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The function $f = \sin x$ is the member of a family of functions closed under the operation $\frac{d}{dx}$. Are there other families of functions that are similarly closed under differentiation? Certainly $e^x$ is a trivial example, and I am aware of a similar relationship for hyperbolic trig functions $\sinh x$ and $\cosh x$.

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These are solutions to homogeneous linear differential equations with constant coefficients.

Consider a function $f(x)$ such that $f(x),f'(x),f''(x),\dots$ eventually loops back on itself or at least no longer introduces "new" functions. What this means is that a some point $f^{(n)}(x)$ is a linear combination of previous derivatives say: $f^{(n)}(x)=c_0f(x)+c_1f'(x)+\cdots+c_{n-1}f^{(n-1)}(x)$ for some constants $c_0,\dots,c_{n-1}$. Then $y=f(x)$ is the solution of the linear differential equation with constant coefficients


It is well known that solutions of such equations are linear combinations of functions of the form $x^k e^{ax}\cos(by)$ and $x^k e^{ax}\sin(by)$ where $k$ is a non-negative integer and $a,b$ real numbers.

Special cases are things like $x^3$, $e^{-2x}$, $\sin(5x)$, $x\cos(3x)$, etc.

Keep in mind that hyperbolic sines and cosines can be written in terms of exponentials so they are represented among this list.

Now if you want $f(x)$ to loop back on itself exactly, you need $f^{(n)}(x)=f(x)$. This is a solution of $y^{(n)}-y=0$ whose characteristic polynomial is $t^n-1$. If $r_1,\dots,r_n$ at the $n^{th}$ roots of unity. Then $f(x)=c_1e^{r_1t}+\cdots+c_ne^{r_nt}$ for some constants $c_1,\cdots,c_n$.

Sorry the last bit involved roots of unity (complex numbers) so that $f(x)$ is written as a linear combination of complex exponentials. I'm not sure how to push this back down into the reals without it becoming really complicated.

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Ah, of course, solutions to differential equations... thanks so much for your help! –  Greg L Nov 9 '11 at 21:05
I've seen functions like these termed as "cyclodifferential functions", in that you recover them after a cycle of $n$ derivatives. –  Guess who it is. Nov 10 '11 at 0:07

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