A function returning to "itself" after p differentiations, and no fewer?

Question

Let D be the differentiation operation restricted to scalar real functions, f:R->R.
So, D(f) = f', the first derivative of f
D^2(f) = D(D(f)) = f'', the second derivative of f, and so forth so that we understand D^p(f) to mean the p^th derivative of f.

What I am wondering about are sort of "irreducible" f that return to themselves after exactly p differentiations for prime p, and *no fewer* than p differentiations.

E.g. Clearly the exponential function f(x) = e^x satisfies D(f) = f and so would be a trivial answer for any p without the restriction of "irreducibility" (I don't know if that's the right word, but it intuitively makes sense). Let me formalize exactly:

Let p be an arbitrary prime number > 1. Define the set of functions which satisfy the following properties:
(i) D^p(f) = f, and
(ii) For any 0 < q < p, D^q(f) does NOT equal f

If such a function f:R->R does not exist for certain p > 1, then why?

Since I am more of a "discrete" mathematics person, I have no idea of what is the hardness of this problem. I apologize in advance if it is either too easy or too hard (either way let me know). Since I asked this question to some mathematics colleagues to no immediate avail, I am guessing it is not trivial. On the other hand, it seems *in my way of thinking* (which admittedly is somewhat different than some others!) that it should be something fundamental in the theory of differential equations, which unfortunately I am not well-versed in. Anyone shed some light, please, and thanks!

Answer 1

Your intuition is very good on this one. e^(x) is the only class of functions that can ever satisfy your constraints. However, you should be considering all functions of the form e^(kx) for all values of k.



Take for example, p = 2.



(k=1) e^x does NOT satisfy condition (ii), but:

(k=-1) e^-x does satisfy condition (i).



D(f) = D(e^-x) = -e^-x.

D²(f) = D²(e^-x) = e^-x = f



Specifically, we are looking for functions such that:

D^p(f) = f, and for the class of functions e^kx,

D^p(e^kx) = e^kx

Differentiating,

k^p e^kx = e^kx

Which leads us to:

k^p = 1.

What you are looking for then, is

k = 1 ^ -p



At first glance, this looks like k=1 is the only solution. But wait! We were able to use k=-1 earlier for p=2. So then do we only care about k = ±1?



If you are versed in complex number theory, you will see that the solution to this problem is straightforward.



For each value of p, there are p solutions to k = 1 ^ -p:

k = cos( 2*pi* n/p ) + i sin( 2*pi* n/p )

is a solution to k = 1 ^ -p for each value of n. Divide up the unit circle in the complex plane into p equal segments. There will be p such unique complex numbers k that satisfy this equation.



Since p is prime, we only need to exclude the case for k = 1 to avoid violating your proprty (ii). Therefore, for each value of p > 1, there are p-1 functions that satisfy (i) and (ii), these functions being specifically:

f = e^kx

f = e^( c + i * s )x

f = exp( x * c ) * exp( x * i * s ) where c = cos( 2*pi* n/p ) and s = sin( 2*pi* n/p ).



And if you want to apply Euler's formula,

f = exp( x * c ) * ( cos( x * s ) + i * sin( x * s ) )

Answer 2

There's lots of functions one can construct that have the property f^n(x) = f(x). For example, for n = 3, we can have the following function:



e^(-x/2) Cos(((√3)/2)x)



The exponential function is the ideal starting point for checking into these kinds of functions because d/dx (e^x) = e^x. Looking at the series expansion of e^x, you can see how one can construct a number of functions that meets f^n(x) = f(x) by dropping all the terms except those which powers increases by n. And then you play around with adding such functions together with arbitrary coefficients to create any number of functions. Hence, you can have a group of functions that meets the condition f^n(x) = f(n) for a given n.

Answer 3

This answer is motivated by the answer of dgr, but sticks strictly to real valued functions of a real variable. These are related to, but are in fact different functions from the real parts of the functions described by dgr. The functions are defined by power series that converge for all real x. For each n, fn(x) returns to itself after n differentiations, but not fewer.



f1(x) = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + . . .



f2(x) = 1 + x^2/2! + x^4/4! + x^6/6! + x^8/8! + . . .



f3(x) = 1 + x^3/3! + x^6/6! + x^9/9! + x^12/12! + . . .



f4(x) = 1 + x^4/4! + x^8/8! + x^12/12! + x^16/16! + . . .



f5(x) = 1 + x^5/5! + x^10/10! + x^15/15! + x^20/20! + . . .



etc. I'm sure you see the pattern. Here is why it works: Take for example f3(x). It is clear that each term, if differentiated 3 times turns into the term in front of it. So the series returns to itself after 3 differentiations. Similarly, for each series, as another example, try f5(x). Each term if differentiated 5 times turns into the term in front of it.



These series can be expressed in closed form. f1(x) = e^x, f2(x) = cosh(x), f3(x) = (1/3)e^x + (2/3)e^(-x/2) cos(sqrt(3) x/2), and after that it gets quite messy.



Finally, although I presented these as real functions of a real variable, they live just as happily as analytic functions of a complex variable.



See also:

https://answersrip.com/question/index?qid=20081012094720AAP4pfA&show=7#profile-info-ovYzQfv0aa

Answer 4

This is an incredible problem that most of the standard books on calculus have omitted. I think the solution involves product of some function and an exponential function ...

any way thank you for you have given me something to think about today...

I will try my best to solve

Answer 5

If w is a primitive nth root of 1, for example w = cos(2pi/n) + i*sin(2pi/n), then the function f(x) = e^(wx) satisfies the equation D^n(f) = f. This is not a real-valued function, but the real part of this function satisfies the same equation.