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If $\{x_n\}$ is a Cauchy sequence then show that $\{\cos x_n\}$ is also a Cauchy sequence.

Let $y_n=\cos x_n$ then for $m>n$, $|y_m-y_n|\leq |\cos x_m-\cos x_n|\leq |\cos x_m|+|\cos x_n|\leq 1+1=2$

Please correct my proof using the definition of Cauchy sequence.

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    $\begingroup$ "correct my proof" I can't find any. $\endgroup$ – drhab May 10 '16 at 11:21
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    $\begingroup$ I don't think your proof is correct. It's better to use $$\cos x_m-\cos x_n=2 \sin \left(\frac{x_n-x_m}{2} \right) \sin \left(\frac{x_n+x_m}{2} \right)$$ which becomes arbitrarily small with $x_n-x_m$ arbitrarily small $\endgroup$ – Yuriy S May 10 '16 at 11:22
  • $\begingroup$ What is $n$ in the statement "for $m>n$"? $\endgroup$ – Jack M May 10 '16 at 11:22
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    $\begingroup$ Your proof is completely wrong, sorry. You must estimate $\cos x_m - \cos x_n$ in terms of $x_m-x_n$. A trivial estimate does not suffice. $\endgroup$ – Siminore May 10 '16 at 11:22
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    $\begingroup$ So... for example, to show that $z_n=1+\frac1n$ is Cauchy, you would write $|z_n-z_m|=\left|1+\frac1n-\left(1+\frac1m\right)\right|\leqslant\left|1+\frac1n\right|+\left|1+\frac1m\right|$? Really? $\endgroup$ – Did May 10 '16 at 11:31
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I think I'll elaborate on my comment. Use the well-known trigonometric indentity:

$$\cos x_m-\cos x_n=2 \sin \left(\frac{x_n-x_m}{2} \right) \sin \left(\frac{x_n+x_m}{2} \right)$$

According to the usual definition of a Cauchy sequence, if we choose $\epsilon>0$ then there exists some $N$ such that for any $n,m>N$ it follows that $|x_m-x_n|<\epsilon$.

Using this definition and the above relation we write:

$$|\cos x_m-\cos x_n|=2| \sin \left(\frac{x_n-x_m}{2} \right) \sin \left(\frac{x_n+x_m}{2} \right)| \leq 2 \frac{|x_m-x_n|}{2} \cdot 1=|x_m-x_n|$$

Obviously from $\{x_k\}$ being Cauchy follows $\{\cos x_k\}$ being Cauchy as well

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  • $\begingroup$ How to get $| \sin \left(\frac{x_n-x_m}{2} \right) | \leq \frac{|x_m-x_n|}{2} $ ? $\endgroup$ – user1942348 May 10 '16 at 13:13
  • $\begingroup$ @user1942348 from the definition of $\sin x$. Either from the series or from geometrical definition. $\endgroup$ – Yuriy S May 10 '16 at 13:18
  • $\begingroup$ $|\sin x|\leq |x|$ $\endgroup$ – Kushal Bhuyan May 10 '16 at 13:45
  • $\begingroup$ @YuriyS The series to show that $|\sin x|\leqslant|x|$? How? $\endgroup$ – Did May 10 '16 at 21:21
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  • Use the fact that the function $f(x) = \cos(x)$ is uniformly continuous. ( Lagrange Mean Value theorem.)
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  • $\begingroup$ The OP is specifically looking for a correction of their proof, not advice on how to write a different proof. $\endgroup$ – Jack M May 10 '16 at 11:21
  • $\begingroup$ @JackM Em. I personally believe that seeing this hint, he can rectify the answer himself. And essentially all proofs of the question are going to based on the same idea. $\endgroup$ – crskhr May 10 '16 at 11:22
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You gave it away. Start with $$|y_m-y_n|=|\cos x_m-\cos x_n|\leq\ ?\ |x_m-x_n|$$ instead, whereby you have to replace the question mark by something meaningful (and correct). A hint: Use the MVT, you will obtain $$|y_m-y_n|\leq|x_m-x_n|\ .$$ The rest is "pure logic".

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  • $\begingroup$ Thank you very much. I also want to get the result of such style. I am unable to detect the ? terms . I would be grateful if you suggest it for me. $\endgroup$ – user1942348 May 10 '16 at 13:10
  • $\begingroup$ Do you mean $|y_m-y_n|=|\cos x_m-\cos{x_n}|\leq\ \sin{ \theta}\ |x_m-x_n|\leq |x_m-x_n|$ as $-1\leq \sin{\theta}\leq 1$ where $x_m<\theta<x_n$? $\endgroup$ – user1942348 May 10 '16 at 14:46
  • $\begingroup$ Sir, Am I correct? $\endgroup$ – user1942348 May 10 '16 at 15:00

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