테일러·매클로린 급수

The factor $(n+1)(n+2)$ reminds us of a derivative. Start with the geometric series.

$$\begin{aligned} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1-x} \right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sum_{n=0}^\infty x^n \right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=0}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^{n-1}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{(1-x)^2} \right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sum_{n=1}^\infty nx^{n-1} \right\}\\ \frac{2}{(1-x)^3}&=\sum_{n=1}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty n(n-1)x^{n-2}\\ &=\sum_{n=0}^\infty (n+2)(n+1)x^n \end{aligned}$$

Let $x=\frac{1}{7}$. Then $|x|<1$, so our series converges.

$$\begin{aligned} \frac{2}{(1-1/7)^3} &=\sum_{n=0}^\infty (n+2)(n+1)\left(\frac{1}{7}\right)^n\\ \frac{2}{(6/7)^3} &=\sum_{n=0}^\infty \frac{(n+2)(n+1)}{7^n} \end{aligned}$$

(a) Substituting $x=-t$ into the known power series $e^x = 1 + x +\frac{x^2}{2!}+\frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$, we see that:

$$\begin{aligned} e^{-t}&= 1 - t +\frac{t^2}{2!}-\frac{t^3}{3!} + \frac{t^4}{4!} - \cdots \\ 1-e^{-t}&= t - \frac{t^2}{2!}+ \frac{t^3}{3!} - \frac{t^4}{4!} + \cdots \\ \frac{1-e^{-t}}{t}&= 1 - \frac{t}{2!}+ \frac{t^2}{3!} - \frac{t^3}{4!} + \cdots \\ \int \frac{1-e^{-t}}{t}\ \mathrm{d}t&= C + x - \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} - \frac{x^4}{4\cdot 4!} + \cdots \end{aligned}$$

Finally, $f(0) = 0$ (since $f(0)$ is an integral from $0$ to $0$) and so $C=0$. Therefore

$$\begin{aligned} f(x) = \int_0^x\frac{1-e^{-t}}{t}\ \mathrm{d}t = x - \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} - \frac{x^4}{4\cdot 4!} + \cdots. \end{aligned}$$

We can also do this calculation entirely in summation notation: $e^{x}= \displaystyle\sum\limits_{n=0}^\infty \frac{x^n}{n!}$, and so

$$\begin{aligned} e^{-t}&= \sum_{n=0}^\infty \frac{(-t)^n}{n!} =1+ \sum_{n=1}^\infty \frac{(-1)^nt^n}{n!} \\ 1-e^{-t}&= - \sum_{n=1}^\infty \frac{(-1)^nt^n}{n!} = \sum_{n=1}^\infty \frac{(-1)^{n-1} t^n}{n!} \\ \frac{1-e^{-t}}{t}&= \sum_{n=1}^\infty \frac{(-1)^{n-1} t^{n-1}}{n!} \\ f(x) = \int_0^x\frac{1-e^{-t}}{t}\ \mathrm{d}t &= \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n\cdot n!} \end{aligned}$$

(b) We set $a_n=A_nx^n = \dfrac{(-1)^{n-1}}{n\cdot n!} x^n$ and apply the ratio test.

$$\begin{aligned} \lim_{n\rightarrow\infty}\Big|\frac{a_{n+1}}{a_n}\Big| &=\lim_{n\rightarrow\infty} \bigg|\frac{{(-1)^{n}}x^{n+1}/{((n+1)\cdot (n+1)!)}} {{(-1)^{n-1}}x^n/{(n\cdot n!)}}\bigg| \\ &=\lim_{n\rightarrow\infty} \bigg( \frac{|x|^{n+1}}{|x|^n} \frac{n\cdot n!}{(n+1)\cdot (n+1)!} \bigg) \\ &=\lim_{n\rightarrow\infty} \bigg( |x|\frac{n}{(n+1)^2} \bigg) \qquad\text{since $(n+1)!=(n+1)\,n!$} \\ &=0 \end{aligned}$$

