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Show that $f(x)\ge 0$ for $0\le x \le 1$, where:

$$f(x) = \arccos(x)^2 -8x(5x^2-2) \sqrt{1-x^2}\arccos(x)+36 x^8-112 x^6+93 x^4-17 x^2$$

The endpoints are $f(0)=\pi^2/4$ and $f(1)=0$. Plotting verifies that the function is positive, but it is not monotonic, or concave. The maximum is the root of a complicated function, so it is not clear how to separate into sub-intervals. There seems to be no way to separate the function into the sum of two simpler parts that are positive.

The problem is equivalent to showing that $\log\hspace{-1pt}\big( h(r)\big)$ is concave on $(0,1]$, where:

$$h(r) = r^{-1}\Big(\pi -2 (1-2r ) \sqrt{r}\sqrt{1-r}-2 \arccos\big(\hspace{-2pt}\sqrt{r}\big)\Big),\quad 0< r \le 1 $$

Update: based on a comment, the problem can be reduced to showing that $g(x) \le 0$ for $1/2 \le x \le 1$, where:

$$g(x) = x\big(x^6-\mbox{$\frac{7}{3}$} x^4 + \mbox{$\frac{103}{72}$}x^2-\mbox{$\frac{25}{144}$}\big)\sqrt{1-x^2}+\mbox{$\frac{5}{9}$}\big(x^4-\mbox{$\frac{19}{20}$}x^2+\mbox{$\frac{7}{80}$}\big)\arccos(x)$$

Since $g$ is monotonic, this is equivalent to showing that $w(x)\ge 0$ on $[1/2,1]$, where:

$$w(x) = x(40 x^2-19)\sqrt{1-x^2}\arccos(x)-144 x^8+378 x^6-323 x^4+93 x^2-4$$

This may seem somewhat obscure, but it turns out to be important for proving a significant theorem on uniqueness of solutions to unmixing linear combinations of independent random variables. One part of the solution I posted in another question. I believe I have a solution for that, though it involves working with a piecewise function and verifying monotonicity of 60 essentially elementary functions. The other part, involving the integral over the radial direction comes down to proving that this function is positive (derived and extracted from a more complicated expression).

Does anyone know how to show that a function like this is positive, or have an idea that I could pursue? I'm asking here because I have no idea how to proceed.

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    $\begingroup$ Ha! I was going to recommend that exact edit to the title. $\endgroup$ Commented Aug 13, 2023 at 2:09
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    $\begingroup$ For $0 \le x \le 1/2$, we have $5x^2-2 < 0$, so you can lower bound $f(x) \ge \arccos(1/2)^2+(36x^8-112x^6+93x^4-17x^2)$ on $0 \le x \le 1/2$. To then establish $f(x) \ge 0$ on $0 \le x \le 1/2$, it just suffices to show $36x^8-112x^6+93x^4-17x^2 > -1$ on $0 \le x \le 1/2$, which should be elementary enough (you can even drop the $36x^8$ term). And it appears, from Mathematica, that $f(x)$ is monotone decreasing on $1/2 \le x \le 1$. $\endgroup$ Commented Aug 13, 2023 at 2:39
  • $\begingroup$ @mathworker21 Thanks, that looks like it works for $0 \le x \le 1/2$. But the monotonicity for $1/2 \le x \le 1$ (an inequality of the derivative) is back to a problem similar to original one isn't it? $\endgroup$ Commented Aug 13, 2023 at 2:45
  • $\begingroup$ I was just addressing your remark in the question: "but it is not monotonic". It sounded like the lack of monotonicity was a negative for you. And it could be that much cruder/dumber bounds work to establish the non-positivity of the derivative. $\endgroup$ Commented Aug 13, 2023 at 2:48
  • $\begingroup$ @mathworker21 I updated the problem based on your response. $\endgroup$ Commented Aug 13, 2023 at 3:02

3 Answers 3

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We have to show that \begin{equation*} g\overset{\text{(?)}}\le0 \text{ on }[1/2,1]. \end{equation*} Note that $p(x):=x^4-\frac{19}{20}x^2+\frac{7}{80}$ is of the same sign as $x-x_*$ for $x\in[1/2,1]$, where $x_*:=\frac{1}{2} \sqrt{\frac{1}{10} \left(19+\sqrt{221}\right)}=0.920\dots$.

