Mathematics - 2014-08-23 (page 4 of 4)

Khallil

8:03 PM

What if you just decide to take the tangent of the equality, @Chris'ssis?

Chris's sis

@Khallil Sure, take it. :-)

Khallil

$a=b \implies a-b=0 \implies \tan \left( a-b \right) = 0 \implies \dfrac{\tan a - \tan b}{1+\tan a \tan b} = 0 \implies \tan a - \tan b = 0 \implies \tan a = \tan b$

Would that be enough?
Or do I have to start from the end (right) and prove the first statement from it ($a=b$), @Chris'ssis?

Chris's sis

@Khallil $$\arctan(\sinh(x))=\displaystyle 2\arctan\left(\tanh\left(\frac{x}{2}\right)\right)$$

you have 2 in front of the expression in the right side. Or I misunderstood your point?

Khallil

That doesn't really change anything does it? I think I haven't understood it properly.

Let $a=\arctan(\sinh(x))$ and $b=2\arctan \left(\tanh \left(\frac{x}{2} \right) \right)$.@Chris'ssis.

Chris's sis

@Khallil There is a formula for $\tan(2\arctan(x))$, maybe you know it well.

Khallil

8:08 PM

Oh, I got it.

$\tan (2\arctan(x)) = \tan (\arctan(x) + \arctan(x))$.

>_>

How did I not see that sooner?

Chris's sis

@Khallil :-)

You can simply start with $$2\arctan \left(\tanh\left( \frac{x}{2 }\right)\right)= \arctan\left(\tanh\left(\frac{x}{2}\right)\right)+\arctan\left(\tanh\left(\frac{‌x}{2}\right)\right)$$

Khallil

I ended up with this @Chris'ssis. $$\dfrac{\sinh (x) - \sinh (x)}{1+\sinh^2 (x)} = 0$$

Which is clearly true as the denominator is $\geqslant 1$ and the numerator is equal to $0$ for all $x\in\mathbb{R}$.

Chris's sis

@Khallil Do you know the formula for $\arctan(x)+\arctan(y)$?

Khallil

I can derive it, but I can't recall it off the top of my head, @Chris'ssis.

Let $f = \arctan(x) + \arctan(y) \implies \tan (f) = \ ... \implies f = \arctan (...)$.

Chris's sis

@Khallil $$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-x y}\right)$$

Khallil

8:16 PM

Yea, that's the one!

I feel good having finished your question. I looked at it yesterday and the compound angle formulae never came to mind, @Chris'ssis.

sigh of relief

Chris's sis

:-)

MrWho

8:28 PM

@Chris'ssis What happened? :-(

Chris's sis

@MrWho What happened?

MrWho

@Chris'ssis Did you solve the problem? I had to leave computer and go out.

Chris's sis

@MrWho hehe, no worry about that. :-)

Here is a cute limit, no need for any asymptotic expansion

MrWho

@Chris'ssis $$2\arctan \left(\tanh\left( \frac{x}{2 }\right)\right)= \arctan\left(\tanh\left(\frac{x}{2}\right)\right)+\arctan\left(\tanh\left(\frac{‌‌x}{2}\right)\right)$$ How did you find this? I didn't know about this.

Chris's sis

$$\lim_{n\to\infty} n\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log(n)-\gamma\right)$$

MrWho

8:34 PM

@Chris'ssis WoW, shame on me :-(

Chris's sis

@MrWho I think it's because of the graphics there above. No worry for that. ;)

MrWho

@Chris'ssis I thought the other $arc$ is the argument of the other $arc$ function, that's why I asked.Simply, I was confused with paratheses.

Chris's sis

@MrWho OK. See my limit above now, maybe you like it. :-)

MrWho

@Chris'ssis Okay, most of your questions have special constants :-O

Chris's sis

@MrWho :D

nosmoking

8:39 PM

for Fourier transform en.wikipedia.org/wiki/Fourier_transform if I replace

$user image$

the 1st in the second, I get nothing: $$f(x)=\int \int f(t)e^{-i2\pi\dzeta(t-x)}$$

MrWho

@Chris'ssis $log(n)$ is natural logarithm or logarithm based $10$ ?

