# Show that $\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)>0$ for $x\geq 4/\pi$

It's a follow up of Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. :

Let $$x>4/\pi$$ then we have :

$$\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)>0$$

My strategy :

For $$x\geq 2$$ we have :

$$\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)\geq 2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\geq \left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)^{\frac{1}{\sqrt{x}}}$$

Remains to show :

$$\lim_{n\to \infty}\cos\left(\frac{2}{x^{2n}}\right)-\left(\frac{x^{2n}-1}{x^{2n}+1}\right)^{\frac{1}{x^{2n}}}=0$$

Lemma 1 :

Let $$x\geq 1$$ $$\frac{\left(x^{2}-1\right)}{x^{2}+1}-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

Proof :

For the RHS we have (using the same approach as in my answer):

$$f(x)=1+\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)+\frac{1}{2}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)^{2}+\frac{1}{6}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{1}{x}-2\right)\left(\frac{x-1}{x+1}-1\right)^{3}\geq\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

And :

$$f(x)-\frac{x^{2}-1}{x^{2}+1}=-\frac{2\left(x-1\right)\left(x^{3}-2x^{2}+7x-2\right)}{3\left(x^{2}+x\right)^{3}\left(x^{2}+1\right)}\leq 0$$

Now we need to show for $$x\geq 2$$:

$$\frac{x^{4}-1}{x^{4}+1}-2\frac{x^{2}-1}{x^{4}+\frac{2}{3}+\frac{1}{3}x^{2}}-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}\geq 0$$

But for $$x\geq 2$$: $$\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}\geq g(x)=\left(\frac{x-1}{x+1}\right)^{-1}\left(1+\left(\frac{1}{x}+1\right)\left(\frac{x-1}{x+1}-1\right)+\frac{1}{2}\left(\frac{1}{x}+1\right)\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)^{2}+\frac{1}{6}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{1}{x}+1\right)\left(\frac{x-1}{x+1}-1\right)^{3}\right)$$

But :

$$\frac{x^{4}-1}{x^{4}+1}-2\frac{x^{2}-1}{x^{4}+\frac{2}{3}+\frac{1}{3}x^{2}}-g(x)=\frac{-2\left(3x^{8}-3x^{7}-x^{6}-3x^{5}+4x^{4}-6x^{3}-4x^{2}-6x+4\right)}{(3x^{3}(x+1)(x^{4}+1)(3x^{4}+x^{2}+2))}<0$$

So the RHS is shown for $$x\ge 2$$

Now the LHS :

$$h(x)=\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)-2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|$$

We have for $$x\in[0,\pi/2]$$:

$$1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\geq \cos(x)\geq 1-\frac{x^{2}}{2}$$

So for $$x\geq 4$$ we have :

$$r\left(\frac{2}{x}\right)-t\left(\frac{2}{\sqrt{x}}\right)-2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\leq 0$$

Where :

$$t\left(x\right)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24},r\left(x\right)=1-\frac{x^{2}}{2}$$

So we are done for $$x\geq 4$$.

Have you another (simpler) proof ?

Bonus/conjecture :

Let $$x>1$$ such that $$x\in[\alpha,\beta]$$ then define :

$$m\left(x\right)=\cos\left(\frac{2}{x}\right)-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

Then it seems $$\exists a,b,c\in[-\infty,\infty]$$ and $$0<\varepsilon$$ arbitrary small such that :

$$b^{-x^{a}}+c-\varepsilon<\sum_{n=1}^{\infty}\left(-1\right)^{n}m\left(x^{n}\right)

All the goal is to determine $$\alpha,\beta$$

• Why $\frac 4 \pi$ ? It should be true for any $x \geq 1$, isn't it ? Feb 7 at 11:42
• @ClaudeLeibovici It's the trivial part ;-) Feb 7 at 12:44

I am going to prove the following inequality :

$$\left(\dfrac{1-t}{1+t}\right)^{\!t}>\cos 2t\quad\forall\,t\in(-1,1)\setminus\{0\}\,.\quad\color{blue}{(1)}$$

Note that by letting $$\;t=\dfrac1x\;$$ for any $$\;x>1\,,\;$$ from $$\,(1)\,$$ we get the OP’s inequality.

