5.19 Maclaurin seriesHL ONLY

1. Definition

The Maclaurin series is a powerful mathematical method used to express complicated continuous functions as an infinite polynomial expansion. For a function $f(x)$ that is infinitely differentiable at $x=0$, the Maclaurin series is defined as

$$ \begin{aligned} f(x) &= f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \dots \\ &= \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n \end{aligned} $$

By calculating more polynomial terms, the series graphically converges closer and closer to the actual curve of $f(x)$ around the origin $(x=0)$.

x y sin(x) P₁(x) P₃(x)

Example 1

Find the Maclaurin series for $f(x) = e^x$ up to the term in $x^3$.

Solution:
We find the derivatives and evaluate them at $x=0$ $$ \begin{aligned} f(x) &= e^x \implies f(0) = e^0 = 1 \\[0.2cm] f'(x) &= e^x \implies f'(0) = 1 \\[0.2cm] f''(x) &= e^x \implies f''(0) = 1 \\[0.2cm] f'''(x) &= e^x \implies f'''(0) = 1 \end{aligned} $$ Substituting these constant values into the Maclaurin formula gives $$ \begin{aligned} e^x &= 1 + 1 \cdot x + \dfrac{1}{2!}x^2 + \dfrac{1}{3!}x^3 + \dots \\ &= \mathbf{1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dots} \end{aligned} $$

2. Standard Maclaurin series

The IB formula booklet provides the following standard Maclaurin series. You are not typically required to derive these from first principles unless explicitly asked to do so.

  • $e^x = 1 + x + \dfrac{x^2}{2!} + \dots = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$
  • $\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dots = \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{x^n}{n}$
  • $\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dots = \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}$
  • $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dots = \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!}$
  • $\arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dots = \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1}$
  • $(1+x)^p = 1 + px + \dfrac{p(p-1)}{2!}x^2 + \dots = \sum_{n=0}^{\infty} \begin{pmatrix} p \\ n \end{pmatrix} x^n$ (General Binomial Theorem)

3. Operations with Maclaurin series

We can cleverly construct the Maclaurin series for far more complex functions by performing algebraic operations on the standard series:

  • Substitution: Replace $x$ entirely with a scaled variable like $2x$, $x^2$, or $-x$.
  • Multiplication: Multiply two infinite series together and expand the resulting brackets (usually gathering terms up to a required degree).
  • Differentiation / Integration: We can cleanly differentiate or integrate the original series term-by-term to yield the expansion of its derivative or integral.

Example 2

Find the Maclaurin series for $f(x) = \cos(2x)$ up to the term in $x^4$.

Solution:
We use the standard series for $\cos x$ from the booklet $$ \cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dots $$ We substitute $x$ with $(2x)$ $$ \begin{aligned} \cos(2x) &= 1 - \dfrac{(2x)^2}{2!} + \dfrac{(2x)^4}{4!} - \dots \\ &= 1 - \dfrac{4x^2}{2} + \dfrac{16x^4}{24} - \dots \\ &= \mathbf{1 - 2x^2 + \dfrac{2}{3}x^4 - \dots} \end{aligned} $$

Example 3

Find the first three non-zero terms of the Maclaurin series for $f(x) = e^x \sin x$.

Solution:
We multiply the known standard series for $e^x$ and $\sin x$ together $$ \begin{aligned} e^x &= 1 + x + \dfrac{x^2}{2!} + \dots \\ \sin x &= x - \dfrac{x^3}{3!} + \dots \end{aligned} $$ Therefore, $$ \begin{aligned} e^x \sin x &= \left(1 + x + \dfrac{x^2}{2} + \dots\right) \left(x - \dfrac{x^3}{6} + \dots\right) \\ &= 1\left(x - \dfrac{x^3}{6}\right) + x(x) + \dfrac{x^2}{2}(x) + \dots \\ &= x - \dfrac{x^3}{6} + x^2 + \dfrac{x^3}{2} + \dots \\ &= \mathbf{x + x^2 + \dfrac{1}{3}x^3 + \dots} \end{aligned} $$

Example 4

Use Maclaurin series to precisely evaluate the limit $\lim_{x\to 0} \dfrac{\sin x - x}{x^3}$.

Solution:
We directly substitute the infinite Maclaurin series for $\sin x$ into the limit expression $$ \sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dots $$ Then, $$ \begin{aligned} \lim_{x\to 0} \dfrac{\sin x - x}{x^3} &= \lim_{x\to 0} \dfrac{\left(x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \dots\right) - x}{x^3} \\ &= \lim_{x\to 0} \dfrac{-\dfrac{x^3}{6} + \dfrac{x^5}{120} - \dots}{x^3} \\ &= \lim_{x\to 0} \left(-\dfrac{1}{6} + \dfrac{x^2}{120} - \dots\right) \\ &= \mathbf{-\dfrac{1}{6}} \end{aligned} $$

4. Maclaurin series via differential equations

Occasionally, a function $y = f(x)$ is purely defined implicitly via a differential equation alongside an initial boundary condition. By sequentially differentiating the equation implicitly, we can construct the needed $y^{(n)}(0)$ values to map out the Maclaurin series.

Example 5

Given that $\dfrac{dy}{dx} = x + y^2$ and $y = 1$ when $x = 0$, logically find the Maclaurin series for $y$ up to the term in $x^3$.

Solution:
It is given that $y(0) = 1$. We evaluate $y'(0)$ using the direct differential equation $$ y'(x) = x + y^2 \implies y'(0) = 0 + (1)^2 = 1 $$ Then, differentiate implicitly with respect to $x$ to find $y''(x)$. $$ y''(x) = 1 + 2y \cdot y'(x) \implies y''(0) = 1 + 2(1)(1) = 3 $$ Differentiate again using the product rule to find $y'''(x)$. $$ \begin{aligned} y'''(x) &= 0 + 2y'(x) \cdot y'(x) + 2y \cdot y''(x) = 2(y')^2 + 2yy'' \\ \implies y'''(0) &= 2(1)^2 + 2(1)(3) = 2 + 6 = 8 \end{aligned} $$ Substituting all constants into the core Maclaurin formula $$ \begin{aligned} y &= y(0) + y'(0)x + \dfrac{y''(0)}{2!}x^2 + \dfrac{y'''(0)}{3!}x^3 + \dots \\ &= 1 + 1x + \dfrac{3}{2}x^2 + \dfrac{8}{6}x^3 + \dots \\ &= \mathbf{1 + x + \dfrac{3}{2}x^2 + \dfrac{4}{3}x^3 + \dots} \end{aligned} $$