no, i need to use sets, but sets of sets also not so nice :D

@Matt: it would be great if you didn't add the (general-topology) to stuff like "is the one point in the one point metric space an open set"? I consider tags like (general-topology) or (ring-theory) or (number-theory) to be on a more advanced level than that what you learn in the first hour about a topic. For instance, I wouldn't add the ring theory tag to a question about expanding (x+y)^2 either.

hmm. is there like an elementary topology tag?

(I know it's hard to draw a clear line but this one question where you insisted is just way below that line).

@anon no, but there is a (metric-spaces) tag for instance.

oh okay

did you see this question? :P

@anon yeah, Wildberger at it again... He has a point, but he would be taken more seriously if he were a bit more moderate in his claims.

engineering? lol

@MarianoSuárezAlvarez May be a "worthless" addition to the list of unanswered questions I guess.

Now, it's calculus time. More correctly, Multivariable...

Hey @TheChaz

@BenjaminLim Hope the gravatar looks atleast a bit different now!

Whew... Another Project Euler problem done. 5 more until level 7.

@Mike Are you interested in group theory? (You asked the room about free products didn't you?)

Interested in it in general? I wanted enough to understand what I was reading. I actually went through a large portion on my brother's old abstract algebra text, which included a lot of group theory. It just didn't bear any mention about free products or such.

Which book, BTW are you looking into?

@Matt Hi

Hi Kannappan.

Pell's equation, I guess ^

I've tried asking questions here (well, not in chat...) but I haven't understood most of the answers. This is actually the second book I checked out trying to figure it all out...

@Mike Probably it helps...

First book I got was on Diophantine equations. It had some info on the topic, but said it involved binary quadrativc forms which were not covered in the book.

@Matt Can you help me with an argument in Linear algebra?

I tried some of the problems anyway. One I got stuck on and asked about here. Someone recommended the text Binary Quadratic Forms by D.A. Buell, which I was fortunate enough to find from a local library.

@KannappanSampath Sorry not now, I'm off to a lecture in a minute.

@MattN :-( Later, then.

I am not into much number theory. I will eventually though. But, if you need some help in groups, I might be able to help you.

@KannappanSampath Just skimmed through quickly, but it looks like it covers only the case x^2-Dy^2=1. That I know how to do, if not all the theory behind it. But apparently, if it equals some other integer, there can be mutiple fundamental solutions. The link I provided does not cover how to find all the fundamental solutions.

@KannappanSampath I'll be back in a few hours. Do you think you'll still need help? I can help you then.

@MattN May be not very long will I take to figure this out. But, I can't say... If I need I'll let you know, Thank you.

Ok!

@MattN Tags should also be useful for those interested in a topic for finding things. "Questions like why is $\mathbb{Z}/n\mathbb{Z}$ a ring?" is not of interest to someone who wants to see stuff about ring theory. Just because a ring is there doesn't make it ring theory. Same in this case: just because a metric space gives a topology (which you can't know if you ask that particular question) doesn't make it belong to the general topology tag in my opinion. The level definitely matters.

Is it always ths quiet here?

@Mariano A site-related help. There is a guy suggesting edits on Arturo's(!) answer. He also asked the question.

I approved it once, because, it is a substantial edit writing out what Arturo sais in more detail.

I rolled it back then because, I was not sure if Arturo would like it.

I asked him to write a new answer.

But he suggested the same edit again.

He should write another answer.

the suggested edit is four times longer than the original

and has a couple of typoes and sub-Arturian English...

@Mariano Precisely that. So, will you please reject the edit?

I did :)

@MarianoSuárezAlvarez Thank you.

I'm here to serve

:)

goodnight!

@Mariano To make sure: Did I handle the situation well? (Assuming I might be put into this, is this the right thing to do?)

@MarianoSuárezAlvarez Good night. See you later.

I don't know if there is another way to preserve the content of the edit without oing what you did

maybe ask on meta?

I've seen a couple of such draconian edits

@MarianoSuárezAlvarez I'll write a question in meta, later today.

