Mathematics - 2014-10-02 (page 2 of 4)

@Committingtoaname: How would you suddenly not be anonymous by saying which university you're attending? Does googling "commiting to a name + uni" immediately yield all of your personal information?

Committing to a name

@Huy Actually yes :)

I won't tell you why so I stay anonymous :)

Basj

8:55 AM

Hi there! I'm a french teacher, and would like to know the english name for this :

2 x^2 + 3 x + 7 = a [ ( b x + c)^2 + 1 ]

or = a (x +b)^2 + c

In french we call this "Mettre sous forme canonique", but in english canonical form is something else

Committing to a name

I speak isiZulu primarily, but I am fairly sure in English it is factorised form

Basj

@Committingtoaname no, factorised form is a (x-x_1) (x- x_2)

this is very different

Huy

@Basj: Maybe completing the square?

@Basj: Oh, I see what you mean. Let me think for a second.

@Basj: A parabola can be represented by $f(x) = a_2 x^2 + a_1 x + a_0$, or $f(x) = a_2(x-x_1)(x-x_2)$ or $f(x) = a_2 (x-u)^2 + v$, the first is the expanded or normal form, the second is the root form and the last is the apex form.

@Basj: We used those notions for parabolas specifically though, not for quadratic equations, so I'm not sure whether you could use the same.

@Basj: en.wikipedia.org/wiki/Completing_the_square

rehband

9:19 AM

@Huy, have you learned any complex analysis?

Huy

@rehband: Of course, I learned some.

rehband

Could you recommend some books/ressources for self study?

Did you enjoy it?

Huy

@rehband: The topic itself is probably the most beautiful in all of mathematics.

rehband

@Huy :)))

Huy

@rehband: Our professor wasn't any good though, and yet everyone enjoyed it a lot.

rehband

9:21 AM

Sweet!

Balarka Sen

@rehband Try Titchmarsh

Huy

@rehband: Ahlfors is one of the popular English books, and Remmert for German, I think. We used Jänich, which I liked because it is very short and compact, but not for everyone.

rehband

@BalarkaSen Ok, I'll google that, merci

Balarka Sen

I learned fundamentals from Titchmarsh and Lipshitz-Siegel

*Lipshultz

rehband

@Huy Ok cool. I'm looking for a chatty book to self-learn with (preferrably in English)

Huy

9:23 AM

@rehband: You can also check out one of our professor's lecture notes: http://www.math.ethz.ch/~salamon/PREPRINTS/cxana.pdf
I didn't check them out yet though.

Balarka Sen

@Huy Those are probably much too advanced

Huy

@BalarkaSen: We used them to start with.

rehband

@BalarkaSen Did you teach yourself with Titchmarsh?

Balarka Sen

@Huy Maybe you had background in differential forms and whatnots.

Huy

@BalarkaSen: Just a bit.

Balarka Sen

9:23 AM

@rehband yes

those are good for fundamentals, @rehband

rehband

@BalarkaSen Ok, thank you and thank you @Huy. I'll go to the library and see if I can find any of these

Huy

Alright, I'm off to uni and to have some lunch. See you around later.

rehband

Take care

Huy

@rehband: If you want my old exercises, I can send you a dropbox link later, I store all of them.

rehband

@Huy Awesome, maybe I'll get back to you later on that!

Balarka Sen

9:26 AM

Titchmarsh is particularly my favorite as he is also the guy who wrote a decent book on zeta functions

rehband

@BalarkaSen Oh looks like Titchmarsh is availabe for free on archive.org

Balarka Sen

Titchmarsh is one of Riemann's students, AFAIK

@rehband "theory of functions"?

rehband

@BalarkaSen Yes

Balarka Sen

cool.

rehband

archive.org/details/TheTheoryOfFunctions

Balarka Sen

9:27 AM

there are a lot of pirated book collection in any case

Basj

@Huy Thanks for completing the square! Is it the same than apex form (never heard of that) ?

rehband

@BalarkaSen Yea. I prefer real life books though anyway. Hate staring at the screen

Balarka Sen

shrugs I like to read books, not to smell them.