Let $$\;f:(-1,1)\to\Bbb R\;$$ be the function defined as

$$f(t)=(t\!-\!2)\ln(1\!-\!t)-(t\!+\!2)\ln(1\!+\!t)\;\;$$ for any $$\;t\in(-1,1)\,.$$

The function $$\,f(t)\,$$ is twice differentiable on the interval $$\,(-1,1)\,,\,$$ moreover $$\; f(0)=f’(0)=0\,.$$

Since $$\;f’’(t)=\dfrac{4t^2}{\left(1-t^2\right)^2}>0\;$$ for all $$\;t\in(-1,1)\setminus\{0\}\;,\;$$ the function $$\,f(t)\,$$ is convex on $$\,(-1,1)\,$$.

Consequently, it results that

$$f(t)>f(0)+f’(0)\,t=0\;\;$$ for any $$\;t\in(-1,1)\setminus\{0\}\,.$$

Hence ,

$$(t\!-\!2)\ln(1\!-\!t)-(t\!+\!2)\ln(1\!+\!t)>0\;$$ for any $$\,t\in(-1,1)\!\setminus\!\{0\},$$

$$t\ln\left(\dfrac{1-t}{1+t}\right)>2\ln\left(1-t^2\right)\;\;$$ for any $$\;t\in(-1,1)\setminus\{0\}\;,$$

$$\left(\dfrac{1-t}{1+t}\right)^{\!t}>\left(1-t^2\right)^2\;\;$$ for any $$\;t\in(-1,1)\setminus\{0\}\,.\quad\color{blue}{(2)}$$

Analogously, we can prove that the function

$$g(t)=\left(1-t^2\right)^2-\cos 2t:(-1,1)\to\Bbb R$$

is convex on the interval $$\,(-1,1)\,,\,$$ moreover $$\,g(0)\!=\!g’(0)\!=\!0\,,$$

consequently,

$$g(t)=\left(1-t^2\right)^2-\cos 2t>0\;\;$$ for any $$\;t\in(-1,1)\setminus\{0\}\;,$$

$$\left(1-t^2\right)^2>\cos 2t\;\;$$ for any $$\;t\in(-1,1)\setminus\{0\}\,.\quad\color{blue}{(3)}$$

From the inequalities $$\,(2)\,$$ and $$\,(3)\,,\,$$ it follows $$\,(1)\,.$$

Maybe not quite simpler, but letting $$a=1/x$$ then $$f(a)=\left(\frac{1-a}{1+a}\right)^{a}-\cos\left(2a\right)=\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)=g(x)$$. Now we consider the Maclaurin series of $$f(a)$$, more specifically note that:

$$\left(\frac{1-a}{1+a}\right)^{a}>1-2a^2+{4\over3}a^4-{2\over5}a^6$$ and $$1-2a^2+{2\over3}a^4>\cos(2a)$$ so that: $$f(a)>1-2a^2+{4\over3}a^4-{2\over5}a^6-\big(1-2a^2+{2\over3}a^4\big)={94\over3}a^4-288a^6$$ which is positive for $$0 making $$g(x)>0$$ for $$x>1$$.

• I understand this are the terms of Maclaurin series, but how fo you get inequalities rigorously? Feb 7 at 15:29
• Not quite rigorous, but you can see it in the graph, and since polynomials are much easier understood, it can be shown there are no other intersections except when $a=0$. An interesting thought that I was trying to generalize, but I have a hard time believing is that depending what term you stop at in the taylor polynomials, the inequalities reverse direction! Feb 8 at 9:44