Is $log(n)=ln(n)$ ?

Alexander Gruber

okay so i'm looking at the roots of degree $m$ polynomials with $n^\text{th}$ roots of unity for coefficients

Chris's sis

@MrWho mathworld.wolfram.com/NaturalLogarithm.html

MrWho

@Chris'ssis Okay, so it's natural logarithm, I think it's better to use $ln$ notation.

nosmoking

$$f(x)=\int \int f(t)e^{-i2\pi \zeta (t-x)}dt d\zeta$$?

Alexander Gruber

8:43 PM

proposition: if $a_0+a_1x+\cdots +a_nx^n$ is one of these then so is $a_0e^{i2\pi j/n}+a_1e^{i2\pi j/n}x+\cdots +a_ne^{i2\pi j/n}x^n$ is too for $j=1,\ldots , n$, so we only need to actually solve one $n^\text{th}$ of the set of these polynomials to find all the roots

Chris's sis

I'm about to fall asleep. :-)

MrWho

@Chris'ssis :D

Daniel Fischer

@MrWho For mathematicians, there is only the natural logarithm. Only engineers and physicists use $\ln$.

MrWho

@DanielFischer Well, I'm physics learner :D

nosmoking

they are all proportional btw

Sab ಠ_ಠ

8:44 PM

What's the best way to read a math book?

Alizter

@Chris'ssis Does $\displaystyle \int_{\Bbb R}\frac1{x^n+1}\mathrm dx$ have a closed form?

Sab ಠ_ಠ

As in Spivak or Apostol?

MrWho

@Sabಠ_ಠ Leave it :D

Alexander Gruber

@Sabಠ_ಠ with a glass of bourbon, laying next to a beautiful woman

nosmoking

best distraction

Sab ಠ_ಠ

8:45 PM

I didn't expect these replies.

MrWho

@AlexanderGruber WoW! what's next? :-O

Sab ಠ_ಠ

:P

nosmoking

who like Fourier tranforms? and could explain me a basic operation of injection the Fourier coefficient expression back in the formula

Sab ಠ_ಠ

@AlexanderGruber I'll do this if I manage to reach 3rd year math

:P

MrWho

@nosmoking I like it, I'm not mature enough for it.

Sab ಠ_ಠ

8:46 PM

I'm not really sure how to go about completing spivak and/or apostol in 2 months starting from the beginning

Chris's sis

@Alizter What happens if $n$ is odd?

Alexander Gruber

alright so back to roots

i've cut out $n$ of the polynomials

but within the roots there should also a repetition of $n$... it should divide the complex plane into $n$ pie slices

(pie slices? is that what you call those?)

Chris's sis

@Alizter $$\int_0^{\infty} \frac{1}{x^n+1} \ dx = \frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)$$ and to get that you first write your integral in terms of beta function.

Alexander Gruber

see, here for example $n=4$, and all we really needed to compute was the upper right quadrant. Then we can just rotate and copy that for the other three and we'll be fine.

also i think this would make a cool rug.

does anybody understand why this set would have that type of symmetry, even after cutting out the duplicate polynomials?

Chris's sis

@Alizter The rest is a piece of cake.

Alexander Gruber

8:54 PM

is it just galois theory?

nabla blah

I have to choose between the classes 'Differential Geometry' or 'Number Systems and the Foundations of Analysis'

Alexander Gruber

@nablablah first one

nabla blah

How come

Alexander Gruber

foundations is boring. manifolds are cool.

Daniel Fischer

@AlexanderGruber If I understand correctly what you've done, the thing is that if $a_0 + a_1 x + \dotsc + a_n x^n$ is one of the polynomials you're looking at, then so is $$a_0e^{i2\pi 0/n} + a_1 e^{i2\pi 1/n}x + a_2 e^{i2\pi 2/n} x^2 + \dotsc + a_n e^{i2\pi n/n}x^n.$$

Or, if $P(x)$ is one, then so is $P(e^{2\pi i/n} x)$.