=P

rehband

Alan Titchmarsh

Alan Fred Titchmarsh, MBE DL (born 2 May 1949) is an English gardener, broadcaster, and novelist. After working as a professional gardener and a gardening journalist, he established himself as a media personality through appearances on gardening programmes. More recently, he has developed a diverse writing and broadcasting career. == Early career == Titchmarsh was born in Ilkley, West Riding of Yorkshire, England, the son of Bessie (née Hardisty), a textile mill worker, and Alan Fred Titchmarsh, Sr., a plumber. After leaving school aged 15, Titchmarsh went to work as an apprentice gardener with...

That must be the author

Balarka Sen

throws table at @rehband

4

It's E. C. Titchmarsh.

Basj

9:31 AM

@BalarkaSen analytic number theorist ?

Balarka Sen

Yes @Basj

Basj

@BalarkaSen are you studying such things ?

Balarka Sen

I have studied bits of it, yes

rehband

Nice, his academic advisor was Hardy

Balarka Sen

Cool @Basj.

@Basj familiar with any good book on ANT?

Basj

9:34 AM

@BalarkaSen oh yes! the book of my advisor : amazon.com/…

it's the most cited reference book in France in ANT

Balarka Sen

wait Tenenbaum is your advisor!?

goodness

Basj

(the link was the English version)

@BalarkaSen yes

Balarka Sen

he is a world famous number theorist, afaik

Basj

i think so, but this world is quite small

Balarka Sen

that's quite cool @Basj, to have a guy with erdos number 1 as an advisor

Basj

9:38 AM

@BalarkaSen yes, then mine has already reached its maximum ;)

Balarka Sen

nice, i'll check that book out

Ice Boy

did you study in Cambridge?

Basj

no in France

Balarka Sen

the only serious book on analytic number theory i ever read was the survey of Iwaniec-Kowalski. never studied it completely, most of what is in there goes above my head.

i am studying some algebra to grasp algebraic nt at the moment

Basj

@BalarkaSen what do you want to study in ANT ? The choice of a good book will depend on this

Sieve methods? Things on zeta function? Things in arithmetic progressions / L functions ?

Balarka Sen

9:41 AM

@Basj not sure. I think algebraic NT suits me better than analytic NT (not sure, as I said. I am a beginner and have only studied the proof of PNT and whatnots about zeta).

Modular forms seems interesting bit of NT but goes above my head ATM

I'd love to learn sieves. Planned it for sometimes, but never went beyond Brun's technique.

Basj

@BalarkaSen ok... Have to go now (teaching soon, just calculus ;) ) see you soon!

Balarka Sen

bu-byes. nice talking with you.

Ice Boy

thanks for dropping in

rehband

@BalarkaSen This Titchmarsh book is so ancient!

Balarka Sen

@rehband I know.

It's good for fundamentals however.

rehband

9:51 AM

@BalarkaSen Ok

Balarka Sen

If you want a newer (but not better, IMO) book, have a look at Spiegel-Lipschultz.

rehband

I think this one will be fine. Maybe I'll skip some of the first parts though (series convergence, uniform convergence,...)

N3buchadnezzar

10:23 AM

I posted le question

Ice Boy

and you got le hint :)

le le

we we

le le

we we

10:30 AM

le le

N3buchadnezzar

we we

Ice Boy

le le

4 hours ago, by Ice Boy

Greetings

Greetings

Greetings

10:37 AM

Greet

GBeau

good morning

Committing to a name

10:57 AM

Greetings

GBeau

Anybody available that I can talk to in order to check my understanding of equivalence relations on a set ($\sim$ on $S$, introductory abstract algebra)?

Committing to a name

Perhaps @GBeau

Though, it has been awhile

GBeau

I can repeat the given definition

Identity: $x\sim x$ for all $x\in S$.
Symmetry: $x\sim y\iff y\sim x$ for all $x,y\in S$.
Transitivity: $x\sim y\wedge y\sim z\implies x\sim z$ for all $x,y,z\in S$.