Same trick as before might give a somewhat shorter solution. Let $$x=2/y$$, $$0. Then $$\cos(2/x)=\cos(y)>0$$ and by monotonicity of $$\ln$$ it suffices to show that $$f(y)=\frac{y}{2}\ln\frac{2/y-1}{2/y+1} - \ln(\cos y)\ge 0.$$ Note that $$f(0)=0$$ so it will be sufficient to show that $$f'(y)\ge 0$$. Now, using Taylor series for $$\ln(1+a)=a-a^2/2+a^3/3+\dots$$, $$\frac{y}{2}\ln\frac{2/y-1}{2/y+1}= \frac{y}{2}\left[\ln(1-y/2)-\ln(1+y/2)\right]=- \sum_{k=1}^\infty \frac{y^{2k}}{(2k-1)2^{2k-1}}$$ and thus, using the expression for the Taylor series of tangents, \begin{align*} f'(y)&=- \sum_{k=1}^\infty \frac{2k y^{2k-1}}{(2k-1)2^{2k-1}}+\tan(y) \\ & =- \sum_{k=1}^\infty \frac{2k}{(2k-1)2^{2k-1}}y^{2k-1}+ \sum_{k=1}^\infty \frac{2\cdot 4^k(4^k-1)\zeta(2k) }{(2\pi)^{2k}}y^{2k-1} \\ & =\frac{y^3}6 + \frac{23y^5}{240}+\frac{227y^7}{5040}+\dots\ge 0 \end{align*} since for $$k\ge 2$$ the coefficient corresponding to $$y^{2k-1}$$ is larger in the second sum: indeed, since $$\zeta(2k)=\sum_{i=1}^\infty i^{-2k}>1$$ $$\frac{\frac{2\cdot 4^k(4^k-1)\zeta(2k) }{(2\pi)^{2k}}}{\frac{2k}{(2k-1)2^{2k-1}}}=\left(1-\frac 1{2k}\right)\frac{16^k-4^k}{\pi^{2k}}\zeta(2k)\ge \min_{k=2,3,\dots}\left(1-\frac 1{2k}\right)\left(1-\frac 1{4^k}\right)\left[\frac{4}{\pi}\right]^{2k} =\left(1-\frac 1{4}\right)\left(1-\frac 1{4^2}\right)\left[\frac{4}{\pi}\right]^{4}>1$$ while for $$k=1$$ the coefficients are the same.

• What is $f(0)=)$?
– Gary
Feb 7 at 12:46
• Probably a typo of $f(0)=0$ Feb 7 at 13:27
• Note that it is enough to show that the function $f(y)$ is convex. But $$f''(y) = \left( {\frac{1}{{\cos y}} + \frac{1}{{1 - \frac{{y^2 }}{4}}}} \right)\left( {\frac{1}{{\cos y}} - \frac{1}{{1 - \frac{{y^2 }}{4}}}} \right) > 0$$ for $0<y<\pi/2$ ($<2$), since $$\cos y < 1 - \frac{{y^2 }}{2} + \frac{{y^4 }}{{24}} = 1 - \frac{{y^2 }}{4} - \frac{{y^2 }}{4}\left( {1 - \frac{{y^2 }}{6}} \right) < 1 - \frac{{y^2 }}{4}$$ for $0<y<\pi/2$ ($<\sqrt{6}$).
– Gary
Feb 7 at 13:39
• Cool solution with factorization of $f''(y)$, Gary! (you might need to add that $f'(0)=0$ as well) Feb 7 at 15:39
• @StasVolkov Thanks. Yes, $f'(0)=0$ is implicitly used in my first sentence.
– Gary
Feb 7 at 23:14

Just a few remarks since other users gave proofs/

$$f(x)=\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)$$ is defined for $$x \geq 1$$ and does not exist in the real domain otherwise.

Let $$x=\frac 1t$$ $$g(t)=\left(\frac{1-t}{1+t}\right)^t-\cos (2 t)$$ Let $$S_n$$ to be the series expansion of $$g(t)$$ to $$O(t^{2n+4})$$. The norm $$\Phi_n=\int_0^1\Big[g(t)-S_n\Big]^2\,dt$$ $$\left( \begin{array}{cc} n & \Phi_n \\ 1 & 5.1643\times 10^{-3} \\ 2 & 2.1261\times 10^{-4} \\ 3 & 6.7153\times 10^{-7} \\ 4 & 2.7507\times 10^{-7} \\ 5 & 8.6409\times 10^{-8} \\ \end{array} \right)$$ So, a low order series represent quite well the function. Moreover, we can show that $$S_{n}$$ does not show any real root.

We could even do much better using, instead of series, the $$[4+2n,2n]$$ Padé approximant $$P_n$$ built around $$t=0$$. Concerning the norms $$\Psi_n=\int_0^1\Big[g(t)-P_n\Big]^2\,dt$$ $$\left( \begin{array}{cc} n & \Psi_n \\ 0 & 5.1643\times 10^{-3} \\ 1 & 2.4784\times 10^{-6} \\ 2 & 9.7386\times 10^{-8} \\ 3 & 2.9313\times 10^{-9} \\ \end{array} \right)$$

In the range of interest, $$S_n$$ and $$P_n$$ are $$\geq 0$$.