Alexander Gruber

9:00 PM

@DanielFischer yes. good. that is clear.

Daniel Fischer

So the set of all zeros is invariant under rotation by $2\pi/n$. Wasn't that your question?

Alexander Gruber

@DanielFischer yeah - i mean that you answered it, thank you

so if i'm generating my polynomials by making tuples of integers $(j_0,j_1,\cdots,j_m)\in \{0,\ldots, n-1\}^m$, i can cut out both $(j_0+j,j_1+j,\cdots,j_m+j)$ for all $j\in \{1,\ldots, n-1\}$, and $(j_0,j_1+1,\cdots,j_m+m)$

Daniel Fischer

Right. The one has exactly the same zeros, the other rotates.

Alexander Gruber

so after unioning the zeroes of that set, i should have one slice that I can then rotate around $n$ times to get the full set of zeroes

Khallil

I've got a quick set theory question. It's more of a notation question.
How would you read this set? $X = \{ (x,y) \in \mathbb{R}^2 : y < x^2 \}$.

Alexander Gruber

9:09 PM

@Khallil the points in the real plane where the second coordinate is less than the square of the first

Daniel Fischer

Not sure what you mean by slice. If you only compute the roots in the angular sector $0 \leqslant \arg z < 2\pi/n$, then you can't throw out the rotations. If you want to omit the rotations, you need to compute all zeros of the polynomials, @AlexanderGruber.

Khallil

Thanks, @AlexanderGruber!

Alexander Gruber

@DanielFischer what I mean by a slice is, I guess, a (minimal) set of points $X$ such that $\bigcup_{j\in\{0,\ldots, n\}} e^{i2\pi j/n}X $ is the full set of roots

Daniel Fischer

Okay. Then it's right.

Alexander Gruber

in other words, I'm trying to compute the whole set, but i want to solve for the fewest number of points.

MrWho

9:16 PM

@Khallil Can you show $X = \{ (x,y) \in \mathbb{R}^2 : y < x^2 \}$ on picture?

Khallil

It's the region of the Cartesian plane under the curve $\ f(x) = x^2$, @MrWho.

MrWho

@Khallil I see, the notation is confusing for me.

Khallil

Yea, it can get kinda confusing at times, @MrWho.

MrWho

@Khallil So $y<0$ right?

Khallil

That's not enough, @MrWho. It's just gotta be every point such that the $y$ coordinates is less than the square of it's respective $x$ coordinate.

So take the point $(3, 6)$ for example.

MrWho

9:23 PM

@Khallil You know, I want to be sure, that's why I ask.

Okay? ...

Alexander Gruber

@DanielFischer yup, it's working. i think that's about the most i can do to speed this up.

MrWho

$(3,6)$ is valid, right? @Khallil

Khallil

If you take the point $(3,6)$ for example, $y>0$ but it's still part of the region under the curve $y=x^2$.

Yea, exactly, @MrWho.

Alexander Gruber

i'm making plots of them ordered by smallest $n^m$ to largest, though i'm not sure if that's really fastest to slowest, with the cuts we made

MrWho

@Khallil So it depends on the points we choose, it's not like a plot or general thing to be specified.

Khallil

9:26 PM

It's the region beneath the curve $\ f(x) = x^2$, @MrWho.
We just took the point $(3,6)$ as a counterexample to your claim that the set $X$ is only satisfied by $y<0$.

MrWho

Yeah

I wanted to draw this

Okay.

Khallil

Sorry, did I spoil it for you?

MrWho

No, thanks @Khallil

Khallil

Here's another one. What's the shape of $A$ on the plane $\mathbb{R}^2$ if $A = \{ (x,y) \in \mathbb{R}^2 : 1 \leqslant x^2 + y^2 \leqslant 4 \}$?

MrWho

@Khallil Circle with radius $\sqrt 3$ ?

no

Khallil

9:31 PM

Nope.

Try and draw $\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \geqslant 1 \}$ and $\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \leqslant 4 \}$. $A$ is the intersection of those two sets of points.

