« first day (1278 days earlier)      last day (4037 days later) » 

12:00
Let me take a look
Why not write that darned thing as $\left(A\setminus B\right ) \cup \left ( B\setminus A \right )$
now why would I write is as $A \backslash B \cup B \backslash A$?
OH! probably to make it easier...
Indeed.
like $A = A $ and $B = B+C$... then I can easily commutative law that thing
@Chris'ssis Taking a look now...
12:03
Ok, I think I have done it.
and then unsub
$A+(B+C) = (A+B) +C$

Starting from the left, let $A = A$ and $B = B+C$ to have $A+B$
@robjohn I wanna show you something marvellous ... :-) (related to that question)
we know that $A+B=(A \cup B) \backslash (A \cap B)$
@Chris'ssis okay :-)
now I'm tempted to use the commutative law
is that possible?
12:05
I don't think so, no.
seeee what are you doing man? lol
@GabrielR. Some great sentences in my paper, including "A central problem in quantum Lie theory is the derivation of fi nitely ultra-empty homomorphisms. "
@robjohn I sent a comment to you.
@usukidoll I will use U for union I for intersection
You got A+(B+C)=A+[BUC\BIC]
from here you get AU[BUC/BIC]\AI[BUC/BIC]
So far so good?
:O I can't see it with latex :(
f.$A+(B+C) = (A+B) +C$\\

