Bilangan e pada Logaritma (Bagian 2)

 D. Lanjutan penentuan nilai e

Perhatikanlah bentuk

$\begin{array}{r}\{\begin{array}{ll}1.& ={(1+{\displaystyle \frac{1}{n}})}^n\\ \\ 2.& ={(1-{\displaystyle \frac{1}{n}})}^{-n}\end{array}\end{array}$.

Menurut Binomial Newton,

$\begin{array}{rl}(a+b{)}^n=& C_0^n{a}^n{b}^0+C_1^n{a}^{n-1}{b}^1+C_2^n{a}^{n-2}{b}^2\\ & +C_3^n{a}^{n-3}{b}^3+\cdots +C_{n-3}^n{a}^3{b}^{n-3}\\ & +C_{n-2}^n{a}^2{b}^{n-2}+C_{n-1}^n{a}^1{b}^{n-1}+C_n^n{a}^0{b}^n\\ & ={\displaystyle \sum _{r=0}^n{C}_r^n{a}^{n-r}{b}^r}\end{array}$.

Bentuk perluasannya, ketika  a=1 dan b=x

$\begin{array}{rl}(1+x)& {}^n\\ =& C_0^n{1}^n{x}^0+C_1^n{1}^{n-1}{x}^1+C_2^n{1}^{n-2}{x}^2\\ & +C_3^n{1}^{n-3}{x}^3+\cdots +C_{n-3}^n{1}^3{x}^{n-3}\\ & +C_{n-2}^n{1}^2{x}^{n-2}+C_{n-1}^n{1}^1{x}^{n-1}+C_n^n{1}^0{x}^n\\ =& C_0^n+C_1^{n}x+C_2^n{x}^2+C_3^n{x}^3+\cdots \\ & +C_{n-3}^n{x}^{n-3}+C_{n-2}^n{x}^{n-2}+C_{n-1}^n{x}^{n-1}\\ & +C_n^n{x}^n\end{array}$

Sehingga

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

Saat  $x={\displaystyle \frac{1}{n}}$,

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

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

$\begin{array}{rl}{U}_n& =1+1+{\displaystyle \frac{1}{2!}(1-{\displaystyle \frac{1}{n}})+{\displaystyle \frac{1}{3!}(1-{\displaystyle \frac{1}{n}})(1-{\displaystyle \frac{2}{n}})}}\\ & +{\displaystyle \frac{1}{4!}(1-{\displaystyle \frac{1}{n}})(1-{\displaystyle \frac{2}{n}})(1-{\displaystyle \frac{3}{n}})}\\ & +...+{\displaystyle \frac{1}{n!}(1-{\displaystyle \frac{1}{n}})(1-{\displaystyle \frac{2}{n}})(1-{\displaystyle \frac{3}{n}})...(1-{\displaystyle \frac{n-1}{n}})}\end{array}$.

Karena bentuk di atas  $(1-{\displaystyle \frac{p}{n}})<1$, 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

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

Dan diketahui pula dari uraian di atas $U_1=2$, maka

$\begin{array}{rl}& U_1<U_n<1+1+{\displaystyle \frac{1}{2!}+{\displaystyle \frac{1}{3!}+{\displaystyle \frac{1}{4!}+...+{\displaystyle \frac{1}{n!}}}}}\\ & \iff 2<U_n<1+1+\underset{\text{deret konvergen}}{\underbrace{\left({\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}}\right)}}\\ & \iff 2<U_n<1+1+\left({\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}}}\right)\\ & \iff 2<U_n<1+1+1\\ & \iff 2<U_n<3\\ & \iff 2<\underset{n\to \mathrm{\infty }}{\text{Lim}}{\displaystyle U_n<3}\end{array}$.

Selanjutnya bentuk  $\underset{h\to 0}{\text{Lim}}{\displaystyle U_n=e}$, 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.