(something is an equivalence relation if it meets these 3 conditions)

(by our given definition)

so I would like to know, for example, if $S = \mathbb{Z}$ and $n\sim m \iff n > m$ for $n,m\in\mathbb{Z}$ is a valid equivalence relation

Committing to a name

Hmmm

I would say not

GBeau

which would be like saying $n > m \iff m>n\text{ for all }m,n\in\mathbb{Z}$ if I'm understanding what an equivalence relation is

which is trivially untrue

Committing to a name

11:12 AM

Yes that is called Reflexivity

Huy

@Basj: Completing the square is the correct term. The apex form just corresponds to parabolas.

Committing to a name

And yes, it is trivially untrue

GBeau

$S = \mathbb{Z}$ and $n\sim m \iff n = 2m$ for $n,m\in\mathbb{Z}$ fails the same test, then

mick

Hi guys

Committing to a name

Oh sorry, yes I meant that reflexivity is what you have called Identity, but yes that symmetric premise is why it was wrong

mick

11:14 AM

Im asking for attention again :(

Committing to a name

Hello @Mick

$n=2m$ fails on $\mathbb{Z}$ again yep

GBeau

ok thank you

I wanted to make sure my understanding was grounded before I attempted $S$ is the set of all differentiable functions on $\mathbb{R}$, and $f\sim g\iff f'(x)=g'(x)$ for all $f,g\in S$, for instance

Committing to a name

@GBeau No problem, happy to help

That may not be true, let me think on it

GBeau

that would make my life easy (I'm asked to prove the truth of any that are, I haven't gotten to this question yet)

(as disproving is typically as easy as a trivial counterexample)

Committing to a name

Sorry, it is actually. Definitely is an equivalence relation

mick

11:18 AM

My latest questions require attention ( only 11 views )

Committing to a name

What is your question @mick

mick

2 questions

0

Q: Finding an optimal path for minimizing an integral.

Let $x,y$ be real numbers. Let the function $f(x,y)$ be real-entire in both $x$ and $y$. Thus $f(x,y)$ is a real-entire Taylor series in the variables $x,y$. How the find a non-intersecting path from $(0,0)$ to $(1,1)$ such that the integral over that path has the minimum absolute value ? $$ ...

and

1

Q: Confused about pochhammer contour?

I know some theorems about complex analysis such as the argument principle. But I do not get the pochhammer contour. I read about it on the wiki page of the beta function , but I do not understand a thing. Why this contour and not another ? Is it based on the argument principle ? Why is it a...

complex-analysis contour-integration gamma-function

Committing to a name

@mick That first one is my favourite sort of math, but weird notation, let me have a look.

GBeau

calculus of variations

now that's something I can do

@mick "occasionally"

(minimizing $T-V$ recovers Newtonian dynamics)

mick

T-V ?

GBeau

11:24 AM

kinetic minus potential energy

see: Noether's Theorem

mick

I do it differently but maybe im too young

GBeau

"Lagrangian Mechanics" if you want the physics usage of this

mick

I only know newton's laws

GBeau

the method is equivalent per noether's theorem

mick

Gotta run guys.
Plz answer my questions
bye :)

GBeau

11:33 AM

oh man this is too early in the morning:

$$S = \mathbb{Z}\text{ and }n\sim m \iff n - m\text{ is prime, for all }n,m\in\mathbb{Z}$$

rehband

Nevermind :)

Huy

@rehband: I think so, why the nevermind?

rehband

Because I just proved it

Huy

@rehband: Good. :D

rehband

:D

GBeau

11:37 AM

I'm making these harder than they are

Chris's sis

@rehband Back. What was it about?

rehband

@Chris'ssis Just some easy inequality

Chris's sis

OK

GBeau

oh this is untrue because $n-n$ is not prime

oh I suspect this one is, in fact, an equivalence relation $S = \mathbb{R}$ and $x\sim y \iff x-y\in\mathbb{Z}$ for all $x,y\in\mathbb{R}$

rehband

11:59 AM

Can we say anything about $$\prod_{k=2}^n \left( 1 - \frac{1}{k-1} \right)$$? It's clear that it tends to $0$ as $n\to\infty$. Does it tend to $0$ like $\frac{1}{n}$?

