Function whose third derivative is itself. 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:


*

*$x\longmapsto c\cdot e^x, c \in R$ are the functions whose derivative is itself.

*$x\longmapsto \cosh(x)={e^x+e^{-x}\over 2}$ and $x\longmapsto \sinh(x)={e^x-e^{-x}\over 2}$ have their second derivatives equal to themselves.

*$x\longmapsto f(x)$, has its third derivative equal to itself.

*$x\longmapsto \cos(x)$ and $x\longmapsto \sin(x)$ have their fourth derivatives equal to themselves.
 A: $y'''-y=0$
$m^3-1=0$
$(m-1)(m^2+m+1)=0$
$m=1$ or 
$m=-0.5\pm \frac{\sqrt{3}}{2}i$
if we take the second term, the solution become
$y=e^{-0.5x}( C_{1}\cos \frac{\sqrt{3}}{2}x+C_{2}\sin \frac{\sqrt{3}}{2}x)$
A: $$f(x)=e^{\omega x}$$
where $\omega$ is a primitive third root of unity.  We have $$f'(x)=\omega e^{\omega x}, ~~ f''(x)=\omega^2 e^{\omega x}, ~~ f'''(x)=\omega^3e^{\omega x}=e^{\omega x}$$
A: Try this: $f(x) = \sum_{m=0}^\infty \frac{x^{3m}}{(3m)!}$. This function is like the exponential function, which can be defined as $e^x=\sum_{m=0}^\infty \frac{x^m}{m!}$, but only taking every third term of the sum. Differentiating term-by-term verifies the desired property.
Generally, the functions $\Lambda_m^k(x) := \sum_{n=0}^\infty \frac{x^{nm+k}}{(nm+k)!}$ for $0\leq k < m$ define the set of $m$ linearly independent solutions to $\frac{d^mf}{df^m} = f$ with $\frac{d^jf}{df^j} \neq f$ for $j < m$. Again, it is easy to see this  using term-by-term differentiation, but actually computing the functions in this form is unfeasible. The theory of ODE's ensures that all solutions are linear combinations of these lambda-functions, but not all linear combinations satisfy our second condition.
Note that all the functions you mention can be realized as linear combinations of these lambda-functions: 
$e^x = \Lambda_1^0(x)$; $\cosh(x) = \Lambda_2^0(x)$; $\sinh(x) = \Lambda_2^1(x)$; 
$\sin(x) = \Lambda_4^1(x) - \Lambda_4^3(x); \cos(x) = \Lambda_4^0(x) - \Lambda_4^2(x)$
PS: If there are an concerns about convergence of these sums, just note that they are essentially the sum for the exponential function but with terms missing. Since the exponential sum is absolutely convergent for all $x$, the sub-sums are also absolutely convergent and may be differentiated term-by-term.
A: The statement means $f''' = f$. We can solve this differential equation using its characteristic equation:
$$f''' - f = 0 \implies r^3-1 = 0 \implies r^3 = 1$$
for $r \in \mathbb{C}$. So $f(x) = Ce^{rx}$. Noting that $f' = Cre^{rx}$, $f'' = Cr^2 e^{rx}$, and $f'''(x) = Cr^3 e^{rx} = Ce^{rx}$, minding that $r^3 = 1$.
A: Let $F=(f,f',f'')^T$ then we have $F'=(f',f'',f)^T$ so we get $F'=AF$ where
$$A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$
hence we get
$$F(t)=\exp(tA)F(0)$$
A: We want to solve $f''' = f$ with $f' \neq f$. Using standard methods for solving ODE with constant coefficients (some of which are reported here) one obtains
$$
f(t) = \frac{1}{3} e^{-t/2} \left(e^{3 t/2}
   \left(f''(0)+f'(0)+f(0)\right)+\sqrt{3} \sin
   \left(\frac{\sqrt{3} t}{2}\right) 
   \left(f'(0)-f''(0)\right)+\cos \left(\frac{\sqrt{3}
   t}{2}\right) \left(-f''(0)-f'(0)+2 f(0)\right)\right)
$$
Now we compute $f'(t)$ :
$$
f'(t) = \frac{1}{3} e^{-t/2} \left(\sqrt{3}
   \left(f''(0)-f(0)\right) \sin \left(\frac{\sqrt{3}
   t}{2}\right)+e^{3 t/2}
   \left(f''(0)+f'(0)+f(0)\right)-\cos \left(\frac{\sqrt{3}
   t}{2}\right) \left(f''(0)-2 f'(0)+f(0)\right)\right)
$$
From which
$$
f(t) - f'(t) = \frac{1}{3} e^{-t/2} \left(3 \left(f(0)-f'(0)\right) \cos
   \left(\frac{\sqrt{3} t}{2}\right)+\sqrt{3} \sin
   \left(\frac{\sqrt{3} t}{2}\right) \left(-2
   f''(0)+f'(0)+f(0)\right)\right)
$$
The solutions are given by the first equation with
$$
f(0)-f'(0)\neq 0 \,\, \mathrm{or }\,\, -2 f''(0)+f'(0)+f(0)\neq 0
$$
A: If you are looking for a real function, following is one:
$$ f(x) = e^{cos(\frac{2}{3}\pi)x} * cos(sin(\frac{2}{3}\pi)x) $$
Plot of f, f', f'', f''':
Plot of f, f', f'', f'''
You can verify it by: derivative-calculator.net
For more common problem: Function whose nth derivative is itself, following is one:
$$ f(x) = e^{cos(\frac{2\pi}{n})x} * cos(sin(\frac{2\pi}{n})x) $$
With n = 1, 2, 4:
$$ f(x)_{n=1} = e^{cos(2\pi)x} * cos(sin(2\pi)x) = e^x * cos(0) = e^x $$
$$ f(x)_{n=2} = e^{cos(\pi)x} * cos(sin(\pi)x) = e^{-x} * cos(0) = e^{-x} $$
$$ f(x)_{n=4} = e^{cos({\pi \over 2})x} * cos(sin({\pi \over 2})x) = e^0 * cos(x) = cos(x) $$
short explain:
$$ 
(e^{wx})' = w * e^{wx} \\
(e^{wx})^{(n)} = w^n * e^{wx} \\
$$
We known $w^n = 1$ has n roots:
$$ w = cos(\frac{2\pi*t}{n}) + isin(\frac{2\pi*t}{n}), t = 0, 1, 2...n-1 $$
let: 
$$ \begin{align}
w_1 & = cos({2\pi \over n}) + isin({2\pi \over n}) \\
w_2 & = cos({2\pi \over n}) - isin({2\pi \over n}) \\
a & = cos({2\pi \over n}) \\
b & = sin({2\pi \over n}) \\
w_1 & = a + bi \\
w_2 & = a - bi \\
\end{align} $$
we got:
$$ \begin{align}
f(x) & = {{e^{w_1x} + e^{w_2x}} \over 2} \\
     & = {{e^{(a+bi)x} + e^{(a-bi)x}} \over 2} \\
     & = {{e^{ax} * e^{bix} + e^{ax} * e^{-bix}} \over 2} \\
     & = {{e^{ax} * (e^{bix} + e^{-bix})} \over 2} \\
     & = {e^{ax} * (cos(bx) + isin(bx) + cos(-bx) + isin(-bx)) \over 2} \\
     & = e^{ax} * cos(bx) \\
     & = e^{cos(\frac{2\pi}{n})x} * cos(sin(\frac{2\pi}{n})x)
\end{align} $$