We need to define the symmetric difference of $A+(B+C) = (A+B) +C$\\

$A+(B+C) = A+[(B \cup C) \setminus (B \cap C)]$\\
$(A \cup [(B \cup C) \setminus (B \cap C)]) \setminus (A \cap [(B \cup C) \setminus (B \cap C)])$
12:09
Yes.
yes so far so good @Studentmath
and then it gets sticky if I distribute the A
Symmetric difference sucks.
Okay, now do you know the rule about A\B=AIB^c?
you mean $A \backslash B = A \cap B'$ ?
FFFFF
12:10
@Chris'ssis which integral?
I don't know how to apply it x.x
@robjohn that one from Mathematica related to Bessel function.
@Chris'ssis Oh, okay...
@robjohn It's very very cute (imo).
12:11
@Studentmath That makes things simple, yes.
yeah but how to apply it since I've never used it before....
@usukidoll First of all apply it on all.
HUH?! OO__OO
For every think with that blackslash, change it to intersection with the ^c (the completing set)
For example
$(A \cup [(B \cup C) \setminus (B \cap C)]) \setminus (A \cap [(B \cup C) \setminus (B \cap C)])$
$A \backslash B = A \cap B'$ using this
12:13
BUC\BIC=(BUC)I(BIC)^c
Erm, let me try with that
Which is just $(A \cap B') \cup (B \cap A')$
Suggestion : Avoid D'Morgan at all costs.
@Chris'ssis I am not familiar with those special functions, so I would have to read up on them to really make sense of your integral.
I am afraid it won't be possible
$B \cup C \backslash \ B \cap C =( B \cup C) \cap (B \cap C)'$
@BalarkaSen
12:14
I just avoided it.
And yes @usukidoll
@usukidoll Do that, and D'Morgan comes to play.
How did you solve it without d'Morgan, Bar?
and you just told me to avoid d morgan
@Studentmath Using the fact that symmetric difference operator is just $(A \setminus B) \cup (B \setminus A)$
And also, I prefer Balarka.
12:16
Question is, can he use that fact..?
@Studentmath Sure, why not?
Can't you, @usukidoll?
maybe let's try it
It's not defined in his question. I would be careful using it without proving it. Anyhow, proving it is rather easy too.
ok then let's prove it
Proving that is toast.
12:17
or which ever is easy
So if you want to avoid d'Morgan it might be wise, yes.. @BalarkaSen
I'm just print screening..did enough latex and it's almost 2:30 in the morning
7 pages worth of latex good gawd
Get to sleep.
Go with Balarka, Usu. It's easier. And yeah, get sleep.
no this and the whole assignment exercises is due on the third
which is now tomorrow
at least guide me towards the end of f
then I can just print screen and latex it tomorrow or today rather
this exercise sucks a through h geez. and since I saw that g is long... h is gonna be hardcore
so how to prove that it's associative?
12:20
Okay, hear me out :
Okay, you need to prove that AUB\AIB=(A\B)U(B\A)
I'll let @BalarkaSen go
$A \cup B \backslash A \cap B = ( A \backslash B ) \cup ( B \backslash A)$
$A + B + C = [[(A\cap B')\cup (B\cap A')]\cap C']\cup[C\cap[(A\cap B')\cup (B\cap A'))]]$
@BalarkaSen I think I found a brilliant solution to your question. I'll probably add the question in my book.
@Chris'ssis Fire it.
You are writing a book?
12:24
@BalarkaSen I'd like to publish a book with problems.
@usukidoll @Studentmath Keep a look out for typos.
@Chris'ssis W.. Wonderful. Just don't give a copy to me.
I might get a heart attack.
@BalarkaSen this will be a bit difficult since I don't have a background in mathematics. I collaborate with one of my former professors. I hope he can help me here.
@Chris'ssis Have you tried writing papers?
Far better than books.
then what happens?
The theory being one at time.
@usukidoll I did it for $(A + B) + C$, do the same for $A + (B + C)$
I'll leave the evil work to you.
12:27
Hi everyone, I would like to understand an exercise that was posted here : math.stackexchange.com/questions/652384/…
ha ha I just start on the left to achieve the right..
by lattice point he means a integer point ?
Thank you in advance for your help
I'll try this again after I sleep... going to bed night guys
Nighto. did much set theory today.
@Chris'ssis Are you collecting real analytic problems only?
@BalarkaSen I mainly create, not collect. Not only.
12:32
Yes, I know that.
Have you considered elliptic modular double/triple/quadruple periodic beasts?
@BalarkaSen no
brb
that's sad, since I would have been much interested in that case.
And in reference [24] and [26], I published something with my father, it seems.
Also, [21] refers to elliptic combinatorics. Much of my likes.
I think it is a integer point, just someone to confirm please.
In [8], there seems to be someone named X. B. Descartes. Odd...
I am seeing a lot of 'groundbraking' in the paper. Is it intentional?
=D
[13] refers to pseudo-finite lagrange primes. What in the name of Merlin is that?
The end of the paper says my derivation of reducible, unique modulli was a milestone in topology.
And also, something odd...
"It is well-knonw that the Riemann Hypothesis holds"
Yeah everybody know that!
Does anybody know of a reference on differential galois theory at the reach of an undergraduate?
12:48
I learned differential galois theory from many books, but I can't recall any of them at the moment.
Mostly I learned it from notes.
But it wasn't really much use to me.
So you know how to prove that e^{x^2} has no "simple" antiderivative ?
Yes, but it's quite a much tough.
Liouville is definitely superior in these works
that differential galois theory.
Yeah that's the gist of it
Are you taking part to IMO?
I remember reading an extension Liouville's works to definite integrals
@GabrielR. No. More work less thinking.
I hear whispers of disagreements.
Hey, @Jasper
@BalarkaSen Hi.
12:56
How many lhfs been killed today?
None, taking a break.
Bye, Precious! gollum, gollum!
Hello, Ramanujan!
Okay, I need to go, bye.
13:37
^ is this kid legit
What is the correct way to label the sides of a rectangle?
For triangles the convention I have been taught is to label the sides by small letters opposite to the corners eg
Is there a simmilar convention for squares?
who cares
lol, call it whatever you want
13:56
Is this true?"Continuous distribution function if and only if continuous probability density function." and "Discrete distribution function if and only if discrete probability density function" ?@robjohn
@Sush I am not sure what you are asking
@N3buchadnezzar with letters
@robjohn In a rectangle with sides ABCD
what should the sides be labeled as?
Can you understand what I ask? If not, please let me know so that I can correct wording, please!@robjohn
Can you understand what I ask? If not, please let me know so that I can correct wording, please!@robjohn
@N3buchadnezzar l and w
14:02
@Sush it seems either obvious, or I am not getting it
