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Given that $f: \mathbb{R} \to \mathbb{R}$ is differentiable and $f'(x)=f(x+\frac{\pi}{2})$ for all $x\in \mathbb{R}$, prove that there exists $x_0$ such that $f(x_0)=0$

I thought that $f(x)=\cos(x)$ but I don't know how to prove it. Moreover is there a solution without using the $\cos(x)$?

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    $\begingroup$ $f(x)$ could be of the form $a\sin x+b\cos x$ $\endgroup$ – Shubham Johri Feb 20 at 9:57
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    $\begingroup$ Stop trying to figure out what $f$ is, and instead think about what you can prove from what you already know about $f$. If $f(x) \ne 0$ anywhere, what would that say about $f$? (Note that $f$ is continuous.). In turn, what does that say about $f'$, given their relationship? And what does that say about $f$ again? Can these two properties of $f$ be made compatible with $f'(x) = f(x + \pi/2)$? $\endgroup$ – Paul Sinclair Feb 20 at 21:34
  • $\begingroup$ @PaulSinclair I see. if $f$ has no zero then $f$ has to be either strictly positive, or strictly negative. Say that it is strictly positive, this means also that $f'$ is strictly positive. So $f$ should be strictly increasing. could I come up with a contradiction from here? $\endgroup$ – karhas Feb 21 at 19:29
  • $\begingroup$ @ShubhamJohri could you describe a way to see this? Or provide me with some reference? $\endgroup$ – karhas Feb 21 at 19:31
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Let us consider the equation $$ f'(x) = f(x+a) $$ with $a \ge 1$ instead. Assume that $f$ has no zeros. Without loss of generality we can assume that $f$ is strictly positive.

It follows that $f$ and $f'$ are strictly increasing, so that for $x < y$ $$ \tag {*} f(y) > f(x) + (y-x) f'(x) \, . $$ In particular, $\lim_{x \to \infty} f(x) = +\infty$.

On the other hand, using $(*)$ again, $$ f'(x) = f(x + a) > f(x) + a f'(x) $$ implies $$ (a-1)f'(x) + f(x) < 0 $$

If $a=1$ then $f(x) < 0$ gives an immediate contradiction. If $a > 1$ then the last inequality shows that $$ e^{x/(a-1)} f(x) $$ is decreasing, so that $$ f(x) \le f(0) e^{-x/(a-1)} \, , $$ contradicting the fact that $\lim_{x \to \infty} f(x) = +\infty$.


Remark: If $0 < a \le 1/e$ then $e^{\lambda a} = \lambda$ has a real solution $\lambda$, and $f(x) = e^{\lambda x}$ satisfies the functional equation and has no zeros (compare How to solve differential equations of the form $f'(x) = f(x + a)$).

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  • $\begingroup$ Bravo and interesting. I was trying to find a proof for the more general equation $f^\prime(x) = f(x+a)$ with $a \gt 0$. What you did can be generalized for $a \gt 1$. The result is also true when $a=1$. What about $a \lt 1$? I don't know for the moment! $\endgroup$ – mathcounterexamples.net 2 days ago
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    $\begingroup$ @mathcounterexamples.net: It is definitely wrong for small $a$. If $e^{\lambda a} = \lambda$ has a real solution $\lambda$ then $f(x) = e^{\lambda x}$ satisfies the given equation. Accordig to math.stackexchange.com/a/61889, this works for $a \le 1/e$. $\endgroup$ – Martin R 2 days ago

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