Taylor series, Maclaurin series

Suppose the function \(f(x)\) is of class \(C^\infty\) in the neighbor of \(x = a\). Then, we can define the following power series:

\[\begin{eqnarray*} \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}t^n &=& \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\\ &=&f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \end{eqnarray*}\]

where \(t = x - a\). If this power series has a positive radius of convergence, and the function defined by it matches \(f(x)\) in the neighbor of \(x = a\), we say the function \(f(x)\) is analytic. \(f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\) is called the Taylor series of \(f(x)\) at \(x=a\). A Taylor series at \(x = 0\) is called a Maclaurin series.

Example. Let us define the function \(f(x)\) on the open interval \((a-r, a+r)\) by the following power series

\[f(x) = \sum_{n=0}^{\infty}a_n(x-a)^n\tag{Eq:eg1}\]

where \(r > 0\) is the radius of convergence of the power series. By Corollary 2 in Calculus of power series, \(a_n = \frac{f^{(n)}(a)}{n!}\) \((n = 0, 1, 2, \cdots)\). Therefore, \(f(x)\) is analytic and (Eq:eg1) gives the Taylor series. □

See also: Calculus of power series

Exponential and trigonometric functions

Lemma

Let \(f(x)\) be a function of class \(C^\infty\) in a neighbor of \(x = 0\) and \(r > 0\). Suppose the following condition is satisfied:

• (\(\dagger\)) There exists an \(M > 0\) such that, for all \(n \in \mathbb{N}_0\) and for all \(x\in \mathbb{R}\), if \(|x| < r\), then \(|f^{(n)}(x)| \leq M\).

Then, \(f(x)\) is analytic at \(x = 0\), and the radius of convergence of its Maclaurin series is at least \(r\).

Proof. Choose an \(x\) such that \(|x| < r\) and consider the finite Maclaurin expansion of \(f(x)\):

\[\begin{eqnarray*} f(x) &=& \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k + R_n,\\ R_n &=& \frac{f^{(n)}(\theta x)}{n!}x^n ~ (0 < \theta < 1). \end{eqnarray*}\]

For the remainder \(R_n\), we have \(|R_n| \leq M\frac{|x|^n}{n!} \to 0\) (\(n \to \infty\)). Therefore, the above finite Maclaurin expansion converges as \(n\to \infty\) and the limit is equal to \(f(x)\). Thus, \(f(x)\) is analytic at \(x=0\) and its radius of convergence is at least \(r\). ■

Example. Let us show that the exponential function \(e^x\) is analytic and its Maclaurin series is given as

\[e^x = \sum_{n=0}^{\infty}\frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. \tag{Eq:Exp}\]

Let \(f(x) = e^x\). For any \(r > 0\), let \(M = e^r\). For any \(x\) such that \(|x| < r\) and for any \(n \in \mathbb{N}_0\), \(|f^{(n)}(x)| = e^x < M\) so that \(f(x)\) is analytic at \(x=0\). Since \(r > 0\) is arbitrary, the radius of convergence is \(+\infty\). For all \(n \in \mathbb{N}_0\), \(f^{(n)}(0) = e^0 = 1\) so that the Maclaurin series is given as in (Eq:Exp). □

Example. It is an exercise to show that the Maclaurin series of \(\sin x\) and \(\cos x\) are given as the following:

\[\begin{eqnarray} \sin x &=& \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots,\\ \cos x &=& \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots. \end{eqnarray}\]

Also, show that the radii of convergence of these series are both \(+\infty\). □

Power functions and the binomial theorem

Given \(\alpha \in \mathbb{R}\) and \(n \in \mathbb{N}_0\), we define the binomial coefficient by

\[\binom{\alpha}{n} = \left\{ \begin{array}{cc} 1 & (n = 0),\\ \frac{\alpha(\alpha - 1)(\alpha - 2)\cdots (\alpha - n + 1)}{n!} & (n \geq 1). \end{array} \right.\]

The numerator for the case when \(n \geq 1\) is the product of \(n\) consecutive numbers from \(\alpha\) to \(\{\alpha - (n-1)\}\). For example,

\[\begin{eqnarray*} \binom{\alpha}{0} &=& 1,\\ \binom{\alpha}{1} &=& \alpha,\\ \binom{\alpha}{2} &=& \frac{\alpha(\alpha - 1)}{2},\\ \binom{\alpha}{3} &=& \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!},\\ &\vdots& \end{eqnarray*}\]

Note that the binomial coefficients are defined for all \(\alpha \in \mathbb{R}\).

