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The following is part of exercise 6.26.21 from Tom Apostol's Calculus Volume 1. I wonder if my proof is correct and if there is a simpler alternative proof.

Prove the following by examining the sign of the derivative of an appropriate function: $$ \frac{2}{\pi}x \lt \sin{x} \qquad \text{if} \qquad 0 \lt x \lt \frac{\pi}{2} \tag{1}\label{1} $$

Let $ f(x)=\sin{x}-\frac{2}{\pi}x $, $ 0 \le x \le \frac{\pi}{2} $ then

$$ f(0)=f\left(\frac{\pi}{2}\right)=0 \tag{2}\label{2} $$

and

$$ f''(x)=-\sin{x} \lt 0 \tag{3}\label{3} $$

From $ \eqref{2} $ and $ \eqref{3} $ we know that $ f $ has a maximum at exactly one point, this together with $ \eqref{2} $ proves $ \eqref{1} $.

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    $\begingroup$ seems correct to me...you can also say by (2) f is strictly concave hence $f(x)=f(1\cdot x+0\cdot y)> 1\cdot0+0\cdot 0=0$ $\endgroup$ – math635 Dec 28 '15 at 1:05
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    $\begingroup$ This is the first part of Jordan's inequality. $\endgroup$ – user258700 Dec 28 '15 at 1:06
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    $\begingroup$ You should consider $f(x)=\sin x-2x/\pi$ for $0\le x\le\pi/2$, just to be picky. $\endgroup$ – egreg Dec 28 '15 at 1:07
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    $\begingroup$ @math635 thanks, was interesting to see how the concavity proves the point by its very definition. Btw, as a comparison, I think my "maximum at exactly one point" is not sufficient as that wording would still allow a minimum at another point, so as you and others pointed out I should've simply used the concavity property. $\endgroup$ – Imre Deák Dec 28 '15 at 10:50
  • $\begingroup$ @math635, one note is that in $ f(x)=f(\lambda x_0+(1-\lambda)y_0) \gt \lambda f(x_0)+(1-\lambda )f(y_0) $ it's not clear to me how you chose $\lambda, x_0, y_0$. I see how it works if we let $ \lambda \in (0,1) \; \text, \; x0=0 \; \text, \; y_0=\frac{\pi}2$. $\endgroup$ – Imre Deák Dec 28 '15 at 12:30
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Your proof is fine: the concavity of the sine function over $\left(0,\frac{\pi}{2}\right)$ gives the wanted inequality in a straightforward way. Anyway, if you like to kill flies with hydrogen bombs, you may consider that: $$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2 \pi^2}\right) $$ hence if $x\in\left(0,\frac{\pi}{2}\right)$ we have: $$ \frac{\sin x}{x}> \prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right) = \prod_{n\geq 1}\frac{2n-1}{2n}\cdot\frac{2n+1}{2n} $$ where the RHS is the reciprocal of the Wallis product, i.e. $\frac{2}{\pi}$ as wanted.

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  • $\begingroup$ The exercise asked for checking derivatives and I never met these products yet (beyond my current level), but nevertheless it's interesting, thanks. $\endgroup$ – Imre Deák Dec 28 '15 at 2:41
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Take $f(x)=\frac{\sin x}{x}$ then $f$ is decreasing as its derivative is negative.

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  • $\begingroup$ That proves that $ \sin{x} \lt x $, but not (1)? $\endgroup$ – Imre Deák Dec 28 '15 at 1:38
  • $\begingroup$ Its decreasing down to $\frac{\pi}{2}$ so this is a lower bound. $\endgroup$ – Rene Schipperus Dec 28 '15 at 1:40
  • $\begingroup$ Ok, I understand now if you meant $ \frac{2}{\pi} $. $\endgroup$ – Imre Deák Dec 28 '15 at 2:16

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