Mathematics - 2015-11-26 (page 1 of 2)

Michael

12:00 AM

@Chris'ssistheartist would a trip integral be volume under a three dimensional surface?

Chris's sis the artist

@Michael aha, you don't seem a high school student at all.

user174558

@Chris'ssistheartist Maybe he is deceiving us. Maybe he is a professor, lol.

Michael

@Chris'ssistheartist haha

@JasperLoy Yes, this was all a test of your analytical skills. No on a serious note, I am actually a good highschool math student. No more

@Chris'ssistheartist but I am interested. So adding integralas adds dimensions?

user174558

@Michael Very strange your school has no HL math.

user174558

@Michael Difficult to answer your random questions like that.

anon

12:07 AM

Interestingly: if you put a unit circle in the plane tangent to the vertical line at the origin, then every line through the origin can be identified with a point on the circle. Rotating the lines in the plane by an angle of θ corresponds to rotating the points on that circle by 2θ.

Michael

@JasperLoy My math right now is legimately a joke. We just finished principles of derivation. Not allowed to use l'hopsitals law. Derivation: not allowed to use any of the cool techniques I saw on this website

user174558

@Michael You may look up multiple integrals and iterated integrals and Fubini's theorem.

Chris's sis the artist

@Michael I think you already know the answer.

anon

volume under a surface can be calculated with double integrals just as areas under curves are calculated with single integrals. a triple integral of density will give mass of a three-dimensional body though @Michael

Michael

@JasperLoy so can you calculate the change in value of switching from integrals? Kind of like differentiating.

user174558

12:09 AM

@Michael Difficult to answer your random questions.

user174558

@Michael I think you already have a lot of reading to do before asking any more.

Michael

@JasperLoy no please mr banana!

Chris's sis the artist

@anon He knows the answers already, but it's kind from you to tell it.

anon

What?

user174558

@Michael Thanks for calling me banana, I love it.

Michael

12:11 AM

@Chris'ssistheartist ok I actually don't know the answer to this one. Since you can calculate the change in x in a graph, can you calculate the change of the value of the function as the number of integrals changes?

anon

the first challenge at hand would be to articulate that in a way that can be understood

user174558

@Michael This is a very confusing statement that does not make sense.

Chris's sis the artist

@anon do you refer to me?

anon

my "What?" refers to you, my "challenge" refers to Michael

Chris's sis the artist

OK

user174558

12:13 AM

I think Michael is anon's sockpuppet account.

Michael

@JasperLoy So so, is there a relation between the changes in values after a function

@JasperLoy is applied by n integrals

user174558

@Michael Your question does not make sense at all. So this means it is time to do more reading and less asking.

Michael

@JasperLoy Yes mr Banana. So what should I read about?

user174558

@Michael Anything you want, I have already given you suggestions above...

Michael

@JasperLoy was it pure mathematics?

user174558

12:15 AM

@Michael That too, and Wikipedia pages.

Michael

@JasperLoy what is a curl in math?

user174558

@Michael It is good to have questions, but too many questions without knowing anything is useless. In the first place, the questions do not make sense, or at least you are not describing it properly

user174558

@Michael You can find all the answers online. No more questions today. =)

Michael

@JasperLoy Oh ok. Thank you so much for dealing with my nonesense! Have a great day

user174558

@Michael When you become a doctor, try to cure as many as you can.

Michael

12:18 AM

@JasperLoy I will I promise. I promise I will try to help as many people as possible so there can be more mathematicians like you to help idiots like me!

user174558

@Michael I am not a mathematician. I am a mental patient.

user174558

Hello @ted.

Ted Shifrin

@anon: Thanks for answering the question, but I don't follow it.

Hi @Jasper.

Jake1234

guys, pls help me out with a question

Let $X=Ult(\mathbb{N})$ be the set of all ultrafilters on $\mathbb{N}$, and consider the topological space $(Ult(\mathbb{N}),\tau)$ with $\tau$ being a topology with a basis $
\left \{ \hat{M}: M \subset \mathbb{N} \right \}$, where $\hat{M}=\left \{ U: U \in Ult(\mathbb{N}), M \in U \right \}$.

morphic

Hi @TedShifrin and @JasperLoy

user174558

12:20 AM

@morphic Bart, Sajindia is Twink but maybe I am wrong

Ted Shifrin

@Michael: I did not allow students to use L'Hôpital early on in calculus, either. For example, when you use L'Hôpital's rule to do the limit $\lim\limits_{x\to 0}\frac{\sin x}x$, you are assuming the limit you're trying to do :)

morphic

@JasperLoy Why did he make a new account

Jake1234

the message is all weirdly formatted here in chat, but it displays fine in the preview where I wrote it (on the main site)

Ted Shifrin

@Jasper: I think you're wrong. Twink is in Montreal, so I would be very surprised if he spoke Spanish rather than French. But maybe ...

Jake1234

anybody have an idea what's wrong?

user174558

12:22 AM

@TedShifrin How do you know he is in Montreal? You met him?

