Are the any non-trivial functions where $f(x)=f'(x)$ not of the form $Ae^x$

This may seem like a silly question, but I just wanted to check. I know there are proofs that if $f(x)=f'(x)$ then $f(x)=Ae^x$. But can we 'invent' another function that obeys $f(x)=f'(x)$ which is non-trivial?

Solution 1:

Observe that if $ f(x)=f'(x) $ then $$ \left(\frac{f(x)}{e^x}\right)'=\frac{f'(x)-f(x)}{e^x}=0 $$ Hence $\dfrac{f(x)}{e^x}$ is constant...

Solution 2:

I think other answers given here assume the existence of a nice function $e^{x}$ and this makes the proof considerably simpler. However I believe that it is better to approach the problem of solving $f'(x) = f(x)$ without knowing anything about $e^{x}$.

When we go down this path our final result is the following:

Theorem: There exists a unique function $f:\mathbb{R}\to \mathbb{R}$ which is differentiable for all $x \in \mathbb{R}$ and satisfies $f'(x) = f(x)$ and $f(0) = 1$. Further any function $g(x)$ which is differentiable for all $x$ and satisfies $g'(x) = g(x)$ is explicitly given by $g(x) = g(0)f(x)$ where $f(x)$ is the unique function mentioned previously.

We give a simple proof of the above theorem without using any properties/knowledge of $e^{x}$. Let's show that if such a function $f$ exists then it must be unique. Suppose there is another function $h(x)$ such that $h'(x) = h(x)$ and $h(0) = 1$. Then the difference $F(x) = f(x) - h(x)$ satisfies $F'(x) = F(x)$ and $F(0) = 0$. We will show that $F(x) = 0$ for all $x$. Suppose that it is not the case and that there is a point $a$ such that $F(a) \neq 0$ and consider $G(x) = F(a + x)F(a - x)$. Clearly we have \begin{align} G'(x) &= F(a - x)F'(a + x) - F(a + x)F'(a - x)\notag\\ &= F(a - x)F(a + x) - F(a + x)F(a - x)\notag\\ &= 0 \end{align} so that $G(x)$ is constant for all $x$. Therefore $G(x) = G(0) = F(a) \cdot F(a) > 0$. We thus have $F(a + x)F(a - x) > 0$ and hence putting $x = a$ we get $F(2a)F(0) > 0$. This contradicts $F(0) = 0$.

It follows that $F(x) = 0$ for all $x$ and hence the function $f$ must be unique. Now we need to show the existence. To that end we first establish that $f(x) > 0$ for all $x$. If there is a number $b$ such that $f(b) = 0$ then we can consider the function $\phi(x) = f(x + b)$ and it will have the property that $\phi'(x) = \phi(x)$ and $\phi(0) = 0$. By argument in preceding paragraph $\phi(x)$ is identically $0$ and hence $f(x) = \phi(x - b)$ is also identically $0$. Hence it follows that $f(x)$ is non-zero for all $x$. Since $f(x)$ is continuous and $f(0) = 1 > 0$ it follows that $f(x) > 0$ for all $x$.

Since $f'(x) = f(x) > 0$ for all $x$, it follows that $f(x)$ is strictly increasing and differentiable with a non-vanishing derivative. By inverse function therorem the inverse function $f^{-1}$ exists (if $f$ exists) and is also increasing with non-vanishing derivative. Also using techniques of differentiation it follows that $f'(x) = f(x)$ implies that $\{f^{-1}(x)\}' = 1/x$ for all $x > 0$ and $f^{-1}(1) = 0$. Since $1/x$ is continuous the definite integral $$\psi(x) = \int_{1}^{x}\frac{dt}{t}$$ exists for all $x > 0$ and has the properties of $f^{-1}$ and it is easy to show that $f^{-1}(x) = \psi(x)$. Clearly the function $(f^{-1}(x) - \psi(x))$ is constant as it derivative is $0$ and hence $$f^{-1}(x) - \psi(x) = f^{-1}(1) - \psi(1) = 0$$ so that $$f^{-1}(x) = \psi(x) = \int_{1}^{x}\frac{dt}{t}$$ Next using inverse function theorem $f(x)$ exists. Thus the question of existence of $f(x)$ is settled.

Now consider $g(x)$ with $g'(x) = g(x)$. If $g(0) = 0$ then we know from argument given earlier that $g(x) = 0$ for all $x$. If $g(0) \neq 0$ then we study the function $\psi(x) = g(x)/g(0)$. Clearly $\psi'(x) = \psi(x)$ and $\psi(0) = 1$ and hence it is same as the unique function $f(x)$. Thus $g(x)/g(0) = \psi(x) = f(x)$ for all $x$. Hence $g(x) = g(0)f(x)$.

The unique function $f(x)$ in the theorem proved above is denoted by $\exp(x)$ or $e^{x}$.