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With looping I mean that the identity is returned after deriving a function $n$ times.

I know the following examples:

n=1:

$$f(x)= e^x = f'(x) $$

n=2:

$$f(x) = e^{-x} = f''(x) $$

n=4:

$$ f(x) = \sin(x) = f''''(x)$$ with $$f''''(x) = f'''(\cos(x)) = f''(-\sin(x)) = f'(-\cos(x)) =\sin(x)$$

What about $n=3$ in particular? Do you know other loops that work?

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3 Answers3

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Your condition is a linear homogeneous differential equation: $$ y^{(n)} - y = 0 $$ It has the characteristic equation $$ z^n - 1 = 0 $$ with the $n$ complex unit roots $\lambda_k = e^{2\pi k i/n}$, $k \in \{0,\dotsc, n-1\}$, as solutions.

As a linear differential equation, the general solution is a linear combination of solutions. In this case $$ y(x) = \sum_{k=0}^{n-1} C_k e^{\lambda_k x} $$

mvw
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The solution to

$$f^{(n)}(x)=f (x)$$

Is given by

$$f (x)=\sum C_ke^{2\pi ki x/n},\ \forall\ k\in\mathbb N,\ 0\le k<n$$

$C_k $ are constants.

Simply Beautiful Art
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  • The real and imaginary parts are already examples of such linear combinations. For example $\cos x = \frac{1}{2} e^{ix} + \frac{1}{2} e^{-ix}$ for the case $n=4$. – Najib Idrissi Nov 20 '16 at 19:02
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$$f(x)=\sum_{k=0}^∞ \frac{x^{kl+m}}{(kl+m)!}, 0â‰Īm<l$$
is equal to all its various order derivatives, including third.

gfppoy
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