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7:25 AM
12
Q: Trigonometric identity

Michael HardyIt is well known (?) that if $\alpha+\beta+\gamma=\pi$ then $4\sin\alpha\sin\beta\sin\gamma = \sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)$ (I think I've seen it in some late-19th-century books, and I read somewhere on the internet (therefore it's true!! right?) that it has repeatedly appeared on the...

I think this question deserves a more descriptive title than just "Trigonometric identity".
The title "Trigonometric identity: If $\alpha+\beta+\gamma=\pi$ then $4\sin^2\alpha\sin^2\beta\sin^2\gamma= (\sin\alpha+\sin\beta+\sin\gamma)(\sin\alpha+\sin\beta-\sin\gamma)(\sin\alpha-\s‌​in\beta+\sin\gamma)(-\sin\alpha+\sin\beta+\sin\gamma)$" is 231 characters, limit is 150.
"If $\alpha+\beta+\gamma=\pi$ then $4\sin^2\alpha\sin^2\beta\sin^2\gamma=(\sin\alpha+\sin\beta+\sin\gamma) (\sin\alpha+\sin\beta-\sin\gamma)(\sin\alpha-\sin\beta+\sin\gamma)(-\sin\alpha+\‌​sin\beta+\sin\gamma)$" is still 209
The RHS alone is 136 characters: $(\sin\alpha+\sin\beta+\sin\gamma)(\sin\alpha+\sin\beta-\sin\gamma)(\sin\alpha-\‌​sin\beta+\sin\gamma)(-\sin\alpha+\sin\beta+\sin\gamma)$
What is your suggestion for better title?
Identity for $4\sin^2\alpha\sin^2\beta\sin^2\gamma$ where $\alpha+\beta+\gamma=\pi$
Identity for $4\sin^2\alpha\sin^2\beta\sin^2\gamma$ expressed as product of terms $(\pm\sin\alpha\pm\sin\beta\pm\sin\gamma)$
Does some of the above two suggestions seem reasonable?
 
 
1 hour later…
8:35 AM
This question gained some attention after (probably) getting into hot questions list and also being mentioned on meta.
-10
Q: How can you simplify $\cos^6x+\sin^6x$ to $1-3\sin^2x cos^2x$?

AlbertHow can you simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$? I know that $\cos^2x+\sin^2x=1$. I thought about trying to use $(\cos^2x+\sin^2x)^3=1$, but that did not lead me anywhere. I found some other questions about the same expression, but they simplify this to another form: Finding $\si...

I wondered how the questions could be salvaged. (So that we do not lose answers posted there.)
One possibility would be to find a duplicate and merge (so that the answers would be moved there). But I did not find a duplicate.
The other possibility would be improve the question.
I know that it is not ideal if "fake context" is added by somebody else than the OP. But seeing that there are already two delete votes, if we want to do something with the post, we are probably running out of time.
So I have edited it. Further improvements are welcome.
 
8:59 AM
The post already went through reopen review.
I wonder whether I will be told off for adding context instead of OP like the last time:

On editing "shown effort" into a question by user other than OP

Jan 21 '15 at 8:55, 1 day 4 hours total – 59 messages, 5 users, 4 stars

Bookmarked Jan 23 '15 at 7:27 by Martin Sleziak

 
 
1 hour later…
10:19 AM
[ SmokeDetector | MS ] Few unique characters in answer, repeating characters in answer: Proof of the Riemann hypothesis based on a precise prime counting function by onepound on math.stackexchange.com
 
@MartinSleziak Or another possibility would be to make a completely new post (with enough context) and then ask moderators to merge/move answers. But let's wait to see whether the above post really gets deleted.
 
10:43 AM
@DanielFischer I am not sure whether you read the above comments. I think that the question was closed deservedly. (And my attempt to add some context was not sufficient to get it reopened from the review.)
Would it be ok if I post a new question about the same identity (and with sufficient context) and then flag for merging?
Main goal is not to lose the answers. (I think that question about this and also answers could be useful for some people.)
As I have mentioned above, I was not able to find a duplicate. (If such question already exists on the site, it would be a better target for merging.)
 
