« first day (2575 days earlier)      last day (2449 days later) » 

12:02 AM
"Hello sign errors my old friend"
 
12:20 AM
hey
 
Hey @LucasHenrique, it's been a while
 
Hello @Daminark!
I've been a bit busy because of my exams
How are you doing?
 
I'm doing alright, thanks! Right now I'm doing summer camp
 
12:38 AM
I'm working on a MathQuill Latex Editor for ChatJax
 
 
3 hours later…
3:54 AM
Guys
Is a male mattress called a matter
 
Absolutely
 
4:19 AM
Whoa, looks like I've had this window open all this time and I've been missing all the fun.
 
ughhh how do I prove that if y' = y + y^3 is zero anywhere then it is zero everywhere?
 
I've got my MathQuill editor working.
 
nice @ATaco
I know that I can use definite integration for this case, but what if the bernoulli differential equation is more complicated...
 
4:39 AM
Oh, I just noticed it's you Leaky. :P Didn't recognize you with an avatar.
 
5:24 AM
[Random to be expanded] Changing lanes and galliean relativity
 
Jon
5:35 AM
That is allowed
Nevermind
Taking linear algebra II this semester, no calculator, and notes allowed on exam? Your thoughts?
 
Hey everyone!
 
Jon
Good morning friend
 
6:29 AM
[Random]
When whole commutative diagrams are functors
ln x + ln y = ln (x+y) => xy = x+y => xy - x - y = 0 => x(y-1) - y = 0 => x(y-1) = y => x = y/(y-1)
 
 
1 hour later…
7:40 AM
@Mr.Xcoder yes, you have two steps: 1) start, and 2)step. Start with $n=0$ or $n=1$ - depending on what you call natural numbers. If the theorem is true for $n=1$ or $n=0$ - good, you've had a perfect start. Now you say that the theorem is true for some $n$ - not for all, just for some. Prove that the theorem is true for $n+1$. Use what you have proven for $n=0$ or $n=1$. Look at some exercises showing you how do to that correctly and try to prove some things yourself.
 
Hey @AlessandroCodenotti and @SteamyRoot!
 
@Daminark Hi
 
Oh hey @TobiasKildetoft! How's it going?
 
Oh btw, when do you have to submit your grant application?
 
7:46 AM
September 15th
 
Ah, hopefully that'll be enough time
Also hey @WillHunting!
 
It should be. The main part of the project description is done now, I just need to tweak stuff
 
@Daminark Hello!
@Semiclassical Thanks, I think the lyrics are fitting in some sense.
 
Sick
How's it going guys?
 
7:51 AM
@Jon It's normal to have no calculator and no notes. In fact, that should be the way an exam goes. All those exams that require calculators and notes are stupid.
 
@Kirill already did that. Thanks for your suggestion!
 
8:49 AM
Hi @LeakyNun
 
Is $S_6 \oplus S_7$ isomorphic to $S_{10}$?
Guten Tag @Mr.Xcoder
 
@LeakyNun You mean those to be the symmetric groups?
 
Bayern? @Mr.Xcoder
@TobiasKildetoft yes
 
And the $\oplus$ should be a $\times$ then?
The first is not correct to use for non-abelian groups
 
@TobiasKildetoft yes
 
8:51 AM
Then no, since the direct product will have more normal subgroups than $S_{10}$.
 
@LeakyNun Umm. What?
 
@TobiasKildetoft can we analyze the order of elements instead?
 
@LeakyNun Sure, that will work too
 
@Mr.Xcoder "i mag ihre Deutsch" looks Bavarian to me
@TobiasKildetoft how?
 
@LeakyNun Ah, yeah, that's it
 
8:53 AM
@LeakyNun The direct product will have elements of larger orders than $S_{10}$
 
so you speak Bavarian German?
@TobiasKildetoft like 5 times 7 = 35?
 
@LeakyNun No, that is the only phrase I know in Bavarian German
 
@LeakyNun Yeah
 
@Mr.Xcoder oh, scheen
@TobiasKildetoft isn't it amazing that the orders of the two groups are the same
 
@LeakyNun What does "scheen" mean?
 
8:55 AM
@Mr.Xcoder schön
 
@LeakyNun Not particularly
 
@TobiasKildetoft heh
 
BTW I learned what "Induction" is
 
@Mr.Xcoder good for you
 
9:00 AM
Is $0$ considered a natural number?
 
@Mr.Xcoder Depends on the context
 
Oh, there is no consensus on that?
Gotta go now. I saw an answer on M.SE on that. Bye o/
 
none at all
 
 
2 hours later…
10:37 AM
@TedShifrin Yes, there is short term evolution. When you fart, you evolve gas for a short time.
 
user302782
Hi
 
user302782
I do not know why I got two down votes for my answer in this question: math.stackexchange.com/questions/2401960
 
user302782
Is there a way that I acn find the reason for the down votes?
 
@VafaKhalighi There are many crazy people on this site who downvote for no reason.
@VafaKhalighi No, there is no way. I have learnt that votes on this site mean nothing. You can ask about your answer, which I haven't read, but don't try to make sense of votes.
 
user84215
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
 
10:46 AM
Just like there is a way to ignore people in chat, there should be a way to ignore people on the main site too, with a simple mouse click!
 
@VafaKhalighi there is only one
 
user84215
This is one of my opinions in my deleted posts on the Meta MSE.
 
user302782
@WillHunting Thanks for your answer. When I got the down votes, I thought that there might have been problems with my answer and I was only curious to see what was the math problem in my answer. I think if the votes become constructive, it helps a lot to improve. Also if one down votes an answer, it would not harm to write a reason for voting down otherwise how does one improve?
 
user302782
@LeakyNun The actual question page shows 1 down vote but my own profile shows -2
 
@VafaKhalighi Exactly. I won't read your answer now, but there are plenty of right answers that get downvotes and wrong answers that get upvotes on SE in general, because almost anyone can cast a vote.
 
user84215
10:52 AM
This fact is also true in the Meta MSE.
 
user84215
Unfortunately
 
@VafaKhalighi A downvote costs you 2 points. That answers the question.
 
@VafaKhalighi your answer can be improved by actually solving for $g^{(n)}(x)$.
 
user302782
@WillHunting Thanks for the explanation.
 
user302782
@LeakyNun Thanks for the feedback. I try to edit my answer with improvements when I get some spare time.
 
user84215
10:56 AM
Please do not pursue these things. Your account may be suspended.
 
user302782
@MathematicsAminPhysics Why does my account get suspended by asking what I did wrong?
 
@VafaKhalighi It doesn't get suspended. He's talking nonsense.
 
user84215
As they have said, some users downvote and upvote for no reason. Suspending your account may be happened for no reason.
 
@MathematicsAminPhysics Now do you mean post ban (question ban or answer ban) or do you mean suspension?
 
user84215
both of them
 
11:02 AM
Because downvotes are related to question ban and answer ban, but they are not related to account suspensions.
 
user84215
How about pursuing unfair downvoting?
 
Many users asked about down-votes on their post on meta and it did not result in suspension.
There is even a separate chatroom which was created for this purpose (among other things) - the rom description says: "You can get feedback for questions, answers and comments."
 
@MartinSleziak His account was suspended for a week for something we don't know. I have decided not to talk to him anymore, because everything he says seems quite trollish and cranky.
 
user84215
You do the best.
 
user84215
Please ignore me.
 
user302782
11:06 AM
@MathematicsAminPhysics I do not get your logic for suspending account. On the other hand, I have had no complaints about the down vote; as you can see, I was only curious to find out what I did wrong so I could improve in my future answers. I can not think why would that make my account being suspended?
 
@VafaKhalighi Like I said, your account won't be suspended for that. He's talking absolute nonsense.
 
user84215
@VafaKhalighi Ok. If you have no complaint, your account will not be suspended.
 
Mods won't suspend any account without a very good reason. Asking why a downvote occurred is a very good question. In fact, anyone that downvotes should comment with their reasoning.
2
 
@Mr.Xcoder In fact, usually, even before a suspension occurs, there will be dialogues with the user.
 
user84215
@Mr.Xcoder ^^^^ To your second sentence.
 
11:10 AM
@WillHunting I know the process.
 
user84215
For me there was no dialogue.
 
They usually always send at least one moderator message before suspending an account, that explains their reasons..
2
 
user84215
No.
 
"Yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
 
@AkivaWeinberger mind=blown
"Yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
> ["Yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation] is false.
> "Yields falsehood when preceded by its quotation" yields truth when preceded by its quotation.
> ["Yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation] is true.
> "Yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
 
11:21 AM
@LeakyNun Hofstadter?
 
@AlessandroCodenotti ?
 
I was guessing the author of that quote
I've seen it before somewhere but I don't remember where
 
how would i know
 
Oh, I wanted to answer to Akiva's message
I hate using this chat from my phone
 
Just a quick conjecture, is any limit point of the set $S = \left\{ \frac{1}{n} + \frac{1}{m} : m,n \in \mathbb N \right\}$ of the form $\frac{1}{k}$ for some $k \in \mathbb N$?
 
11:30 AM
I second your conjecture
 
11:41 AM
@AlessandroCodenotti @LeakyNun
Incidentally, a "quine" is a computer program that outputs its own code.
 
@TimTheEnchanter [Approach0] says yes. Find the derived set of $\{\frac{1}{n} + \frac{1}{m}: m,n \in \mathbb{N}\}$ and prove it is such. and find the number of limit points of the set{$ \frac{1}{m} +\frac{1}{n}:m,n \in \Bbb N$}
But I do not doubt that there are many other posts about this on the main.
 
I have a problem and I have absolutely no idea how to go about it: Let $A = \{\sqrt{3}-\sqrt{2}\}+\{\sqrt{4}-\sqrt{3}\}+\dotsb+\{\sqrt{200}-\sqrt{199}\}$, where $\{x\}\text{ is the fractional part of }x$. What is $\lfloor A\rfloor$? We also have choices: a) 50, b) 12, c) 13, d) 49
@LeakyNun Can you help?
 
"is a sentence fragment" is a sentence fragment
- Wikipedia
 
RIP me
 
11:45 AM
@Mr.Xcoder inconsistent sign
 
Oh sorry, correcting
@LeakyNun Fixed
Absolutely no clue
 
@Mr.Xcoder prove that each term is less than 1
 
@LeakyNun Well, of course each term is smaller than one since it's the fractional part
 
(non-maths) student solves the differential equation $\frac{1}{\sin(2t)} x'(t) + x(t) = \frac{1}{2}x^2(t)$ by substituting $y = x^{-1}$, and finds (correctly) that $y(t) = \frac{1}{2}\left(1 + c\exp\left( -\frac{1}{2} \cos(2t)\right)\right)$
 
@Mr.Xcoder I mean, inside the fractional part
 
11:48 AM
@Mr.Xcoder No, the stuff inside it
Snaped
 
OOhhh.... Let me think
 
"Therefore $x(t) = \frac{1}{y(t)} = 2 + \frac{2}{1+c\exp(...)}$"
WHYYYYYYYYY
 
@LeakyNun How?
 
@Mr.Xcoder prove that $\sqrt3-\sqrt2<1$
@SteamyRoot I call that "not even wrong"
to be fair, you did say that said student is non-maths
 
Wait, I typoed it again. It should've been 1/2 + 1/c exp(...)
whatever
Well, yeah, but the idea is like... You solve a pretty damn difficult differential equation...
 
11:51 AM
Oh I see a quite simple way to do that (re prove it's less than 1)
 
@AkivaWeinberger of course it's simple
 
Without resorting to concavity or anything
which was equivalent to my first idea
 
And then you go all 1/(a+b) = 1/a + 1/b
 
@AkivaWeinberger in that case it would be too complicated
 
$\sqrt{3}-\sqrt{2}<1 | \text{^2}$, $(\sqrt{3}-\sqrt{2})^2<1, \implies 3-2\sqrt{6}+2<1$
 
11:53 AM
Squaring it was an idea too but I couldn't finish the thought in my head so I abandoned it
(I, uh, don't have pencil and paper handy)
 
@Mr.Xcoder begging the question fallacy
@AkivaWeinberger you just have to square it the right way
 
@LeakyNun Yeah, but it's positive, so that's really an $\Leftrightarrow$
so it's technically OK
 
Hint to get on the right track?
 
@LeakyNun (Do we actually have the same idea?)
 
@AkivaWeinberger he explicitly wrote $\implies$
 
11:54 AM
@LeakyNun Yeah but still
Whatever
 
@AkivaWeinberger no, I don't need to square it to prove it
 
$\sqrt{4} > \sqrt{3} > \sqrt{2} > \sqrt{1}$ and $\sqrt{4} - \sqrt{1} = 2- 1 = 1$
 
Well, $|\sqrt{3}-\sqrt{2}|=\sqrt{3}-\sqrt{2}$
 
I mean I'll say the thing if you're OK with that @LeakyNun
It's just one word really
 
11:55 AM
it's one word
but don't say it
 
OK sure
 
Godammit, I have to go. Will be back shortly
 
How was the eclipse?
Actually I don't remember where you are
 
@AkivaWeinberger hong kong
 
If you're in Hong Kong you didn't see an eclipse
 
11:56 AM
user image
2
 
@AkivaWeinberger right
 
(Sniped)
 
@SteamyRoot hairgenesis?
$\begin{array}{rcl}
\dfrac1{\sin(2t)} \dfrac{\mathrm dx}{\mathrm dt} + x &=& \dfrac12x^2 \\
\dfrac1{\sin(2t)} \dfrac{\mathrm dx}{\mathrm dt} &=& \dfrac12x^2 - x\\
\dfrac2{x^2-2x} \dfrac{\mathrm dx}{\mathrm dt} &=& \sin(2t)\\
\displaystyle \int \dfrac2{x^2-2x} \dfrac{\mathrm dx}{\mathrm dt} \ \mathrm dt &=& \displaystyle \int \sin(2t) \ \mathrm dt\\
\displaystyle \int \dfrac2{x^2-2x} \ \mathrm dx &=& \displaystyle \int \sin(2t) \ \mathrm dt\\
\displaystyle \int \left(\frac1x-\frac1{x+2}\right) \ \mathrm dx &=& \displaystyle \int \sin(2t) \ \mathrm dt\\
@SteamyRoot amirite
 
Looks like it definitely could be
 
How do I create olympiad combinatorics simulator in python ? e.g To solve IMO 2010 P5 (See here), you create a python applet like this: where you can push buttons and gain intuition ?
 
12:01 PM
inb4 definite integrals
 
The absolute values seem like they might be troublesome, though. I use a different approach
 
@AlexKChen just use console :P
 
What's console ?
 
@AlexKChen cmd
 
12:03 PM
Graphical handing are bit better.
 
use tkinter for graphical
for a demonstration of console interaction:
while 1:n=int(input("Enter a number: "));print(n**2)
if you're mostly doing that for yourself, then a console-interaction would be easier to program
 
Oh, so learning tkinter is sufficient for doing the grpahics stuff ?
 
Also, can I do most Simon Tatham's game on tkinter ?
 
@AlexKChen I think so
I've never used TkInter, as a disclaimer
 
12:06 PM
Oh OK.
 
@LeakyNun I'm back
Can you give me a hint?
 
@Mr.Xcoder try more first
 
@LeakyNun Ok. So I should abandon the "square approach"?
Or square it another way?
 
@Mr.Xcoder all roads lead to rome
@AlexKChen here's a console-interaction program I wrote:
 
Fancy poetry
 
12:13 PM
coins = [1]*6
while True:
	print("coins: %s" % coins)
	allowed = [(1,i) for i in range(1,6) if coins[i-1] != 0] + [(2,i) for i in range(1,5) if coins[i-1] != 0]
	print("allowed operations: %s" % allowed)
	op = 0
	while op not in allowed:
		try: op = eval(input("input operation: "))
		except: pass
	if op[0] == 1:
		coins[op[1]-1] -= 1
		coins[op[1]] += 2
	else:
		coins[op[1]-1] -= 1
		coins[op[1]], coins[op[1]+1] = coins[op[1]+1], coins[op[1]]
as you see, it can be done in very few lines of code
 
Let me do it by hand, thinking in MathJax is hard
 
@LeakyNun Why it's showing "No module named tkinter ?"
 
@Mr.Xcoder you may not be able to find the answer. that isn't the point. at least you have explored in the vast world of mathematics, instead of relying on others to give you the answer
 
@AlexKChen Because you must install it first
 
@AlexKChen what did you do?
@AlexKChen you need to do python -m pip install tkinter
 
12:20 PM
I wrote import tkinter as tk in IDLE
@LeakyNun Like what I did the other day ?
 
@AlexKChen yes
 
@LeakyNun What do you mean by at least you have explored in the vast world of mathematics, instead of relying on others to give you the answer?
 
@Mr.Xcoder exploration...
trying out different methods
failing
as they say in Chinese
 
I have no other methods in mind, but I'll seek one
 
failure is the mother of success
mathematics is not about the answer
not about finding the correct answer
maths isn't about finding the correct answer, it's about finding the correct answer
 
12:24 PM
unlike physics
 
Mr.Xcoder is surly a troll.
 
@AlexKChen i don't think so
 
@Mr.Xcoder I don't think that, too.
@LeakyNun It's stuck at "downloading/unpacking tkinter"
Now showing:
No distribution at all found for tkinter
In red ink.
 
@AlexKChen no, he isn't
 
OK I was joking :P
Why the error ^^
 
12:27 PM
@LeakyNun I have proven that $\sqrt{x+1}-\sqrt{x}<1, \forall x \ge 2$.
I think it's ok
 
@AlexKChen wait a minute, I'm figuring it out
@Mr.Xcoder how did you prove it?
 
by squaring
wait a bit
 
@AlexKChen you don't need to install it; it's already inside python
what is your python version?
 
2.7.9
 
@AlexKChen OS?
 
12:32 PM
@LeakyNun I found a mistake in my proof... Rewriting
 
Windows 8.1
 
for python 2.7.9 it is Tkinter not tkinter apparently
 
Damn $h!t
Yeah works now.
 
@LeakyNun I must specify from the beginning that $x\ge 1$. $\sqrt{x+1}-\sqrt{x}<1\leftrightarrow (x+1)-2\sqrt{x^2+x}+x<1\leftrightarrow \\ 2x+1-2\sqrt{x^2+x}<1\leftrightarrow 2x-2\sqrt{x^2+x}<0\leftrightarrow 2x<2\sqrt{x^2+x}\implies x<\sqrt{x^2+x}$. $\sqrt{x^2+x} > x, (\forall x\ge 1)\implies x<\sqrt{x^2+x} \text{ is true.}$ That implies our initial relation, $\sqrt{x+1}-\sqrt{x}<1$ holds as well $\forall x\ge 1$. Correct? I think I have been struggling a bit, not 100% sure it's correct.
 
@Mr.Xcoder do you want to know two more methods to prove it?
 
12:43 PM
@LeakyNun Yes, I'd be interested in a shorter and maybe more complete approach. I'd also like to know if mine is valid.
 
yes, your approach is valid
 
Thanks. I'll let you write / hint to the other 2.
 
first method: move $\sqrt2$ to the right hand side before squaring both sides
second method: conjugate
 
I don't know what conjugates are though
 
@Mr.Xcoder Take $\sqrt3-\sqrt2<1$ (what you're trying to prove)
Multiply both sides by $\sqrt3+\sqrt2$
What do you get
 
12:47 PM
@LeakyNun $\sqrt{3}<\sqrt{2}+1\leftrightarrow 3<2+2\sqrt{2}+1\leftrightarrow 0<2\sqrt{2}\leftrightarrow 0<\sqrt{2}$. Correct?
 
@LeakyNun (Re: the first one) Oh nice
 
@AkivaWeinberger Hm, I proved that in 2 different ways with help from Leaky. Can we get back to the question?
 
Yeah OK so they're all less than 1 so
 
@Mr.Xcoder the conjugate of a-b is a+b
 
@LeakyNun Now I know all are smaller than $1$. No idea what to do next
 
12:49 PM
@Mr.Xcoder first part: prove they are all less than 1 (done)
 
$\rm\{thing\}=thing$
 
second part: prove they are all greater than or equal to 0
@AkivaWeinberger wait, not so fast
 
@LeakyNun Well that's obvious
Can we pull up the question again though for a second
 
1 hour ago, by Mr. Xcoder
I have a problem and I have absolutely no idea how to go about it: Let $A = \{\sqrt{3}-\sqrt{2}\}+\{\sqrt{4}-\sqrt{3}\}+\dotsb+\{\sqrt{200}-\sqrt{199}\}$, where $\{x\}\text{ is the fractional part of }x$. What is $\lfloor A\rfloor$? We also have choices: a) 50, b) 12, c) 13, d) 49
 
Thanks
Yeah OK so they're all between 0 and 1, so the floor of the thingies are the thingies
 
12:50 PM
@LeakyNun Well, $(\forall x\ge 1): \sqrt{x+1}>\sqrt{x}\implies \sqrt{x+1}-\sqrt{x}>0$
 
@LeakyNun Why greater than or equal to 0 and not greater than 0?
 
@Mr.Xcoder because of what we are going to do next
third part: prove that each term is equal to what it inside its bracket
 
And now, obviously, since ($\sqrt{x+1}-\sqrt{x})\in(0;1)\implies \{\sqrt{x+1}-\sqrt{x}\}=\sqrt{x+1}-\sqrt{x}$
@LeakyNun Proven ^?
 
@Mr.Xcoder yes
now continue on your question
 
12:54 PM
Mhmmm
@LeakyNun Thanks for all your help!
 
@Mr.Xcoder no problem. thanks to @AkivaWeinberger also
 
oH yes
@AkivaWeinberger Thanks for all your help!
 
You're welcome
(I don't know why you wanted to drag out that last step but OK)
 
$\require{mhchem}$
$\ce{H_3O^+}$
It works
 
@Secret Hydronium Ion?
 
12:59 PM
yup
in The h Bar, 4 hours ago, by Slereah
Oh can we import libraries now?
Time to begin the chemistry invasion lolololololol
 

« first day (2575 days earlier)      last day (2449 days later) »