Then ... dayum, @mathh.

I'm not sure. I think that's similar to an unsolved problem.

The P versus NP problem comes to mind.

You misunderstood my problem @Khallil. What I asked for is using math notation to inform everyone that the statement is true. Ok, let me show you an example.. Say that the statement Q is $ab=0 \implies a=0 \text{or} b=0$.. We clearly know this is true, but is there a better way than "Q is true" to inform everyone that Q is true?

Oh, I get what you're referring to now. Words really are the best way to do justice to math.

Unless you have a good reason to pack your stuff with notation, don't.

It would be great, though ...

I'd really like to use notation for that.

People reading what you write wouldn't.

@DanielFischer

@robjohn I mean it's not enough to invoke the Cauchy product. The infinite versions are caught by some theorems.

@PedroTamaroff Si?

@Chris'ssis Okay...

@DanielFischer I am trying to figure out how to metrize (?) a normal second countable space.

@Chris'ssis Spivak proves that more generally, I think.

I wouldn't, @Pedro

It could be bad for your health.

@Chris'ssis That is why I was a bit more careful.

@robjohn @DanielFischer or anyone else, perhaps you know it?

@PedroTamaroff Do you remember that result that is more general?

@Chris'ssis I am wrong, sorry.

Is there a way of claiming "Q is true" without using words, i.e. just by using math notation?

@PedroTamaroff Take a look at the proof of the Bing-Nagata-Smirnov theorem. I don't think you can have a nicer general construction.

@mathh know what?

Regarding your theorem, consider $A(t)=\sum a_nt^n$, $B(t)=\sum b_nt^n$ and $C(t)=A(t)B(t)$ and let $t\to 1^{-}$, I guess.

@robjohn my last comment.

@mathh $Q \iff (0 = 0)$

But don't, please don't.

@DanielFischer DANG. I was trying to come up with an idea myself. Is it too complicated? Convoluted?

@PedroTamaroff Yeah, that is the Abel's original proof to the theorem. :D

@DanielFischer That's what I thought too.. too bad there's no better way.

@mathh you mean this comment?

@PedroTamaroff Normal second countable is pretty abstract. I think you can simplify the general construction, but basically, it's still ugh, unless I'm missing something.

@robjohn well, yes. My other comment later summarized the problem.

@mathh Does Q have a name that would be recognizable by people?

@robjohn Q can be any statement. E.g., we could let Q be the statement $0=0$. Is there a way of informing others that it is true without using words, i.e. just by using math notation? We could have $Q \iff 1=1$ as Daniel said, but is there a better way?

@JasperLoy Sup, long time no see.

@KarlKronenfeld You should say hi to me instead, lol.

@mathh This sounds highly contextual. It depends on what Q is. If it is applicable, you could prove it as a theorem or lemma and cite it as such. This seems to be quite a broad question. Can you give an example?

@JasperLoy I should... say hi to you instead of you..

idgi

@Karl Confusing, for sure.

@KarlKronenfeld I am confused.

@JasperLoy confucius?

@N3buchadnezzar Confucius is a silly man, don't associate me with him

Confucius still lives.

@JasperLoy Man who drop watch in toilet have shitty time - Confucius.

Only Buddha is wise.

Life is suffering - Buddha

Man who run in front of car gets tired, man who run behind car get exhausted.

@DanielFischer is pretty wise.

@N3buchadnezzar No meat, no pudding, lol

One of the rappers in the group "Die Antwoord" has a tattoo of the phrase "pretty wise".

@JasperLoy Confucius says many smart things in broken english.

"$Q(x,y)$ is true $\forall x,y$, thus Q(1,2) is true." We could denote this as "$( ( Q(x,y)\iff 1=1 ), \forall x,y)\implies Q(1,2)$". @robjohn But I'm searching for better ways of showing this.

Obese people are pretty wide.

@KarlKronenfeld Hey! What are you up to?

@mathh The best way to write this is "Q is true", period. Notation like you're using solely obfuscates this, and the point of notation is for clarity and brevity.

@PedroTamaroff He must be up to no good.

@MikeMiller Germans are pretty spies

@N3b I like to wear pretty ties

white man speak with fork tongue :D

@MikeMiller I prefer the allies?

@PedroTamaroff As I was telling mike the other day, I am working on the infinitely prime twins conjecture

@KarlKronenfeld I am currently working on my mental problems.

@KarlKronenfeld Wow, alone?

The conjecture relating to the number of twin primes is too difficult though, so I stick to prime twins

@N3buchadnezzar I find most philosophers silly.

It would be good if the person being ignored can see who is ignoring him.

Let us propose this new chat feature, shall we?

@PedroTamaroff I am working with EnjoysMath

I find it silly to ignore people in this chat. When you do that, the conversation does not make sense.

@KarlKronenfeld Are you in the same university as he?

@Karl You're quite lucky to work with someone so skilled.

@MikeMiller How do you know he is skilled?