@MarianoSuárezAlvarez They scare the hell out of me. : )

@KannappanSampath You claimed that we don't have an isomorphism $W \cong W^{\ast*}$. But if the map you defined is injective, by rank nullity wouldn't it be surjective too since $\dim W = \dim W^{\ast}$?

@Ilya Hi!

Hi guys

Oops.......looks like the weekend is still on!

@BenjaminLim Yes, I realised I goofed. Sorry!

I'd like to discuss when you're free. (I had to go to combinatorics lecture.)

Hey @Kannappan

Hi @RajeshD How do you do?

fine thanks

whats up with you

Not much. I am in the exam phase.

I am having difficult time following multivariable calculus.

yes....i used to think it is a trivial(atleast an easy) thing...but when i came to know about it...Ah....I found it very difficult.....

good grasp of basics in linear algebra is necessary for it i guess

I am looking for a good place to begin learning about them.

well for the exams...i suggest sticking to the notes

@RajeshD Yes, but, I would not say much. Upto the understanding what a linear function is; what the bases of a vector space are. Pretty much that.

@RajeshD True dat. I agree with this.

(My teacher asks us to prove propositions right from the notes. )

@KannappanSampath I am no expert in this but Ah...here you go I have downloaded them a while ago...but too lazy to view them

@RajeshD Thank you, for that.

do you have implicit function theorem and the stuff like that in your syllabus?

Hello :-)

@RajeshD No.

@robjohn Hello!

@Kannappan: this paper says that the non-existence of the projective plane of order 10 was proven by computer, much the same as the 4-color map theorem; checking a multitude of cases.

@robjohn Yes that is right. But, long before this proof appeared, it was shown that the automorphism group of the projective plane of order 10 is trivial.

Combination chapter is making me cry , I need help please

That is why I asked for which you wanted a reference. so you want a reference for the triviality?

@robjohn Yes. For the triviality of Automorphism group.

@x4d33746153706c306974 do you have a particular question?

@robjohn yes please , lemme ask

(All known constructions, well, there are not many, that lead to projective plane also lead to one with trivial automorphism group.)

@robjohn No of ways to make selection of four letter words from "Calcutta"

Do you allow letters to repeat? For instance, is cctt a word?

probably the two t's

@KannappanSampath That's not given in question :(

and the two c's and two a's

@KannappanSampath but I guess it's allowed

answer is 26 but I'm unable to figure anwer

Inclusion Exclusion should help you I guess.

Consider the two c's, two t's two a's distinct.

So, there are $\binom 8 4 $ words of length 4...

Hi @Ben

Now, you have over counted some words...

What are they?

@KannappanSampath cc tt aa

@x4d33746153706c306974 No, I am asking you how does a word look if it was overcounted?

For instance, a word with two $c\ '$s would be overcounted...

@KannappanSampath I'm not sure how would I count those words :(

OR, perhaps, this is getting complicated.

@KannappanSampath I think the intent is that there are 6 words coming from the two c's and two a's

So, we'll count word with 4 distinct letters; three distinct letters, two distinct letters, 1 distinct letter...

@robjohn Yes, I realised this. I complicated it unnecessarily,

Will the above thing work ^?

Hi @robjohn

In permutation I can easily find this by : n! / (a! b!...) : a,b.. are number of same letters in word

the question I have is is the capital C different from the lower case c?

@robjohn no , they are same , words are case-insenstive

Strange:

I am getting 26 words only.

@KannappanSampath yeah that's the answer

How did you solve it? 0_o

Oh, good.

Let me explain it to you:

In a four letter word:

1. There can be four distinct letters.

2. There can be three distinct letters.

3.There can be two distinct letters.

4. There can be one letter repeated four times.

@x4d33746153706c306974 Do you agree?^

@KannappanSampath yes

aah , I got it now

So, let's get rid of trivial case: I claim (4) cannot happen. Do you see why?

@x4d33746153706c306974 Nice.

So, you know the argument now, right?

