AHL 5.19—Maclaurin series

A Maclaurin series is a way of representing a function as an infinite polynomial centred at $x=0$. The motivation is simple: polynomials are among the easiest functions to differentiate, integrate, and evaluate , so if we can express a complicated function like $\sin x$ or $e^x$ as a polynomial, we gain a powerful computational tool.

Maclaurin Series: The Maclaurin series of a function $f(x)$ is the infinite polynomial expansion:
$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\cdots ={\sum }_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$
where $f^{(n)}(0)$ denotes the $n$th derivative of $f$ evaluated at $x=0$.

The key requirement is that the polynomial $P(x)$ must agree with $f(x)$ at $x=0$ in every derivative , meaning $P(0)=f(0)$, $P^{\prime }(0)=f^{\prime }(0)$, $P^{\prime \prime }(0)=f^{\prime \prime }(0)$, and so on for all orders.

Suppose we want to find a polynomial $P(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$ that matches every derivative of $f(x)$ at $x=0$. We determine the coefficients $a_n$ one at a time by substituting $x=0$ after each differentiation.

Step 1: Set $x=0$ in $P(x)$:
$P(0)=a_0⟹a_0=f(0)$

Step 2: Differentiate and set $x=0$:
$P^{\prime }(x)=a_1+2a_{2}x+3a_3{x}^2+\cdots ⟹P^{\prime }(0)=a_1=f^{\prime }(0)$

Step 3: Differentiate again and set $x=0$:
$P^{\prime \prime }(x)=2a_2+6a_{3}x+12a_4{x}^2+\cdots ⟹P^{\prime \prime }(0)=2a_2=f^{\prime \prime }(0)⟹a_2=\frac{f^{\prime \prime }(0)}{2!}$

Step 4: Differentiate once more and set $x=0$:
$P^{\prime \prime \prime }(x)=6a_3+24a_{4}x+\cdots ⟹P^{\prime \prime \prime }(0)=6a_3=f^{\prime \prime \prime }(0)⟹a_3=\frac{f^{\prime \prime \prime }(0)}{3!}$

The pattern is clear. In general:
$a_n=\frac{f^{(n)}(0)}{n!}$

Substituting back gives the Maclaurin series formula:
$f(x)={\sum }_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$