CURED - 2021-10-31

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Lalit Tolani

4:34 AM

@amWhy Yes I wanted to point out to question only, the way OP framed the question confused me :-)

2 hours later…

Jitendra Singh

6:49 AM

Consider closing. Also present in HNQ

3 hours later…

José Carlos Santos

10:00 AM

delete Although correct, this answer adds nothing new to other answers posted many years ago.

1 hour later…

user21820

11:16 AM

No, that wasn't what the question was. Peter asked why we can no longer see whether we have voted before trying to delete-vote again (and before it is actually deleted). Currently, we cannot, but the SO meta post I linked says what Peter remembered and is dated 2020. I am unaware of any other place besides that tooltip that this information can be seen (certainly not in the timeline).

SmokeDetector

[ SmokeDetector | MS ] Bad keyword in link text in body (69): Optimal play in Strings & Coins by malinoal on math.SE

user21820

But I wrongly assumed that there was no way to see it because I did not realize that the tooltip even existed. It takes a while to show up after hovering over the "delete" link, but whenever I want to delete a post I immediately click without hesitation so I never saw that tooltip. Hence my mistake. @amWhy: Sorry forgot to reply to your message.

Lalit Tolani

12:03 PM

Q: Inequality involving Sequences

I am learning sequences and series and I came across with this problem but have no idea to solve it. I tried using AM-GM inequality but it did not work out. Any idea how to solve this one? Given a sequence $a_{1}, a_{2}, a_{3}, \cdots$ of non-negative real numbers satisfying (i) $a_{n}+a_{2 n} \g...

2 hours later…

amWhy

2:32 PM

@user21820 No problem. That has been my understanding too: the tool tip. Perhaps it does not work in review queues? I'll test it out.

user21820

@Peter That factor-for-me question is up for deletion now.

@amWhy Maybe the change was at the same time as the one-vote limit.

That reminds me, @XanderHenderson, my last flag was declined with the reason that I already voted before "on Apr 25 '19 at 17:06", but the timeline shows no such vote from me at all. Isn't that a bug?

islamm

Peter

2:53 PM

@user21820 gone

user21820

@Peter Great!

amWhy

Consider deletion. Note also that one of the answers has two delete votes, but has been upvoted since.

Please delete (answer): needs one downvote, then a delete vote.

^^^^ The question answered by the linked answer needs closure and deletion. Two of the answers received an upvote after two delete votes were cast. I only linked one of the answers.

@user21820 So close to @Peter The Great! ;D

user21820

@amWhy not? (≈ "and why not?")

Xander Henderson

@user21820 On which question?

(And I apologize if I am slow to get back to you---I am at the AMATYC conference this weekend, and will be driving home this afternoon).

user21820

3:09 PM

@XanderHenderson Here is the timeline.

@XanderHenderson Sure, no problem.

Xander Henderson

There is no error. You and one other person voted to close the question two years ago. The question was then reopened, thus clearing your votes. Votes which do not lead to a change of state in the question don't show up on the timeline.

amWhy

@user21820 I suspect you are referring to no evidence of haved voted to delete? Because you voted to close on the given date.

^^^^I can't know what your flag was about, but I see no delete vote from you on the timeline.

user21820

@XanderHenderson I see, but why should I be prevented from voting again, since my first vote was never effected?

Xander Henderson

@user21820 A totally useless answer is "because that is the way the software is set up". You can only vote once, whether or not your vote is eventually effected.

I would argue that this is not how things should be, but that is the way that the software is set up. I would suggest that this is something which should be brought up in the main meta.

islamm

user21820

3:21 PM

@XanderHenderson Oh. Ok.

amWhy

@user21820, @Peter, I can't ping you from the Cafe, but we have A HALLOWEEN PARTY there. All are welcome!

user21820

D1, D2, D3.

D4, D5, D6.

D7, D8, D9.

D10, D11, D12.

C1, C2, C3.

C4, C5, C6.

C7, C8, C9.

C10, C11, C12.

@amWhy Ok will drop by!

amWhy

3:57 PM

C/D. One answer is worse than low quality.

amWhy

4:39 PM

This question is really quite poor, despite the asker claiming to show effort; it strikes me as an effort only to occupy space..

user21820

5:28 PM

@amWhy Gone.

amWhy

5:47 PM

@user21820 Thanks!

Hi, @Joe !

amWhy

6:28 PM

-6

Q: Prove if $f(x+y)=f(x)f(y)$then $f(x)=xf(1)$.

Let $f:R→R$ be a monotone function such that $f(x+y)=f(x)f(y)$ , $∀x,y∈R$. Prove: there exists constant a such that $f(x)=xf(1)$. My attempt: if $x=y=0$ $\implies f(0) = 0$ if $y=0$ $\implies f(x+1) = f(x) + f(1)$ if $x=x+1\implies f(x)= xf(1)$ Is my proof correct? If it isn't, then I am looking ...

calculus continuity

^ commented earlier. Still at only two close votes, and now three poor answers. Please that there is no sincere "attempt", despite "attempt" being written in bold by the OP.

2 hours later…