In response to a Quora question, I wrote the following:

What is the function that is its own derivative?

To eliminate functions that are too exotic, let us restrict this question to functions that can be written as power series. So, we seek the function $f(x)=f'(x)$ where $f(x)=a_0+a_1x+a_2x^2+...$.

But the derivative of such a function is given by $f'(x)=a_1+2a_2x+3a_3x^2+...$. Thus, if $f(x)=f'(x)$, then we must have


If we choose $a_0=1$, this means that $a_i=1/i!$. (Any other choice of $a_0$ just means trivially multiplying $f(x)$ by that number.)

So what is the function, $f(x)=1 + x + x^2/2! +x^3/3! + ...$?

Let me try something. Let me see if I can find $f(x+y)$ in terms of $f(x)$ and $f(y)$:

$$f(x+y)=1 + x + y + (x+y)^2/2! + ...,$$


$$f(x)f(y) = 1 + x + y + x^2/2 + y^2/2 + xy + ...,$$

It doesn't take too long to show that the pattern continues, and $f(x+y)=f(x)f(y)$.

Now if I know $f(1)$, I know $f(2)=f(1)^2$. I also know $f(0.5)=\sqrt{f(1)}$. In other words, I can calculate $f(x)$ for any rational number. Or any number, for that matter: it is clear from the above that $f(x)=f(1)^x$.

So all that remains is calculating $f(1)$, the number that is at the core of all these magical properties. This number is defined as $f(1)=1+1+1/2!+1/3!+...$. This sum converges, giving the value $f(1)=2.718281846...$. In other words, $f(1)=e$, and the magic function I sought was the exponential function $e^x$.