@Simply me more Mutivarible-Calc(anything I missed) Contour Integration, Real Analysis stuff, and working on more diffclut problems and using Computer Algebra Systems to automate tedious steps
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I have a copy but never really learned from it, but after learning from other, not-so-satisfactory places, I realised Spivak's is probably the best for someone wanting to learn about calculus. Too busy studying logic and programming things now to bother with math. :P
@SimplyBeautifulArt Hey, @SimplyBeautifulArt Will you provide, here, a comment or perspective as to whether you think my most recent edit to my answer here: math.stackexchange.com/questions/2289098/… is more confusing than it is helpful? (Do not upvote, though, since that's not what I'm seeking. It's just nice to have an extra pair of eyes.) The edit (addition) appears below the "horizontal line"
@projectilemotion, @EricStucky, Please see my request immediately above, since I'd appreciate any/all feedback, here. Not looking for upvotes; I simply want to know if my "my edit addend* today, below the horizontal "line", makes my post more complete, or more confusing, or both :-/
Of course... still hoping @Eric and/or @Simply might add their thoughts too. What I was initially thinking of was that we have a composite number as being $\neq -1, 0, 1$, but the product $\pm 1$ times the product of two primes.
Just corrected my mistaken "two" when it should have been "too". Usually people make mistakes/typos between "to" vs. "too". Only those immersed in math will sometimes write "two" instead of "to" and "too"!
Hi, @TheGreatDuck. I do understand about the mod chat, and I was, for a time, chatting too much there, too. As for the other user I addressed: ....it seems the "new user" has been a member of MSE chatrooms for only three days.
If in a group $G$, $xy^n=y^{n+1}x$ and $ yx^n=x^{n+1}y$ for some $n \in \mathbb{N}$, show that $x=y=e$, where $e$ is the identity of $G$.
I found this problem from the book "Modern Algebra" written by Surjeet singh and Qazi Zameeruddin. This problem is on the page 59. I've been trying to...
@TheGreatDuck You are a big boy and know better, and could have cut the conversation off (or redirect. You wrote the last comment there (did not need to), introducing a mathematical question that does not belong in the Mod chat, nor was it solicited. So Don't act so indignant.
Hey, @Theoretical! Welcome to the chat room!
@TheG "Im trying to figure out the maximum number of integers divisible by any number in [1,n] within the range [n^2,(n+1)^2] basically an integer b in [n^2,(n+1)^2] is a composite if and only if an integer 1 < a < n divides b combinatrics should reveal the answer the trick is how to avoid counting duplicates
@amWhy: I think I might have gone about that a slightly different way, and made the observation that sometimes you'll get negative numbers, following that up with the definition that negative numbers require at least five (or six) factors to be composite. But what you have written seems fine.
You are welcome to email me, @SBM. Let me know if you'd like my email address, and if so, when you're ready to copy it (so I can post in a comment briefly) after which I will follow up to delete the address)!
Mod's know, or can contact by email, even if they don't know it, so I'm not worried about what they might be able to read. I just don't want to leave my email address so it's too available to too many people. @Mithrandir is here, and can help me delete a comment with my email address, if it persists beyond the window of time we each have to delete a comment. @SBM Are you ready?
@amWhy also, if necessary I can purge it so that even people who can see deleted messages won't be able to see it (if it's edited to not show it before it's deleted)
OK ... I read the blerb ... is it like code golf ? ... no you explain me ? ... you say it is easy to learn in a few days ... can you refer me to the stuff ? ... I am interested ...sell it me ... I will but anything ! Kind Regards DS
@TheGreatDuck ... I was trying to remember what Legendre's conjecture was ... thats right ... there is a least one prime between n & n^2 ... probably easier than Bertrands postulate !
So, trivial observation: the inductive structure you're proposing works no matter which range you take, as long as it's at least three. I mean: if you wanted to prove that there's always a prime strictly between $n$ and $n+3$, you would be able to provide a proof by induction as long as it were true for prime numbers. (It's not, which is why the statement is wrong, of course.)
So it seems that your question is, in some essential way, about distribution of prime gaps, i.e., can we put an upper bound on how far it is between a prime and some other one. Which, yes, you can, but you wouldn't expect that to be very straightforward.
@SimplyBeautifulArt I'm surprised I didn't get the little message in my message post, to the effect that my "post has been edited." But thanks, I think?
@SimplyBeautifulArt But what's the question, or puzzle? I see "Read this carefully ... .. .. " and a hint. Is the puzzle to translate? Decode? Otherwise, any one who "Read your title and post(hint) carefully" has succeeded.
Looks good so far, but usually I would do it in the reverse manner. Your integrand should be the derivative of something... anyways, what are you integrating with respect to?
i'm integrating with respect to r, usually when apply ferynman's trick with an expotient in their integrand one would integrate with respect to that exponent
@Simply I should have mentioned that I introduced a parameter
@Simply this integral i'm taking come from when I using the estimation lemma, skiping over details but in order to find my upper bounds I have to simplify the integral
@Simply also here's an alternate approach that I considered with Feynman's Integration trick for that integral I mentioned earlier
$$\frac{\partial(I)}{\partial{\theta}}=\int_{0}^{\pi}(-1)^{er}-1 \frac{-(1)e^{ir \theta} - 2 \theta i }{r}$$
^ Set the partial derative with respect to theta equal to the integral
then move the partial derative on the RHS next to our integrad
@Simply is that also a proper way of doing it in this case also ?
Back in April of 2010, Joel shared our assumptions about the role of small sites in the newly minted Stack Exchange network:
If a site does not have enough activity at the end of 90 days, it will be closed down. Any existing Q&A will be archived and made available for download, but the site...
If in a group $G$, $xy^n=y^{n+1}x$ and $ yx^n=x^{n+1}y$ for some $n \in \mathbb{N}$, show that $x=y=e$, where $e$ is the identity of $G$.
I found this problem from the book "Modern Algebra" written by Surjeet singh and Qazi Zameeruddin. This problem is on the page 59. I've been trying to...
https://repl.it/01001001010011000110110001011000
Hint:
This is the Realm of Simply Beautiful…
This is the Realm of Simply Beautiful…