(You're on the right lines with it being to do with circles.)

MrWho

@Khallil The first circle contains the second circle, so the intersection is the circle with radius $2$ .

Khallil

Not quite, @MrWho.

Is $\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \geqslant 1 \}$ really a circle?

MrWho

:-*

No

But what is it?

The notation is annoying :(

Khallil

Try and sketch both of them on the same graph.

Oh, if the notation is the problem, then I'll just write it out in words.

$1) \qquad \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \geqslant 1 \}$ is the set of all points in the real plane such that $x^2 + y^2 \geqslant 1$.
$2) \qquad \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \leqslant 1 \}$ is the set of all points in the real plane such that $x^2 + y^2 \leqslant 4$.

MrWho

I know that the second set is the one with radius $2$

I interpret that the first set is a circle with radius more or equal to $1$

probably the first one is the set of all points on the plane outside the circle with radius one.

:-(

I've hanged :O

Khallil

9:38 PM

Yes, that's right!

MrWho

@Khallil That's why I haven't been successful tackling analysis, because these kinds of set notations :(

Khallil

So if $1)$ is the set of all points on and outside of the circumference of a circle with radius $1$ and if $2)$ is the set of all points on and inside of the circumference of a circle with radius $2$. Then what is the intersection between those two?

Is it really that common in Analysis, @MrWho?

MrWho

@Khallil Yeah, I've seen alot of such stuff, haven't you read Analysis?you should know better than me.

Khallil

I haven't studied any Analysis yet.

I'm a newbie. I just started doing set theory, @MrWho.

I really only discovered that math is fun about two years ago.

MrWho

@Khallil I haven't studied too, but I have seen and checked stuff in it.

@Khallil Not seriously, I've done decent math, but didn't practice for some time and I'm like :-(

However, I haven't read set theory for Analysis

Khallil

9:42 PM

I totally understand. Once you get into the hang of thinking about something to do with math everyday, you'll get better!

MrWho

Yeah :)

Khallil

Oh, I'm not reading it for Analysis. I'm just reading it out of interest. I want some sort of foundation in math before I try to go into any 'higher level' stuff and I've read that set theory is pretty fundamental.

MrWho

@Khallil Yeah, but it's boring for me.

Khallil

I'd highly recommend looking through Book of Proof by Richard Hammack, @MrWho.

@Alizter recommended it to me and I'm enjoying it at the moment.

MrWho

@Khallil I'll absolutely check it out.Indeed, you recommended it on time, because I'm going to pick books for this year.

Khallil

9:46 PM

Nice! Are you in college, @MrWho?

I'm guessing the average college student is about 18 in their first year. Would that be right?

MrWho

@Khallil Yeah :D

I'm about to be in college :D

Khallil

I can only go by the UK standard which is 18 by university which I think, is equivalent to college elsewhere.

MrWho

Are you 18?

Khallil

Oh, me too! I took a year out but I'll be starting in the beginning of October. I'm 19 now.

MrWho

Yeah the same then :D

Khallil

9:48 PM

I think the UK system is pretty terrible.

We study derivatives and integrals but we aren't given a formal introduction to them.

All the teaching is done so that you can perform algorithms that the exams test.

The only real thinking that's done is during the applied modules to do with mechanics.

MrWho

@Khallil Yeah, I know what you're talking about, the same happened for me.We have a lot in common :D

@AlexanderGruber Any supplementary book for Linear Algebra course? ( The book volume is important, I want to finish it by studying simultaneously both course and the book)

Khallil

With the decline in teaching quality in secondary schools, sixth forms and high schools, it's not a surprise that we're in the same situation. Many others are too which is a great shame, @MrWho.

Alexander Gruber

@MrWho what level of linear algebra?

MrWho

@Khallil The point is, don't care about them, even in the university, the only thing they care about is money.

@AlexanderGruber Introduction.

Alexander Gruber

@MrWho get gilbert strang's book and watch his lectures on MIT opencourseware concurrently

MrWho

9:54 PM

@AlexanderGruber Plus, some cool tricks and ground for the advanced one.

