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This is just a curious question, but is the following true?

$$f(x) = f'(x)\iff f(x) = e^x.$$

I can prove that $\dfrac{\mathrm d}{\mathrm dx}\left(e^x\right) = e^x$ from using the the formula, $e^x := \operatorname*{\lim}\limits_{n\to 0}(1+n)^{1/n}.$ For those who are not familiar with the proof, it can be found here. Or, you can go here for a similar approach.

However, my question is asking whether or not $f(x) = e^x$ is the only function equal to its own deriv. I suspect it is true, but how can we prove that $e^x$ is the only value equal to its derivative, for any $x$?

I consider it very likely that there exists another question out there, perhaps exactly like this. If so, please comment the link below, and I will go straight to it, delete this post, and give you a muffin. I do not intend on trolling or wasting anyone's time.

Thank you in advance.

Mr Pie
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  • No, see $f(x)=0$. In general, $f(x)=Ke^x$ is a solution. –  Mar 21 '18 at 13:14
  • There's also $f(x)=2e^x$. I think there are a few more. – Aweygan Mar 21 '18 at 13:14
  • We do have $f(x) = Ce^x$ for some constant $C$, though. – Ben Grossmann Mar 21 '18 at 13:14
  • I'm sure this is a duplicate. If $f$ is another function that satisfies the differential equation, differentiate $f(x)/e^x$ to show you get a constant. Google https://www.google.com/search?q=uniqueness+of+exponential+function&ie=utf-8&oe=utf-8&client=firefox-b-1 – Ethan Bolker Mar 21 '18 at 13:15
  • Ah yes, I didn't think of that. I kind of left out the case $f(x) = 0$ since that is trivial. Is there a way of proving so? – Mr Pie Mar 21 '18 at 13:16
  • Actually $f(x)=0$ is just another case of $f(x)=Ke^x$, the one with $K=0$. –  Mar 21 '18 at 13:17
  • There's no need to delete your post, by the way. It's actually useful to have duplicate posts like yours to make the original easier to find. – Ben Grossmann Mar 21 '18 at 13:21
  • @Omnomnomnom ok. I know for next time. But how do people find these other links anyway, do you know? Do they have them listed in their favourites or something? – Mr Pie Mar 21 '18 at 13:23
  • @user477343 Sometimes somebody will remember answering a similar question. Sometimes it's a matter of noticing the "related" questions on the right. Sometimes it's a matter of googling exactly the right phrase. However it happens, it's a lot more likely to happen when a lot of people are trying at the same time. – Ben Grossmann Mar 21 '18 at 13:26
  • @EthanBolker, applying the product rule $\dfrac{\mathrm d}{\mathrm dx}\big(u(x)\cdot v(x)\big) = u'(x)\cdot v(x) + u(x)\cdot v(x)$ I get that $$\begin{align} \frac{\mathrm d}{\mathrm dx}\big(f(x)e^{-x}\big) &= f(e^{-x} - xe^{-x}) \ &= -f(x-1)e^{-x}.\end{align}$$ Thank you very much :P – Mr Pie Mar 21 '18 at 13:29
  • @user477343: $f(xy)\ne f(x),y$ in general (actually, setting $x=1$ shows that there is a very limited set of functions that satisfy this equation). – robjohn Mar 21 '18 at 13:32
  • @robjohn $f(e^{-x} - xe^{-x}) = fe^{-x} - fxe^{-x} = -f(x-1)e^{-x}$ ..... ? Is that right? Oh, and I made a typo in the product rule. The last summand should have a $v'(x)$ instead of $v(x)$. – Mr Pie Mar 21 '18 at 13:34
  • @user477343: no, that is the whole point of my previous comment. – robjohn Mar 21 '18 at 13:36
  • @robjohn so then how do we show $-$ oh wait... ahhhh hahah sorry about that. Silly mistake faceslap – Mr Pie Mar 21 '18 at 13:37

1 Answers1

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If $f=f'$, then let $g=\frac f\exp$. Then$$g'=\frac{\exp\times f'-f\times\exp}{\exp^2}=0.$$Therefore, $g=C$, for some $C\in\mathbb R$. In other words, $f=C\times\exp$.

José Carlos Santos
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  • I am familiar with the $\text{exp}$ notation, so I understand. I would have upvoted if I did not reach my daily voting limit, hahah. Thank you very much though :) – Mr Pie Mar 21 '18 at 13:31