If \(\alpha\) is a non-negative integer, then \(\binom{\alpha}{n}\) is the number of combinations when we choose \(n\) elements out of \(\alpha\) ("\(\alpha\) choose \(n\)") for \(n = 0, 1, 2, \cdots, \alpha\), or \(\binom{\alpha}{n} = 0\) for \(n > \alpha\).

Lemma 

\[\binom{\alpha - 1}{n-1} + \binom{\alpha - 1}{n} = \binom{\alpha}{n}.\tag{Eq:binsum}\]

Proof. If \(n = 1\), we have

\[\binom{\alpha - 1}{0} + \binom{\alpha - 1}{1} = 1 + (\alpha - 1) = \alpha = \binom{\alpha}{1}.\]

Thus, (Eq:binsum) holds.

If \(n > 1\), 

\[\begin{eqnarray*} \binom{\alpha - 1}{n-1} + \binom{\alpha - 1}{n} &=& \frac{(\alpha-1)(\alpha - 2)(\alpha - 3)\cdots (\alpha -n + 1)}{(n-1)!}\\ && + \frac{(\alpha-1)(\alpha - 2)(\alpha - 3)\cdots (\alpha - n)}{n!} \\ &=&\frac{\{n + (\alpha - n)\}(\alpha-1)(\alpha - 2)(\alpha - 3)\cdots (\alpha - n + 1)}{n!}\\ &=&\frac{\alpha(\alpha-1)(\alpha - 2)(\alpha - 3)\cdots (\alpha - n + 1)}{n!}\\ &=& \binom{\alpha}{n}. \end{eqnarray*}\]

■

For \(\alpha \in \mathbb{R}\), consider the power series

\[\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n.\tag{Eq:binpow}\]

If \(\alpha \in \mathbb{N}_0\), then this is a polynomial of degree \(\alpha\). Otherwise, \(\binom{\alpha}{n} \neq 0\) for all \(n\in\mathbb{N}_0\), and as \(n \to \infty\),

\[\left|\frac{\binom{\alpha}{n+1}}{\binom{\alpha}{n}}\right| = \left|\frac{\alpha - n}{n+1}\right| \to 1\]

so that the radius of convergence of the power series (Eq:binpow) is 1. 

See also: Power series (Theorem (Radius of convergence)).

Theorem (Binomial Theorem)

For \(|x| < 1\) and \(\alpha \in \mathbb{R}\),

\[(1 + x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n.\tag{eq:binser}\]

Proof. Let \(f(x)\) be the function on \((-1, 1)\) defined by the right-hand side of (Eq:binser). By term-wise differentiation, we have

\[f'(x) = \sum_{n=1}^{\infty}n\binom{\alpha}{n}x^{n-1} = \sum_{n=1}^{\infty}\alpha\binom{\alpha-1}{n-1}x^{n-1}.\]

Using the above Lemma,

\[\begin{eqnarray*} (1+x)f'(x) &=& \alpha\sum_{n=1}^{\infty}\left\{\binom{\alpha-1}{n-1}x^{n-1} + \binom{\alpha-1}{n-1}x^{n}\right\}\\ &=&\alpha\left[1 + \sum_{n=1}^{\infty}\left\{\binom{\alpha-1}{n-1} + \binom{\alpha-1}{n}\right\}x^{n}\right]\\ &=& \alpha\left\{1 + \sum_{n=1}^{\infty}\binom{\alpha}{n}x^n\right\}\\ &=& \alpha f(x). \end{eqnarray*}\]

Let \(g(x) = (1+x)^{\alpha}\). We have

\[(1+x)g'(x) = (1+x)\alpha (1 + x)^{\alpha - 1} = \alpha g(x).\]