@KannappanSampath "I claim (4) cannot happen" ??

So, to check your answers, I'll break it up:

1. $\binom 5 4$

2. $3 \cdot \binom 4 2$

3. $\binom 3 2$

4. $0$

@x4d33746153706c306974 The fourth case of the four choice I listed cannot happen.

^This cannot happen

@KannappanSampath How did that come I'm wondering

Well, if you want to have three distinct letters, it means, you have one letter repeating twice and two letters different from each other and different from the repeating letter.

The repeating letter can be $c$ or $a$ or $t$.

Didn't get

The remaining is chosen from four of the letters of "calut"...

@x4d33746153706c306974 Read again please...

I'll have vada and be back.

@x4d33746153706c306974: There are 54 4-letter words that can be made using just the c's a's and t's

@KannappanSampath I don't think this case can occur.

@robjohn Shouldn't it be 15 ? 6 C 4

hmm.......vada and sambhar!

I'm damn more confused now :(

@KannappanSampath hello

yes

so now you know that $\dim W = \dim W^{\ast}$?

@robjohn I can't believe I got like 14 upvotes on one question on the tensor product

Hey @Ben can you give a link to it?

@robjohn How did you come up with that?

@RajeshD math.stackexchange.com/questions/135208/…

@x4d33746153706c306974 there are 6 words with just the c's and a's: ccaa caca caac acac acca aacc

@RajeshD All I did was calculate some kernel

there are 12 more with just the a's and t's and just the c's and t's

Ah...pardon me I know nothing about abstract algebra or what ever that is there

@RajeshD It's ok

Then if you count the number of 4 letter words with all three, there are 36.

@robjohn you're using any formula to compute that?

@x4d33746153706c306974 Yes, but a different formula for specific cases

@robjohn so can you explain "There can be three distinct letters" case ? I'm still not understanding it

3⋅(4 C 2)

There are two kinds of letters, the c-a-t of which there are two each and the l-u of which there are only one.

@robjohn Yes, right, I put $0$ against $4$ in my break up I gave in case you missed that. : )

@BenjaminLim Yes. :)

@robjohn and what does that 4 and 3 represent in that formula ?

@x4d33746153706c306974 what 4 and 3 in what formula? I have not mentioned any formulas yet.

@RajeshD Right. Institute is an heaven. Who gives you snacks these days at the institute? chuckles

@robjohn 3⋅(4 C 2)

@x4d33746153706c306974 That is the 18 that is the number of 4-letter words that can be made from the c's, a's, and t's.

@robjohn then what about u and l ?, we're not going to consider them in word? 0_o

oh my mother , combination is really way harder :(

@x4d33746153706c306974 have patience. we have to count many cases here.

@robjohn So can we restart with my problem? :)

"No of ways to make selection of four letter words from "Calcutta""

@tb I disagree with you very much. For example set-theory and elementary-set-theory is a case where it's been solved differently. But instead of having elementary-X we could introduce an elementary tag and then tag accordingly. In any case, I think you know what you're doing so I'm happy to conform with whatever you think is right.

@KannappanSampath Ok. Still need help with linear algebra? (Be warned though, I'm not sure I can help: ))

@MattN Thanks for liking my answer on tensor products

@BenjaminLim That was well deserved : ) I saw you got a lot more upvotes too.

@MattN hahahahahahaha thanks :D :D :D

@MattN I don't think it deserved that many upvotes though

All I did was calculate some kernel

@MattN Hmmm, I figured it out a while ago. Thanks again.

@KannappanSampath Ok, very good : )

Ok bye guys!!

Bye @Ben. Take care.

We had a user while ago, with a combinatorics problem. Robjohn and I solved it. Robjohn and I attempted to explain. The user just left without leaving a word!!! Crazy!

@robjohn you there ?

@RajeshD yes

Someone to comment on my gravatar?

@KannappanSampath did we solve it completely?

@robjohn : Seems like DayLateDon is too late to respond here

@robjohn I mean, I gave all the case-wise break up no?

