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I have the inequality $\cos^2(−\frac{\pi}{4}+\frac{\pi}{p})>\frac{1}{2}+\frac{π}{p^2}$​​ and have found that it holds for $p>p^*$, where $p^*$ is some positive number (around ~2.8). I'm looking for a way to determine $p^*$ and to prove the inequality for $p$ greater than this threshold. Any suggestions on approaches or relevant theorems would be greatly appreciated.

Thank you!

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    $\begingroup$ You can differentiate the difference, show it's positive for positive $p$, and solve $\cos^2(\pi(\frac1x-\frac14))=\frac12+\frac\pi{x^2}$ numerically. It doesn't seem like this root has a closed form. $\endgroup$
    – Daniel Weber
    Commented Nov 3, 2023 at 11:00
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    $\begingroup$ Actually the derivative is not always positive. It becomes negative for $p>5.3$ approximately. Asymptotically the function is $\pi/p$. $\endgroup$
    – Brendan McKay
    Commented Nov 3, 2023 at 12:05
  • $\begingroup$ You may also write it as $\sin(\frac{2\pi}{p})>\frac{2\pi}{p^2}$ or $\sin(2\pi x)>2\pi x^2$, seeing their graphs could be clearer. $\endgroup$
    – Toni Mhax
    Commented Nov 3, 2023 at 13:00
  • $\begingroup$ Would a series solution work? $\endgroup$
    – Тyma Gaidash
    Commented Nov 4, 2023 at 0:32

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Sketch:

Write $p=1/x$, so that you want to show $f(x) = \sin(\pi/4+\pi x)-1/2-\pi x^2\ge 0$ for $0\le x\le 0.355$ approximately.

First, $f''(x)<0$ in $[0,1/2]$ (by a large margin), so $f'(x)$ is strictly decreasing in $[0,1/2]$. Moreover, $f'(0)>0$ and $f'(1/2)<0$, so $f(x)$ is strictly increasing up to a maximum, then strictly decreasing. Since $f(1/2)<0$, there is a single point in $(0,1/2)$ where it equals 0. You have to find that point numerically as there is almost certainly no closed expression. I get $x=1/2.8185102203577819$.

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  • $\begingroup$ Why do you (confidently) say there is no closed expression? $\endgroup$
    – mathworker21
    Commented Nov 3, 2023 at 13:21
  • $\begingroup$ I guess it is equivalent to find $X$ in $\sin(X)=aX^2$ for some $a$ see my comment. It seems to be in an approximative way. $\endgroup$
    – Toni Mhax
    Commented Nov 3, 2023 at 14:39
  • $\begingroup$ @mathworker21 I added "almost certainly". Equations like this only have closed-form solution in special cases, but admittedly I'm only 99% sure this is not one of them. $\endgroup$
    – Brendan McKay
    Commented Nov 4, 2023 at 0:30

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