# Taylor's Theorem: Remainder Formula and Error Bound Examples

> Taylor's theorem with the Lagrange remainder: how close a Taylor polynomial is to f(x). Statement, error bound, graph and worked examples with e and sin x.

Source: https://practicecalculus.com/theorems/taylors-theorem
Updated: 2026-11-03

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

## At a glance

- **Hypotheses:** $f$ has $n + 1$ derivatives on an open interval $I$ containing $a$, and $x$ is in $I$
- **Taylor polynomial:** $P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k$
- **Remainder (Lagrange form):** $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$ for some $c$ between $a$ and $x$
- **Error bound:** $|R_n(x)| \le \frac{M}{(n+1)!}|x - a|^{n+1}$ where $|f^{(n+1)}| \le M$
- **Case n = 0:** The mean value theorem
- **Course:** Calculus 2

## Key takeaways

- Taylor's theorem says $f(x)$ equals its degree $n$ Taylor polynomial plus a remainder you can bound.
- The Lagrange remainder looks like the next term of the series, with the derivative evaluated at an unknown $c$.
- Bound $|f^{(n+1)}|$ by $M$ on the interval to get the error bound $\frac{M}{(n+1)!}|x - a|^{n+1}$.
- A Taylor series equals $f$ exactly when the remainder goes to $0$ as $n \to \infty$.

**Taylor's theorem** says that a smooth function equals its Taylor polynomial plus an error term, and it gives that error term an exact form. That turns "the Taylor polynomial is a good approximation" into a guaranteed number of correct decimal places.

## Statement of Taylor's theorem

Suppose $f$ has derivatives up to order $n + 1$ on an open interval $I$ containing $a$. Let

$$P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n$$

be the degree $n$ Taylor polynomial of $f$ at $a$. Then for every $x$ in $I$ there is a number $c$ **between $a$ and $x$** such that

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

This is the **Lagrange form of the remainder**. When $a = 0$ the polynomial is called a Maclaurin polynomial.

## Taylor's inequality (the error bound)

You almost never know $c$, so you bound the derivative instead. If $|f^{(n+1)}(t)| \le M$ for every $t$ between $a$ and $x$, then

$$|f(x) - P_n(x)| = |R_n(x)| \le \frac{M}{(n+1)!}|x - a|^{n+1}$$

The bound gets small fast because $(n+1)!$ grows faster than any power.

## What Taylor's theorem means

The remainder looks exactly like the next term of the Taylor series, except the derivative is evaluated at some point $c$ instead of at $a$. For $n = 0$ the statement is $f(x) = f(a) + f'(c)(x - a)$, which is the [mean value theorem](/theorems/mean-value-theorem). Taylor's theorem is the higher-order version.

```graph
title: 'Taylor''s theorem: Taylor polynomials of sin(x) at 0 hug the curve near 0'
caption: 'The degree 1, 3 and 5 Maclaurin polynomials of $\sin x$. Near $x = 0$ the error is tiny; it grows with $|x|$ and shrinks as the degree rises.'
x: [-4, 4]
y: [-2, 2]
functions:
  - expr: sin(x)
    label: y = sin(x)
  - expr: x
    label: degree 1, y = x
    dashed: true
  - expr: x - x^3/6
    label: degree 3, y = x − x³/6
    dashed: true
  - expr: x - x^3/6 + x^5/120
    label: degree 5
    dashed: true
```

## Why the hypotheses matter

**Enough derivatives.** The remainder formula uses $f^{(n+1)}$. For $f(x) = |x|$ at $a = 0$, $f'(0)$ does not exist, so there is not even a degree 1 Taylor polynomial.

**A remainder that does not shrink.** Taylor's theorem always holds when its hypotheses do, but it does not promise the error goes to $0$. The function $f(x) = e^{-1/x^2}$ (with $f(0) = 0$) has every derivative equal to $0$ at $x = 0$, so every Taylor polynomial at $0$ is the zero polynomial. The remainder $R_n(x) = f(x)$ never shrinks, and the Taylor series converges to $0$, not to $f$. A function equals its [Taylor series](/blog/taylor-series) only when $R_n(x) \to 0$.