Is there anything we can do about it

@robjohn Completely forgot to ask you! Did the thing I gave you interest you? The t-designs and such...

Hi!

Hi Daniil

????

@KannappanSampath you mean the symmetric design paper you sent?

@robjohn Yes. That one.

I'd be happy if you can comment on how you liked it; of course, it's structure and the appearance also need improvement.

@KannappanSampath I haven't gotten a chance to read much of it. I was having trouble finding motivation for parts and so it was taking at while to push through on definitions alone. The definition of the incidence structure was a bit confusing; given $\{X,\mathbb{B},I\}$, is the $B$ used to define $I$ just an arbitrary subset of $2^X$?

@robjohn Yes, a kind of. Actually, I should admit, I made it look simple. The usual way of defining incidence, as done in most texts is rather confusing.

Let me try to elaborate:

What we are given is, $(X, \Bbb{B})$

hello

@AOAOne Hello. : )

Now, the $\Bbb{B}$ is the block set, consisting of blocks, $B$, which are arbitrary subsets of $X$. @robjohn

So, you have, $X \supseteq B \in \Bbb{B}$.

Yes, I got that.

Now, I define the incidence $I$ to be $I=\{(x,B) \in X \times \mathbb{B} \mid x \in B\}$

@KannappanSampath wait, I thought that $B\subset2^X$

can anyone help me with a counting problem

@robjohn That is a serious blunder. Let me fix that for you. Can you please give me a moment?

It seems that $B\in\mathbb{B}$, but that is only mentioned in the definition of $I$

Yes, you're right.

That is a bit confusing. I mean that I would have made mention of that separately.

So, the model is really this:

You have a set of points, $X$.

Say, $X=\Bbb{R}^2$

Then, the blocks $B$ are collection of some points.

sorry, dumb question - I=E/R, dI/dt=((dE/dt)R-(dR/dt)I)/(dR/dt)?

@JoeStavitsky do you know the quotient rule?

@KannappanSampath So $B\subset X$ not $2^X$?

yea, (f/g)'=(f'g-g'f)/g'

@robjohn Yes, so, $B$ in the specific case can be a circle or line or really any weird collection of points.

@JoeStavitsky no. $(f/g)'=(f'g-fg')/g^2$

o, d'oh

@KannappanSampath okay, so two confusing points in the beginning of the paper (one an error).

@robjohn Yes, sorry about that. But, please do not give up. I promise you, I did not write very badly.

sorry, still a little stuck. I was given E, R, and dR/Dt - they want me to solve for dI/dt

So, I=E/R

right

I,E and R are obvious

dI/dt={R(dE/dt)-E(dR/dt) }/R^2

got that

So, you need dE/dt...

right

exactly

So, this could be $0$, because you may be working with a dry cell or something...

"could be" is alas insufficient

@JoeStavitsky "could be" is a case that either needs to be considered or shown not to be possible.

can anyone help me fine tune my recursive counting problem

@Henning, can you do combinatorics?

(I have a question.)

@MattN Sometimes.

Apparently, the correct answer is 26.

Actual words or just strings of 4 letters each? :-)

Hey did I not solve it @Matt?

@KannappanSampath I have no idea, did you try? This guy invited me to another room to help him.

@HenningMakholm strings of 4 letters each...

@HenningMakholm well the person who asked me didn't know. Maybe we can do strings first and see if we get 26.

My instinct would be to start with 8! and then try to correct for double-counting the ones that have two a's, c's or t's.

Yes. The problem I tripped over was the duplicate removal : )

That would end up with some kind of three-level inclusion/exclusion tree, which feels like it should be manageable.

@MattN Look here please...

Does that help?

Hm. We also have to consider double counting of words with just one A, C, or T.

@HenningMakholm I have one slightly more specific question: a different thought I had was to first count all 4 letter words with all different letters. I have 5 different letters so I thought I get 5! different 4 letter words with all different letters. The problem is that that's already far more than 26. Is my counting wrong or do they mean actual words?

@KannappanSampath Let me see.