@robjohn, NP :)
Are you asking what the relationship is between the distribution function
and the probability density function?
hi @Charlie
@skullpatrol, I am asking the relationship between CONTINUOUS distribution function
and CONTINUOUS probability density function. As well as the relationship between DISCRETE distribution function
and DISCRETE probability function.
Can you please help?
@skullpatrol
What are the definitions of each function?
By the way the name BADASS was better than skullpatrol. I think you changed it because there is porn site named as badass, right?
14:12
No.
@skullpatrol, definition? I will have to LaTeX!
@skullpatrol And then the sides $AD$, $DC$ ... should be labeled as?
A clockwise labeling seems very strange since everything that has to do with angles and alike is measured counterclockwise
Just as you have said AD, ...
or AB, ...
@skullpatrol The verticies in a polygon are labeled with lowercase letters, which you should know
14:18
I am annoyed. Thomas/Finney Calculus doesn't contain a proof for the triangle inequality
@BalarkaSen I have an answer to this question, but it is too long for chat. If you ask a question, I can post the answer there.
Fermat couldn't have said it better :-)
@Alizter They probably expect that to have been done in high school Algebra.
@skullpatrol I proved it. Now for the general case of n numbers
@skullpatrol The answer I have is 695 characters of LaTeX and chat won't let me post it
@robjohn Wow
14:33
@skullpatrol I added a line to make a step clearer, so now it is up to 756
@robjohn Put it on mathbin
@Chris'ssis: have you worked on BalarkaSen's question?
@robjohn Just a bit. I have guests in my house now. Did you work on it?
@Chris'ssis I have an answer written up, but it is 756 characters of LaTeX. I guess I can try to split it up.
@robjohn Really? That's awesome.
14:37
@Chris'ssis I see that BalarkaSen has not asked any questions. I said if he posted it as a question I could post my answer.
@robjohn I'd be very glad to see you answer.
Ah, I don't get the too long message if I post a multiline answer...
@BalarkaSen Here it is:
Using
$$
(1-x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}4^{-n}x^n
$$
and
$$
\int_0^1\log(x)x^n\,\mathrm{d}x=-\frac1{(n+1)^2}
$$
we get
$$
\begin{align}
\sum_{n=1}^\infty\binom{2n}{n}\frac{H_n}{4^n n}
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\binom{2n}{n}\frac1{4^n n}\left(\frac1k-\frac1{k+n}\right)\\
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\binom{2n}{n}\frac1{4^n k(k+n)}\\
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\binom{2n}{n}\frac1{4^n k}\int_0^1x^{n+k-1}\mathrm{d}x\\
&=\int_0^1\sum_{k=1}^\infty\frac{x^k}{k}\frac{(1-x)^{-1/2}-1}{x}\,\mathrm{d}x\\
2
@robjohn Geezzzz, that's why I say you're absolutely brilliant! Very nice!
Nice, thanks, @robjohn.
@BalarkaSen @robjohn is a math god. (it's good to know that)
14:45
Correction : Integral and series god.
Am I right?
@BalarkaSen no. He has brilliant answers in many areas.
@robjohn thanks for sharing this.
@BalarkaSen not a god, but I do answer a lot of analysis questions
You haven't yet answered by $k$-height double factorial sum question $$\sum_{n\geq 0} \left [\frac{(2n-1)!!}{(2n)!!}\right]^k x^n$$
@BalarkaSen I think people must be appreciated for what they have already done, not for what they didnt' do yet.
(I am appreciating it, I just wanted robjohn to answer this one. Don't get me wrong)
14:50
@BalarkaSen there are always questions where we need to think of.
I have heard of robjohn in many of the forums I am in, and I respect his mathematical abilities. That is why I am asking him.
@Chris'ssis I know, but I am sure he would think more if I ping him more =D
@BalarkaSen with answer or not, he's definitely a very rare piece here. (sorry @robjohn, I don't wanna create an embarrassing situation but I need to say some things)
@BalarkaSen I think he is definitely in the top of MSE users. (I'd say more than that but it's ok to let things this way)
He is a very rare piece of mathematician, indeed, I don't disagree.
@BalarkaSen Let me show you something
I would be glad if I ever got even a bit of a googolplex part of his mathematical ability.
14:54
@BalarkaSen I don't know if I've ever seen a nicer proof than this one for a triple series - mathematica.stackexchange.com/questions/39228/…
@BalarkaSen that is highly difficult and it's done in a very brilliant way!
@BalarkaSen for me was a BIG lesson to learn.
@BalarkaSen: you are aware that $\frac{(2n-1)!!}{(2n)!!}=\binom{2n}{n}4^{-n}$, right?
@robjohn Indeed.
@robjohn I've understood my error and decided to delete the answer because it would need to be rewriten almost completely. I saw your answer: you've managed to write the antiderivative using only two cases. I've found it nice!
@Chris'sis robjohn is a brilliant mathematician, as I have said before, and it doesn't surprise me to see such brilliant answer from him.
@BalarkaSen yeah, that's true.
14:58
Actually, the most brilliant thing I have ever learned from robjohn is evaluating finite sum using residues.
@AméricoTavares I needed the second case just because things get weird when $\sin(x/2)=0$
@AméricoTavares Your answer is discontinuous, right?
@BalarkaSen I'm glad such persons exist. (unfortunately too rare)
@robjohn Yes.
brb, my guests...
@Chris'ssis Yes.
Even the members of I&S uses his methods from time to time.
The evaluating-finite-sum-using-residues is one of those methods which RV used while solving a problem.
He learned it from robjohn too.
15:03
@AméricoTavares The problem is that you are integrating a piecewise defined function (pieces of functions whose antiderivatives we know).
@AméricoTavares the constants of integration do not need to be the same on each piece of $\mathbb{R}$
@AméricoTavares they should be chosen so that the final antiderivative is continuous
(Whew, @sis went like fire when I didn't appreciate @robjohn enough!)
@robjohn Yes, that was the problem. As it was the antiderivative would be valid only in an interval not including any jump. I've found two of those constants to make it valid in the interval $]-\pi,\pi[$.
@robjohn Thanks for your explanations.
@AméricoTavares Since the integral over $[2n\pi,2(n+1)\pi]$ is $4$, the jump needs to be $4$ each $2\pi$. That is why I have $4\left\lfloor\frac{x}{2\pi}\right\rfloor$
Now, I need to walk an impatient dog :-)
BBL
15:39
@robjohn You have dogs?
Hi!
Were $\limsup$ and $\liminf$ introduced because they always exist?
@Dave
I asked a statistic question yesterday that didn't get much love, can anyone check it ?
0
Q: Chebychev to get at least 90 %

