Taylor expansion of e^x

យើងឃើញថា
$$\int_0^xe^t\mathrm{d}t=e^x-1 \Rightarrow e^x-\int_0^xe^t\mathrm{d}t=1 $$
តាង$\varphi:C(\mathbb{R})\to C(\mathbb{R});f(x)\mapsto \int_0^xf(t)\mathrm{d}t$
នាំឲ្យ
\begin{align} e^x-\varphi(e^x)&=1\\ \Rightarrow (1-\varphi)(e^x)&=1\\ \Rightarrow e^x&=(1-\varphi)^{-1}(1)\\ &=(1+\varphi+\varphi^2+\varphi^3+\cdots)(1)\\ &=1+\varphi(1)+\varphi^2(1)+\varphi^3(1)+\cdots\\ \end{align}
ដោយ
\begin{align} \varphi(1)&=\int_0^x1\mathrm{d}t=x\\ \varphi^2(1)&=\varphi(\varphi(1))=\varphi(x)=\int_0^x t \mathrm{d}t=\frac{x^2}{2!}\\ \varphi^3(1)&=\varphi(\varphi^2(1))=\varphi\Bigl(\frac{x^2}{2!}\Bigr)=\int_0^x \frac{t^2}{2!} \mathrm{d}t=\frac{x^3}{3!}\\ &\vdots \end{align}
ដូចនេះ យើងបាន
$$ \large e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots $$