Should I spam today?

A substitution is going to crack it, I know it.

Spam maths problems? :P

@Sawarnik

@KhallilBenyattou No, spam nonsense :)

And substitution? No need at all, its Verrry simple!

@Sawarnik That would be a terrible use of your day!

lurks

@meer2kat Greetings sir.

@Sawarnik sup

@KhallilBenyattou Hint: You don't need to evaluate the integral!

How am I supposed to get the c/31 if I don't integrate it directly? @Sawarnik

@KhallilBenyattou Just think. :P :P

It also looks like a square if I'm not mistaken, that is the area looks like a square.

@meer2kat You are up early again

@KhallilBenyattou $1 < \dfrac{1+x^{30}}{1+x^{60}} < 1+x^{30}$ for $x \in (0,1)$

@Sawarnik Check out my earlier post!

You could have asked help from the calc teacher @meer2kat .

@KhallilBenyattou Yes, I see. Just writing that properly.

@meer2kat Morning :)

@KhallilBenyattou So where is the substitution? :P

@KhallilBenyattou mornin

@WillHunting work

I have a piece of Mathematics I would like to include in Wikipedia. However they won't let me since they are saying it is original research. How should I proceed? I have no experience in publishing in Journals. And the result is very elementary so it is not sure it would be accepted in a Journal. Yet I think because it is elementary it should be included in Wikipedia.

@WillHunting Will you remain Will Hunting?

@meer2kat I read what you wrote in the poetry room and figured you are Christian

@Sawarnik I don't know

@WillHunting i am

@WillHunting why?

I just watched The Amazing Spider-man, nice movie.

@meer2kat Nothing, just making small talk as usual, lol

@WillHunting how about you?

@WillHunting After the first 3, I gave up on the series.

Watch Batman.

@meer2kat I used to be Christian for a while. Now I consider myself a Buddhist. But I know that the words Christian and Buddhist can have many meanings

@WillHunting very true

@KhallilBenyattou This is a new series, much better to me

I stopped watching spider-man and batman when I was a kid

@meer2kat Oh, and you didn't email me, but that's OK, lol

@skullpatrol Did you watch the Batman movies? The Dark Knight?

@skullpatrol To solve Riemann? :P

lol

@skullpatrol I managed to solve 20 per cent of my mental problems last month

@WillHunting nice work

@sawarnik I saw what you wrote, lol

@WillHunting let me know if you need any help, pal

@KhallilBenyattou Not good enough for Alyssa, I think. :P

@Sawarnik I look like Justin Bieber

I am wondering if any universal algebraist is out here ?

how about a local arithmetician?

:P

how about an irritating kid?

How about a mental arithmetician?

:P

I think Sawarnik is a pretty irritating kid

to each their own

their

thanks

I wonder why this mistake is so common

@WillHunting lack of understanding

I should irritate somebody now, getting bored.

like all common mistakes

You can't irritate me

@WillHunting How come?

@Sawarnik I choose not to be irritated by you

@WillHunting I choose to irritate you.

@Sawarnik You are powerless

@WillHunting Yes, powerless enough to clutter your inbox :P

Quiet suddenly :P Ok, bye.

Hi @r9m

@skullpatrol hello :)

Guys, I need help with an integral: $$ \displaystyle \begin{aligned} \int \dfrac{e^{x} + 1}{e^{x} - 1} \text{ d}x & \overset{u=e^{x} -1}= \int \dfrac{u+2}{u(u+1)} \text{ d}u \\ & = 2 \int \dfrac{\text{d}u}{u} - \int \dfrac{\text{d}u}{u+1} \\ & = 2\ln |u| - \ln|u+1| + \mathcal{C} \\ & = 2\ln |e^{x} - 1 | - \ln |e^{x}| + \mathcal{C} \\ & \quad \because e^{x} > 0 \text{ for all } x \in \mathbb{R}, \\ & = 2\ln | e^{x} - 1 | - x + \mathcal{C} \end{aligned} $$

Wolfram gets a different answer so I'm not sure if this is an alternate form. Anyone?

@r9m How are you doing today? :)

@KhallilBenyattou sleepy .. :O (yawns)

@r9m Me too. I did too much maths yesterday!

@KhallilBenyattou I'm pretty sure they are alternate forms

@AmericanLuke Thank you! I was getting worried there as a friend was adamant that I was wrong.

$$|e^x-1|=|1-e^x|$$

Try any number for x. It'll still be equal

@AmericanLuke Hold on for a moment, Wolfram had the form on the right hand side but without the absolute value parentheses ...

> alternate forms assuming $x>0$

@WillHunting email?

