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A posted answer was soon withdrawn after my comment showed a fundamental flaw in it. The comment saved the answerer some embarrassment, and perhaps provided a little illumination for him. Also it spared readers from having their time wasted, or even being misled, by a false argument. However, there is no trace of this event. Possibly the answerer was grateful for being rescued from his folly, as I would have been, but it seems that there is no way to express thanks in this situation.

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  • $\begingroup$ If you direct me to the answer, I can undelete it and request that the answerer edit it to indicate that it's wrong instead of deleting it. $\endgroup$ – Qiaochu Yuan Jun 14 '11 at 16:58
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    $\begingroup$ @Qiaochu: Dear Qiaochu, I think people should be able to delete answers that have been indicated to be wrong, for both reasons indicated above: to save embarrassment, and to save everybody else's time. Thus I'm not sure that you should undelete the answer, even if it is identified. While I can understand the OP's desire for thanks, it may be that they have to be satisfied with the knowledge that they've done a good deed. Regards, $\endgroup$ – Matt E Jun 14 '11 at 19:06
  • $\begingroup$ @Qiaouchu: Thank you for the offer. For the record, the answer was to the recent question about a limit of the iterated sine function: math.stackexchange.com/q/45283/875 . However, I request clemency in this case, as the unfortunate poster thinks himself saved from public shame and would be doubly smitten by his blunder being published after all. I enjoy credits---but not at such expense to someone else's ego! In general, thought could still be given to how a comment pointing out an unpatchable flaw in a (consequently withdrawn) answer might be acknowledged. $\endgroup$ – John Bentin Jun 14 '11 at 19:07
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    $\begingroup$ @Matt: I appreciate your comment, and consider myself thus well enough rewarded! $\endgroup$ – John Bentin Jun 14 '11 at 19:40
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I was the one who retracted the answer, but not because of your comment. Sorry :) It was another comment by Didier Piau, which pointed out the flaw. My answer was:

Here is another answer, which essentially rewords the answer of @Didier. The sine is a contraction mapping on interval $(−1,1)$, i.e. it satisfies

$$|\sin x−\sin y|<\beta |x−y|$$

where $\beta<1$. Now for contraction mapping $T$ with the fixed point $v$ and any point $v_0$ we have that

$$|T_kv_0−v|≤\beta^k|v_0−v|$$ Take $v_0=\sin n$ and $v=0$ and we get that

$$|\sin^{(n)}n|≤\beta^{n−1}|\sin n|≤\beta^{n−1}$$ Taking the limit we get the zero.

And the comments were:

I don't see your first statement. For example, let y=0 and β=1−ϵ, where 0<ϵ<1. Then we can easily find δ>0 such that sinx−x>−ϵx for all x>0 such that x<δ. (Taking δ=ϵ would do.) – John Bentin 17 hours ago

@John, β is not chosen freely, the statement says that such β exists. It follows from the mean-value theorem: sinx−siny=cosα⋅(x−y) with α∈(−1,1), take β=sup{cosα,α∈(−1,1)}. – mpiktas 17 hours ago 3

Precisely: This β is 1 (and a geometric rate of convergence cannot hold, for good reasons). Sorry. – Didier Piau 16 hours ago

@Didier, thanks for catching that. I will retract my answer. – mpiktas 16 hours ago

I do not know whether Didier got my message. I thought about commenting separately, but since all that happened close to my working day end, I never did.

My intentions for the answer were I hope good, I thought that there exists a simple answer relating to standard result about contraction operators. For some reason I figured $\cos$ is less than one in interval $(-1,1)$ :) These things happen, and for that reason I always thoroughly recheck my results, usually the next day, with the fresh mind. This time somebody else found the mistake before me. Since it was major mistake, I immediately withdrew my answer.

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  • $\begingroup$ I showed that $\beta$ cannot be less than $1$. In your reply to me (which I missed), you said that one could take β=sup{cosα,α∈(−1,1)}, and Didier pointed out that this is $1$. $\endgroup$ – John Bentin Jun 15 '11 at 8:32
  • $\begingroup$ @John, I did not say that your comment was incorrect. It is, though I thought that I found the flaw in it at that time, hence my comment. But I still retracted the answer because of the comment of @Didier, though naturally without your initial comment, I would not have commented and subsequently Didier would not have submitted the comment which was the reason of my retraction. Phew :) $\endgroup$ – mpiktas Jun 15 '11 at 9:05
  • $\begingroup$ Thanks. All settled, I hope! Just on a point of English: "It is", referring to the preceding sentence, means "it is incorrect"; but I gather from the context that you mean "it is correct". Phew also :) $\endgroup$ – John Bentin Jun 15 '11 at 10:21
  • $\begingroup$ @John, yes I meant it is correct :) $\endgroup$ – mpiktas Jun 15 '11 at 10:37

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