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Basically most of us know that $\frac {\textrm{d}}{\textrm{d}x} \cos x = -\sin x$ . Also $ \cos (x+\frac{\pi}{2})=-\sin x$
That makes $$ \frac {\textrm{d}}{\textrm{d}x} \cos x = \cos (x+\frac{\pi}{2}) $$ So, out of curiosity I wondered what other functions could have this property. It's pretty interesting that by translating the graph of a function by some vector we get the graph of its derivative.
Therefore, here's my attempt:
I tried to find the defining property of this function. Here we go: $$ f'(x)= f(x+\frac{\pi}{2}) $$ That makes $$ f''(x)= \left( f'(x)\right)'= f(x+\frac{\pi}{2})'=f'(x+\frac{\pi}{2})=f(x+\pi)$$ Generally $$f^n(x)=f(x+n\frac{\pi}{2})$$ With $f^n(x) $ being the $n$-th derivative of $f$. And I'm stuck at this step. What makes cosine adapt to this property is that $ \cos (x+\pi)=-\cos x$ which allows it to be the solution of the following differential equation: $$f''(x)+f(x)=0$$ Though I don't want to admit that $f(x+\pi)=-f(x) $

So, could you please help me? Isn't there any way to determine all the functions that satisfy the previously acknowledged property? Are they infinite?

Carw Lucas
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  • Interesting observations. Did you also notice that $sin(x)$ is a solution for $f''(x)+f(x)=0$? This means that sine may have analogous properties. – mucciolo Mar 05 '17 at 16:39
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    Just to point out: $F(x)=e^{\lambda x}\implies F'(x)=\lambda e^{\lambda x}=e^{\lambda (x+\frac {\ln \lambda}{\lambda})}=F(x+\frac {\ln \lambda}{\lambda})$. Of course, there is no $\lambda \in \mathbb R$ such that $\frac {\ln \lambda}{\lambda}=\frac {\pi}2$ – lulu Mar 05 '17 at 16:45
  • Note that if $f$ satisfies this, then $f(x+a)$ satisfies this. In particular, then $\cos (x+\pi/2)=\sin x$ satisfies this equation. – Thomas Andrews Mar 05 '17 at 16:52
  • Certainly $f(x)=k$ where $k$ is some constant would have his property but I admit this isn't a very interesting example. – Karl Mar 05 '17 at 17:02

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We have:

$$\sin(x+\pi/2)=-\sin(\pi/2-x)=\cos x=(\sin x)'$$

Actually, $\cos(x+a)$ for any $a$ will work.

Applying Fourier transform rules, your original equation becomes:

$$i\omega \hat f(\omega)=e^{i\omega \pi/2}\hat f(\omega)$$

So, when $i\omega \neq e^{i\omega\pi/2}$, $\hat f$ must be zero. The set of solutions to $i\omega =e^{i\omega\pi/2}$ is discrete, and contains only $-1$ and $1$ on the real line. Are there complex solution? I don't know.

The result is that $\hat f$ must be a linear combination of delta functions at the roots $\omega_1,\omega_2,...$, an thus that $f$ is a linear combination of $e^{i\omega_k x}$, The cases $\omega=1,-1$ give $\sin x$ and $\cos x$.

(This corresponds to the comment given above, where $\frac{\ln \lambda}{\lambda}=\frac{\pi}{2}$, by setting $\lambda =i\omega$.)

Thomas Andrews
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