@PedroTamaroff What classes are you taking this term?

@JasperLoy So?

@N3buchadnezzar Then at least he doesn't have to talk to him.

@KarlKronenfeld ORLY

@PedroTamaroff I have a feeling that this is all a joke.

@PedroTamaroff Yeah, we're trying to get topos theory involved.

@MikeMiller If you're being sarcastic, I don't want to hear it.

I am going to say for the millionth time I am confused about whether Karl Kronenfeld = Ed Gorcenski, lol.

@JasperLoy lol, what could possibly lead you to think that?

@PedroTamaroff How about you?

@KarlKronenfeld I can't remember the 3 names involved in this. Two avatars look similar and one of them had 2 usernames.

@JasperLoy I was referring to something else

sorry for the confusion

I used to be ironic, and sometimes I'm still facetious, but never sarcastic.

@KarlKronenfeld I now remember that the other name is Arkamis, lol.

@MikeMiller until you took an arrow to the knee?

@Karl What did you find out about TPC?

@JasperLoy As per Gauss, they are.

@BalarkaSen nothing it's (I)PTC bro

If a philosopher says something true, it's trivial

If he says something not trivial, it's false

@KarlKronenfeld what's with the I?

Hm, why not?

oh, you are referring to that paper

infinitely twin prime conjectures.

haha

@skullpatrol Wait there is even a category called $\mathbf{knee}$?

:D

I am pretty sure you didn't get the joke.

Hi Skullpatrol

hi

John Jack

Are you familiar with the result "An increasing unbounded sequence has no convergent subsequence"? Does that seem valid to you?

it seems valid to me

@KarlKronenfeld I'm studying commutative algebra, trying to study combinatorial topology, and with some finals ahead.

We believe in you, @Pedro!

So, do we feel that beer helps or hinders maths learning?

It depends how much

Never tried it. Don't intend to.

how can alcohol help learning?

I'm not talking about getting wasted! Just slogging through a lot of questions and the stress is building up. Thought about a (small) beer but concerned it might make me fall asleep.

@skullpatrol Yes, more accurately I would have asked if it hinders or not. I don't expect to become Hawking after a Budvar.

I enjoy a drink or two sometimes but usually never more than that

Funny quotes, @N3!

How are you, @Huy?

@Khallil: I'm fine, how are you?

I'm a bit more than fine, @Huy. ^_^

@r9m: It's not a Dutch name, so Huy isn't the same as Hi.

@Khallil: How so?

Meeting up with friends you haven't seen in a while is really invigorating!

(y)

Meeting new people brings even more happiness.

Both happened to me today!

That really depends.

Hey everyone. Is the $\left(\exists \,\,x\in\mathbb{R}\right) \left[x^2 =x\right]$ notation correct to denote $\exists \,\,x\in\mathbb{R}$ such that $x^2 =x$, and why ? This is the notation a commenter in this question used.

I thought it was more common to use a $|$ or a colon.

@Khallil Today I put a comment there showing that they can both be ambiguous.

Can a statement belong to a set?

Yes it can.

Anything can be ambiguous without context.

@DanielFischer Chat guidelines expired

I swear more than half my mistakes are careless ones. Not sure if that's better or worse than clueless ones!

@Alizter Any algebra lately? Or is it all integrals?

Chat guidelines | $\LaTeX$ in chat

@BalarkaSen Hmm many integrals but I have also been looking at solutions of cubics/quartics

Trying to make symmetrical substituations to solve them

Symmetrical substitutions?

For example $x^2-2x+5$ then $x\mapsto 1-x$ and $x^2+4$

For a simple quadratic case

Of course, linear transformations are dumb while working with appropriate cubics.

by symmetrical substitutions I mean subs that reveal symmetry

Though there exists quadratic transformations that reduces a cubic $x^3 + ax^2 + bx + c = 0$ to $z^3 + k = 0$

@Alizter have you ever studied Lagrange resolvents?

Nope never heard of it

It's in Dummit-Foote. Do you have a copy?

Nope unfortuantely

Bring yourself one.

@BalarkaSen I am poor at the moment I might ask my library to get one

@Alizter Just download it.

But Prof. Ted said that was bad!

ok

got a copy

sorry ted

eww photocopied

Is $x^2+(\log\cos x)^2$ analytic on $[0,\frac\pi2)$

Is this answer correct people? Give a hand please!

@Darksonn Yes

thanks

@BalarkaSen Was your link photocopied?

@Alizter photocopied?

what d'you mean?

@BalarkaSen The pdf is pictures of the pages of the book

rather than the book iteself

maybe, maybe not.

[blah]

just search it there.

yeah its photo copied

oh well

they all seem to be the same

don't be picky.

@BalarkaSen No it is slower to scroll

I need help finding the a series representation for $$f(x)=\frac{x}{x^2+(\cos\log x)^2}$$

and download

@Alizter don't you have DjVu?