DaveIm am given a list of integers and asked to give the interval that contains at least 90 % of my values. Values : $62,56,72,83,66,77,62,71,50,58, 74,81,76,67,70,70,69,67,80,81, 74,53,73,55,66,88,73,61,63,70, 72,63,75,68,78,75,61,69,80,82, 87,57,74,74,85,68,75,63,81,73$ At First, I found the fol...

 
1 hour later…
16:58
How was tennis @TedShifrin?
Is it valid to multiply one side of an inequality? For example $a>b\implies 3a>2b$
Nope
$a=-1, b=-1.1$
@Mike Let's say for a positive $a$ and $b$
In that case it IS true, but not for the reasons you want
$a>b \implies 2a>2b$
Since $a$ is positive $3a > 2a$
So $3a > 2a > 2b \implies 3a > 2b$
Any 'other' inequalities you get will just be weakenings of the ones you have
Or... false
@Mike I found a work around
i need to prove $2^n<n!,\qquad\forall n\ge 4$
Let $n=4$
17:13
Ah
That's using the trick I told you ;)
We have $2^n=16<4!=24$
So without donig any 'both-side' multiplication, let's look at the inductive ste
So holds for $n=4$
$2^n < n! \implies 2^{n+1} < 2 \cdot n!$
@Mike I got it but I need a check
17:14
Yep
Assume the theorem holds for some $k-1>4$
then clearly $2^{k-1}<(k-1)!$
therefore
$2^k<2(k-1)!$
and we have $2(k-1)!<k!$
because $k-1>4$
Therefore by induction the therorem holds?
That's not quite what you want at the end there
You want $k>2$
Since $k!$ is $k \cdot (k-1)!$
But obviously you have that.
Hmmm would it be unrigorous to not state $2<5$?
Are you talking about the base case?
Oh
I see what you mean
Nah, you can just say $k>2$ because $k-1>4$
If someone docks you points for that they're a dingus
Alright thanks :)
I am having trouble finding an example of where induction cannot be used and strong induction has to be used instead.
17:27
3 weeks ago I could give one to you, easy :P gimme a sec
OK
Imagine you have a $2 \times n$ chessboard
As a function of $n$, how many tilings does it have by dominoes (each domino face coveres two squares)?
Prove your formula
There's also a much more important example I can give after this
Whats that?
Oh, you didn't like that one? :P
It's hard to imagine for me :P
(The answer, perhaps surprisingly, is the fibonacci sequence :))
@Mike: Misted out. :( Need to go for a brisk walk.
17:33
(to prove it you need to know that it's true for $n$ and $n-1$ - weak induction won't do it)
@Ted It's never a :( when one goes for a walk
@Alizter The other one is the fundamental theorem of arithmetic: every positive integer has a unique prime factorization.
@Alizter: Given $a+1/a\in \Bbb Z$ prove same is true for all $(a+1/a)^n$.
@Ted I think you may have meant to write differently there
No, @Mike, the :( was for the mist.
The idea is, if $n+1$ is prime, clearly it has a unique prime factorization; if not, it's divisible by some number $a$, and we can factor $a$ uniquely and $\frac{n+1}{a}$ uniquely by strong induction
@Ted You mean to say the same is true for $a^n+ 1/a^n$
What you wrote is rather trivial
Oops, duh, yeah.
17:35
Howdy folks. Is this right?

