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This conjecture is false. A nice counter example is given by David Speyer in the comments.

Yesterday I imagined up a nice little conjecture on real trigonometric polynomials. Much to my surprise, I couldn't find any such thing on googling the net. I don't know what are the implications or importance (if any) of this conjecture, if it were correct. I'd like to know whether the conjecture is true, and also a proof. In case its already treated in the literature, please give a pointer to a source.

Conjecture

Let $f_N(x)$ be a real trigonometric polynomial of degree $N$. Let $$P = \{x_p|x_p \in (0,2\pi) \land (f^'_N(x_p) = 0 \lor f^''_N(x_p) = 0)\}$$ Let $$B = \{(x_i,x_j)| x_i,x_j\in P \wedge x_i\ne x_j\}$$ Then $$\min\limits_{(x_i,x_j)\in B} |x_i-x_j| \ge \frac{\pi}{2N}$$


Note : $f^'_N(x)$,$f^''_N(x)$ are respectively, the first and second derivatives of $f_N(x)$


PS : Please let me know if you find this theorem interesting or useful in any part of Mathematics.


EDIT 1 : Adding a short summary of the statement in words as requested by Chris Taylor to avoid any possible confusion to the reader while parsing the symbols.

Let $P$ be the set of all points where either the first derivative or second derivative of a given real trigonometric polynomial of degree $N$ vanish. The conjecture is that the shortest of the distances between any two distinct points in $P$ is at least $\frac{\pi}{2N}$. I hope this summary in words is accurate. If not please let me know the mistakes.

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    One does not call it a theorem unless it had been proved.2012-05-23
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    @Sasha : theorem changed to conjecture.2012-05-23
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    or if it was the last unproved claim written by a famous amateur mathematician.2012-05-23
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    @Dustan Levenstein : lol. my bad, sorry for the mistake.2012-05-23
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    Is your conjecture "The smallest distance between distinct maxima, minima or inflection points of an $n$th degree trigonometric polynomial lying in $(0,2\pi)$ is at least $2\pi/n$"?2012-05-23
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    @ChrisTaylor : yes, the points where either first or second derivative is zero, but its $\frac{\pi}{2N}$. please check the question.2012-05-23
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    It's often considered good style to write out the meaning of your symbolic expressions in words, rather than leave them as symbols - it avoids your readers from having to parse the symbols as words themselves (and potentially make mistakes, as I did). Of course, you need to ensure that the human-readable form is unambiguous...2012-05-23
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    @ChrisTaylor : I agree. I'll try and add a summary.2012-05-23
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    Am I confused? This seems very false. Consider $(4 - \epsilon) \cos \theta - \cos (2 \theta)$. This is a trig. polynomial of degree $2$ with critical points at $0$, $\pi$ and $\pm \cos^{-1} (1-\epsilon/4)$. By choosing $\epsilon$ sufficiently small (and positive) I can make the three critical points near $0$ as close as I want.2012-05-23
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    @David Speyer : Thank you very much for your nice counter example. I am really sorry for the goof up and for wasting your time. I'd like to delete this question after a while with your permission.2012-05-24
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    @RajeshD Please don't delete it. Other people might make the same conjecture and find this post useful.2012-05-24
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    @David: if Rajesh decides to not delete his question, you may want to post that comment as an answer.2012-05-24
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    @David Speyer : I do not want to delete this question.Please consider adding your comment as an answer so that I could accept it.2012-05-27

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Adding this answer by popular request:

The degree $2$ trignometric polynomial $(4-\epsilon) \cos \theta - \cos (2 \theta)$ is a counter-example. It has critical points at $0$, $\pi$ and $\pm \cos^{-1} (1-\epsilon/4)$; by choosing $\epsilon$ small enough, I can make the zeroes near $0$ as close as I like.

Found by thinking first "there is usually no obstacle making roots of a polynomial close together" and then by thinking "well, where are the roots of $\frac{d}{d \theta} (a \cos (2 \theta) + b \cos (\theta))$?"