Bilangan e pada Logaritma (Bagian 2)

 D. Lanjutan penentuan nilai e

Perhatikanlah bentuk

$\begin{aligned}\begin{cases} 1. & =\left ( 1+\displaystyle \frac{1}{n} \right )^{n} \\\\ 2. & =\left ( 1-\displaystyle \frac{1}{n} \right )^{-n} \end{cases} \end{aligned}$.

Menurut Binomial Newton,

$\color{blue}\begin{aligned}(a+b)^{n}\color{black}=\, &\color{red}C_{0}^{n}a^{n}b^{0}+C_{1}^{n}a^{n-1}b^{1}+C_{2}^{n}a^{n-2}b^{2}\\ &+C_{3}^{n}a^{n-3}b^{3}+\cdots +C_{n-3}^{n}a^{3}b^{n-3}\\ &+C_{n-2}^{n}a^{2}b^{n-2}+C_{n-1}^{n}a^{1}b^{n-1}+C_{n}^{n}a^{0}b^{n}\\ &\color{black}=\displaystyle \sum_{r=0}^{n}C_{r}^{\color{red}n}a^{\color{red}n\color{black}-r}b^{r} \end{aligned}$.

Bentuk perluasannya, ketika  a=1 dan b=x

$\color{blue}\begin{aligned}\color{black}\quad(1+x)&^{n}\\ \color{black}=\, &\color{red}C_{0}^{n}1^{n}x^{0}+C_{1}^{n}1^{n-1}x^{1}+C_{2}^{n}1^{n-2}x^{2}\\ &+C_{3}^{n}1^{n-3}x^{3}+\cdots +C_{n-3}^{n}1^{3}x^{n-3}\\ &+C_{n-2}^{n}1^{2}x^{n-2}+C_{n-1}^{n}1^{1}x^{n-1}+C_{n}^{n}1^{0}x^{n}\\ =\, &\color{red}C_{0}^{n}+C_{1}^{n}x+C_{2}^{n}x^{2} +C_{3}^{n}x^{3}+\cdots \\ &+C_{n-3}^{n}x^{n-3} +C_{n-2}^{n}x^{n-2}+C_{n-1}^{n}x^{n-1}\\ &+C_{n}^{n}x^{n}\\  \end{aligned}$

Sehingga

$\begin{aligned}(1+x)^{n}&=1+nx+\displaystyle \frac{n(n-1)}{2!}x^{2}+\displaystyle \frac{n(n-1)(n-2)}{3!}x^{3}\\ &+... +\displaystyle \frac{n(n-1)(n-2)...(n-r+1)}{(r-1)!}x^{r-1}+... \end{aligned}$.

Saat  $x=\displaystyle \frac{1}{n}$,

$\begin{aligned}\left ( 1+\displaystyle \frac{1}{n} \right )^{n}&=1+\displaystyle \frac{n}{1}.\left (\frac{1}{n} \right )^{1}+\displaystyle \frac{n(n-1)}{1.2}\left ( \displaystyle \frac{1}{n} \right )^{2}\\ &+\displaystyle \frac{n(n-1)(n-2)}{1.2.3}\left ( \displaystyle \frac{1}{n} \right )^{2}\\ &+...+\displaystyle \frac{n(n-1)(n-2)...1}{1.2.3...n}\left ( \displaystyle \frac{1}{n} \right )^{n} \end{aligned}$.

Jika  $U_{n}=\left ( 1+\displaystyle \frac{1}{n} \right )^{n}$, maka didapatkan

$\begin{aligned}U_{n}&=1+1+\displaystyle \frac{1}{2!}\left ( 1-\displaystyle \frac{1}{n} \right )+\displaystyle \frac{1}{3!}\left ( 1-\displaystyle \frac{1}{n} \right )\left ( 1-\displaystyle \frac{2}{n} \right )\\ &+\displaystyle \frac{1}{4!}\left ( 1-\displaystyle \frac{1}{n} \right )\left ( 1-\displaystyle \frac{2}{n} \right )\left ( 1-\displaystyle \frac{3}{n} \right )\\ &+...+\displaystyle \frac{1}{n!}\left ( 1-\displaystyle \frac{1}{n} \right )\left ( 1-\displaystyle \frac{2}{n} \right )\left ( 1-\displaystyle \frac{3}{n} \right )...\left ( 1-\displaystyle \frac{n-1}{n} \right ) \end{aligned}$.

Karena bentuk di atas  $\color{red}\left ( 1-\displaystyle \frac{p}{n} \right )<1$, dengan  $p,n\in \mathbb{N}$, maka akan diperoleh

$U_{1}<U_{n}<1+1+\displaystyle \frac{1}{2!}+\displaystyle \frac{1}{3!}+\displaystyle \frac{1}{4!}+...+\displaystyle \frac{1}{n!}$

Serta diketahui bentuk

$\begin{aligned}&\displaystyle \frac{1}{2.3}<\frac{1}{2.2}\\ &\displaystyle \frac{1}{2.3.4}<\frac{1}{2.2.2}\\ &\displaystyle \frac{1}{2.3.4.5}<\frac{1}{2.2.2.2}\\ &...\\ &\displaystyle \frac{1}{2.3.3.4...n}<\frac{1}{2^{n-1}}\\ \end{aligned}$.

Dan diketahui pula dari uraian di atas $U_{1}=2$, maka

$\begin{aligned}&U_{1}<U_{n}<1+1+\displaystyle \frac{1}{2!}+\displaystyle \frac{1}{3!}+\displaystyle \frac{1}{4!}+...+\displaystyle \frac{1}{n!}\\ &\Leftrightarrow \: 2<U_{n}<1+1+\underset{\color{blue}\textrm{deret konvergen}}{\underbrace{\left ( \displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{n}} \right )}}\\ &\Leftrightarrow \: 2<U_{n}<1+1+\left ( \displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}} \right )\\ &\Leftrightarrow \: 2<U_{n}<1+1+1\\ &\Leftrightarrow \: 2<U_{n}<3\\ &\Leftrightarrow \: 2<\underset{n\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle U_{n}<3 \end{aligned}$.

Selanjutnya bentuk  $\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle U_{n}=e$, dengan e adalah bilangan irasional dengan bentuk desimal e = 2,71828....

DAFTAR PUSTAKA

1. Koesmantoro, Rawuh (Ed.). 2001. Matematika Pendahuluan (Seri Matematika). Cet. VII. Bandung: ITB.