Example 5.4.1 $\exp x=\sum _{n=0}^{+\mathrm{\infty }}\frac{x^n}{n!}$. Evaluate the radius of convergence. When $x\ne 0$, $$\underset{n\to +\mathrm{\infty }}{lim}\frac{\left|\frac{x^{n+1}}{(n+1)!}\right|}{\left|\frac{x^n}{n!}\right|}=\underset{n\to +\mathrm{\infty }}{lim}\frac{|x|}{n+1}=0<1$$ So $\mathrm{\forall }x\in \mathbb{C}$, $\exp x$ is convergent, meaning the domain of convergence is $\mathbb{C}$ and the radius of convergence is $R=+\mathrm{\infty }$.

Evaluate $\exp x$. Notice that $$(\exp x{)}^{\prime }=\sum _{n=1}^{+\mathrm{\infty }}{\left(\frac{x^n}{n!}\right)}^{\prime }=\sum _{n=1}^{+\mathrm{\infty }}\frac{x^{n-1}}{(n-1)!}=\exp x,\mathrm{\forall }x\in \mathbb{R}$$ So $\exp x$ is the solution to $$\{\begin{array}{l}{y}^{\prime }=y\\ y(0)=1\end{array}$$ Since ${\text{e}}^x$ is also the solution to the equation above, according to the uniqueness of the solution, we have $\exp x={\text{e}}^x,\mathrm{\forall }x\in \mathbb{R}$.

Similarly, we could prove that $$S(x)=\sum _{n=0}^{+\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}=\sin x,C(x)=\sum _{n=0}^{+\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n}}{(2n)!}=\cos x,\mathrm{\forall }x\in \mathbb{R}$$ After simple observation we could get famous Euler formula: ${\text{e}}^{\text{i}x}=\cos x+\text{i}\sin x$.

Example 5.4.2 Evaluate ${\int }_0^1{x}^x\text{d}x={\int }_0^1{\text{e}}^{x\ln x}\text{d}x$.

According to the uniform convergence of $\exp$, we have $$\begin{array}{rl}I& ={\int }_0^1\sum _{n=0}^{+\mathrm{\infty }}\frac{(x\ln x{)}^n}{n!}\text{d}x=\sum _{n=0}^{+\mathrm{\infty }}\frac{1}{n!}{\int }_0^1{x}^n(\ln x{)}^n\text{d}x=\sum _{n=0}^{+\mathrm{\infty }}\frac{1}{(n+1)!}{\int }_0^1(\ln x{)}^n\text{d}{x}^{n+1}\\ & =\sum _{n=0}^{+\mathrm{\infty }}\frac{1}{(n+1)!}[x^{n+1}(\ln x{)}^n{|}_0^1-n{\int }_0^1{x}^n(\ln x{)}^{n-1}\text{d}x]=\sum _{n=0}^{+\mathrm{\infty }}\frac{(-1{)}^n}{(n+1{)}^{n+1}}\end{array}$$

Example 5.4.3 Seek the power series of $(1+x{)}^{\alpha }=\sum _{n=0}^{+\mathrm{\infty }}{a}_n{x}^n$.

Notice that $y=(1+x{)}^{\alpha }$ is the solution to $$\{\begin{array}{l}(1+x)y^{\prime }=\alpha y\\ y(0)=1\end{array}$$ Assuming $S(x)=\sum _{n=0}^{+\mathrm{\infty }}{a}_n{x}^n$ is the solution to the equation above, then $a_0=1$, and $$(1+x)\left(\sum _{n=1}^{+\mathrm{\infty }}n{a}_n{x}^{n-1}\right)=\alpha \sum _{n=0}^{+\mathrm{\infty }}{a}_n{x}^n\Rightarrow (n+1)a_{n+1}+n{a}_n=\alpha a_n$$ Therefore, we get a recursion formula: $a_{n+1}=\frac{\alpha -n}{n+1}{a}_n$, i.e. $$a_n=\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!},\mathrm{\forall }n\ge 0$$ Terminally, $$(1+x{)}^{\alpha }=\sum _{n=0}^{+\mathrm{\infty }}\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}{x}^n$$ Evaluate its radius of convergence. According to D’Alembert’s test, $$\underset{n\to +\mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to +\mathrm{\infty }}{lim}\frac{|\alpha -n||x|}{n+1}=|x|$$ Hence, when $|x|<1$, series is convergent; when $|x|>1$, series is divergent.

When $x=1$, $S(1)=\sum _{n=0}^{+\mathrm{\infty }}{a}_n=\sum _{n=0}^{+\mathrm{\infty }}\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}$. According to Raabe’s test, $$n(\frac{|a_n|}{|a_{n+1}|}-1)=n(\frac{n+1}{n-\alpha }-1)\stackrel{n\to +\mathrm{\infty }}{\to }\alpha +1$$ So when $\alpha >0$, $S(\pm 1)$ is absolutely convergent; when $\alpha <0$, $S(\pm 1)$ is not absolutely convergent.

When $\alpha <0$, $$a_n=(-1{)}^n\frac{(-\alpha )(-\alpha +1)\cdots (n-\alpha -1)}{n!}>0$$ So $S(-1)$ is divergent. ... Then it’s left as exercise. :)

Terminal results. Assuming $D$ is the domain of convergence, then $$D=\{\begin{array}{ll}(-1,1)& \alpha \le -1\\ (-1,1]& \alpha \in (-1,0)\\ [-1,1]& \alpha >0\end{array}$$ meaning $\mathrm{\forall }x\in (-1,1)$, we have the general binomial theorem $$(1+x{)}^{\alpha }=\sum _{n=0}^{+\mathrm{\infty }}\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}{x}^n$$

Example 5.4.4

1.

Take $\alpha =-1$, we have $$\frac{1}{1+x}=1-x+x^2-x^3+\cdots ,x\in (-1,1)$$ Integrate it term by term, we have $$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots +\frac{(-1{)}^{n-1}{x}^n}{n}+\cdots ,x\in (-1,1)$$ Particularly, the series is also convergent when $x=1$, so $$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$ In numerical mathematics, we often use $$\ln \frac{1+x}{1-x}=\ln (1+x)-\ln (1-x)=2x+\frac{2}{3}{x}^3+\frac{2}{5}{x}^5+\cdots$$ which is much faster.

2.

When $x\in (-1,1)$, $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots$$ $$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots ,x\in (-1,1)$$ Particularly, the series is also convergent when $x=\pm 1$, so $$\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$

3.

When $x\in (-1,1)$, $$(\arcsin x{)}^{\prime }=(1-x^2{)}^{-\frac{1}{2}}=1+\frac{(-\frac{1}{2})}{1!}(-x^2)+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}(-x^2{)}^2+\cdots$$ Then $$\arcsin x=x+\frac{1}{6}{x}^3+\frac{3}{40}{x}^5+\cdots ,x\in (-1,1)$$

Example 5.4.5 Not all functions of $C^{\mathrm{\infty }}$ could be expressed as a convergent power series. Example, $$f(x)=\{\begin{array}{ll}{\text{e}}^{-\frac{1}{x^2}}& x\ne 0\\ 0& x=0\end{array}$$ then we have $f^{(n)}(0)=0,\mathrm{\forall }n$, so the power series is constantly $0$, wrong!