## Worked examples

**Example 1: approximate $e$ with $P_4$ and bound the error.**

**Step 1.** For $f(x) = e^x$ at $a = 0$, every derivative is $e^x$ and equals $1$ at $0$. So

$$P_4(1) = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} = \frac{65}{24} \approx 2.70833$$

**Step 2.** On $[0, 1]$, $f^{(5)}(t) = e^t \le e < 3$, so take $M = 3$.

**Step 3.** $|R_4(1)| \le \frac{3}{5!}\cdot 1^5 = \frac{3}{120} = 0.025$.

The true error is $e - 2.70833 \approx 0.00995$, comfortably inside the bound.

**Example 2: $\sin(0.1)$ to high accuracy.**

The Maclaurin polynomial $P_3(x) = x - \frac{x^3}{6}$ gives $\sin(0.1) \approx 0.1 - \frac{0.001}{6} \approx 0.0998333$. Since the $x^4$ coefficient of sine is $0$, $P_4 = P_3$, so we may use $R_4$. Every derivative of $\sin$ is bounded by $M = 1$:

$$|R_4(0.1)| \le \frac{1}{5!}(0.1)^5 = \frac{0.00001}{120} \approx 8.3 \times 10^{-8}$$

So $0.0998333$ is correct to at least six decimal places.

**Example 3: how many terms for $e^{0.5}$ within $0.001$?**

On $[0, 0.5]$, $e^t \le e^{0.5} < 2$, so $|R_n(0.5)| \le \frac{2\,(0.5)^{n+1}}{(n+1)!}$.

| $n$ | Bound $\frac{2\,(0.5)^{n+1}}{(n+1)!}$ |
| --- | --- |
| $2$ | $\frac{2(0.125)}{6} \approx 0.042$ |
| $3$ | $\frac{2(0.0625)}{24} \approx 0.0052$ |
| $4$ | $\frac{2(0.03125)}{120} \approx 0.00052$ |

The bound first drops below $0.001$ at $n = 4$, so $P_4(0.5)$ is accurate enough.

## Common mistakes

1. **Using $(n)!$ instead of $(n+1)!$.** The remainder after a degree $n$ polynomial uses the $(n+1)$st derivative and $(n+1)!$.
2. **Bounding the derivative at $a$ only.** $M$ must bound $|f^{(n+1)}|$ on the whole interval between $a$ and $x$.
3. **Assuming the Taylor series always equals $f$.** It does only when the remainder tends to $0$.

## Related theorems and guides

Taylor's theorem generalizes the [mean value theorem](/theorems/mean-value-theorem). For building the polynomials and series themselves, see [Taylor series](/blog/taylor-series) and [power series](/blog/power-series). In several variables, the second-order Taylor expansion uses mixed partial derivatives, which [Clairaut's theorem](/theorems/clairauts-theorem) says can be taken in either order.

## Frequently asked questions

### What does Taylor's theorem say?

If $f$ has $n + 1$ derivatives on an interval containing $a$ and $x$, then $f(x) = P_n(x) + R_n(x)$, where $P_n$ is the degree $n$ Taylor polynomial at $a$ and $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$ for some $c$ between $a$ and $x$.

### What is the Lagrange error bound?

If $|f^{(n+1)}(t)| \le M$ for all $t$ between $a$ and $x$, then the error in using $P_n(x)$ to approximate $f(x)$ is at most $\frac{M}{(n+1)!}|x - a|^{n+1}$. It is the same as Taylor's inequality.

### How do you know how many terms of a Taylor polynomial you need?

Write the error bound as a function of $n$ and increase $n$ until the bound is smaller than the accuracy you need. The factorial in the denominator makes the bound shrink quickly.

### What is the difference between Taylor's theorem and a Taylor series?

A Taylor series is the infinite sum of all the terms. Taylor's theorem is about a finite polynomial and gives an exact formula for the leftover error, which is how you prove a Taylor series converges to the function.