.01) = 1 + 0.05 - 0.00125 = 1.04875$$

The actual value is sqrt(1.1) = 1.04881…, so the error is about 0.00006. The third derivative of sqrt(x) is 3/(8x^(5/2)), which at x = 1 gives 3/8. The remainder bound is

$$|R_2| \leq \frac{3/8}{3!}(0.1)^3 = \frac{0.375}{6}(0.001) = 0.0000625$$

The actual error fits within this bound.

Real-World Application

Taylor approximations are used constantly in computing:

• Game engines approximate sin, cos, and exp with degree-3 or degree-5 polynomials for speed
• Physics simulations linearize equations near equilibrium using degree-1 Taylor polynomials
• Machine learning uses Taylor expansions to analyze how loss functions behave near optima
• Financial models approximate option pricing formulas with polynomial expansions
• Scientific computing uses error bounds to determine how many terms are needed for a given precision

You’ve Got This

Taylor series turn complicated functions into polynomials that are easy to evaluate. The error bound tells you exactly how good your approximation is. For values close to the center, even a few terms give remarkable accuracy. The factorial in the denominator is your friend: it makes the error shrink fast. This is one of the most practical tools you’ll take away from Calculus 2.

Quiz

Using 5 terms of the Maclaurin series for $e^x$ at $x = 1$ gives approximately

• A.3.0
• B.2.5
• C.2.708
• D.2.718

The Lagrange remainder helps you

• A.find the exact function value
• B.bound the error of the polynomial approximation
• C.determine the radius of convergence
• D.compute derivatives

For small x, the approximation $\sin x \approx x$ is accurate because

• A.sin is always equal to x
• B.the next term $x^3/6$ is tiny when x is small
• C.the series diverges
• D.cos x equals 1

Adding more terms to a Taylor polynomial

• A.always makes it worse
• B.has no effect on accuracy
• C.reduces the error near the center
• D.changes the center point

The factorial in the remainder denominator means the error

• A.grows with more terms
• B.stays constant
• C.shrinks rapidly with more terms
• D.depends only on the center