Given a group G, K a subgroup of G, and x,v elements of G... we can introduce multiplication of cosets of K as uK o vK = (uv)K but then we have to prove its well-defined ( if aK=cK and bK=dK then abK = cdK ), and we show thats only true if K is normal subgroup..

Now, we can introduce the multiplication of cosets of K based on product of group subsets ( en.wikipedia.org/wiki/Product_of_group_subsets ) , and from that we can easily prove that if K is normal in G, then (xK)(vK) = x(Kv)K = xvKK= xvK , but now ( in this second multiplication ) after we proved that K normal yields closure, do we still have to prove its well-defined ?

@AmericanLuke That explains it!

In the first introduction, we assumed multiplication of cosets was closed in the Quotient Group, and had to prove it was also well-defined in case K is normal. In the second definition, we didn't assume it was closed, proved it's closed if K is normal but now we don't need to prove it's well-defined ?

If $x>0, e^x-1=|1-e^x|$

@Sawarnik hello .. I am watching a movie (Oblivion)

Ok, watch in peace :)

@DanielFischer To give context about the product I was looking to expand, I was asked to prove that given $(a_n)$ a sequence of complex numbers all not equal to $-1$, if $\sum_{k=1}^{\infty}|a_k|$ converges then $\prod_{k=1}^{\infty}(1+a_k)$ converges.

You mean \prod_{k=1}^{\infty} (1+a_k) :)

@GabrielR.

@DanielFischer I have a rather ugly proof. It is easy to prove that that given $(a_n)$ a sequence of positive real numbers, the convergence of $\sum_{k=1}^{\infty}a_k$ implies the convergence of $\prod_{k=1}^{\infty}(1+a_k)$.

@Waffle'sCrazyPeanut Is that Joostan from TSR? You have the same avatar :)

@GabrielR. Okay. The trick is to relate $\lvert \log (1+a_n)\rvert$ and $\lvert a_n\rvert$, where you take the principal branch of the logarithm (assuming $n$ large enough so that $\lvert a_n\rvert < 1$).

@KhallilBenyattou Are you interested in geometry?

@DanielFischer It follows here that $\prod_{k=1}^{\infty}(1+|a_k|)$ converges. It gets ugly when I then prove that the sequence $\prod_{k=1}^{n}(1+a_k)$ is Cauchy (the huge sum appears)

@GabrielR. the absolute convergence of $\sum\limits_na_n$ implies the convergence of $\prod\limits_n(1+a_n)$

@GabrielR. Well, don't do that ;) Use the logarithms, they were invented for such things.

@KhallilBenyattou Why would they cover geometry in university? Do you mean non-Euclidean or axioms or something like that ..

@DanielFischer I don't know what a branch is

@Sawarnik I think so yea. Geometric proofs too :)

Although it isn't as in depth as you might think.

Oh. I don't know anything about them!

@DanielFischer Wikipedia seems to say it's complex analysis

math.stackexchange.com/questions/754417/… .

@robjohn yes, it is indeed what I attempt to prove :)

@GabrielR. Yes. Though you don't need much. You just need to know that $$\log (1+z) = \sum_{n=1}^\infty (-1)^{n-1}\frac{z^n}{n}$$ gives you a function with $\exp(\log w) = w$ for $\lvert w-1\rvert < 1$.

Well, you don't really need much of the formula, the existence and a bound on the remainder is what you need.

@GabrielR. The second order terms can cause either the sum or the product to diverge even though the other converges.

@KhallilBenyattou Prove that there are infinitely many rational points for an equilateral triangle of length 1. A rational point is a point within the triangle such that its distances to the vertices are rational numbers. Its very tough, you can spend a day on it :)

@robjohn But here we have $\sum \lvert a_n\rvert < \infty$ as a premise, so here that cannot happen.

@Sawarnik At first glance, I'm guessing you can use proof by contradiction?

@DanielFischer yes. I didn't see that in the original statement, so I wanted to make it clear that absolute convergence of one implies absolute convergence of the other. Conditional convergence is not really useful here.

@KhallilBenyattou You can try it :D

Park time. BBL

@KhallilBenyattou ... then somehow derive a result which is a ... How? Clarify the somehow!

@KhallilBenyattou Meh... My gravatar is stuck there for over an year or something :D

@Sawarnik Ah I see. I'll try it later. Am I on the right lines?

@DanielFischer Wow... I discovered another short form -- All this time, I've been using brb :D

@KhallilBenyattou Ok, take a look. math.stackexchange.com/questions/485755/…

wow you guys talk a lot

@meer2kat Maybe you don't talk enough?

Haha!

@KhallilBenyattou Right!

@meer2kat It's chat, what do you expect?

@KhallilBenyattou See the link. BBL.

@DanielFischer I can surely write $|log(1+a_n)| \leq |a_n| + O(|a_n|^2)$ but when summing these $O$ I'm not granted convergence

@DanielFischer actually $|log(1+a_n)| \leq |a_n| + o(|a_n|)$ should do ?

@KhallilBenyattou touche. hey i talked a lot yesterday

@DanielFischer touche

@GabrielR. You can have an estimate in both directions, e.g. $$\frac{2}{3}\lvert u\rvert \leqslant \lvert \log (1+u)\rvert \leqslant \frac{4}{3}\lvert u\rvert$$ for $\lvert u\rvert \leqslant \frac{1}{4}$. That gets you to the goal nicely.

@DanielFischer I agree. Thanks!

my project that i've been working on for two months just got submitted. i'm a nervous wreck

@meer2kat I'm sure you'll be fine!

Never thought I'd have such a popular question. 2500 views, goodness

@meer2kat Whichie which?

:14969139 acronymslist.com/cat/chatting-acronyms.html

@robjohn That's too many!

@Sawarnik You can find it there

Ok :)

@meer2kat Which q?

nobody uses 90% of those acronyms

Greeetings

@robjohn is it familiar to you? $$\int_0^1 \log(s) \sin(2 \pi n s) \ ds$$

@Chris'ssis It should be an application of $\int_0^1\log(x)x^a\,\mathrm{d}x=-\frac1{(a+1)^2}$

DYJHIW FUBAR IYKWIM

@meer2kat It's probably your lovely avatar :)

@Chris'ssis Limit of n to infinity?

@N3buchadnezzar what's the matter?

@robjohn I just tried to use all the acronyms...

@N3buchadnezzar ikwym

@N3buchadnezzar $n$ is a natural number

oh my god.

@Mike ?

@Chris'ssis Well if you let n to infinity you could use Rieman-lebegue lemma

@N3buchadnezzar I don't need to let $n \to \infty$

@KhallilBenyattou haha!

@Sawarnik I can't help it :P

@robjohn I missed to think of that.

@N3buchadnezzar That looks like an application of Integration by Parts

@N3buchadnezzar Thank, but I need another form without involving $\text{Ci}(x)$

@Chris'ssis If it involves $\mathrm{Ci}$, how are you going to get a simpler form without it?

@Chris'ssis $$-\frac{2 (\log (2 \pi n)+\gamma )+\Gamma (0,2 i n \pi ,-2 i n \pi )}{4 \pi n}$$ hehe

I don't claim that this is correct, but I don't guarantee it's not either

@robjohn I think I approached things here in a wrongly way. (this is just a small part of a harder question)

I'll reconsider my thinking way here. Thanks.

@MickLH It looks interesting. Thanks! :-)

hello

hi

so, what is buzzing in here?

@AwalGarg Not much, how are you? :)

@AwalGarg I'm on a gamma function binge the last couple days as I've been finishing up my arbitrary precision calculator

@AwalGarg Hi.

@KhallilBenyattou i am intelligent as always :)

I've just been listening to an awesome instrumental piece of music. It starts at 17:20 youtube.com/watch?v=XF1MXOtEsQk

@meer2kat I have changed my mind. You are very very pretty!

@AwalGarg There's not a speck of arrogance in your post, at all! :P

When will it be my turn to be pretty?

@AwalGarg Are you an undergrad?

@KhallilBenyattou ha ha

@user127001 Maybe next life.

@KhallilBenyattou yes

:(

@Sawarnik I think @meer2kat is mad at me and won't come say hi because I'm here ;(

and i well always be one

@AwalGarg At which university? I can't see why you'll always be one if you're intelligent.

When will it be my turn to be intelligent

@KhallilBenyattou i don't like studies

and i have just passed high school

@user127001 Maybe next life.

:(

@AwalGarg Who does? :P

i am not into college yet

@KhallilBenyattou Right.

@KhallilBenyattou yeah, right

I'm guessing you're waiting to graduate? @AwalGarg

@KhallilBenyattou i am waiting to avoid graduation as it comes

@KhallilBenyattou Anyways you probably don't know of Indian universities.

@Sawarnik thumbs up

@Sawarnik My mind is open, enlighten me!

@KhallilBenyattou i would like to do the holy thing

@KhallilBenyattou Ok, the most enlightening fact is that you re intelligent.

@Sawarnik I'll be back in 20 minutes. Got to make some maths up and solve Riemann.

@AwalGarg What holy thing?

@MickLH Why do you think so?

@KhallilBenyattou Solving Riemann may take more than 20 mins, I think.