Solve for $f(x)$ and for $n$ related to consecutive derivatives

Question

Seeing that the sine derivative returns to be the same function every $4$ derivatives, it occurred to me to ask myself the following 2 problems

Statement 1: Get all the functions such that $\large{\frac{d^n}{dx^n}f } =f(x),$ with $n = 4$

For this statement i know that for $f(x) = \sin x$ holds, but how i can get others functions that holds? Is there a theorem related to this? And what about of the functions for other values of $n$ ?

Analogous statement 2: If is possible, get $n$ such that $\large{ \frac{d^n}{dx^n}f = f(x)}$ with $f(x) = \sin x$

Obviously, without derivative consecutively and for other possible functions.

So, are some theorems or ways to solving these problems?

@SamuelA.Morales It should be multiplied by a $c\in \mathbb{R}$, though. — Sage Stark, Jul 05 '20 at 00:33

score 1 · Answer 1 · answered Jul 05 '20 at 00:29

This is just solving a fourth order ODE: $$y^{[4]}(x)-y(x)=0$$ We guess a solution of the form $y(x)=Ce^{rx}$: $$Cr^4e^{rx}-Ce^{rx}=0$$ $$r^4=1$$ We are looking for the fourth roots of unity in $\mathbb{C}.$ They are $1,-1,i$, and $-i$. Therefore a general solution is a linear combination of these four linearly independent solutions. $$y(x)=c_1e^x+c_2e^{-x}+c_3e^{ix}+c_4e^{-ix}$$

Now we use Euler's formula $e^{i\theta}=\cos(\theta)+i\sin(\theta)$: $$y(x)=c_1e^x+c_2e^{-x}+c_3(\cos(x)+i\sin(x))+c_4(\cos(x)-i\sin(x))$$ In order for $y$ to be real valued, it must be the case that $c_4={c_3}^*$. Let $c_3=\frac{a}{2}+i\frac{b}{2}$. Then, $$y(x)=c_1e^x+c_2e^{-x}+a\cos(x)-b\sin(x)$$ Renaming our variables, $$y(x)=Ae^x+Be^{-x}+C\sin(x)+D\cos(x)$$ @Samuel A. Morales did mention that these four fundamental solutions are the only linearly independent solutions that exist to these fourth order ODE, but any linear combination of these four fundamental solutions will also have the property you seek. It's also fairly easy to see how this generalizes to higher derivatives.

@Eduardo S. Well it worked didn't it? I found four linearly independent solutions. Since this is a fourth order ODE, there cannot be any others. — K.defaoite, Jul 05 '20 at 20:04

score 0 · Answer 2 · answered Jul 04 '20 at 23:58

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Notice how in order for this to work the derivatives must belong to a cycle, from my knowledge the only ones that do this are:

$$f(x)=e^x\\ j(x)=e^{-x}\\ g(x)=\sin(x)\\ h(x)=\cos(x)$$

I do not know any theorem regarding this, I came up with the solutions using intuition, such as noting that $x^\alpha$ will not work, neither any $\log_\alpha{x}$, nor $\alpha^x$

answered Jul 04 '20 at 23:58

Samuel A. Morales

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what about a combination of these functions? – ESCM Jul 05 '20 at 19:55
You are totally right, this makes the question have infinetly many solutions – Samuel A. Morales Jul 05 '20 at 20:06

Solve for $f(x)$ and for $n$ related to consecutive derivatives

2 Answers2