Do three distinct functions f, g and h exist such that f'=g, g'=h and h'=f? First year math student here taking Analysis II. Today we learned about sinh, cosh and tanh and how to take their derivatives. Our teacher, in his magnanimity, wrote on the board after fiddling with sinh and cosh that sinh'=cosh and cosh'=sinh. I noticed the oscillatory nature of this, and after commenting on it, quickly asked the question you see up above. He wasn't able to provide an answer; a friend of mine inquired on it and we ended up learning that functions whose nth derivative is itself have a special name. Yet we are no closer to finding three distinct functions that fit the bill. I would also kindly ask that if you are able to generalize this to the nth derivative please do so!
 A: The equation can be written $f'''(x) = f(x)$, that is to say, $f$ is its own third derivative. It is a case of a linear differential equation with constant coefficients. These are solved by looking for exponential solutions $f(x) = e^{\lambda x}$. This leads to the algebraic equation $\lambda^3 = 1$, hence $\lambda$ is a cubic root of unity in ${\mathbb C}$. These roots are
$1, (-1 \pm i\sqrt{3} )/2$, hence the general solution
\begin{equation}
f(x) = A e^x + \left(B e^{i x \sqrt{3}/2} + C e^{- i x \sqrt{3}/2}\right) e^{-x/2}
\end{equation}
A: The other answers show how to find solutions using roots of unity. Once you have those, linear combinations of them are also solutions.
Since
$$
e^{ix\sqrt3/2}=\cos\frac{\sqrt3}2\,x+i\sin\frac{\sqrt3}2\,x,
\qquad 
e^{-ix\sqrt3/2}=\cos\frac{\sqrt3}2\,x-i\sin\frac{\sqrt3}2\,x,
$$
taking the sum and difference of these is also a solution. We get solutions
$$
e^{-x/2}\,\cos\frac{\sqrt3}2\,x,\qquad e^{-x/2}\,\sin\frac{\sqrt3}2\,x.
$$
This allows us to write the full real general solution as
$$
f(x)=Ae^x+B\,e^{-x/2}\,\cos\tfrac{\sqrt3}2\,x+C\, e^{-x/2}\,\sin\tfrac{\sqrt3}2\,x.
$$
As long as $B$ or $C$ is not zero, this $f$ will satisfy $f'\ne f$, $f''\ne f,f'$, $f'''=f$.
A: Yes, such functions do exist. One construction uses $n$'th roots of unity. Let $\zeta_n=e^{\frac{2\pi i}{n}}$. We see then that $\zeta_n^n = 1$, and that $n$ is minimal with this property. Then the function
$f(x) = e^{\zeta_n x}$
satisfies your desired property. Indeed,
$f^{(k)}(x) = \zeta_n^k e^{\zeta_n x} = \zeta_n^k f(x) \: ,$
thus $f^{(n)}(x) = f(x)$, and all previous are distinct, as $\zeta_n^i \neq \zeta_n^j$ for $i,j<n$, $i\neq j$.
Any constant times $f$ satisfies the same property.
