Taylor series, Maclaurin series

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

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(a)}{n!}{t}^n& =& \sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(a)}{n!}(x-a{)}^n\\ & =& f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+\cdots \end{array}$$

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. $f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(a)}{n!}(x-a{)}^n$ 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

$$\begin{array}{}\text{(Eq:eg1)}& f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(x-a{)}^n\end{array}$$

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 $C^{\mathrm{\infty }}$ in a neighbor of $x=0$ and $r>0$. Suppose the following condition is satisfied:

• ($†$) 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 $|f^{(n)}(x)|\le M$.

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 $f(x)$:

$$\begin{array}{rcl}f(x)& =& \sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}{x}^k+R_n,\\ R_n& =& \frac{f^{(n)}(\theta x)}{n!}{x}^n(0<\theta <1).\end{array}$$

For the remainder $R_n$, we have $|R_n|\le M\frac{|x{|}^n}{n!}\to 0$ ($n\to \mathrm{\infty }$). Therefore, the above finite Maclaurin expansion converges as $n\to \mathrm{\infty }$ 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

$$\begin{array}{}\text{(Eq:Exp)}& e^x=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{x}^n=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots .\end{array}$$

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 $+\mathrm{\infty }$. 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:

$$\begin{array}{rcl}\sin x& =& \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots ,\\ \cos x& =& \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots .\end{array}$$

Also, show that the radii of convergence of these series are both $+\mathrm{\infty }$. □

Power functions and the binomial theorem

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

$$(\genfrac{}{}{0ex}{}{\alpha }{n})=\{\begin{array}{cc}1& (n=0),\\ \frac{\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -n+1)}{n!}& (n\ge 1).\end{array}$$

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

$$\begin{array}{rcl}(\genfrac{}{}{0ex}{}{\alpha }{0})& =& 1,\\ (\genfrac{}{}{0ex}{}{\alpha }{1})& =& \alpha ,\\ (\genfrac{}{}{0ex}{}{\alpha }{2})& =& \frac{\alpha (\alpha -1)}{2},\\ (\genfrac{}{}{0ex}{}{\alpha }{3})& =& \frac{\alpha (\alpha -1)(\alpha -2)}{3!},\\ & ⋮& \end{array}$$

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

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

Lemma 

$$\begin{array}{}\text{(Eq:binsum)}& (\genfrac{}{}{0ex}{}{\alpha -1}{n-1})+(\genfrac{}{}{0ex}{}{\alpha -1}{n})=(\genfrac{}{}{0ex}{}{\alpha }{n}).\end{array}$$

Proof. If $n=1$, we have

$$(\genfrac{}{}{0ex}{}{\alpha -1}{0})+(\genfrac{}{}{0ex}{}{\alpha -1}{1})=1+(\alpha -1)=\alpha =(\genfrac{}{}{0ex}{}{\alpha }{1}).$$

Thus, (Eq:binsum) holds.

If $n>1$, 

$$\begin{array}{rcl}(\genfrac{}{}{0ex}{}{\alpha -1}{n-1})+(\genfrac{}{}{0ex}{}{\alpha -1}{n})& =& \frac{(\alpha -1)(\alpha -2)(\alpha -3)\cdots (\alpha -n+1)}{(n-1)!}\\ & & +\frac{(\alpha -1)(\alpha -2)(\alpha -3)\cdots (\alpha -n)}{n!}\\ & =& \frac{\{n+(\alpha -n)\}(\alpha -1)(\alpha -2)(\alpha -3)\cdots (\alpha -n+1)}{n!}\\ & =& \frac{\alpha (\alpha -1)(\alpha -2)(\alpha -3)\cdots (\alpha -n+1)}{n!}\\ & =& (\genfrac{}{}{0ex}{}{\alpha }{n}).\end{array}$$

■

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

$$\begin{array}{}\text{(Eq:binpow)}& \sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{n})x^n.\end{array}$$

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

$$\left|\frac{(\genfrac{}{}{0ex}{}{\alpha }{n+1})}{(\genfrac{}{}{0ex}{}{\alpha }{n})}\right|=\left|\frac{\alpha -n}{n+1}\right|\to 1$$

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 $\alpha \in \mathbb{R}$,

$$\begin{array}{}\text{(eq:binser)}& (1+x{)}^{\alpha }=\sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{n})x^n.\end{array}$$

