# e (數學常數)

(由自然底數跳轉過嚟)

${\displaystyle \mathbb {N} \subset \mathbb {Z} \subset \mathbb {Q} \subset \mathbb {R} \subset \mathbb {C} }$

${\displaystyle e}$自然指數同埋自然對數函數嘅底數。有時又叫做自然底數歐拉數Euler's number），個名來自瑞士數學家歐拉；佢嘅數值大約係（小數點後20位）：

${\displaystyle e=2.71828182845904523536\ldots }$

## 定義

${\displaystyle e}$ 可以用導數嚟定義。如果試吓對隨便一個指數函數 ${\displaystyle f(x)=a^{x}}$ 求導，根據基本原理：

${\displaystyle f'(x)={d \over dx}a^{x}=\lim _{h\to 0}{a^{x+h}-a^{x} \over h}=\lim _{h\to 0}{a^{x}a^{h}-a^{x} \over h}=a^{x}\lim _{h\to 0}{a^{h}-1 \over h}}$

{\displaystyle {\begin{aligned}\lim _{h\to 0}(e^{h}-1)&=\lim _{h\to 0}h\\\lim _{h\to 0}e^{h}&=\lim _{h\to 0}(1+h)\end{aligned}}}

${\displaystyle e=\lim _{h\to \infty }\left(1+{1 \over h}\right)^{h}=\lim _{h\to \infty }\sum _{k=0}^{\infty }\left((_{k}^{h}){1 \over h^{k}}\right)}$
{\displaystyle {\begin{aligned}&=\lim _{h\to \infty }\left((_{0}^{h}){1 \over h^{0}}+(_{1}^{h}){1 \over h^{1}}+(_{2}^{h}){1 \over h^{2}}+\dots \right)\\&=\lim _{h\to \infty }\left({1 \over 0!}+{h \over 1!}{1 \over h}+{h(h-1) \over 2!}{1 \over h^{2}}+{h(h-1)(h-2) \over 3!}{1 \over h^{3}}+\dots \right)\\&=\lim _{h\to \infty }\left({1 \over 0!}+{1 \over 1!}+{1-{1 \over h} \over 2!}+{(1-{1 \over h})(1-{2 \over h}) \over 3!}+\dots \right)\\&=\lim \left({1 \over 0!}+{1 \over 1!}+{1 \over 2!}+{1 \over 3!}+\dots \right)\\&=\lim _{h\to \infty }\sum _{n=0}^{h}{1 \over n!}\end{aligned}}}

${\displaystyle {d \over dx}a^{x}={d \over dx}e^{x\ln(a)}={d \over d(x\ln(a))}e^{x\ln(a)}\times {d \over dx}(\ x\ln(a)\ )=a^{x}\ln(a)}$

{\displaystyle {\begin{aligned}\lim _{h\to 0}(a^{h}-1)&=\lim _{h\to 0}h\ln(a)\\\lim _{h\to 0}a^{h}&=\lim _{h\to 0}\ (\ 1+h\ln(a)\ )\\a&=\lim _{h\to 0}\ (\ 1+h\ln(a)\ )^{1 \over h}\end{aligned}}}

${\displaystyle e^{\ln(a)}=\lim _{h\to \infty }\left(1+{\ln(a) \over h}\right)^{h}=\lim _{h\to \infty }\left(1+{1 \over h}\right)^{h\ln(a)}}$

${\displaystyle e^{x}=\lim _{h\to \infty }\left(1+{x \over h}\right)^{h}=\lim _{h\to \infty }\left(1+{1 \over h}\right)^{hx}}$

${\displaystyle e^{x}=\lim _{h\to \infty }\sum _{n=0}^{h}{x^{n} \over n!}}$

## 無理數

${\displaystyle e}$ 係一個無理數。可以用反證法證明。

${\displaystyle e={P \over Q}=1+{1 \over 1}+{1 \over {1\cdot 2}}+{1 \over {1\cdot 2\cdot 3}}+\dots }$
${\displaystyle Qe=P=Q+{Q \over 1}+{Q \over {1\cdot 2}}+{Q \over {1\cdot 2\cdot 3}}+\dots \in \mathbb {Z} }$
${\displaystyle k!\cdot Qe=k!\cdot Q+{k!\cdot Q \over 1}+{k!\cdot Q \over {1\cdot 2}}+{k!\cdot Q \over {1\cdot 2\cdot 3}}+\dots \in \mathbb {Z} }$，當中嘅 ${\displaystyle k}$ 係大於 ${\displaystyle Q}$ 嘅任意正整數。
${\displaystyle k!\cdot Qe=L+{Q \over {k+1}}+{Q \over {(k+1)(k+2)}}+{Q \over {(k+1)(k+2)(k+3)}}+\ldots }$，入面 ${\displaystyle L}$ 係整數。
${\displaystyle 0<{Q \over {k+1}}+{Q \over {(k+1)(k+2)}}+{Q \over {(k+1)(k+2)(k+3)}}+\dots <{Q \over {k+1}}+{Q \over {(k+1)^{2}}}+{Q \over {(k+1)^{3}}}+\dots ={Q \over k}<1}$