@BalarkaSen Whats that?

a software.

well, pdflite is also cool

it renders books much faster

This is a big book

Scroll to the Galois theory chapter.

Believe me, you need only pieces of information about groups and vector spaces to study GT.

I still don't get how vector spaces are relevent

Well, field extensions are essentially realized as vector spaces, no?

Nope never seen that analoogy

It's not an analogy. It's a fact.

@BalarkaSen My other book skips field extensions and goes straight into galois groups

galois groups without the notion of field extensions... doesn't really make sense

@BalarkaSen This is alot to think about for the moment. My school starts tomorrow so I must be getting ready for that stuff. I will keep this in mind to study when I have time. Need to get my physics in gear

@BalarkaSen As is they aren't studied in depth too much

well, galois groups are really corresponding groups of field extensions not fields

do you know what a vector space is?

@BalarkaSen Oh well. I gotta study other stuff for now. But this is really interesting!

yeah, OK, read it up anytime you want. just think less about those integrals.

Hey, @Darksonn. Have you considered finding the Taylor/Maclaurin series expansion of your function? $$ f(x) = \displaystyle \underbrace{\sum_{r=0}^{\infty} \dfrac{f^{(r)}(a)}{r!} \ (x-a)^r }_{\text{Taylor Expansion}} \ \overset{a=0}= \underbrace{\sum_{r=0}^{\infty} \dfrac{f^{(r)}(0)}{r!} \ x^r }_{\text{Maclaurin Expansion}} $$

@BalarkaSen I do them for fun nothing serious ;)

plus analysis needs to be studied

for me

more specifically differential equations

@Alizter integrals are absolutely not about all of analysis

because they have interesting properties that i want to explore with algebra

@Alizter Told ya, differential GT.

@BalarkaSen I am not even sure how to solve ODE's at the moment so I need some work on this.

@Khallil Sure, the hard part is the $r$'th derivative

a lot of work.

Oh, right. Let me have a crack at it, @Darksonn.

=)

@Khallil Thanks

@Khallil @Darksonn That function is not easily differentiable

don't bother

@Alizter I know

It can be done with 2 nested Faà di Bruno's

@Darksonn Try and find some other power series for it

also @Khallil stop showing off with your $\LaTeX$ skills ;)

What's a Faà di Bruno?

Faà di Bruno

I'll try, @Alizter!

Oh yea, how'd your results go?

@Khallil 5a* 3a 3b and a c

Then I got an e in d1

I hate that module

That's awesome!

Oh, forget D1. D1 sucks.

2 of my a's are borderline a* so i'm gonna have them remarked

Did you do C1-4?

@Khallil Next year

Yea, do that! It'd be a shame to go on regretting not having them remarked.

1 ums off of a* in dt electronics

You'll be so ready for them. They are a walk in the park compared to the stuff discussed on here!

1 UMS?!?!?!

Are you on Edexcel for math?

@Khallil yes

Yep, the modules are über easy.

@Khallil FP1-3 and C1-4 are gonna be good

I am thinking of doing the extra AS with some more modules

FP3 and C4 are the good ones. FP2 doesn't go into much depth in terms of explanations.

maybe by skipping d1

Yea. The mechanics modules are good fun.

The stats modules are also nice, but I have a particular distaste towards them.

@Khallil I got an A in stats gcse and this is more mathsy

I think I will enhoy them

:-)

Are you on TSR?

(The Student Room)

@Khallil I lurk occasionally

Does anyone know of a closed form expression for exp(aD^n)f(x), just as exp(aD)f(x) = f(x+a) [where D is the differential operator]?

Gotta get my physics in order though

You must've seen some of my posts if you were in the math threads. I was on there for a while. I hit 4000 posts and stopped it.

Physics is really enjoyable at A-Level. Discussing concepts has always been more appealing to me than memorising random facts.

Almost everything links together. It's nice!

I also am taking computing.

One I missed a* by 8ums at gcse

anyway gotta sleep first day of sixthform tomorrow

Ah, cool! Have fun! ^_^

Are there any requirements for the sum for the following to be allowed, other than the sum not diverging? $$\frac d{dx}\sum f(x)=\sum\frac d{dx} f(x)$$

I think that's the only requirement. Did you find that in a book, @Darksonn?

(I'm asking because I've only ever seen a handful of people on the internet using that result to find sums.)

hi all

anyone got an apostol book near him?

Which one?

1

Calc?

yep

one variable cal with linear algebra

I do.

it's about mathematical induction

What's up?

can you check page 33 pls

Something to do with a parabolic segment?

this gives an equality

now he says to obtain A(k+1) as a consequence it suffices to show

Let me read through it for a second!

this is the part where I'm stuck, at this point I'm confused. it seems like he's showing A(k+1) < A(k+1)

Ok.

I'm not sure why I see it like this It's weird

So we've assumed $A(k)$ has been proven to be true for some $k \geq 1$.

yep

I think I know where your confusion lies.