 
2 hours later…
12:30 PM
I think that Sum $\sum_{x=0}^{\infty} \frac{x}{2^x}$ is a duplicate. (It is possible that the duplicate target I was able to find was not the best.)
The post already went through the review: math.stackexchange.com/review/close/716546
Am I missing some reason why it should not be closed as a duplicate?
And if it should be closed, one vote is missing.
@MartinSleziak It is closed now. Thanks to Daniel Fischer.
 
Found one myself.
 
@DanielFischer I think you could leave the message - in case somebody peeks in here and has another suggestion for a dupicate.
And I have edited this one to be at least a bit better readable - it now looks almost the same as the one you linked: mean of two consecutive number helps proving both number equals..
 
12:48 PM
@MartinSleziak Since I don't need to wait for four other dupe-votes (hmm, wonder when we'll get the first gold badge), that's not necessary.
 
I see you have already found a duplicate and closed.
I wonder why the duplicate is not shown among linked questions. Let's wait a bit, it might take some time.
 
I see it in the linked question list. shouldiblamecaching.com
 
@DanielFischer Still I thought that it might be reasonable to leave it here. There are users who enjoy searching for questions. And, moreover, if somebody else tries to search, we might find a few more duplicates.
 
I didn't know whether you were already looking, and in case you hadn't been, wanted to un-ping, to not cause pointless work for you. Could have edited the @Martin out, but deleting is quicker.
 
BTW I still do not see the link in the sidebar of 4=5. Is this possible? But since you see it, it's probably caching.
When I try direct link math.stackexchange.com/questions/linked/2021178 then it is shown there.
BTW Where's the problem in this equation? Resulting in $4 = 5$ is slightly different, so probably not a duplicate. But they are similar - so maybe linking each to the other one might be useful.
 
1:02 PM
Oh, you mean in the sidebar of the duplicate, a link to the target? I see none there either. Caching, always caching.
 
Well probably not always.
Some questions linked in the comments are not shown in the sidebar here: How to find the sum of series $\sum_{i=1}^{\infty}\frac{i}{2^i}$?. Direct link shows only two questions as well: math.stackexchange.com/questions/linked/683863
And I am sure I saw a few other examples. Maybe it would be reasonable to collect them and ask on meta.SE.
 
 
1 hour later…
2:21 PM
What about these two questions? Merge? Or only duplicate (if so, in which direction)?
6
Q: $X$ is homeomorphic to $X\times X$ (TIFR GS $2014$)

Praphulla KoushikQuestion is : Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $X\times X$. Then which of the following is true: $X$ is not connected. $X$ is not compact $X$ is not homeomorphic to a subset of $\mathbb{R}$ None of the above. I guess first two options are fa...

2
Q: What can be said about topological properties of a space $X$ that is homeomorphic to $X\times X$ $?$

user118494 Given that $X$ is a topological space with infinite cardinality and $X$ is homeomorphic to $X\times X$. (A)   $X$ cannot be connected. (B)   $X$ cannot be compact. (C)   $X$ cannot be homeomorphic to a subspace of $\mathbb R$. (D)   None of the above. If I take $\mathbb Q$...

 
 
2 hours later…
3:52 PM
@MartinSleziak I have voted to close and also flagged for merging.
I chose the newer question as the duplicate target, since I think the answers there are simpler.
 
 
1 hour later…
4:57 PM
@MartinSleziak I asked first.. how can this be a duplicate of that question — Praphulla Koushik 6 mins ago
@PraphullaKoushik The time is not the main thing in order to decide which of the two question should be duplicate target, see meta. Of course, if you disagree, feel free to vote to close the other one. — Martin Sleziak 59 secs ago
 
 
1 hour later…
5:58 PM
These three questions are also very similar to each other:
67754 requires non-discrete space, 722412 requires connected space, the only requirement in 1384455 is that the space is infinite.
As far the answers go, in my opinion the simples answer is the space of the form $X=A^{\mathbb N}$ for some $A$ (countable power). Both 722412 and 1384455 have such answer.
 
 
6 hours later…

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