Chris's sis

@rehband This is precisely 0.

rehband

Sorry, it was meant to start at $k=3$ (cant edit anymore)

Huy

@rehband: Then it is $\frac{1}{n-1}$.

rehband

@Huy Ok, good. Why is that so?

Huy

@rehband: Write it out.

rehband

12:03 PM

Okay

I see, thank you :P

GBeau

can somebody nudge me in the right direction here; I'm trying to prove: $$x-y\in\mathbb{Z}\wedge y-z\in\mathbb{Z}\implies x-z\in\mathbb{Z}\text{ for all }x,y,z\in\mathbb{R}$$
It's true to me logically, but I'm being asked to think about it in terms of equivalence classes (this is an introductory abstract algebra course)

Huy

@GBeau: Let $x-y = a \in \mathbb{Z}$, and $y-z = b \in \mathbb{Z}$. Then $x-z = x + b - y = a + b \in \mathbb{Z}$.

GBeau

oh, I see

Committing to a name

How are you doing @GBeau?

GBeau

ok, I think I finished that section, I'm back to the other one that had stumped me

Committing to a name

12:18 PM

The differentiable real valued?

GBeau

no, prove or disprove this statement: "There exists a prime $p>3$ such that $p+2$ and $p+4$ are also prime numbers."

it is my guess that it can be shown untrue by somehow showing $3\mid p+2\vee 3\mid p+4$

(for all $p>3$)

and that guess is based on the context hinting in the question

Huy

@GBeau: Yes, one of the three numbers will always be divisible by 3.

GBeau

which is where I got stuck; I had started by attempting to see if $3\mid(p+2)(p+4)$ could be shown, but I haven't yet made progress on it

Committing to a name

Hint: Think of consecutive integers

GBeau

oh I guess this is obvious in modular arithmetic

I guess I need to formalize that

thank you that helped actually

Committing to a name

12:26 PM

I hope so :)

robjohn

@rehband The $k=2$ term is $1-\frac1{2-1}=0$

Committing to a name

$ 2\pmod 3, 1 \pmod 3, 0 \pmod 3$, and since you start on any of these, etcetc

robjohn

@rehband okay, then the product is $\frac1{n-1}$

Huy

Is the trace of operators cyclic as well?

robjohn

@rehband Oh, I see Huy answered already

The product $\displaystyle\prod_{k=3}^n\frac{k-2}{k-1}$ telescopes

Huy

12:32 PM

@robjohn: He knows. :P

Will Hunting

@DanielFischer Is your secret condition improving?

Huy

@robjohn: But I'm a bit confused with something else. How do I get from $$\operatorname{Tr} \left( e^{-\beta H} e^{\tau H} e^{\beta H} A e^{-\tau H} e^{-\beta H} B \right)$$ to $$\operatorname{Tr} \left( e^{-\beta H} B e^{\tau H} A e^{-\tau H} \right),$$ I thought cyclicity would help but I'm a bit helpless because I don't see how I could cycle the product such that the additional terms cancel. $A,B$ are operators and $H$ is a Hamiltonian (self-adjoint operator).

Oh, I think I can commute the first three operators because they are just differ by a constant and then use cyclicity and then I'm done, right?

GBeau

12:48 PM

oh wow I think I was doing something like this the other day (humbler)

was being asked to show if $AB-BA=0$ (over matrices), then $e^Ae^B=e^{A+B}$ holds

(otherwise it doesn't)

Huy

@GBeau: Yes, I used that fact a bit earlier in my calculations.

John Jack

1:03 PM

Hi Huy

robjohn

1:21 PM

@Chris'ssis: I thought I was going to get to use my integrals of $\int_0^{\pi/2}\log(1+\cos(x))\,\mathrm{d}x$ and $\int_0^{\pi/2}\log(1-\cos(x))\,\mathrm{d}x$ in this answer, but it came out much simpler. Mathematica has a little bit of trouble verifying the result though. I needed to up MaxRecursion before it would compute 20 places

@Huy I would need to look a bit at that... it has been a while since I've looked at the trace of exponentials...

However, I am taking my dog to the park and then my bunnies to the vet for a snip-snip.

