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How do I prove $\sin x$ is uniformly continuous on $\mathbb R$ with delta and epsilon?

I proved geometrically that $\sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|\sin x_1 - \sin x_2|\le|\sin x_1|+|\sin x_2|<|x_1|+|x_2|$$

But this doesn't help me much finding a delta...

Thanks for any help!

P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).

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I changed $sinx_1$, etc., to $\sin x_1$. That is standard TeX usage. –  Michael Hardy Jan 1 '13 at 19:35
    
This is a particular case of (at least) two more general results. First: a periodic continuous function on $\mathbb{R}$ is uniformly continuous on $\mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $\sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$. –  1015 Jan 1 '13 at 19:41
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4 Answers

up vote 5 down vote accepted

Let $\epsilon>0$ and $x,y\in \mathbb{R}$. We want $$\left|f(x)-f(y)\right|<\epsilon\implies \left|\sin x-\sin y\right|<\epsilon\implies \left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|$$ Because $$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$ it suffices $$2\left|\sin\frac{x-y}2\right|<\epsilon$$ when $$\left|x-y\right|<\delta\implies \left|\frac{x-y}2\right|<\delta$$ SInce $\left|\sin x\right|\le \left|x\right|$, $$2\left|\sin\frac{x-y}2\right|\le 2\left|\frac{x-y}2\right|<2\delta$$

Choosing $\delta=\frac{\epsilon}{2}>0$ will do the trick. Because $\delta$ doesn't depend on $x,y$, the continuity is uniform

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Hi, thanks for the fast response! Why $$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again! –  Harold Jan 1 '13 at 19:32
    
because $\cos \alpha \leqslant 1 $, $\forall \alpha \in \mathbb{R}$. –  Amihai Zivan Jan 1 '13 at 19:34
    
Because $\left|\cos(\bullet)\right|\le 1$. –  Michael Hardy Jan 1 '13 at 19:34
    
Oh, right! And why $|sinx|<|x|$? –  Harold Jan 1 '13 at 19:35
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@Harold You want $\left|\sin y\right|\le \left|y\right|\iff -\sin y\le -y\iff \sin y\ge y$ for $y<0$ near $0$. But $\sin x\le x$ for $x>0$. Multipliying by $-1$ gives $-\sin x\ge -x\iff \sin (-x)\ge -x\iff \sin y\ge y$ –  Nameless Jan 1 '13 at 20:10
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By Mean Value Theorem,

$$ |\sin{x}- \sin{y}| \leq |x-y| |\cos{\xi}| \leq |x-y|, \quad x\leq\xi \leq y.$$

Hence, you may choose $\epsilon=\delta$.

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Since $\sin x$ is a periodic continuous function with a period $2\pi$, it suffices to prove that it is uniformly continuous on $[0, 2\pi]$. Since $[0, 2\pi]$ is compact, this follows from the well-known theorem.

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There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, y\in(-\pi, \pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|\sin(x) - \sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case $$|\sin(x) - \sin(y) | \le |x - y|.$$

This gives us uniform continuity on $(-\pi, \pi]$, so by periodciity the sine function is uniformly continuous on the entire line.

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Another way to prove this identity: math.stackexchange.com/questions/620305/… –  GinKin Dec 28 '13 at 10:08
    
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions. –  ncmathsadist Dec 29 '13 at 19:56
    
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense. –  GinKin Dec 29 '13 at 20:03
    
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique. –  ncmathsadist Dec 30 '13 at 0:46
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