What functions have the property that $\frac{d}{dx}f(x) = c \cdot f(x+1)$?

Question

If we are allowed to pick any real-valued constant $c$ that helps, when does

$$\frac{d}{dx}f(x) = c \cdot f(x+1)$$

In other words, when does the derivative of a function $f(x)$ equal some constant times $f(x)$?

EXAMPLE

If we set $f(x) = \sin{ (\frac{\pi}{2}x) }$, then

$$\frac{d}{dx} \left( \sin (\frac{\pi}{2}x) \right) = \frac{\pi}{2}\sin (\frac{\pi}{2}(x+1)) $$

This also works for $f(x) = \cos (\frac{\pi}{2}x) $:

$$\frac{d}{dx} \left( \cos (\frac{\pi}{2}x) \right) =\frac{\pi}{2} \cos (\frac{\pi}{2}(x+1)) $$

WHAT I'M AFTER

Is there some way of finding formulas that possess this property?

Perhaps there is a way to formalize a method. I'd like to know if there's an easy way to find formulas that work.

Answer 1

If we let $f\in \mathcal{C}^{\infty}$, then, we can use $f(x+1)$'s Taylor expansion around $1$ to get $$f'(x)=\sum_{k\ge 0}f^{(k)}(1)\frac{x^k}{k!}=\sum_{k\ge 0}f(k+1)\frac{(cx)^k}{k!}\\ \Rightarrow f(x)=\sum_{k\ge 0}f(k)\frac{(cx)^k}{k!}$$ So, for example taking $$f(k)=\left\{\begin{array}{rl} 0 & k\ \mbox{even}\\ (-1)^k& k\ \mbox{odd} \end{array} \right.$$ gives $f(x)=\sin(cx)$ similarly you'll get $f(x)=\cos(cx)$ by taking $$f(k)=\left\{\begin{array}{rl} (-1)^k & k\ \mbox{even}\\ 0& k\ \mbox{odd} \end{array} \right.$$ with $c=\pi/2$.

Answer 2

For fixed $c,$ if $a / e^a = c,$ then $e^{ax}$ works. Linear ODE, so you can add solutions with the same $c.$

Errands to run, look at it later. Possible benefit to allowing complex $a.$ Yes, try $a=i \pi / 2$ or similar. Right, next try $a=-i \pi / 2.$ You can add or subtract these, divide by $i$ if needed, no harm in constant multiple. I think sums of these give all; just a guess for now.

EEDDIITT: The holomorphic function $$g(z) = \frac{z}{e^z}$$ is far from injective. for example, $g(i \pi / 2) = \pi/2$ and $g(-i \pi / 2) = \pi/2.$ So, to get the original value of $c = \pi/2,$ we can take $$ A e^{i \pi x / 2} + B e^{-i \pi x / 2} $$ with real or complex constants $A,B.$ As $$ e^{i \pi x / 2} = \cos \frac{\pi x}{2} + i \sin \frac{\pi x}{2}, $$ we can get your original examples as $$ \left( e^{i \pi x / 2} + e^{-i \pi x / 2} \right) / 2, $$ then $$ \left( e^{i \pi x / 2} - e^{-i \pi x / 2} \right) / (2i). $$

There should be a countably infinite set of solutions to $$ \color{magenta}{ \frac{z}{e^z} = \frac{\pi}{2}}, $$ occurring in conjugate pairs. Information about these can be gathered from the Lambert W Function, because $$ \color{magenta}{ -z e^{-z} = -\frac{\pi}{2}}, $$ so we can take $v=-z$ and solve $$ \color{magenta}{ v e^v = -\frac{\pi}{2}}. $$ Compare the first example in VALUES. Evidently MATLAB allows you to demand the non-principal branches, one at a time. I suspect that if you actually own Mathematica you can evaluate ProductLog[1, -Pi/2] but it does not seem to cooperate on Alpha. Oh, well. Once i made a Newton's method thing for tyhe complex plane, maybe i can find the first non-principal (conjugate pair of) value myself, numerically at least. Matter of carefully finding an approximate location first...

EEDDIITT worked this time ONE, $$ v_1 = -1.60429 \pm 7.64719 i ; \; \; \; z_1 = 1.60429091344801 \pm 7.64719227612459 i $$ TWO $$ v_2 = -2.19834 \pm 13.981208 i ; \; \; \; z_2 = 2.1983426299819 \pm 13.981208306240 i $$

Answer 3

The infinitesimal generator corresponding to the translation operator is differentiation:

$$ e^{t \frac{d}{dx}} f(x) = f(x+t). $$

That is, if we write $D = d/dx$ then

$$ D f = c e^{D} f. $$

So at least formally, $f$ is an eigenfunction of the operator $ D - ce^{D} $ corresponding to the eigenvalue $0$.

Instead, we begin by making an ansatz. That is, let us try $f = e^{kx}$. Then what we want to solve is now the following equation.

$$ 0 = (D - ce^{D})e^{kx} = (k - ce^{k}) e^{kx} \quad \Longrightarrow \quad k - ce^{k} = 0. $$

This equation can be solved using the Lambert W-function, and any complex solution to this equation leads to a solution to our original equation. This is the very approach that Will Jagy adopted.

Edited. I should correct my answer in that functional analysis does not seem to provide enough power to analyze the solution. Sorry about this confusion.

Answer 4

By way of an extended comment, consider the following. By putting $f(x)=g(cx)$ we can convert the equation to $c g'(cx) = c g(cx+c)$ or with $cx \to x$, $$g'(x) = g(x+c).$$

See this question for how to find all solutions of $g'(x)=g(x+1)$, a very similar problem.

Here's a more convincing reason why Will Jagy's ansatz is probably the only solution. Take the Fourier transform: $$\mathcal{F}[g'(x)] = \mathcal{F}[g(x+c)]$$ implying $$-2 \pi \xi i G(\xi) = e^{2 \pi i \xi c} G(\xi)$$ implying $G(\xi) = 0$ except for when $\xi = -e^{2 \pi i \xi c}/{2 \pi i}$. For one such solution $\xi = a$, then $$G(\xi) =A \delta(\xi - a)$$ which goes back to $$g(x) = A e^{2 \pi i a x}$$ as a solution. So $f(x) = g(cx) = A e^{2 \pi i c a x}$.