This question already has an answer here:

(counting $\sin x$ as a variation of $\cos x$).

They are self-similar in that their derivative is also a function of themselves. Crucially, this sort of feedback means that higher derivatives are also similar, resulting in curves that are smooth in what seems to me to be a unique way. For the examples in the title:

$f'(x) = f(x)$ is true for $f(x)=e^x$

$f'(x) = f(x+\pi/2)$ is true for $f(x)=\cos x$

EDIT: That's for the two functions given in the title. The question is: are there any other functions that are similar to their derivative in some other way? Just one example is enough (the "duplicate" doesn't address this).

BTW It seems to me that these are the only solutions to these differential equations, and e.g. a different offset for the second one just changes the period of the $\cos$ function. But I don't see how to show this - or even how to think about it.

BTW this was inspired by Euler's equation, relating $\cos, \sin,$ and $e$. I think the fundamental connection is that they all are self-similar, and the rest is clever bit-twiddling (like how integers alternate even/odd, and $f(x) = (-1)^x$ alternates positive/negative).

marked as duplicate by Ross Millikan algebra-precalculus Sep 16 at 1:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $f=f'$ is an ode so you can use the uniqueness part of the ode theorem. idk about the ones where you translate $x$. good question – Tim kinsella Sep 16 at 1:47
  • this looks relevant… – Tim kinsella Sep 16 at 1:50
  • The duplicate is not perfect because it is just for the second part, but that is the hard one. There is also a rescaling, but the logic is the same. – Ross Millikan Sep 16 at 1:59

Well, they are the only ones. Including of course $\sin{x} , \sinh{x}, \cosh{x}$. That is why when you have a linear differential equation with constant coefficients, the solution is a linear combination of these functions, they are similar to their derivatives. As you said, trigonometric functions are exponentials too (but complex), so you can tell why these functions are the solutions to these particular type of equations, when you are relating a function with its derivatives in a linear way.

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