Will Hunting

1:41 PM

@robjohn What are you snipping off?

Committing to a name

Pathways to the testicles

Will Hunting

Might be just a haircut, lol.

Committing to a name

That is very true Will, I just assumed based on being bunnies

Will Hunting

Have you watched the movie Good Will Hunting?

Committing to a name

I have, but not since very long ago

I understand the name reference

@Will Do you have any books or textbooks you would recommend to your dearest colleagues?

Apologies on typo, I am tired, please forgive me

Will Hunting

1:56 PM

@Committingtoaname I recommend my 12 holy math books, I think I mentioned them to you already.

Committing to a name

Oh that was you indeed, apologies once again

I do ask everyone that question

Will Hunting

@Committingtoaname Very good, I like to ask people about books too, as I am always in search of the best books on any given topic.

Committing to a name

Would you recommend any books outside of Mathematics?

Will Hunting

@Committingtoaname I am currently trying to come up with a list of books for learning some languages, but that list is not finalised yet... I need to think maybe one more year...

Committing to a name

Languages are hard by book. I find applications are much more friendly and rewarding. Have you tried the website Duolingo?

Will Hunting

2:00 PM

@Committingtoaname I prefer books with audio, I do not like learning from websites.

Aditya

hi

Committing to a name

Hello @aditya

Aditya

hello @Committingtoaname

what do you mean by "committing to a name"?

Committing to a name

@WillHunting I understand not wanting to learn from websites, there was actually an interesting result of research about there being a real and substantial difference between reading on paper and on screen. I can't find the actual source at the moment, so you can read somewhat about it here if interested: http://www.pri.org/stories/2014-09-18/your-paper-brain-and-your-kindle-brain-arent-same-thing .

That being said, I do still recommend Duolingo, which works mainly via audio.

:17959345 Well I had the user1230123123 name originally, and I didn't want to change it, since that would be committing and altering cognitive bias based on peoples feelings towards a certain name. So I went with one that would be mostly neutral!

Aditya

where do you study?

Committing to a name

2:07 PM

In Australia :)

Aditya

college/school/research/etc?

Committing to a name

Anonymous University

Aditya

r u jokin'?

Committing to a name

If I tell people what university, I could lose my anonymous nature, refer to starred posts.

Aditya

why are you trying to become the invisible man?

Committing to a name

2:09 PM

So I don't have emotional investment into my action. It allow me to be laid back :)

Aditya

just went over my head

tell me something else about ye?

Committing to a name

Read Second sense: Decision to not connect emotionally : en.wikipedia.org/wiki/Emotional_detachment

Aditya

do you know about me?

Committing to a name

No I don't, tell me about yourself

Aditya

I'm here studying at highschool

in India

Committing to a name

2:12 PM

What school exactly by name

Aditya

step by step

Semiclassical

i feel similarly: one can deduce my location from my profile (and perhaps guess something about my math tastes) but not my identity.

Aditya

how would you know my school, why are you asking

Committing to a name

I was asking because you asked me, and thought it odd that I not answer ;)

Aditya

ah! @Semiclassical the genius who answers most questions?

Semiclassical

2:14 PM

psh

i'm rather middle of the road, all things considered.

Aditya

middle of the road, in sense?

Semiclassical

and my rigor is in particularly not that consistent. (which i guess goes nicely with my name here, lol)

Committing to a name

Goodnight friends!

It is 12:16AM here now

Aditya

it 7:46 PM

here

don't be awake so late?

Committing to a name

Triangulating your location as we speak

Aditya

2:16 PM

you

'll

get

Committing to a name

India

Bye :)

Aditya

noooooo!

someone here?

Semiclassical

nope. totally not. nope.

Aditya

any you? a mod? a bot?

Semiclassical

(self-refuting statements, how i love thee)

i am here, i'm just being a bit silly

John Jack

2:22 PM

Hi @WillHunting

Will Hunting

@JohnJack Hi! Do I know you?

Aditya

where were you hiding @WillHunting?

ah! ignoring me :(

Will Hunting

@Aditya Nowhere.

