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Suppose $f(x)$ is a smooth periodic function with period $T$ and suppose the function crosses the $x$-axis at a finite number of points $x_i \in [0,T)$, $i=1,2,\ldots,N$, within a single period $T$, where $f(x_i) = 0$ and $f'(x_i) \neq 0 \ \forall i$.

It seems intuitively obvious to me that $N$ should be even: as the function is periodic, when it cross below the $x$-axis it must cross back at some point within the period, however I do not know how to show this mathematically.

Moreover, the number of crossings from above should equal the number of crossings from below, i.e.

$$ \sum_{i=1}^N \mathrm{sgn}(f'(x_i)) = 0.$$

Again, this is intuitive, but I do not know how I could calculate this mathematically. Any hints would be appreciated as I would like to try and derive this myself.

Hermitian_hermit
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  • In general $N$ could be infinite. For example, if you paste properly some copies of the function given in this answear https://math.stackexchange.com/questions/2205980/smooth-function-with-infinite-oscillation to get a periodic function. – Wiktor Vacca Feb 08 '22 at 19:00
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    A couple of technicalities: say $x_i\in [0,T)$ rather than $[0,T]$. Otherwise $f(x)=\sin(x)$ is a counterexample. Also, $N$ need not be finite. Consider, for example $f(x)=e^{-1/\sin^2 x}\sin(1/\sin x).$ – Plutoro Feb 08 '22 at 19:01
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    Also, $f(x)=2\sin(x)+\sin(2x)$ is technically a counterexample because while $f$ crosses the $x$ axis at $0$ and $\pi$, $\pi$ does not meet your definition for $x_i$ because $f'(\pi)=0$. Therefore, $N=1$. However, the essence of your hypothesis still holds: $f$ crosses the $x$-axis twice. You just need a definition for a crossing point that does not rely on derivatives. – Plutoro Feb 08 '22 at 19:15
  • I think a bigger problem is to define "cross". What if comes from below and touches in then goes back down. What if it hits the $x$ axis and stays equal to $0$ for an interval. I'd say the definition to "cross" must be something like. The graph crosses at $x$ if $f(x)$ and there is a $\delta>0$ where for all $w\in(x-\delta,x)$ and $y\in(x,x+\delta)$ then either $f(w) < 0 < f(y)$ or $f(w)> 0 > f(y)$. By intermediate value theorem you can prove everything you need. (Induction wouldn't hurt). – fleablood Feb 08 '22 at 19:18
  • @fleablood well I stated in my question that the function is smooth and that the derivative at the intersection is non-zero, so I think this would avoid the two examples you have provided. – Hermitian_hermit Feb 08 '22 at 19:29
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    @Matt0410 Did you mean my examples? If so, my first example is smooth and has infinitely many crossing points. In my second example, this is exactly my point. It is possible that $f(x)$ crosses the $x$ axis at a point where the derivative is zero, so this point does not get counted among the $x_i$ and so $N$ is odd when the number of crossings is even. – Plutoro Feb 08 '22 at 19:41
  • Oh... good point. Still use intermediate value to prove that for all values of $w\in (x_i, x_{i+1}$ that all values of $f(w)$ must have the same sign (do a proof by contradiction). Thenas $f'(x_i)\ne 0$ that means that for $x_{i-1} < u < x_i < w < x_{i+1}$ that $f(u)$ and $f(w)$ have opposite signs. Then do proof by induction that if $x_{i}< w<x_{i+1}$ and $x_{i+ W}< v<x_{i+W}$ then $f(w)$ and $f(v)$ must have the same sign if $W$ is even and opposite signs if $W$ is odd. – fleablood Feb 08 '22 at 19:47

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This answer assumes the following version of the question: there are $N\in\mathbb{N}$ points $x_1,\ldots,x_N\in[0,T)$ where $f(x_i)=0$ and $f'(x_i)\ne 0$, and there are no other zeros (so in particular there is no $x$ such that $f(x)=0$ and $f'(x)=0$). As mentioned in the comments to the question, if we do not make these assumptions then there are counterexamples, since a function can "cross the $x$-axis" (being first above it then below it) while having derivative $0$ at the crossing point, which messes up the count so that we may get $N$ that is odd.

Assume without loss of generality that $f(0)>0$. (The case $f(0)=0$ follows similarly; I'm leaving out the detail because you said you want to derive this yourself. Let me know if you need more detail.)

At the first crossing point $x_1$ we have $f'(x_1)\ne 0$ so by smoothness there is a neighbourhood around $x_1$ such that $f'(x)>0$ or $f'(x)<0$. Since $f(0)>0$ and this is the first crossing point we have $f'(x)<0$ for all $x$ in a neighbourhood of $x_1$. This means $f$ is decreasing there and so there is some point $p>x_1$ such that $f(p)<0$ and such that there is no zero between $x_1$ and $p$. (Such $p$ exists since we are assuming there is only a finite number $N$ of zeros in $[0,T)$.)

By periodicity $f(T)>0$, so by the intermediate value theorem there must exist some $p<x_2<T$ such that $f(x_2)=0$, so that there are at least two zeros. If there is a third one $x_3$ you repeat the argument.

This same explanation also shows that your second claim is also true, that is, that the function alternates between being decreasing and increasing at the points $x_i$.

Snaw
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  • If $f(0)=0$, or in general $f$ is $0$ on the boundary of a period then this doesn't work because you can't assume $f(0)>0$. – Wiktor Vacca Feb 08 '22 at 19:07
  • @WiktorVacca That is true, but the OP only asked for a hint. The case $f(0)=0$ follows similarly. – Snaw Feb 08 '22 at 19:09