Alexander Gruber

linear algebra done right by axler is also real good

MrWho

@AlexanderGruber Do I need prerequisite for Linear Algebra?

Alexander Gruber

@MrWho nah, you should be able to jump into those just fine. anybody can do linear algebra.

Balarka Sen

Phew. Just finished the writing about RH.

MrWho

@AlexanderGruber Is there any easy book to go with, for geometry? I think we need Euclidean geo in linear algebra!

Balarka Sen

9:56 PM

Eulid? Who is that?

Never heard of 'em.

Alexander Gruber

@MrWho i'm not sure, i never took geometry. I don't think you need it for linear algebra.

MrWho

@BalarkaSen I made a typo!Your eye just spots flaws :D

Balarka Sen

@MrWho I was joking.

Eyes of a bird, ya know.

Alexander Gruber

linear algebra is all about vectors and matrices, you really just need to be able to multiply and add numbers together in a pattern. It's very easy at introductory levels.

Balarka Sen

@AlexanderGruber yuck linear algebra

Alexander Gruber

9:58 PM

it gets harder maybe 6 months or so in, but not much harder. I think it's definitely easier than calculus, that's for sure.

MrWho

@AlexanderGruber Can I get to advanced level in one year?considering the other stuff?

Balarka Sen

I got a sudden distaste for LA after reading Artin do the LA-overflow throughout his proofs

eugghhhh

Alexander Gruber

@MrWho yeah, definitely.

Khallil

Everything that isn't number theory and combinatorics is yuck to you, @BalarkaSen.

>_>

Balarka Sen

No, @Khallil

I like complex analysis

MrWho

9:59 PM

@Khallil Agree with you !

Khallil

Alexander Gruber

i like complex numbers. i don't think i really like complex analysis though. :p

Balarka Sen

@AlexanderGruber that 'cause you are an algebraist.

MrWho

@AlexanderGruber Everything with Analysis in it is just the domination of boredom!

Alexander Gruber

@BalarkaSen Really I just don't like inequalities

Khallil

10:01 PM

Enough of this discrimination. We are all mathematicians! (Apart from the statisticians. Not them, eww.)

Alexander Gruber

That's what it comes down to

MrWho

@AlexanderGruber Yeah, euggh inequalities.

Balarka Sen

@AlexanderGruber I like ineqs. I just don't like ineqs at infinitesimal level.

Alexander Gruber

@BalarkaSen maybe you're a finitist.

Balarka Sen

I am a hard analyst.

Alexander Gruber

10:02 PM

try it sometime. You can belligerantly refuse to do anything involving infinite sets and nobody will say anything about it because they think you're crazy.

it's worked out very well for me

MrWho

:))

Balarka Sen

=p

user161954

Hm, if we have the absolute galois group of some field where the following males sense (say Q, if you so wish), then complex conjugation surely can be lifted to the absolute galois . But how do we extend complex conjugation? For example, we might have stuff like sqrt(a +bi) - where does that go under the conjugation on the absolute galois group?

MrWho

I want to black out

For a while

:)

Balarka Sen

@user161954 i am not sure what you are saying.

are you asking how is conjugation in the absolute galois groups?

@AlexanderGruber What did you think of my groupified galois theory?

Alexander Gruber

10:10 PM

@BalarkaSen sounded pretty groupy

Balarka Sen

Alexander Gruber

would you like a problem?

Balarka Sen

@AlexanderGruber depends

can't know until you post it

user161954

I am asking how it extends, yes.

Alexander Gruber

Let $\Gamma_0=S_n$ and $\Gamma_i=\operatorname{Aut}\left(\Gamma(\Gamma_{i-1}\right))$. here $\Gamma$ (by itself without an index) is the prime graph. What's the smallest $m$ such that $\Gamma_m=1$?

Balarka Sen

10:13 PM

dunno anything about graphs, really.

