Is $e^x$ the only non-trivial function for which the differential operator is the identity operator?

Question

Given that the derivative of a function is defined as:

$$\frac{d}{dx}f(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\tag{1}\label{1}$$

and:

$$\frac{d}{dx}e^x=e^x\tag{2}\label{2}$$

Can we get $e^x=f(x)$ from the following condition?

$$f(x)=\frac{d}{dx}f(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\tag{3}\label{3}$$

Could the differential operator be interpreted as the identity operator for $f(x)=e^x$?

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Is $e^x$ the only non-trivial function for which the differential operator is the identity operator?

Answer 1

Using elementary methods for "solving a differential equation" we can find all such functions.

Suppose $f(x)$ is a differentiable function $\mathbb R \to \mathbb R$ such that $$f'(x) = f(x)\quad\text{for all }x.\tag{1}$$ Consider the function $g(x) = e^{-x}f(x)$. Use the product rule to compute \begin{align} g'(x) &= \frac{d}{dx}\big(e^{-x}f(x)\big) = f(x)\frac{d}{dx}\big(e^{-x}\big)+e^{-x}\frac{d}{dx}\big(f(x)\big)\\ &= -f(x)e^{-x} + e^{-x} f(x) = 0 \end{align} But if $g'(x) = 0$ for all $x$, it follows that $g(x) = c$ for some constant $c$. (Proof from the mean value theorem.) Therefore \begin{align} g(x) &= c \\ e^{-x}f(x) &= c \\ f(x) &= c e^{x} \end{align} The functions $ce^{x}$ are the only solutions for $(1)$. There are of course infintely many solutions, one for each constant $c$.

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The same method shows that the solutions for functions $f : \mathbb C \to \mathbb C$ that satisfy $f'(z)=f(z)$ for all $z \in \mathbb C$ are the functions $f(z) = c e^{z}$. One for each complex constant $c$.

Answer 2

Operators aren't equal somewhere but unequal somewhere else; that's sloppy terminology. What we'd usually say is functions of the form $ke^x$ are the only eigenfunctions of $\frac{d}{dx}$ with eigenvalue $1$, or that they're the only functions on which $\frac{d}{dx}$ has the same action as the identity operator does, or they're its only fixed points, or they're the only elements of the kernel of $\frac{d}{dx}-I$.

Actually, I think that talk of "action" will probably raise eyebrows too; I should say these are the functions with the same image under both operators.

Answer 3

If consider functions of the form $ke^x$, then YES.

Assume $f(x)>0$. As $$\mathrm{\frac{d(\log(f(x)))}{dx}}=\frac{1}{f(x)}\cdot\mathrm{\frac{df(x)}{dx}}=1,$$ We have $$\int{\mathrm{\frac{d(\log(f(x)))}{dx}}}\mathrm dx=\int 1\mathrm dx=x+C,$$ where $C\in \mathbb{R}$ is a constant number. Now The left hand side is $\log(f(x))$ and the right hand side is $x+C$. Thus $$f(x)=e^{\log(f(x)}=e^{x+C}=e^C\cdot e^x=ke^x,$$

If $f(x)<0$, then consider $\mathrm{\frac{d(\log(-f(x)))}{dx}}$. In a similar way, we have $$f(x)=-e^C\cdot e^x=ke^x,$$ where $k=-e^C<0$. where $k=e^C\ge 0$ is a constant number too.