14

My problem is to show that $$\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$$ for all $x\in\Bbb R$.

I was thinking of first finding the max and then show that its less than $1/\pi$. But it is hard to find it. I get that the series is equal to $$f(x)=\frac{x-\sin x}{x^2}.$$ Then, $$f'(x)=\frac{1-\cos x}{x^2}-\frac{2(x-\sin x)}{x^3}=0$$ if and only if $$2\sin x=x(1+\cos x),$$ which I am unable to solve, appart from the obvious solutions $x=0$ and $x=\pi$. But if $x=\pi$ is the max, then we are done because $f(\pi)=1/\pi$.

Martin Sleziak
  • 53,687
Pas gêné de poser une question
  • 141
  • You can rearrange that trig equation by using $\sin x = 2\sin(x/2)\cos(x/2)$ and $\cos x = \cos^2(x/2) - \sin^2(x/2)$, resulting in $\tan(x/2) = x/2$. You then might be able to deduce something about the other solutions. – wltrup Aug 17 '15 at 17:52
  • The function $g(x)=tan(x)-x$ has derivate $tan^2(x)$, so in the interval $[0,\frac{\pi}{2})$, $g(x)$ is strictly increasing. This shows that the only root of $g(x)$ in the given interval is $0$. Maybe this helps. – Peter Aug 17 '15 at 18:33
  • Possibly relevant: Prove that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$ – Bart Michels Aug 17 '15 at 18:39
  • I just searched for 'sin inequality' hoping to find some questions about inequalities of that kind. There may be others. – Bart Michels Aug 17 '15 at 18:40
  • For $x\ge 4$, we have $f(x)<\frac{x+1}{x^2}<\frac{5}{16}<\frac{1}{\pi}$ because the function $\frac{x+1}{x^2}$ strictly decreases for $x>0$. – Peter Aug 17 '15 at 18:48
  • For $x<-1$, it is clear that $f(x)$ is negative. That means, we need only consider the interval $[-1,4]$ and together with my argument for $g(x)=tan(x)-x$, it should be no problem to check it. – Peter Aug 17 '15 at 18:51
  • For $0\leq x$ we have $\sin x>x-\frac{x^3}6$ so for $x\leq\frac\pi6$: $\frac{x-\sin x}{x^2}<\frac x6\leq\frac1\pi$. For $x\in[0,\frac\pi2]$ we have $\sin x\geq\frac2\pi x$ so for $x\in[\pi-2,\frac\pi2]$: $\frac{x-\sin x}{x^2}\leq\frac{2-\pi}{\pi x}\leq\frac1\pi$. For $x\geq\pi$ we have $\sin x\geq\pi-x$ so $\frac{x-\sin x}{x^2}\leq\frac{2x-\pi}{x^2}\leq\frac1\pi$ (The last inequality is just $(x-\pi)^2\geq0$.) It remains to do $[-1,0]$, $[\frac\pi6,\pi-2]$ and $[\frac\pi2,\pi]$. – Bart Michels Aug 17 '15 at 18:58

3 Answers3

4

Using $f(x)=\frac{x-\sin(x)}{x^2}$, the inequality is equivalent to: $$ \frac{x-\sin(x)}{x^2}≤\frac{1}{\pi}\iff x\left(1-\frac{x}{\pi}\right)≤\sin(x) $$ Thus, define $g(x):=x\left(1-\frac{x}{\pi}\right)$.

Note that $g(x)=g(\pi-x)$ and $\sin(x)=\sin(\pi-x)$. Therefore it suffices (by symmetry) to prove the inequality for $x\in\left(-\infty;\frac{\pi}{2}\right]$.

Firstly, we prove it for $x\in\left[0;\frac{\pi}{2}\right]$:

We have $g(0)=\sin(0)=0$, thus it is sufficient to prove, that $g'(x)≤\sin'(x)$ for $x\in\left[0;\frac{\pi}{2}\right]$. Now: $g'(x)=1-\frac{2x}{\pi}$ and $\sin'(x)=\cos(x)$. So it remains to prove: $$ 1-\frac{2x}{\pi}≤\cos(x) $$ Since $\cos''(x)=-\cos(x)≤0\space\forall x\in\left[0;\frac{\pi}{2}\right]$, $\cos(x)$ is concave on the interval. As $g'(0)=\cos(0)=1$ and $g'\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0$, $g'(x)$ connects two points on the graph of $\cos(x)$, and together with the concavity, this implies the above inequality.

Secondly, we prove it for $x\in\left(-\infty;0\right]$:

By the same argument as above, we just have to show, that on this interval $g'(x)≥\sin'(x)\iff 1-\frac{2x}{\pi}≥\cos(x)$ holds. This is true because $1-\frac{2x}{\pi}≥1≥\cos(x)$ for all $x\in\left(-\infty;0\right]$

So the inequality is indeed true.

Redundant Aunt
  • 12,030
  • 2
  • 22
  • 66
0

Note that $f$ is odd. Thus, it suffices to look at the interval $[0, \infty)$.

Lemma: For $x$ > $\pi$, we have $$\sin(x) > \frac{(\pi^2-x^2)x}{\pi^2+x^2}.$$

Proof: See here.

Thus, for $x > \pi$, we have

$$ \frac{1}x - \frac{\sin(x)}{x^2} < \frac{1}x + \frac{\pi^2-x^2}{x(\pi^2+x^2)} := g(x).$$

For $x > \pi$ (or even $x > 0$), the RHS is monotonically decreasing. Thus, we can conclude that for $x \in (\pi, \infty),$ we have $$f(x) < g(\pi) = \frac{1}{\pi}.$$

Now we will find the maximum of $f$ on $[0, \pi]$. First, we check the endpoints, $$\lim_{x \rightarrow 0} f(x) = 0$$ and $f(\pi) = \frac{1}{\pi}$.

Now we find all the extremas on our desired interval. Taking the derivative and setting it equal to $0$ gives $$2\sin(x)-x \cos(x) - x = -2 \cos(x/2)(x \cos(x/2)-2\sin(x/2)) = 0.$$ The only root of the first factor in our interval is at $x = \pi$. Now we consider $$x \cos(x/2)-2\sin(x/2) = \sqrt{x^2+4}\sin(x/2-\tan^{-1}(2/x)) = 0.$$ The roots of $\sin(\theta)$ are at multiples of $\pi$. Note that $$(x/2-\tan^{-1}(x/2))' = \frac{x^2+8}{2x^2+8}$$ so this function is strictly increasing for $x \ge 0$. Furthermore, $$\lim_{x \rightarrow 0^{+}} \frac{x}2- \tan^{-1}(2/x) = - \frac{\pi}2$$ and $$\frac{\pi}2 - \tan^{-1}(\pi/2) \approx 1$$ so $$\sin(x/2-\tan^{-1}(2/x))$$ has no zeroes in this interval. Thus, the global maximum is at $x = \pi$ which givse $f(\pi) = \frac{1}{\pi}$ as desired.

Community
  • 1
Sandeep Silwal
  • 8,135
-1

That is o.k. for obtaining which solution get max and which min use second derivative test: if $f^"(a)<0$ then $a$ get max and if $f^"(a)>0$ then $a$ get min.

Ali
  • 2,277