How many ways are there to pair off 10 women at a dance with 10 out of 20 available men?

Starting with the first woman, we can pair her with one of 20 men. Once we choose one pairing, we can then pair the second with one of 19 men. This continues until we get to the last woman, who can be paired with one of 10 men. So the total number of pairings will be given by 20!/9!
Though perhaps indicative of a method of proof :)
@agent154 Last one will be paired with one of 11 men
@Mike AHHH Good one. Of course.
Oh yes, my mistake
so $20!/10!$
Perhaps a better way to think about it is to give every woman a number (what a terribly misogynistic proof) and then assign each man either a number or nothing
In other words, we've picked - in order - 10 men
So that the number is $P(20,10) = 20!/10!$
Could someone help me retag this question?
0
Q: How to show that alle graphs share a point

Max AdaloudisI'm currently preparing for my exams and I do not this question, could someone explain it to me? $f(x)= \sqrt{px-4p+4}$, the graph has $5$ 'branches' left, and $5$ on the right. Now these all come together at a certain point, and the question is what point exactly. Here's what the answer sheet t...

17:41
Hi @ted
I always wondered it's very nice having formulae such as $\sum^n_{k=1}k=\frac{n(n+1)}2$ that you can prove and all. But how does one discover them?
Hi @Charlie
I'm pretty sure none of the tags are appropriate, but I can't understand the question well enough to figure out what it's asking
How is it going @ted?
@Alizter Look at the output and see if you recognize it.
17:42
fine, thanks, @Charlie, and you?
Here's a problem I'm not sure how to attack:

How many ways can a committee be formed from four men and six women with four members, at least two of whom are women, and Mr. and Mrs. Baggins cannot both be chosen?

How do I think about two very specific elements from two different sets?
Difference equations @Alizter.
Would I have to consider 5 cases, one of which is all women, and the other four are either 2 women with/without Mrs. Baggins, and 3 women with/without Mrs. Baggins?
@Alizter check this PDF out
L-L
L-L
@BalarkaSen How did you learn so much so early ? Tell me about your experiences, your motivation, and the guidance you received. Also how were you able to grasp such advanced topics so early(easily?) ?
17:45
So it'd be

$${6\choose 4}+{3\choose 1}{6\choose 3}+{4\choose 1}{5\choose 3}+{3\choose 2}{6\choose 2}+{4\choose 2}{5\choose 2}$$?
ugh
@agent154 Surely one considers only making a 4-member committee of 3 men and 5 women? If one can't pick Mr. or Mrs. Baggins then you're just scrapping two candidates
Unless you meant they can't both be in at the same time
(also the ugh wasn't at you, I just realized it might seem that way)
Yeah, they can't be in the same time
Ugh$^2$ @Mike :)
@agent154 Surely you just want inclusion-exclusion then. Total committes minus committees they're both in
@Ted I should probably finish up simplicial = singular today.
Simplicial approximation?
17:51
@BalarkaSen I have one dog, yes.
I'm not actually there yet, to be frank :P
Did you ever do that quaternion exercise?
Nope!
bad boy.
That's correct.
17:54
Shakes head forlornly.
It's okay, @TedShifrin, I'll just start tiling $\Bbb R^3$ with those cubes...
LOL
OK, i'm outta here ...
See ya, time to read
18:12
hey @robjohn
would be much appreciated if you took a look at my Chebychev question
@L-L Too much of a broad question.
18:27
119 users currently talking in 50 rooms. That's an average of 2.38 users in each room. Two guys are having a conversation per room and a third guy is merely listening or so.
2
18:39
Hi @Charlie!
Hi @jasper
 
2 hours later…
20:13
hey, i've got a pretty elementary terminology question. when people refer to "the Jacobian" (like, in calculus) do they generally mean the matrix of functions $$\left(\begin{array}{cc}\partial_x g_1(x,y) & \partial_y g_1(x,y) \\ \partial_x g_2(x,y) & \partial_y g_2(x,y) \end{array}\right)$$ or the matrix of values of these functions evaluated at a certain point,
i.e. $$\left(\begin{array}{cc}\partial_x g_1(x,y)\mid_{(x_0,y_0)} & \partial_y g_1(x,y)\mid_{(x_0,y_0)} \\ \partial_x g_2(x,y)\mid_{(x_0,y_0)} & \partial_y g_2(x,y)\mid_{(x_0,y_0)} \end{array}\right)$$?
the first
20:56
@AlexanderGruber The total derivative, usually.
But I call the Jacobian the determinant of ${\bf Df}$.

« first day (1278 days earlier)      last day (4037 days later) »