This is smaller than $1$ no matter what $x$ is. So the series converges for all $x$.

$$\begin{aligned} &e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \ge 1+x\qquad\text{for all $x\ge 0$}\\ &\implies e^x-1\ge x \\ &\implies \frac{x^3}{e^x-1}\le \frac{x^3}{x} = x^2 \\ &\implies \int_0^1\frac{x^3}{e^x-1}\,\mathrm{d}x \le \int_0^1 x^2\,\mathrm{d}x=\frac{1}{3} \end{aligned}$$

(a) We know that $e^x = \sum\limits_{n=0}^\infty \frac{x^n}{n!}$ for all $x$. Replacing $x$ by $-x$, we also have $e^{-x} = \sum\limits_{n=0}^\infty \frac{(-x)^n}{n!}$ for all $x$ and hence

$$\begin{aligned} \cosh(x) =\frac{1}{2}\big[e^x+e^{-x}\big] =\frac{1}{2}\Big[\sum_{n=0}^\infty \frac{x^n}{n!} + \sum_{n=0}^\infty \frac{(-x)^n}{n!}\Big] = \sum_{\genfrac{}{}{0pt}{}{n=0}{n\text{ even}}}^\infty\frac{x^n}{n!} = \sum_{n=0}^\infty\frac{x^{2n}}{(2n)!} \end{aligned}$$

for all $x$. In particular, the interval of convergence is all real numbers.

(b) Using the power series expansion of part (a),

$$\begin{aligned} \cosh(2) &= 1 +\frac{2^2}{2!}+\frac{2^4}{4!} +\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} = 3\frac{2}{3} +\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} \end{aligned}$$

So it suffices to show that $\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!}\le 0.1$. Let's write $b_n = \frac{2^{2n}}{(2n)!}$. The first term in $\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!}$ is

$$b_3 = \frac{2^6}{6!} = \frac{2^6}{6 \times 5 \times 4 \times 3\times 2} = \frac{4}{45}$$

The ratio between successive terms in $\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!}$ is

$$\begin{aligned} \frac{b_{n+1}}{b_n} = \frac{2^{2n+2}/2^{2n}}{(2n+2)!/(2n)!} =\frac{4}{(2n+2)(2n+1)} \le \frac{4}{8\times 7} =\frac{1}{14}\qquad \text{for all }n\ge 3 \end{aligned}$$

Hence

$$\begin{aligned} \sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} &\le \overbrace{\frac{4}{45}}^{b_3} + \overbrace{\frac{4}{45}\times \frac{1}{14}}^{b_4\le} + \overbrace{\frac{4}{45}\times \frac{1}{14^2}}^{b_5\le} + \overbrace{\frac{4}{45}\times \frac{1}{14^3}}^{b_6\le} + \cdots\\ &=\frac{4}{45}\,\frac{1}{1-\frac{1}{14}} =\frac{4}{45}\,\frac{14}{13} =\frac{56}{585}<\frac{1}{10} \end{aligned}$$

(c) Comparing

$$\cosh(t) = \sum_{n=0}^\infty\frac{t^{2n}}{(2n)!} = \sum_{n=0}^\infty\frac{{(t^2)}^n}{(2n)!} \qquad\text{and}\qquad e^{\frac{1}{2}t^2} = \sum_{n=0}^\infty\frac{{(\frac{1}{2}t^2)}^n}{n!} = \sum_{n=0}^\infty\frac{{(t^2)}^n}{2^n n!}$$

we see that it suffices to show that $(2n)! \ge 2^n n!$. Now. for all $n\ge 1$,

$$\begin{aligned} (2n)! &=\overbrace{1\times 2\times\cdots\times n}^{n\text{ factors}} \overbrace{(n+1)\times (n+2)\times\cdots\times 2n}^{n\text{ factors}} \\ &\ge\overbrace{1\times 2\times\cdots\times n}^{n\text{ factors}} \overbrace{2\times 2\times\cdots\times 2}^{n\text{ factors}} \\ &=2^n\, n! \end{aligned}$$