@MattN Good point. So they must mean actual words. Or whoever thought 26 was correct is just wrong.

@MattN "I have 5 different letters, so I have $\binom 54 $ 4 lettered strings..."

@HenningMakholm Well I wasn't sure because the other possibility is that there aren't 5! 4 letter strings... but I'm quite sure of it.

@KannappanSampath No, strings are ordered.

@MattN I thought he was not thinking about orders.

@MattN I agree on 5! -- each 4-letter string corresponds exactly to one permutation of the 5 letters (just tack the single unused letter onto the back).

@HenningMakholm But isn't counting actual words a boring and tedious question?

From a mathematical viewpoint: Yes, certainly.

My thoughts. And then seeing as this is a maths site I discarded the possibility that he might be counting actual words.

Words are those present in a dictionary according to you people?

Yes. Well sort of. But if it's a combinatorics homework question then no.

Which raises the question of which language. All of them?

So, I did not care about order; (s)he told me not care I guess!

@KannappanSampath Yes, the question stated "words" which implies ordered selections.

Now I'm still confused.

@MattN Then, there are way too many.

@KannappanSampath Unless they mean actual words. But then the question isn't mathematical.

I think the question implicitly assumes what I did assume.

Which is?

that strings are unordered collection of letters.

Then you get ${8 \choose 4} = 70$, no?

Right and then we have to remove the dupes.

How did you do that?

I approached this by looking at the number of distinct letters in the four lettered string as in the bookmarked conversation.

Ok. And did you get 26?

@MattN Yes, I did.

BTW, in which room are you discussing that?

@KannappanSampath In this one.

@HenningMakholm Mystery resolved: Kannappan's solution is right. They wanted selections without order and not real words : )

I agree with this but why is it that everyone is oblivious to this idea when the answer in question is a Bill answer?

can someone help me with my counting function

Only one way to find out.

@tb PS: I'd love to book you in for an argument about it but I know you don't have time. : )

@AOAOne does this have to do with Calcutta?

no?

calcutta?

india?

@AOAOne someone asked how many four letter words could be made from the letters in calcutta earlier

oh

no it is about dice

then ask already :-)

It's not about transitive dice, is it?

@HenningMakholm if die_1 = die_2 and die_2 = die_3...

Oops. "non-transitive", of course.

@HenningMakholm The die is cast

Oh, that is intransitive dice...

@AOAOne: did you have a question?

It is the passive voice, which requires that the verb is transitive.

yes, nontransitive dice

hey guys... what happens here to 2xdx? youtube.com/…

i asked in SE and got an answer but technically for a very different problem so i am trying to understand how it works and adjust it to my problem

Is this advanced roshambo?

kinda

basically counting valid sets of three dice where no matter which die you pick, i can pick one of the remaining two and beat you >50% of the time in terms of rolling a higher value

e.g. 112555 124444 223337

anybody?

@Joe it gets wrapped up in the du term

@Joe if you let u = x^2+4 then du is 2xdx. In your original integral, you then replace all your terms. x^2+4 turns to u and you replace 2xdx with du, leaving you with u^100 du

but is du not equal to x^2+4?

n, sorry, thats u

A group $G$ is isomorphic to $(\mathbb{Z},+) iff $G$ is cyclic and of infinite order. Does that answer this one?

the OP is looking for a simple characterization

u is x^2+4 yes. derivative of u is du. so derivative of u=x^2+4 is du=2x + 0 dx = 2xdx

that's very simple

also, where is +c from?\

@JoeStavitsky when you differentiate a constant, you get 0

+C is a term you append when you take the integral of something. it represents an arbitrary constant

ah got it

because the derivative of, say, x^2+4 and x^2+0 and x^2+242343 are all the same -- 2x

thus, when you take the anti-derivative of a function, you only know it up to an additive constant

so when you are asking for the integral of 2x, you say x^2+c

because c could be any constant (including 0)

so that is accurate u substitution right?

@AOAOne yes

@robjohn would you happen to know why my current counting function doesnt seem to work properly?