I'm looking for a function $f$, whose third derivative is $f$ itself, while the first derivative isn't.
Is there any such function? Which one(s)? If not, how can we prove that there is none?
Notes:
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$x\longmapsto c\cdot e^x, c \in R$ are the functions whose derivative is itself.
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$x\longmapsto \cosh(x)={e^x+e^{-x}\over 2}$ and $x\longmapsto \sinh(x)={e^x-e^{-x}\over 2}$ have their second derivatives equal to themselves.
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$x\longmapsto f(x)$, has its third derivative equal to itself.
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$x\longmapsto \cos(x)$ and $x\longmapsto \sin(x)$ have their fourth derivatives equal to themselves.
Best Answer
$$f(x)=e^{\omega x}$$ where $\omega$ is a primitive third root of unity. We have $$f'(x)=\omega e^{\omega x}, ~~ f''(x)=\omega^2 e^{\omega x}, ~~ f'''(x)=\omega^3e^{\omega x}=e^{\omega x}$$