# Maclaurin and Taylor series (HL only)

> IB Mathematics AA HL · IB AA HL Calculus
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-maclaurin-and-taylor-series/

This module covers Taylor and Maclaurin series, which represent infinitely differentiable functions as infinite polynomial expansions. You will learn to derive expansions, use them for approximation, and solve complex limit problems.

**Prerequisites:** [Differentiation of elementary functions](https://www.owlsprep.com/study/ib-math-aa-hl-u3-rules-of-differentiation/); [Infinite series convergence](https://www.owlsprep.com/study/ib-math-aa-hl-u5-infinite-sequences-and-series/)

## Learning objectives

- State the general formula for Taylor and Maclaurin series
- Derive and recall standard Maclaurin series expansions
- Find Taylor series expansions about any non-zero center
- Apply series to approximate function values and evaluate indeterminate limits

## Core Definitions

**Taylor Series** — For a function $f(x)$ that is infinitely differentiable at $x=a$ (called the center of expansion), the Taylor series is given by:

*Example:* Expands $f(x)$ around any point $x=a$

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

**Maclaurin Series** — A special case of Taylor series where the center of expansion is $a=0$, given by:

*Example:* Expands $f(x)$ around the origin

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$

**Worked example:** Find the first 3 non-zero terms of the Maclaurin series for $f(x) = e^x$

1. Calculate derivatives and evaluate at $a=0$:
2. $f(x) = e^x \implies f(0) = 1$ 
$f'(x) = e^x \implies f'(0) = 1$ 
$f''(x) = e^x \implies f''(0) = 1$
3. Substitute into the Maclaurin formula:
4. $n=0$ term: $\frac{1}{0!}x^0 = 1$ 
$n=1$ term: $\frac{1}{1!}x^1 = x$ 
$n=2$ term: $\frac{1}{2!}x^2 = \frac{x^2}{2}$
5. Final result (first 3 non-zero terms):
6. $$e^x \approx 1 + x + \frac{x^2}{2}$$

> **tip**
>
> Always remember the $n!$ denominator in all series terms — this is the most frequently penalized mistake in exams.

> **Exam tip:** You can be asked to derive the general formula for a Taylor or Maclaurin series from first principles, so memorize the general form.

## Standard Maclaurin Series

There are 6 standard Maclaurin series you should memorize for exams, along with their intervals of convergence:

| Function | Maclaurin Series | Interval of Convergence |
| --- | --- | --- |
| $e^x$ | $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ | $(-\infty, \infty)$ |
| $\sin x$ | $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ | $(-\infty, \infty)$ |
| $\cos x$ | $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ | $(-\infty, \infty)$ |
| $\ln(1+x)$ | $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$ | $(-1, 1]$ |
| $(1+x)^k$ | $\sum_{n=0}^{\infty} \binom{k}{n} x^n$ | $(-1, 1)$ |
| $\frac{1}{1-x}$ | $\sum_{n=0}^{\infty} x^n$ | $(-1, 1)$ |

**Worked example:** Use the standard $\cos x$ series to find the first 4 non-zero terms of the Maclaurin series for $\cos(2x)$

1. Recall the standard expansion for $\cos u$:
2. $$\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + ...$$
3. Substitute $u = 2x$:
4. $$\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + ...$$
5. Simplify each coefficient:
6. $$1 - \frac{4x^2}{2} + \frac{16x^4}{24} - \frac{64x^6}{720} = 1 - 2x^2 + \frac{2x^4}{3} - \frac{8x^6}{45} + ...$$

**Check your understanding**

1. What is the interval of convergence for the Maclaurin series of $\ln(1+x)$?

   - $(-\infty, \infty)$
   - $[-1, 1]$
   - $(-1, 1]$
   - $(-1, 1)$

   *Answer:* $(-1, 1]$

   *Why:* The series $\sum \frac{(-1)^{n+1}x^n}{n}$ converges when $-1 < x \leq 1$, diverges at $x=-1$.

> **Exam tip:** Memorizing standard series saves 5-10 minutes per question compared to deriving them from scratch in an exam.

## Taylor Series About Non-Zero Centers

To expand a function around any center $a \neq 0$, we use the general Taylor series formula. A useful shortcut is substitution: let $u = x - a$, expand $f(a+u)$ as a Maclaurin series in $u$, then substitute back $u = x-a$.