Alexander Gruber

well it's time to learn buddy!

here you go

Balarka Sen

not sure if i want to

yuck wikipedia

2 days ago, by Balarka Sen

I think I have an allergy for the math in wikipedia. They disfigure some facts in a sophisticated manner in the process of briefing the theories; usually hard to catch but doesn't fool the eyes of the one who studied the topic rigorously.

oops bed time. if you have an interesting problem in group theory, let me know by ping.

Alexander Gruber

@BalarkaSen fortunately you have not studied graph theory rigorously, so your eyes can be fooled

(that is, by the way, a problem in group theory.)

Khallil

10:48 PM

Hey, @TRiG. I haven't seen you in these parts. How're you doing?

TRiG

@Khallil I pop by now and again.

Doing pretty well, actually. You?

Khallil

Largely the same, @TRiG. It's pretty late but I don't feel tired which is refreshing.

TRiG

@Khallil I'm going on a protest march tomorrow. I really should be abed.

Or at least off the computer.

Khallil

Now I can tell why you have an English stack exchange account, @TRiG!

I've never heard the word abed before. It's pretty cool!

TRiG

in V'dibarta Bam, Mar 4 '13 at 19:43, by TRiG

I sometimes make the effort to carefully phrase things for a euphonious use of language in ephemeral written communication.

;)

Actually, I don't find the word abed that unusual; it forms a fairly standard part of my vocabulary.

Khallil

10:53 PM

Pulls out dictionary

Hahaha! It's admirable that you can incorporate words like that into your lexicon.

I've been trying to do the same, but not for the purpose of showing off. Just to improve my ability to describe things without obscurity.

TRiG

Maths and English were my best and my favourite subjects in school, respectively.

Khallil

That makes sense. If you're good at something, it's likely that you'd enjoy it too. It's not a necessary condition, but it does helpful.

TRiG

@Khallil I have a shortcut set up in Firefox so that if I type d word into the address bar I am brought directly to the OneLook Dictionary search results.

Khallil

I found English to be a bore in school. I think that was down to the morbid focus on exams as opposed to learning more about the language and it's roots.

That's pretty cool, @TRiG. I use Safari mainly because Chrome and it's extensions are so power-draining.

TRiG

@Khallil While I was good at maths in school, I've abandoned it since. I can barely follow quite a lot of this site. I used to offer grinds (as far as I know, this word exists only in the English language as used in Ireland: it means extra tuition separate to school) in maths, but I don't any more because I've forgotten a lot of the syllabus.

@Khallil Where are you from?

Khallil

10:59 PM

Oh, I see. I can hardly follow most of the stuff on here either, @TRiG. I'm just a newbie to be honest.

I'm from the UK, @TRiG.

TRiG

@Khallil Country of origin, really. Where did you go to school?

Ah. Just across the sea.

I was in Manchester a week or so ago. Discworld Convention. Great fun.

Khallil

Indeed. My parents and their parents and their parents etc. are all Moroccan. I live in England however.

What's the Discworld convention, @TRiG?

(Also, apt username.)

TRiG

@Khallil I'm a bit of a mongrel meself. Born and raised in Ireland, but I still have an English accent, despite having lived here all my life. I blame BBC Radio 4 (and my English parents, but mainly BBC Radio 4).

Firefox spell checker doesn't like meself. Firefox spell checker doesn't understand Hiberno English.

Khallil

That's pretty cool. That explains your reference to 'grinds', too. I hardly listen to the radio, unless I inadvertently overhear somebody else listening to it.

What's Hiberno English, @TRiG?

Hibernian rings a bell, but that's a Scottish football team, haha!

TRiG

@Khallil The Discworld Convention is a Con dedicated to the Discworld books of Sir Terry Pratchett. The Irish Discworld Convention is next year.

@Khallil The Latin name for Ireland is Hibernia.

Khallil

11:06 PM

Ah, I see. I wonder what a Scottish team is doing with an Irish name.

Was your name inspired by trigonometry, @TRiG?

