My Mendelson book is missing. AHHHHHHHHH!

@PeterTamaroff Is that where you learned about limits via directions?

@peoplepower Yep!

I really liked that.

Brian Scott told me about yet another approach which I cannot remember.

Filters?

He notes in the preface what I noted at the time, that it is related to (ultra)filters.

@peoplepower Shilov does? I never read the preface.

Brian told me what a filer was, I think, but I have completely forgotten about it.

@PeterTamaroff Yes, in a parenthetical remark.

@peoplepower Where? Can't find it,.

Page x of the 1996 edition. He refers to "'filters' of Cartan"

Oh, there it is.

@peoplepower I found it nice how directions apply to Riemann Integrals.

@PeterTamaroff I don't know this application (like I don't know most of analysis).

@peoplepower Well, say I give you an interval $[a,b]$

Then let $\mathscr P$ be the set of all partitions of $[a,b]$.

Right.

We define the mesh a the partition (finite) $P=\{a=x_0,x_1,\dots,x_n=b\}$ as $$\Delta(P)=\max_{1\leq i \leq n}|x_i-x_{i-1}|$$

Is the author's description on page 275?

Kinda.

Yes.

Then if we define $E_\delta =\{P\in\mathscr P:\Delta(P)<\delta\}$

The set of $\{E_\delta \, ;\delta >0\}$ defines a direction.

Yes, that is interesting.

@peoplepower Yeah.

@peoplepower Does this feel long-winded?

I really like the $\Leftarrow$ direction of Hausdorff's criterion proof.

Never thought one could give a proof by arguing an "algorithm" should terminate.

what implications can I deduce from the simplicity of some group?

do you know any nice examples?

if it's finite, you can deduce ... that it is one of the groups in the Classification

I meant something outside of group theory

and yes finite

@anon The Holy Classification.

@PeterTamaroff I like it, though I cannot judge it very well.. you know because of my lack of knowledge in that area. I like the $\Leftarrow$ direction particularly well too.

@peoplepower Right.

@JacobBlack OK?

@JacobBlack I like peach jam.

how can I visualize projective line over F_7?

the projective space of a 1-dim space is just a single point no?

yes, but projective line means we start with 2D space and quotient it

that's just an oddity with the notation people use

right, right

what is a projective line?

ultimately, I think finite structures are where geometry degenerates into combinatorics and have no obvious truly visual meaning.

but I want it to :(

maybe there's a different more combinatorial way to see it

13 more days...

Wow, posting completely different answers is scary. There's an uncertainty of how they'd be responded to.

commons.wikimedia.org/wiki/Projective_line_over_a_finite_field

@ethereal :)

"Intuition on fundamental theorem of arithmetic"

sigh

Lol. And both answers seem to strap it on some other result our OP may not have intuition for.

@jorge hello

@charlie wasssaaap

@JorgeFernández everything fine, you?

I totally struck out with a girl in the bus

but apart from that everything is great, oh and I might have overdone it at the gym

I'm so tired I can talk properly

@JorgeFernández fascinating

is it standard practice to downvote Meta Questions that you disagree with?

@Lepidopterist Usually, yes.

by the way I mean the school bus, so I was talking to her and she took out her ipad and started playing this game (candy crush saga) and I started to hel her out. And I sort of made her lose. And it's one of those games where you have a number of lives and you need to wait half an hour to play again. And I ran out of things to say. So it was awkward silence the rest of the way.

@Lepidopterist I once got a question $-8$-ed.

isn't it standard practice to explain downvotes?

@JorgeFernández she can't play and that's your fault?come on

@Lepidopterist In meta, I don't think so. They usually mean "I a disagree" or "I'm not in favour of this".

i see

tough crowd

i've activated the ego-defenses, though

i suspect i'll lose and beeping and booping will drown me out

@Charlie no, she's kind of good, It's sort of like bejeweled but without time, so you have to think a lot about a move apparently, I just saw a regular move, went for it and the GAME OVER title showed up on the screen and I made my "oops" face

I'm not really an expert with the ladies, one of the things I do sometimes is pretend to be an idiot and get them to explain class stuff to me.