**Worked example:** Find the first 3 non-zero terms of the Taylor series for $f(x) = \ln x$ centered at $a=1$.

1. Calculate derivatives and evaluate at $a=1$:
2. $f(x) = \ln x \implies f(1) = 0$ 
$f'(x) = \frac{1}{x} \implies f'(1) = 1$ 
$f''(x) = -\frac{1}{x^2} \implies f''(1) = -1$ 
$f'''(x) = \frac{2}{x^3} \implies f'''(1) = 2$
3. Substitute into the Taylor formula:
4. $$f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + ...$$
5. Simplify to get the first 3 non-zero terms:
6. $$\ln x \approx (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}$$

> **warning**
>
> Functions like $\ln x$ have no Maclaurin series, because they are undefined at $a=0$. Always use Taylor series centered at a point where the function is defined.

## Applications: Approximation and Limits

Two of the most common exam applications of Maclaurin/Taylor series are approximating function values near the center, and evaluating indeterminate limits that would be cumbersome with L'Hospital's rule.

**Worked example:** Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ using Maclaurin series.

1. Substitute the Maclaurin expansion of $e^x$ into the numerator:
2. $$e^x - 1 - x = \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\right) - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + ...$$
3. Divide the entire expression by $x^2$:
4. $$\frac{e^x - 1 - x}{x^2} = \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} + ...$$
5. Take the limit as $x \to 0$: all terms containing $x$ approach 0, so the limit is:
6. $$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{1}{2}$$

**Check your understanding**

1. What is the center of the Taylor series: $\sum_{n=0}^{\infty} \frac{(-1)^n (x-2)^n}{n!}$

   - 0
   - 2
   - -2
   - n

   *Answer:* 2

   *Why:* The general term of a Taylor series is $\frac{f^{(n)}(a)}{n!}(x-a)^n$, so $a=2$ here.

## Common pitfalls

- **Wrong:** Forgetting the $n!$ denominator in series terms
  - Why it fails: This is the most common avoidable mistake, and examiners always penalize missing factorials
  - Correct: Always write out the general formula with $n!$ before substituting derivative values
- **Wrong:** Trying to find a Maclaurin series for a function undefined at 0, e.g. $\ln x$
  - Why it fails: All derivatives will be undefined at 0, so no valid expansion can be found
  - Correct: Use Taylor series centered at a point where the function is defined, e.g. $a=1$ for $\ln x$
- **Wrong:** Forgetting to raise the substitution constant to the power of the term, e.g. writing $(kx)^n = kx^n$
  - Why it fails: This leads to incorrect coefficients for all terms after the first
  - Correct: Simplify $(kx)^n = k^n x^n$ for every term when substituting into a standard series
- **Wrong:** Using a series outside its interval of convergence
  - Why it fails: The series does not equal the function outside this interval, so results are meaningless
  - Correct: Always confirm your $x$ value lies within the interval of convergence before using the series
- **Wrong:** Treating the binomial series for non-integer $k$ as a finite expansion
  - Why it fails: Only for positive integer $k$ is the binomial expansion finite; for all other $k$, it is infinite
  - Correct: Use the infinite binomial series with interval of convergence $(-1, 1)$ for non-integer $k$

## Cheatsheet

| Type | Formula | Key Note |
| --- | --- | --- |
| Maclaurin (center 0) | $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$ | Special case of Taylor |
| Taylor (center $a$) | $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$ | Works for any center $a$ |
| $e^x$ | $\sum \frac{x^n}{n!}$ | All real $x$ |
| $\sin x$ | $\sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ | All real $x$ |
| $\cos x$ | $\sum \frac{(-1)^n x^{2n}}{(2n)!}$ | All real $x$ |
| $\ln(1+x)$ | $\sum \frac{(-1)^{n+1}x^n}{n}$ | $-1 < x \leq 1$ |
| $(1+x)^k$ | $\sum \binom{k}{n}x^n$ | $-1 < x < 1$, all $k$ |

## What's next

Maclaurin and Taylor series are foundational tools for advanced calculus, used in error estimation, solving differential equations, and numerical analysis. In IB AA HL, the next core topic related to this sub-topic is Lagrange error bounds, which teaches you to calculate the maximum error of a Taylor polynomial approximation. Understanding series expansions also makes evaluating complex indeterminate limits much easier, and is used to solve many types of differential equations that cannot be solved with standard algebraic methods.

- [Fundamental theorem of calculus](https://www.owlsprep.com/study/ib-math-aa-hl-u5-fundamental-theorem-of-calculus/)

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