For $x\in[1/2,1]\setminus\{x_*\}$, let \begin{equation*} h(x):=\frac{g(x)}{p(x)}. \end{equation*} It suffices to show that \begin{equation*} h\overset{\text{(?)}}\ge0 \text{ on }[1/2,x_*) \tag{10}\label{10} \end{equation*} and \begin{equation*} h\overset{\text{(?)}}\le0 \text{ on }(x_*,1]. \tag{20}\label{20} \end{equation*} For $x\in[1/2,1)\setminus\{x_*\}$, \begin{equation*} h''(x)=\frac{x h_2(x)}{28800 p(x)^3\,\sqrt{1-x^2}}, \end{equation*} where \begin{equation*} h_2(x):=-345600 x^{14}+1266240 x^{12}-1823120 x^{10}+1301740 x^8-478566 x^6+85096 x^4-5777 x^2-497<0 \end{equation*} for $x\in[1/2,1]$.

So, $h$ is convex on $[1/2,x_*)$ and concave on $(x_*,1]$.

We have $h(1)=0=h'(1)$. So, \eqref{20} follows.

Also, $h'(3/4)=0.119\ldots>0$ and hence for all $x\in[1/2,x_*)$ we have $h(x)\ge h(3/4)+h'(3/4)(x-3/4) \ge h(3/4)+h'(3/4)(1/2-3/4)=0.102\ldots>0$. So, \eqref{10} follows as well. $\quad\Box$

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  • $\begingroup$ Iosif and @mathworker12 if you send me your info I will acknowledge you in publication. $\endgroup$ Commented Aug 13, 2023 at 4:30
  • $\begingroup$ My comment seems to have been deleted somehow ... I had posted "Nice! Thanks!" $\endgroup$ Commented Aug 13, 2023 at 13:33
  • $\begingroup$ Iosif, I got your info from your page. I'll acknowledge your solution in publication. I don't understand why the @ function works sometimes and sometimes it doesn't. $\endgroup$ Commented Aug 13, 2023 at 13:35
  • $\begingroup$ @japalmer : I think there is a (strange, to me) policy against comments such as "nice, thanks". $\endgroup$ Commented Aug 13, 2023 at 16:24
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If you can bound $f'(x)$, you only need to numerically evaluate finitely many points to show that the function is positive everywhere. (Given $-A < f'(x) < A$, if $f(x_0) = C$, take $x_1 = x_0 + C/A$.)

This doesn't work at $f(1) = 0$, but showing the negativity of $f'(x)$ around that point (between the last x-value you check and x=1) would suffice. Unfortunately, $f'(1) = 0$, so you'd need to show the positivity of $f''(x)$ near $1$. $f''(x)$ is positive at $x=1$, so the original strategy will work for it. It just unfortunately means you'd have to bound $ f'''(x)$.

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  • $\begingroup$ Ah, I guess this would work for the modified problem $g(x)$ (updated above) since it is monotonic and the derivative can be evaluated at $x=1$ and bounded? $\endgroup$ Commented Aug 13, 2023 at 3:18
  • $\begingroup$ Not sure I can bound the derivative since I think that's basically the original problem $\endgroup$ Commented Aug 13, 2023 at 4:03
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Proof.

We split into two cases.

Case 1: $5x^2 - 2 \le 0$

Using $\arccos x \ge \sqrt{1 - x^2}$ for all $x\in [0, 1]$ (equivalently $y \ge \sin y$ by simply letting $x = \cos y$), we have \begin{align*} &f(x)\\ \ge{}& (1 - x^2) - 8x(5x^2 - 2)\sqrt{1 - x^2} \cdot \sqrt{1 - x^2} + 36x^8 - 112 x^6 + 93x^4 - 17x^2\\ ={}& \left( x+1 \right) \left( 36\,{x}^{5}+36\,{x}^{4}-40\,{x}^{3}+17\,x+ 1 \right) \left( x-1 \right) ^{2}\\ \ge{}& 0. \end{align*}

Case 2: $5x^2 - 2 > 0$

Using $\arccos x \ge \sqrt{1 - x^2}$ for all $x \in [0, 1]$ and $\arccos x \le \frac{\sqrt{1 - x^2}}{x}$ for all $x \in [0, 1]$ (equivalently $y \le \tan y$ by simply letting $x = \cos y$), we have \begin{align*} &f(x)\\ \ge{}& (1 - x^2) - 8x(5x^2 - 2)\sqrt{1 - x^2} \cdot \frac{\sqrt{1 - x^2}}{x} + 36x^8 - 112 x^6 + 93x^4 - 17x^2\\ ={}& \left( 36\,{x}^{4}-40\,{x}^{2}+17 \right) \left( x-1 \right) ^{2} \left( x+1 \right) ^{2} \\ \ge{}& 0. \end{align*}

We are done.

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