# Functions with the same derivatives [duplicate]

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Functions with the same derivatives

Besides, $f(x)=e^x$, what other non-zero functions are the same as their derivatives no matter how many times the derivative is taken?

## marked as duplicate by Pedro Tamaroff♦, user61527, Potato, Seirios, Dominic MichaelisAug 25 '13 at 8:18

• @Chris Culter Thanks for pointing this out! – Dilippard Aug 25 '13 at 4:49

If $f:\mathbb{R}\to\mathbb{R}$ satisfies $f'(x)=f(x)$ for all $x\in\mathbb{R}$ then $f(x)=Ae^{x}$.

To see this suppose $g$ is such a function. Then take $h(x)=g(x)e^{-x}$. Then

$h'(x)=g'(x)e^{-x}-g(x)e^{-x}=g(x)e^{-x}-g(x)e^{-x}=0$. So $h$ is constant.

So $g(x)=Ae^{x}$ for some real constant $A$. So after requiring the function to satisfy this for the first derivative it is already determined. Further derivatives don't matter.

It was noted by kahen that if we allow the function to be defined on an open domain $A\subset\mathbb{R}$ which has more than one connected component then the function is locally of the form $Ce^{x}$ (where the constants may differ from component to component). This can be seen by applying the proof to each of the connected components. Recall that if $f'(x)=0$ for all $x\in B$ and $B\subset\mathbb{R}$ is open and connected then $f=C$ on $B$ where $C$ is a real constant.

• I can't take credit for the proof. I believe I saw this in Spivak. – user71352 Aug 25 '13 at 4:43
• @user71352 Thanks! – Dilippard Aug 25 '13 at 4:50
• Your welcome! Glad to help. – user71352 Aug 25 '13 at 4:51
• A tiny correction: You can only conclude that $h$ is locally constant. Consider for example $\mathbb R \setminus \{0\} \owns x \mapsto \begin{cases} e^x & \text{if }x < 0 \cr 2e^x & \text{if }x > 0\end{cases}$ – kahen Aug 25 '13 at 4:52
• @kahen That example isn't continuous, hence not entire. – user61527 Aug 25 '13 at 4:53

You can see $A e^x$ is the only such function by looking at the power series. Assume $g(x)$ is such a function with $g(x)=g'(x)=g''(x)=\ldots$. Then since $g(x)$ has infinite derivatives it also has a power series: $$g(x) = A + a_1 x + a_2 x^2 + \ldots = g'(x) = a_1 + 2a_2 x + 3 a_3 x^2+\ldots$$ $$\to \quad A = a_1 =2!a_2 = 3!a_3 = \ldots n!a_n$$ So we have: $$g(x) = A\left( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right)=Ae^x$$ Thus proving $A e^x$ is the only such function.

As user71352's very nice answer shows, $e^{x}$ is the unique function that does this, up to rescaling. A similar function is

$$f(x) = \sin{x}$$

which satisfies $f^{(4)} = f$. The cosine function also does this.