@JorgeFernández geez I could talk to ya play mortal kombat against shao khan and m make fatality

@Charlie well girls in my school seem to be more interested in shoes and other stuff I'd be considered gay if I started talking about

@JacobBlack Women only seem to be attracted to me for my extremely handsome looks and my rock hard body. It pisses me to no end (jking)

@JorgeFernández that's why I don't talk to girls, I don't have what to talk about

@charlie but you are a girl, doesn't that make things easier

?

@Lepidopterist Meh, try not to worry much.

I have sort of like a Raj complex, my voice starts coming out really weird.

@JorgeFernández no, makes it unbearable

no, shaky and higher pitched than normal.

That's life

It doesn't bother me a lot, After a while my voice goes back to normal, but that's only if I really get interested in the conversation which rarely happens. However at least I try to get a conversation going.

if {.} is the fractional part of x, how can I show for all integers a, $$0<{\{\frac{x}{a}}\}-\frac{{\{x}\}}{a} $$

x not an integer

@Ethan what is the fractional part?

modulo 1 operator, it removes the integer part of x

oh, so like $x-\lfloor{x}\rfloor$

so {5/2} would be 1/2

yes

well I am trying to show for all integers a, $[\frac{[x]}{a}]=[\frac{x}{a}]$

it just suffices to show the last quantity is greater then zero

@JacobBlack Go to Jonas' starred message, I used lol around there too.

Slow night tonight, on main...Anything noteworthy happening here?

@user58512 yes, I see that for prime p the projective lines are integer multiples of a vector "wrapped around," yet I don't see how to do this for higher powers of primes, it is not easy to work with mentally, and is infeasible by hand when p is large.

hi

im having trouble in number theory when they ask questions like if a square of an odd number is of the form 8k+1. Questions that ask about number form. Im wondering if there are other resources that could help me understand it.

what exactly is the trouble?

Hello

Have you seen some software to teach propositional calculus?

To teach how to simplify formulas following the rules etc

...you could ask @robjohn about that

@anon about what?

@anon Ah...

@leo There is Logic 2010 from UCLA.

@leo Barwise and Etchemendy (sp?) have a text which includes software ("Tarski'w World"), and other programs for proofs, propositional logic, predicate logic, natural deduction. I TA'd for a course using that text.

@robjohn How goes it? Are we going to see a green mean face at St. Patrick's Day?

@amWhy I've tried green. It doesn't work. Perhaps I will experiment with other shades...

@robjohn I can imagine, it might be hard to see the frown and mean eyes, meanie!

@amWhy Does this look okay?

@robjohn Looks great! As discernible as the current face! ;-)

@amWhy It wasn't the visibility, it just didn't look right.

@robjohn It's funny, since to me your gravatar seems to portray an "alter ego", in the sense that you come across as so likable...

@robjohn So the contrast is rather funny, though there's something to be said for not letting too many people know one's a really nice person...preemptive defense, so to speak.

@amWhy I am a mean square >8(

@robjohn Okay, we'll pretend ;-)

)8<

proofs that involve divisibility I have trouble with, such as 24|(n)(n+1)(n+2)(n+3)(n+4) or if a number is of the form 3n+2 then it has prime divisor is of the form 3n+2. Im using burton's book and he seems to put alot of these type f questions in there. I really don't get how to approach these type of questions.

are you familiar with modular arithmetic?

a bit yes not too well

is that how you approach these type of questions?

yes

in the first case, {n,n+1,n+2,n+3,n+4} are five consecutive terms, so one of them must be a multiple of four, at least two of them must be even, and at least one must be a multiple of three - can you see these things?

yeah i can.

in the second case, you assume otherwise, write 3n+2 as a product of things equal to 1 mod 3, then reduce mod 3 to get 2=1

yeah how did approach that? because I don't see that

Also, rule out the case of a factor being =0(mod 3).

@user60887 if both a & b are 1 mod 3, then a*b=1*1=1 mod 3, and hence multiplying any number of things congruent to 1 mod 3 will give you another thing that is 1 mod 3

im not sure why I have trouble see these things.

yeah i understand that now.

@user60887 That is a nice book, good!

yeah i know its a good book. Hey @anon when you first were learning to solve problems like these I pointed out what did you do to get better at them?