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

$$f^{\prime }(x)=\sum _{n=1}^{\mathrm{\infty }}n(\genfrac{}{}{0ex}{}{\alpha }{n})x^{n-1}=\sum _{n=1}^{\mathrm{\infty }}\alpha (\genfrac{}{}{0ex}{}{\alpha -1}{n-1})x^{n-1}.$$

Using the above Lemma,

$$\begin{array}{rcl}(1+x)f^{\prime }(x)& =& \alpha \sum _{n=1}^{\mathrm{\infty }}\{(\genfrac{}{}{0ex}{}{\alpha -1}{n-1})x^{n-1}+(\genfrac{}{}{0ex}{}{\alpha -1}{n-1})x^n\}\\ & =& \alpha [1+\sum _{n=1}^{\mathrm{\infty }}\{(\genfrac{}{}{0ex}{}{\alpha -1}{n-1})+(\genfrac{}{}{0ex}{}{\alpha -1}{n})\}{x}^n]\\ & =& \alpha \{1+\sum _{n=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{n})x^n\}\\ & =& \alpha f(x).\end{array}$$

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

$$(1+x)g^{\prime }(x)=(1+x)\alpha (1+x{)}^{\alpha -1}=\alpha g(x).$$

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

$$(1+x)y^{\prime }=\alpha y.$$

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

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

Then,

$$h^{\prime }(x)=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{\{g(x){\}}^2}=\alpha \cdot \frac{f(x)g(x)-f(x)g(x)}{(1+x)\{g(x){\}}^2}=0$$

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 $\alpha \in {\mathbb{N}}_0$, $(\genfrac{}{}{0ex}{}{\alpha }{n})=0$ for $n>\alpha$. Therefore, in this case, we have

$$(1+x{)}^{\alpha }=\sum _{n=0}^{\alpha }(\genfrac{}{}{0ex}{}{\alpha }{n})x^n.$$

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

Exercise. Show that $(\genfrac{}{}{0ex}{}{-1}{n})=(-1{)}^n$. □

Example. With $\alpha =-1$, we have

$$\frac{1}{1-x}=\sum _{n=0}^{\mathrm{\infty }}{x}^n=1+x+x^2+x^3+\cdots$$

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 $(\arctan x{)}^{\prime }=\frac{1}{1+x^2}$, we have

$$\begin{array}{}\text{(Eq:atanps)}& \frac{1}{1+x^2}=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{x}^{2n}.\end{array}$$

By term-wise integration,

$$\arctan x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}{x}^{2n+1}+C$$

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

$$\begin{array}{}\text{(Eq:atans)}& \arctan x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}{x}^{2n+1}.\end{array}$$

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.

$$\frac{1}{1-x}=\sum _{n=0}^{\mathrm{\infty }}{x}^n(|x|<1).$$

$$(1+x{)}^{\alpha }=\sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{n})x^n(|x|<1)$$

where

$$(\genfrac{}{}{0ex}{}{\alpha }{n})=\{\begin{array}{cc}1& (n=0),\\ \frac{\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -n+1)}{n!}& (n\ge 1).\end{array}$$

$$\log (1+x)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}{x}^n(|x|<1).$$

$$\sqrt{1+x}=1+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}(2n-3)!!}{(2n)!!}{x}^n(|x|<1)$$

where the double factorials are defined as

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

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

$$\frac{1}{\sqrt{1-x}}=\sum _{n=0}^{\mathrm{\infty }}\frac{(2n-1)!!}{(2n)!!}{x}^n(|x|<1).$$

$$\frac{1}{\sqrt{1-x^2}}=\sum _{n=0}^{\mathrm{\infty }}\frac{(2n-1)!!}{(2n)!!}{x}^{2n}(|x|<1).$$

$$\arctan x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}{x}^{2n+1}(|x|<1).$$

$$\arcsin x=\sum _{n=0}^{\mathrm{\infty }}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}(|x|<1).$$

$$\begin{array}{rcl}{e}^x& =& \sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}(x\in \mathbb{R}),\\ \sin x& =& \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}(x\in \mathbb{R}),\\ \cos x& =& \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n)!}{x}^{2n}(x\in \mathbb{R}).\end{array}$$