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2 hours later…
8:50 AM
0
Q: Mechanical pencils for math notes: Most suitable lead type?

kdbPreviously in my PhD and now at my job I do a lot of handwriting of mathematics, with accompanying text notes, and I have found mechanical pencils to be the most suitable tool for the job, as they combine the comfort of ergonomically fitting products, low effort for the wrist and repeatedly corre...

 
 
2 hours later…
10:28 AM
reopen: OP provided attempt in the comments
 
 
2 hours later…
12:44 PM
@darijgrinberg An attempt in the comments does not count unless somebody edits the attempt into the question body. The editor does not have to be the asker. I have "rescued" several questions by relocating such an attempt into the question body when I felt that the question should be preserved/reopened. I believe that this is the best way. Practice what you preach and all that.
 
1:10 PM
@JyrkiLahtonen fine, done
 
2:00 PM
@darijgrinberg Voting to reopen.
 
@darijgrinberg Reopened
 
@TheSimpliFire When did you get that shiny lozenge next to your name?
 
Congratulations.
And good luck with the new beta site.
 
Thanks, the site stats seem promising - for now :P
 
2:46 PM
@XanderHenderson It was a valiant effort (in gentle) but it really looks now to be a case of self-sustaining intransigence. Let's leave the room to its fate and actually get some things done here :)
 
 
2 hours later…
4:26 PM
0
Q: Failing to visualise Stiener's argument - Isoperimetric Theorem

hargun3045Hi stackexchange community, My question is regarding Steiner's (incomplete) proof for the Isoperimetric problem Before I ask, I've done the following: Read the proof as presented in 'What is Mathematics' Re-read the proof Searched the stackexchange for similar questions Now, my question: I...

I think that the question is overly chatty. Would anyone object to my editing it down a bit? e.g. the list of things done by the asker seems extraneous.
 
5:13 PM
I made some edits.
 
 
1 hour later…
6:30 PM
D1, D2, D3.
D4, D5, D6.
D7, D8, D9.
 
7:00 PM
0
Q: Textbook Error? $\big [e^{\int_{0}^{x} e^{-t^2} dt} \big ]' = \int_{0}^{x} e^{-t^2} dt \cdot e^{-x^2} $

user_hello1Textbook's Answer $ \big [e^{\int_{0}^{x} e^{-t^2} dt} \big ]' = \int_{0}^{x} e^{-t^2} dt \cdot e^{-x^2} $ My Answer $ \big [e^{\int_{0}^{x} e^{-t^2} dt} \big ]' = e^{\int_{0}^{x} e^{-t^2} dt} \cdot e^{-x^2} $

MSE is not for errata. :\
 
7:22 PM
As I understand it, the predominant opinion was that – given sufficient context (!) – such a question can be on-topic.
 
Thank you, @MartinR. It seems that I upvoted this answer at some point in the past. According to that answer, the question I linked above should be closed on the grounds that it lacks context.
 
 
3 hours later…
10:20 PM
I like this answer. The subsequent edits don't improve it. :(
 

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