Jake1234

Let $f:Y \rightarrow \mathbb{R}$ be a bounded continuous function, where $Y$ is a subset of $X$, which contains only the principal ultrafilters - $\left \{A: A \subset \mathbb{N}, \left \{n \right \} \subset A \right \} $ for some $n \in \mathbb{N}$.

Show that this function can be continuously extended to $X$.

Ted Shifrin

Jake, I've never played with ultrafilters in my entire life, and I'm not going to start now.

Jake1234

I guess it looks normal now.

Ted Shifrin

No, haven't met him, but there were discussions about the geometry course he was taking. Maybe it was all a farce, but I dunno.

morphic

Don't you usually cover ultrafilters in introductory point-set topology

Jake1234

12:22 AM

:) no problem

Ted Shifrin

NOOOOOO @morphic.

user174558

@morphic No.

Ted Shifrin

Mr eyeglasses, you have the strangest ideas of what is "basic" or "elementary" based on your school curriculum!

user174558

@morphic The sections on filters in topology books are usually starred.

Chris's sis the artist

@Michael your try is pretty funny in a way, but I assure you that for my questions it might not be enough a life to answer them, with PhD or not.

@Michael Let's take one

Calculate by series manipulation only $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3$$

Ted Shifrin

12:24 AM

Not even in Munkres, and I recall Kelley's book covered nets, but I don't remember ultrafilters.

I no longer have all the books, so I can't look.

Chris's sis the artist

@Michael you're free to use any resources existing on earth. I mean I have nothign against if you call another 1000 PhD people like you and thing together of my question.

user174558

@TedShifrin Will you be celebrating Thanksgiving with family?

Ted Shifrin

Nope, haven't done that in about 35 years, @Jasper.

user174558

@TedShifrin Oh, maybe you should, this Christmas or something.

morphic

I went to a Thanksgiving party at my school and that will be the extent of my celebration

Ted Shifrin

12:26 AM

I decided long ago that cooking for good friends was a lot more fun than dealing with family, although my sister and her husband and I get along great.

For most of the last 30 years, I invited friends and lost souls to my house and spent 3 days cooking.

Chris's sis the artist

@Michael calculate in closed form by real methods $$\int_{-1}^1 \frac{\log^2(1-x)}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx$$ in the next 10 years.

morphic

Several TV show scenes are being filmed in some our buildings this week so it's kind of crowded here

user174558

@TedShifrin Aha. I hope your mum is OK.

Ted Shifrin

She hasn't known who I am in about 4-5 years, @Jasper. I will visit her in the nursing home when I go back to ATL in January.

Chris's sis the artist

@Michael let me give you one to think of for 20 years.

morphic

12:28 AM

:(

Michael

@Chris'ssistheartist Oh. My. God. My brain hurts. Ok I will clip mark that and I'll be back in a quick 10.

user174558

@morphic Maybe they are looking for you Mr Handsome.

Ted Shifrin

wow, @morphic, what TV show?

morphic

@TedShifrin Three of them: Elementary, The Blacklist, and Blindspot. I have seen them filming Blindspot near my church before a couple months ago

Ted Shifrin

Wow, I'll have to do serious googling.

morphic

12:29 AM

@TedShifrin They are young people TV shows, I think

Ted Shifrin

LOL ... you mean younger than you?

morphic

They said there will be fake gunfire noises on campus lol

Ted Shifrin

Maybe they make money on campus this way.

morphic

Well Elementary is about a modern version of Sherlock Holmes. It has Jonny Lee Miller (he was the main character in the movie Hackers but I don't know what else he's been in), and also Lucy Liu is in it

user174558

I want to join the show Mr Bachelor, lol.

Chris's sis the artist

12:31 AM

$$\sum_{n=1}^{\infty} \left(\left(\zeta(2)-1-\frac{1}{ 2^2}-\cdots -\frac{1}{n^2}\right)} \left(\zeta(3)-1- \frac{ 1}{2^3}-\cdots -\frac{1}{n^3} \right)\left(\zeta(4)-1-\frac{1}{2^4}-\cdots - \frac{1}{n^4}\right)\right)^2$$

morphic

Blindspot is based on a book series of the same title I think. It has the main character from the new 300 movie and the lady in the Thor movies that is in Thor's gang

Ted Shifrin

that doesn't sound like young persons' stuff, mr eyeglasses. And Blacklist is FBI drama. Although I used to love James Spader.

morphic

Oh, I only know James Spader from The Office (US version) lol

Too bad I didn't get any photos today. Too busy with schoolwork

user174558

I love The Wonder Years!!!

Ted Shifrin

I remember Spader from movies like Sex, Lies, Videotape, plus he was in Boston Legal for years.

morphic

12:33 AM

Oh, I've never heard of any of those

Michael

Is there a way I can show someone an equation I used in word and they can check if it is right?