John Jack

@WillHunting I don't know, possibly. Could I ask a question. Could you say that the only time the limit of a function and the limit of the absolute value of the function can be said to always coincide is if the limit of either is found to be zero?

Aditya

most likely!

Will Hunting

2:25 PM

@JohnJack No, take f=1.

Aditya

where is the limit 0? @WillHunting

John Jack

But the limit of f = 1 is not zero.

Will Hunting

@JohnJack I am trying to interpret your question the way you asked it.

John Jack

The limit at some point is zero.

Aditya

@WillHunting atleast read it whole?

Will Hunting

2:27 PM

Anyway, the sentence is very ambiguous, so I cannot answer the question. =)

Aditya

excuses

bye

Semiclassical

Right now, any $f(x)>0$ in the nbhd of a given point will provide a counterexample if if has limit at that point

John Jack

If the limit at some point is zero for a function then the limit of the absolute value is always zero and the converse holds as well. Is that result always true?

Will Hunting

@JohnJack Yes.

John Jack

Okay kewl thanks @WillHunting

Will Hunting

2:30 PM

@JohnJack Try to prove it yourself, very simple.

John Jack

Yeah it is.

Will Hunting

I am surprised you ask this simple question...

John Jack

You first ask if you know me then are surprised at my questions...does that make sense?

Darksonn

I'm surprised WA can't solve this $$\int\left[\frac{\sin x}{\sqrt{1-x^2}}+ \arcsin x\cdot\cos x\right]\,\mathrm dx$$

Chris's sis

@Darksonn It can do it easily if you have Pro version. It's a very easy integral eventually.

Darksonn

2:44 PM

@Chris'ssis I know it's easy, it's just $\arcsin x\cdot\sin x$. That's why I'm suprised

Chris's sis

@Darksonn They want you to pay for the version that can compute such an integral. W|A servers are very powerful, but pretty limited as regards the free version.

Darksonn

@Chris'ssis The free version has done some integrals which seem much more advanced

Chris's sis

@Darksonn This is less important since all is related to the computational time required and it is based on the way W|A approaches these problems.

Darksonn

@Chris'ssis I guess so, it probably splits the integral up by the sum, which isn't a way of solving that makes much sense in this case

rehband

@robjohn Yep, thank you! :)

Chris's sis

2:51 PM

@rehband Does it seem to me that you have difficulties solving very easy problems, but you understand the harder ones? It doesn't make much sense to me. :-)

rehband

@Chris'ssis I'm missing a lot of basic knowledge. Haven't done any of this stuff in uni yet :P I'm still a total newbie!

Btw. have you seen this series? $$\sum_{n=2}^{\infty} \prod_{k=2}^n \left( 2 - e^{1/k} \right)^a$$

Chris's sis

What about it?

rehband

An awesome exercise in Furdui is to see for which values of $a>0$ it converges. Working on it at the moment

Chris's sis

At first sight it looks like $$\prod_{k=2}^n \left( 2 - e^{1/k} \right)^a \approx \frac{1}{n^a}$$

rehband

@Chris'ssis Yep, how did u see that so fast?

Chris's sis

3:03 PM

@rehband I had luck.

rehband

@Chris'ssis Ok :)

Chris's sis

@rehband This is related to your previous question, the very easy one.

rehband

@Chris'ssis Indeed

Chris's sis

3:25 PM

@rehband I can do that in more ways.

$$\prod_{k=2}^n \left( 2 - e^{1/k} \right)=\exp\left(\sum_{k=2}^{n}\log\left(1+1 - e^{1/k}\right)\right)\approx \exp\left(\sum_{k=2}^{n}1 - e^{1/k}\right)$$
$$\approx \exp\left(-\sum_{k=2}^{n}\frac{1}{k}\right)\approx \exp\left(-H_n\right)\approx \exp\left(-\log(n)\right)= \frac{1}{n}$$

Alizter

3:43 PM

eek number theory hurting my brain

rehband

@Chris'ssis Oh wow. Learned a new technique there, thank you! Short and sweet.

Chris's sis

@rehband Welcome ;)

@robjohn Yeap, nice.

Transcript for

$Mathematics$ Mathematics