导数是自身的函数

如何找出这么特殊的函数?

原文链接:Finding the function that is its own derivative

设函数为$f(x)$,满足

\begin{equation}
\frac{d}{dx} f(x) = f^\prime (x) = f(x)
\label{df}
\end{equation}

如果$f(x)=x$行不行?

\begin{equation}
f(x) = x \implies f^\prime (x) = 1
\label{fx}
\end{equation}

$f^\prime (x) \ne x$,不满足\eqref{df}。

再试$f(x)=1+x$,

\begin{equation}
f(x) = 1+x \implies f^\prime (x) = 1
\label{fx1}
\end{equation}

接近答案了,看到“1”了。什么的导数会是$x$?

\begin{equation}
f(x) = 1+x+\tfrac{1}{2}x^2 \implies f^\prime (x) = 1 + x
\label{fx2}
\end{equation}

什么的导数是$\frac{1}{2x^2}$?

\begin{equation}
f(x) = f(x) = 1+x+\tfrac{1}{2}x^2 + \tfrac{1}{2 \cdot 3}x^3 \implies f^\prime (x) = 1 + x + \tfrac{1}{2}x^2
\label{fx3}
\end{equation}

以此类推,如果有无穷多项,

\begin{equation} \begin{split} f(x) &= 1+x+\tfrac{1}{2}x^2 + \tfrac{1}{2 \cdot 3}x^3 + \cdots \tfrac{1}{n!}x^n \cdots \\ &\implies f^\prime (x)=1 + x + \tfrac{1}{2}x^2 + \cdots \tfrac{1}{(n-1)!}x^{n-1} \cdots \end{split} \label{fxinf} \end{equation}

能不能把$f(x)$写得更精致些?

显然,$f(0)=1$,而$f(1)$是某个数,设此数为$e$:

\begin{equation}
f(1) = 1+1+\tfrac{1}{2} + \tfrac{1}{2 \cdot 3} + \cdots = e
\label{e}
\end{equation}

计算前面聊聊几项,就看出$e$收敛到$e \approx 2.717\dots $。

对于$f(2)$,达到$e$同样精度,需要多算几项,结果为$7.382\approx 2.717\times 2.717\approx e^2$

\begin{equation}
f(2) = 1+2+\tfrac{1}{2}4 + \tfrac{1}{2 \cdot 3}8 + \cdots = e^2
\label{e2}
\end{equation}

以此类推,

\begin{equation}
f(x) = 1+x+\tfrac{1}{2}x^2 + \tfrac{1}{2 \cdot 3}x^3 + \cdots \tfrac{1}{n!}x^n + \cdots = e^x
\label{ex}
\end{equation}

即函数$f(x)=e^x$的导数是自身:

\begin{equation}
\frac{d}{dx} e^x = e^x
\label{dself}
\end{equation}

标签: 导数, 微分, 指数函数, e

添加新评论