@anon Hmm I am pretty sure the right hand side with the $v$ term is non-zero

@anon The right hand side of the last equation here

@user60887 practice

i will. i just hate giving up. I tend to struggle on one problem for an hour till I dont get it then I give up

@user60887 with elementary number theory, you could easily get away with seeing most of the clever tricks at least once before you do exercises where they're required. so, perhaps instead of trying your hand at problems at this stage, simply read up and see the overarching themes and approaches in the solutions that other people have already done.

while doing math is the best way to learn math, it is not always the most efficient way to begin learning it

oh ok thanks. so basically look up similar problems? ill do that then

Anyone active?

Hello?

Hello

You have any familiarity with tridiagonal systems?

No sorry.

huhu

maybe you still get the great answer badge for the one on (-1)^2=1 thread

Hello everyone.

@JacobBlack What is your definition of lhf?

@awllower hi :)

what is something like C(8,3) ?

chose (8,3) ?

I think it means the number of ways of choosing 3 objects from 8 ones.

I t is my guess...

Let U be a set with eight elements, $C(8,3)=|\{S\subset U:|S|=3\}|=\binom 83$.

@anon What is an efficient way to begin learning it? Reading about it?

@JacobBlack Ah. I'd say I focus on lhf too then, but just not as frequently.

hello

Hi

@awllower, I posted a counter-example for the galois thing, but it is quite ugly compared to the conceptual solutions

I tried once to construct counter-examples as well, but failed to do so, due to some bad choice of elements and orderings.

now im thinking about the fano plane

unrelated

Hehe

it's really got me hooked

What a pity that I do not know what it is. :(

ill show you :)

i have these two different diagrams of it

and it has the same symmetry group as en.wikipedia.org/wiki/Klein_quartic

@Jonas Hello! Are you around? :)

Or, @robjohn too...

@kan Hey there!

Hi @robjohn!

@kan How are things going?

How are you doing? Things are on well!

Really glad that I have learnt some linear algebra finally.

@user58512 It seems intriguing!

it is very.. I was dreaming about it :D

:D

@robjohn I was going to ask something in connection to weirstrass approximation theorem.

@kan sounds interesting :-)

@robjohn I have a continuous function on a locally compact metric space, [0, oo).

So, now, I want to approximate that function by functions of the type e^{-x} p(x). where p is a polynomial.

Wait, I think I got it now....

let's see...

@kan :-)

these are so strange en.wikipedia.org/wiki/…

how about approximating e^x by polynomials also...

I think that works.

Quite a surprising result indeed. !!

bye!

@kan I don't know what your question is...

@JacobBlack That is the intent, yes.

> PSL(2, 7) can be thought of geometrically as a group of symmetries of the projective line P^1(7);

can you help me see this?

from here en.wikipedia.org/wiki/PSL%282,7%29

@robjohn Hah! the question is to find a polynomial p(x), given eps> 0, such that |f(x) - e^{-x}p(x)| < eps.

f is a continuous function on [0,oo).

@kan, that seems really difficult as x grows

especially if f(x) does not tend towards zero rapidly

@user58512 you have a polynomial that can grow arbitrarily large.

it can only grow polynomially large

@user58512 So, I think take polynomials h_n(x) ---> e^x. And, p_n(x) ---> f(x). Now, notice you have e^{-x}h_n(x)p_n(x) ---> f(x).

@robjohn ^^. Did I make some sense ? @user58512

I doubt that h_n(x) ---> e^x exists

Weirstrass. e^x is a continuous function too...

doesn't weierstrass only hold for a bounded interval?

Taylor gives you one.

@user58512 locally compact is good enough.

oh, I didn't know that

I mean, isn't Taylor giving you one? 1, 1+x, 1+x+x^2/2,...?

The error of that diverges as x grows

Hmm, quite. What should I be thinking of then?

I really don't believe such a polynomial exists, even though you're telling me it's a theorem

@robjohn might have some pointers and comments about the mess I have just created.

@user58512 could you please look up stone weirstrass theorem?

why? you told me it already

@user58512 Well, if you believed what I said, I am lost then.

are you letting the degree of the polynomial depend on epsilon?

