Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\cosh x$.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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

$$y^{(n)}-c_{n-1}y^{(n-1)}-\cdots-c_1y'-c_0y=0$$

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.

share|improve this answer
    
Ah, of course, solutions to differential equations... thanks so much for your help! –  Greg L Nov 9 '11 at 21:05
1  
I've seen functions like these termed as "cyclodifferential functions", in that you recover them after a cycle of $n$ derivatives. –  J. M. Nov 10 '11 at 0:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.