Maclaurin Series Examples: Expanding exp, sin, cos and Deriving Euler’s Formula

Table of Contents

Introduction
Maclaurin Series Expansion of \(\mathrm{e}^x\)
Maclaurin Series Expansion of \(\sin{x}\)
Maclaurin Series Expansion of \(\cos{x}\)
Euler’s Formula

Introduction

Last time, I wrote an article about Taylor series expansion.

テイラー展開の理論を解説〜なぜ多項式で表現できるのか〜

テイラー展開をゆるーく導出しましょう。テイラー展開の素晴らしさは、関数を多項式で表現できることにあると思います。

To recall the Taylor series expansion:

Taylor Series Expansion

When a function \(f(x)\) is infinitely differentiable in an interval containing \(x = a\), then \(f(x)\) can be expressed as

\begin{eqnarray*}
f(x) &=& f(a) + f^{\prime}(a)\, (x – a) + \frac{1}{2!}f^{\prime\prime}(a)\, (x-a)^2 + \cdots + \frac{1}{n!}f^{(n)}(a)\, (x-a)^n + \cdots \\
&=& \sum_{k = 0}^{\infty} \frac{1}{k!}f^{(k)}(a)\, (x-a)^k
\end{eqnarray*}

This is called the “Taylor series expansion around \(a\)”.

In particular, the Taylor series expansion at \(x = 0\), that is, around the origin, is called the “Maclaurin series expansion“.

Maclaurin Series Expansion

\begin{eqnarray*}
f(x) &=& f(0) + f^{\prime}(0)\, x + \frac{1}{2!}f^{\prime\prime}(0)\, x^2 + \cdots + \frac{1}{n!}f^{(n)}(0)\, x^n + \cdots \\
&=& \sum_{k = 0}^{\infty} \frac{1}{k!}f^{(k)}(0)\, x^k
\end{eqnarray*}

In this article, as concrete examples of this, we will calculate the Maclaurin series expansions of \(\mathrm{e}^x,\ \sin{x},\ \cos{x}\). First, let me present the results upfront.

\begin{eqnarray*}
\mathrm{e}^{x} &=& 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots = \sum_{k=0}^{\infty} \frac{x^{k}}{k !} \\
\\
\sin x &=& x-\frac{x^{3}}{3!} + \frac{x^5}{5!}\, – \frac{x^7}{7!} + \cdots = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2 k+1}}{(2 k+1) !} \\
\\
\cos x &=& 1-\frac{x^{2}}{2!} + \frac{x^4}{4!}\, – \frac{x^6}{6!} + \cdots = \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{(2 k) !}
\end{eqnarray*}

Let’s actually calculate and verify these.

Maclaurin Series Expansion of \(\mathrm{e}^x\)

To calculate the Maclaurin series expansion, let’s compute the derivatives of \(f(x) = \mathrm{e}^x\).
From the properties of the exponential function \(\mathrm{e}^x\),

\begin{eqnarray*}
f^{\prime}(x) &=& \mathrm{e}^x \\
\\
f^{\prime\prime}(x) &=& \mathrm{e}^x \\
\\
f^{(3)}(x) &=& \mathrm{e}^x \\
\\
&\vdots&
\end{eqnarray*}

the function form remains unchanged no matter how many times we differentiate. Therefore,

\begin{eqnarray*}
f^{\prime}(0) = f^{\prime\prime}(0) = f^{(3)}(0) = \cdots = 1
\end{eqnarray*}

Substituting this into the Maclaurin series expansion formula,

\begin{eqnarray*}
\mathrm{e}^{x} &=& 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots \\
&=& \sum_{k=0}^{\infty} \frac{x^{k}}{k!}
\end{eqnarray*}

is confirmed.