Thus, both \(y = f(x)\) and \(y = g(x)\) satisfies the same differential equation 

\[(1+x)y' = \alpha y.\]

Let us define the function \(h(x)\) on \((-1, 1)\) by

\[h(x) = \frac{f(x)}{g(x)}.\]

Then,

\[h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{\{g(x)\}^2} = \alpha\cdot\frac{f(x)g(x) - f(x)g(x)}{(1+x)\{g(x)\}^2} = 0\]

so that \(h(x) = c\) (constant) on \((-1, 1)\). But \(c = h(0) = \frac{f(0)}{g(0)} = 1\). Thus, \(f(x) = g(x) = (1+x)^{\alpha}\). ■

Remark. When \(\alpha \in \mathbb{N}_0\), \(\binom{\alpha}{n} = 0\) for \(n > \alpha\). Therefore, in this case, we have

\[(1 + x)^{\alpha} = \sum_{n=0}^{\alpha}\binom{\alpha}{n}x^n.\]

This is the binomial theorem you might have learned in high school. □

Exercise. Show that \(\binom{-1}{n} = (-1)^n\). □

Example. With \(\alpha = -1\), we have

\[\frac{1}{1 - x} = \sum_{n=0}^{\infty}x^n = 1 + x + x^2 + x^3 + \cdots\]

for \(|x| < 1\). This is the formula of geometric series. □

Example. Let us show that \(\arctan x\) is analytic and find its Maclaurin series and radius of convergence. Applying the binomial theorem to \((\arctan x)' = \frac{1}{1 + x^2}\), we have

\[\frac{1}{1 + x^2} = \sum_{n=0}^{\infty}(-1)^{n}x^{2n}. \tag{Eq:atanps}\]

By term-wise integration,

\[\arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n + 1}x^{2n+1} + C\]

where \(C\) is a constant. But \(\arctan(0) = 0\) so \(C = 0\). Therefore

\[\arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n + 1}x^{2n+1}.\tag{Eq:atans}\]

The radius of convergence of the right-hand side of (Eq:atans) is equal to that of (Eq:atanps), which is 1 (verify!). Thus, \(\arctan x\) is analytic, and its radius of convergence is 1. □

List of frequently used Maclaurin series

It comes in handy if you memorize the following Maclaurin series. Make sure you can derive them.

\[\frac{1}{1 - x} = \sum_{n=0}^{\infty}x^n ~~~ (|x| < 1).\]

\[(1 + x)^{\alpha} = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n ~~~  (|x|< 1)\]

where

\[\binom{\alpha}{n} = \left\{ \begin{array}{cc} 1 & (n = 0),\\ \frac{\alpha(\alpha - 1)(\alpha - 2)\cdots (\alpha - n + 1)}{n!} & (n \geq 1). \end{array} \right.\]

\[\log(1 + x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n ~~~ (|x| < 1).\]

\[\sqrt{1 + x} = 1 + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(2n-3)!!}{(2n)!!}x^n ~~~ (|x| < 1)\]

where the double factorials are defined as

\[n!! = \left\{ \begin{array}{cc} n(n-2)(n-4)\cdots 2 & (\text{$n$ is even}),\\ n(n-2)(n-4)\cdots 1 & (\text{$n$ is odd}),\\ \end{array} \right.\]

and \(0!! = (-1)!! = 1\).

\[\frac{1}{\sqrt{1 - x}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}x^{n} ~~~ (|x| < 1).\]

\[\frac{1}{\sqrt{1 - x^2}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}x^{2n} ~~~ (|x| < 1).\]

\[\arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1} ~~ (|x| < 1).\]

\[\arcsin x = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1} ~~ (|x| < 1).\]

\[\begin{eqnarray*} e^x &=& \sum_{n=0}^{\infty}\frac{x^n}{n!} ~~~ (x\in \mathbb{R}),\\ \sin x &=& \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} ~~~(x\in \mathbb{R}),\\ \cos x &=& \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n} ~~~(x\in \mathbb{R}). \end{eqnarray*}\]