@user58512 the polynomial depends on epsilon. Given an epsilon, I should get you a polynomial. That is all.

ok then it's possible

Heya

@N3buchadnezzar Hello!! :)

Long time no see Kanapanananana

True! How are you? How is university?

It is good, tough though.

And you?

@N3buchadnezzar I am doing fine enough. But, not great! Let's see how this pans out.

BRB.

How many 6 digit numbers contain at least 3 equal digits?

I was thinking $$N = 9 \cdot 10^5 \, - \, \left[ {6 \choose 0} \cdot 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \, + \, {6 \choose 1} \cdot 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \, + \, {6 \choose 2} \cdot 9 \cdot 9 \cdot 8 \cdot 7 \right]$$, correct, wrong, maybe?

Hi. I remember that someone posted a few days ago the name of a site like MSE, but I cannot recollect it right now. If any of you recollect something about that pls let me know. Thanks.

@Chris'ssisterandpals, hi did youg et my message

@N3buchadnezzar Ahem.

@kan Since they are fried in a pan! It is a joke (yoik), laugh I say, laugh!

Haha

@N3buchadnezzar :)

Man!

@kan Woman!

@user58512: hi. I'm here.

@user58512: which message do you refer at?

@kan Yes.

@JonasTeuwen Wow!! Well, I had a question I resolved wrongly. So, I am still thinking about it.

@Jonas Do you want me to describe that or you had looked at it?

@Chris'ssisterandpals, I said you are the best so if there's a problem with the site we should change the site instead of losing you

also I had a limit to show you but I've forgot it :(

bye

guys, does anyone know about the Banker's discount problem?

@user58512: are you around? You can let the limit here and read it when I have time. Now I'm a bit busy with a new class of limits (discovered by me).

I saw some days ago (not more than 2 weeks) the name of a site that is very similar to SE (I don't remember the user that posted it). I appreciate if you can let me know which this site is. Thanks!

@kan I don't think this can be done without some extra constraints on $f$

@robjohn Hmph. No more is given... :(

@JacobBlack Hi, Jacob! How's it going?

@JacobBlack Yes.

@kan Take $f=1$, then I don't think you can uniformly approximate it on $[0,\infty]$ by any $p(x)e^{-x}$.

@kan Hi

@JayeshBadwaik Hi!

I am in Bangalore right now. Will leave tomorrow morning though.

Had come for the TIFR CAM interview.

@JayeshBadwaik Oh, wow! Should have told me sometime before!! :(

@JayeshBadwaik Hope you did a good job!

@robjohn Thinking about it...

@kan I think I did, lets see.

Bleh the OP of this question unaccepted my answer to accept an answer which says "feed the problem to Mathematica" >8(

@kan Yeah, I could have. I did not have much time anyway. May be next time.

)8<

@JayeshBadwaik Sure.

@kan Take $\epsilon=\frac12$

@robjohn I was with 1.

@kan You can do it with $1$, but you need to be able to do it with any $\epsilon$

@robjohn That's a shame. Your answer seems superior.

Hmm, how'd you prove that? The polynomial shoots off to infinity, although $e^{-x}$ decays to $0$.

so, your argument should be something different from taking limit of that inequality as $x \to \infty$ and concluding that $\epsilon > 1$.

@robjohn Am I making sense?

Oh, yeah, may be it is legitimate too. Since $\frac{p(x)}{e^{x}} \to 0$ as $x \to \infty$...

@robjohn, so, I'd presume this means, if $f$ can be approximated by a $p(x)e^{-x}$, then, $f(x) \to 0$ as $x \to \infty$?

hi

hi

Oh dear, it is dinner time. @robjohn, would you still be around later?

half an hour from now?

@kan for any polynomial, $\displaystyle\limsup_{x\to\infty}|1-p(x)e^{-x}|=1$

@kan I am taking Lilly for a walk now, I will be back later

Hah, sure thing.

@JacobBlack I do you see "all" higher than "week", "month", "quarter" and "year", but it may mean over all.

The limit, I'll think now.

@GustavoBandeira Punjab Protest Pics?

@JayeshBadwaik Yep. Reuters just posted.

Why post it here?

@skullpatrol Answer.