Graph of \(\mathrm{e}^x\) (blue) and its Maclaurin series expansion

Maclaurin Series Expansion of \(\sin{x}\)

Let’s compute the derivatives of \(f(x) = \sin{x}\).

\begin{eqnarray*}
f^{\prime}(x) &=& \cos{x} \\
\\
f^{\prime\prime}(x) &=& -\sin{x} \\
\\
f^{(3)}(x) &=& -\cos{x} \\
\\
f^{(4)}(x) &=& \sin{x}\\
\\
&\vdots&
\end{eqnarray*}

After taking the fourth derivative, we return to the original function. Therefore, the \(k\)-th derivative is

\begin{eqnarray*}
f^{(k)}(x) &=& \left\{
\begin{array}{ll}
\cos{x}\ \ &(k = 1, 5, 9, \dots) \\
-\sin{x}\ \ &(k = 2, 6, 10, \dots) \\
-\cos{x}\ \ &(k = 3, 7, 11, \dots) \\
\sin{x}\ \ &(k = 4, 8, 12, \dots) \\
\end{array}
\right. \\
\\
&=& \sin{\left(x + \frac{k\pi}{2} \right)}\ \ (k = 1, 2, 3, \dots)
\end{eqnarray*}

Therefore, at \(x = 0\),

\begin{eqnarray*}
f^{(k)}(0) &=& \left\{
\begin{array}{ll}
1\ \ &(k = 1, 5, 9, \dots) \\
0\ \ &(k = 2, 6, 10, \dots) \\
-1\ \ &(k = 3, 7, 11, \dots) \\
0\ \ &(k = 4, 8, 12, \dots) \\
\end{array}
\right.
\end{eqnarray*}

it has a value only when \(k\) is odd. Substituting into the Maclaurin series expansion formula,

\begin{eqnarray*}
\sin x &=& x-\frac{x^{3}}{3!} + \frac{x^5}{5!}\, – \frac{x^7}{7!} + \cdots \\
&=& \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2 k+1}}{(2 k+1) !}
\end{eqnarray*}

is confirmed.

Graph of \(\sin{x}\) (blue) and its Maclaurin series expansion

Maclaurin Series Expansion of \(\cos{x}\)

Let’s also compute the derivatives of \(f(x) = \cos{x}\).

\begin{eqnarray*}
f^{\prime}(x) &=& -\sin{x} \\
\\
f^{\prime\prime}(x) &=& -\cos{x} \\
\\
f^{(3)}(x) &=& \sin{x} \\
\\
f^{(4)}(x) &=& \cos{x}\\
\\
&\vdots&
\end{eqnarray*}

After taking the fourth derivative, we return to the original function. Therefore, the \(k\)-th derivative is

\begin{eqnarray*}
f^{(k)}(x) &=& \left\{
\begin{array}{ll}
-\sin{x}\ \ &(k = 1, 5, 9, \dots) \\
-\cos{x}\ \ &(k = 2, 6, 10, \dots) \\
\sin{x}\ \ &(k = 3, 7, 11, \dots) \\
\cos{x}\ \ &(k = 4, 8, 12, \dots) \\
\end{array}
\right. \\
\\
&=& \cos{\left(x + \frac{k\pi}{2} \right)}\ \ (k = 1, 2, 3, \dots)
\end{eqnarray*}

Therefore, at \(x = 0\),

\begin{eqnarray*}
f^{(k)}(0) &=& \left\{
\begin{array}{ll}
0\ \ &(k = 1, 5, 9, \dots) \\
-1\ \ &(k = 2, 6, 10, \dots) \\
0\ \ &(k = 3, 7, 11, \dots) \\
1\ \ &(k = 4, 8, 12, \dots) \\
\end{array}
\right.
\end{eqnarray*}

it has a value only when \(k\) is even. Substituting into the Maclaurin series expansion formula,

\begin{eqnarray*}
\cos x &=& 1-\frac{x^{2}}{2!} + \frac{x^4}{4!}\, – \frac{x^6}{6!} + \cdots \\
&=& \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{(2 k) !}
\end{eqnarray*}

is confirmed.