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I have to prove this inequality:

$$\cos^{10}x\le 1-x^2 \quad \text{for all}\quad x \in [0,0.5]$$

Is there any easy and/or elegant way to do this? I can do this with Taylor, but it's really a mess. :(

Thanks in advance

Blue
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Logic_Problem_42
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  • "Elegant" might be hard to get, since $0.5$ (radians) isn't a nice angle range. – coffeemath Apr 17 '18 at 09:58
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    It actually holds for all $x$ from $[-0.99,0.99]$, which are even uglier numbers :) – Logic_Problem_42 Apr 17 '18 at 10:02
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    The right hand side of the equations may suggest trying to compare the cosine to its Taylor truncation, for example $f(x) = 1-\frac{x^2}{2}$. Then we see that $1-x^2 \leq 1- \frac{x^2}{2} \leq \cos{x}$ ... – Matti P. Apr 17 '18 at 10:06
  • Can't we use MVT? – John Glenn Apr 17 '18 at 13:40
  • It suffices to prove that $\cos^{3}x\le 1-x^2$, which on the surface seems easier. – lhf Apr 17 '18 at 14:01
  • @Logic_Problem_42 Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 19 '18 at 19:37

4 Answers4

2

Note that

$$\cos(x)^{10}\le 1-x^2\iff 10\log\cos x\le\log (1-x^2)$$

and

  • $10\log\cos x\le 10 \log (1-\frac12x^2+\frac1{24}x^4)\le-5x^2+\frac5{12}x^4\quad$ by $\log(1-x)<x$
  • $\log (1-x^2)\ge-x^2-x^4$ to be proved

and since

$$-5x^2+\frac5{12}x^4\le -x^2-x^4 \iff4x^2-\frac{17}{12}x^4\ge 0\iff x^2(4-\frac{17}{12} x^2)\ge 0\\\iff -\sqrt{\frac{48}{17}}\le x\le \sqrt{\frac{48}{17}}$$

the inequality is proved.

To prove

  • $\log (1-x^2)\ge-x^2-x^4$

let consider

  • $g(x)=\log (1-x^2)\ge+x^2+x^4$

and note that

  • $g(0)=0$
  • $g'(x)=\frac{2x^3-4x^5}{1-x^2}$ and $g'(x)>0$ for $x\in(0,1/2)$

and thus $g(x)\ge0$ for $x\in[0,1/2]$.

user
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2

Here is a partial answer.

We have $\cos x \leq 1 - \frac{x^2}{2} + \frac{x^4}{24}$ for all $x$ because the cosine series is alternating.

Therefore, to suffices to prove that $\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)^{10} \le 1 -x^2$ for $x \in [0,0.5]$.

It is easy to verify that $f(x)=1 -x^2-\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)^{10} $ has a local minimum at $x=0$ and that $f(0)=0$.

The hard part is to prove that $f(x)\ge 0$ for $x \in [0,0.5]$. As we can see in the graph below, the local maximum is after $x=0.5$, but this seems harder to prove.

enter image description here

lhf
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  • Actually, it suffices to prove that $\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)^{3} \le 1 -x^2$, or equivalently, that $x^{10} - 36 x^8 + 504 x^6 - 3456 x^4 + 12096 x^2 - 6912\le0$ for $x \in [0,0.5]$. – lhf Apr 18 '18 at 00:55
2

In order to prove $$ 10\log\cos x\leq \log(1-x^2)\qquad \text{for }x\in\left[0,\tfrac{1}{2}\right]$$ it is enough to show $$ 5 \tan x \geq \frac{x}{1-x^2} \qquad \text{for }x\in\left[0,\tfrac{1}{2}\right] $$ then use termwise integration. On the other hand $5(1-x^2)\tan(x)$ is a (log-)concave function on the given interval, hence its graph lies above the secant line through $(0,0)$ and $\left(\frac{1}{2}, \frac{15}{4}\tan\frac{1}{2}\right)$: $$ 5(1-x)^2 \tan(x) \geq \frac{15}{2}\tan\left(\frac{1}{2}\right) x\geq\frac{15}{4}x\geq x $$ and the original inequality (which turns out to be pretty loose far from the origin) is proved.

Jack D'Aurizio
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1

Partial answer.

It equal to $(1-\sin^2x)^5\leq 1-x^2$.

$LHD≒1-5\sin^2x<1-x^2$, at $x≒0$.

Since $\sin x \leq x $, it must hold at $x=0.5$.

$LHD≒0.27$

$RHD=0.75$

Takahiro Waki
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