TRiG

Hiberno-English

Hiberno‐English (sometimes referred to as Irish English) is the dialect of English written and spoken in Ireland (Latin: Hibernia). It comprises a number of sub-dialects, such as Ulster English, Dublin English and Cork English. English was brought to Ireland as a result of the Norman invasion of the late 12th century. Initially, it was mainly spoken in an area known as the Pale around Dublin, with Irish spoken throughout the rest of the country. By the Tudor period, Irish culture and language had regained most of the territory lost to the colonists: even in the Pale, "all the common folk… for the...

Khallil

Also, do you go to university, @TRiG?

TRiG

Jul 11 '13 at 13:43, by TRiG

@skullpatrol Hi. But my name predates any knowledge I have of trig.

Jul 11 '13 at 13:43, by TRiG

> Timothy Richard Green

@Khallil Not any more. I have a degree in chemistry (in which group theory turned out to be surprisingly useful) and now work in web development.

Khallil

Where did you go graduate from, @TRiG? Really? I don't know too much about group theory, but I didn't think it'd be involved with chemistry of all things.

TRiG

@Khallil Athlone IT. Just down the road.

The more symmetrical a molecule is, the more stable it is (as a general rule). So thinking about symmetries (including glide reflections, rotational symmetry, and the like) is useful to chemistry. If you have the sort of brain that can visualise a 3d object and subject it to transformations (as I have), this can be quite good fun.

Khallil

11:13 PM

I've never heard of that university before. I'm guessing you lived at home whilst attending the university. Did you feel like you were missing out on the social side of things, @TRiG?

TRiG

@Khallil Lived at home. Commuted by bus.

r9m

@Chris'ssis :D

@Chris'ssis COOL !! :D

Khallil

That's pretty sweet, @TRiG! I get bus sick after long journeys, so I'm not sure if I could've coped with that.

TRiG

@Khallil Meh. It wasn't that far. I live in Tullamore. Not far at all.

Khallil

Quick question on drawing Venn diagrams. When told to draw things like $B-A$ and $(A-B) \cap C$, do you always assume the sets intersect?

r9m

11:25 PM

@Khallil If they don't intersect its trivial :P .. so draw 'em like they intersect :)

Khallil

Thought so, @r9m!

Also, long time no see!

TRiG

@r9m Intersecting sections could be empty.

r9m

@TRiG yes :) .. but we make it look like they intersect ..

AbstractionOfMe

Anyone know about theory of computation?

r9m

@Khallil :)

JEET TRIVEDI

11:28 PM

Hello

Khallil

That's very true, @TRiG. I think @r9m is just saying that it's not exactly noteworthy since they don't intersect.

(I think I'm siding with @r9m because I'm so lazy.) ^_^

TRiG

Woo. That article on Hiberno English lists some words I use all the time, some I'd never heard of, and some I had no idea were unique to Ireland.

r9m

@Khallil not because it is not noteworthy ... but basically what @TRiG said .. the intersecting sections being empty is a particular case of the general scenario :)

JEET TRIVEDI

In my opinion, When representing sets which are not completely described in Venn diagrams, it's always safer to assume intersection than to assume no intersection (unless explicitly stated to do so)

r9m

@Khallil ya bro .. me lazy too .. very lazy :P

Khallil

11:32 PM

Ah, I see. That makes a load more sense!

Hey, @JEETTRIVEDI!

JEET TRIVEDI

Hey @Kha

*@Khallil

TRiG

> Get youse your homework done or you're no goin' out!

user1534664

Basic question, guys. I don't feel like its worth starting a topic on this. I want to factorize x²-64. That's the same as x² + 0x - 64. Now I have to find 2 numbers that add up to 0, and are the product of -64. Therefore I set up the following system of equations:
a - b = 0
a * b = -64

a-b=0 would be the same as a = b, now substitute that into second equation:
b² = - 64. How can something squared equal -64? I already know the two numbers I'm looking for are -8 and 8, but if we say a = b, then thats false since -8 does not equal 8... Can someone clear my confusion?

TRiG

a + b = 0

> I have to find 2 numbers that add up to 0

user1534664

oh haha

thanks dude

TRiG

11:36 PM

@user1534664 No problem.

See, I haven't forgotten everything! I feel good now.

Khallil

How did you get that vertical line in your post, @TRiG?