Chris's sis the artist

@Michael $$\int _0^{\infty }\int _0^{\infty }\int _0^{\infty }\frac{\left(x \cot \left(\frac{x}{2}\right)-2\right) \left(y \cot \left(\frac{y}{2}\right)-2\right) \left(z \cot \left(\frac{z}{2}\right)-2\right) e^{-x y z}}{xyz}dx dy dz$$

Ted Shifrin

We hate Word here. :)

user174558

But we love words. =)

Ted Shifrin

One movie, morphic, Sex, Lies, and Videotape, I think.

Chris's sis the artist

12:34 AM

@Michael come here with all you best people you know and give me a lesson in my area

morphic

@TedShifrin Do you cringe whenever a math paper is written in Word

Chris's sis the artist

another one

$$\sum_{n=1}^{\infty} \left(\left(\zeta(2)-1-\frac{1}{ 2^2}-\cdots -\frac{1}{n^2}\right) \left(\zeta(3)-1- \frac{ 1}{2^3}-\cdots -\frac{1}{n^3} \right)\left(\zeta(4)-1-\frac{1}{2^4}-\cdots - \frac{1}{n^4}\right)\right)^2$$

Ted Shifrin

No one does it any more, seriously, morphic.

Michael

@Chris'ssistheartist is that a challenge to the god of highschool integrals himself?

Ted Shifrin

Some old fuddy duddies type homework assignments in Word.

Michael

12:35 AM

How can I upload an image?

Ted Shifrin

Hit "upload," @Michael.

Michael

where is it good sir?

Ted Shifrin

Next to "Send" to the right of the text box you type in.

user174558

@Michael The bottom.

Chris's sis the artist

@Michael Look, I just changed my mind: show me a brilliant idea, not a whole solution and I count it as if you had a whole solution.

Michael

12:36 AM

@JasperLoy do you need certain reputation to use it?

morphic

When a student was submitting math homework via e-mail here, the professor asked him to attach it as a .doc file lol...

user174558

@Michael I don't know. Ask the papaya.

Ted Shifrin

@morphic: Here you go.

morphic

Wow, 1989

The only movie I know from that year is Batman

Ted Shifrin

LOL, I was already tenured by then.

morphic

12:36 AM

Wow!!!

user174558

I love all the 4 Indiana Jones movies.

Chris's sis the artist

@Michael I don't know what stuff you created solved so far, but you talk to someone that created thousands of problems, not baby questions.

@Michael but it would be so great to give me your brilliant solutions! To learn from you or from your other 1000 friends of yours with PhD.

morphic

Harrison Ford kind of looks like he'd be a grumpy guy in real life

user174558

@Chris'ssistheartist I remember I called Sayan a baby and he was upset, ROFL.

morphic

@JasperLoy Who is Sayan

user174558

12:38 AM

@morphic I bet he can run faster than me.

Chris's sis the artist

@Michael I come in a TV show with you all, I mean to make a nice challenge with much fun.

Michael

@Chris'ssistheartist I just solved the twin prime conjecture.

user174558

@morphic Remember Me

morphic

@JasperLoy I haven't seen him in a while

user174558

@Michael Are you a troll?

Michael

12:39 AM

@JasperLoy it was a joke

Ted Shifrin

LOL, you shouldn't ask such questions, @Jasper, since you troll with the best of 'em.

Chris's sis the artist

@Michael If you wanna challenge me, then let's do it. So, tell me a word about any of the stuff above. What does it go better, tan or hyperbolic functions?

user174558

=)

Michael

@Chris'ssistheartist easy. you see, there is a. there is a... There is a 2^2. Aha. I saw that. Checkmate. Hyperbolic

user174558

@Chris'ssistheartist I don't think he is able to answer you.

Chris's sis the artist

12:40 AM

@Michael What would you do in front of this stuff, really, other than admire it profoundly? It's the only thing you can do, to admire it much and hope one day I might like to show you some ways to calculate them.

user174558

Wait, is Sajindia male or female? I cannot tell from the pic @TedShifrin and @morphic.

Chris's sis the artist

@Michael but I'm not sure I wanna show you anything since you seem to know so much stuff.

Michael

@Chris'ssistheartist well first I need to get this chat to look like normal and not dollar signs. Then maybe give it 20 years. THen I might try to understand a part of it.

imgur.com/CZLc35C

Ted Shifrin

May identify as female, @Jasper, but cis male, as far as I can tell.

user174558

@Michael You might take 20 years to get this chat to look normal, LOL.

Michael

12:42 AM

@Chris'ssistheartist imgur.com/CZLc35C Will this function work for arc length of a quadratic?

Jake1234

my god, I hate it when I get stuck on a problem and can't let it go.

morphic

@JasperLoy I think male; I saw he made a video about coming out as gay or something

Jake1234

I'm starting to think my OCD tendencies and math don't go together very well.

user174558

@Sajindia Are you the user called Twink?

user174558

@Jake1234 Sorry to hear. I know it is hell.

Michael

12:43 AM

imgur.com/CZLc35C . WOuld this function be a good way to calculate the arc length of a function between two points?

Excluding the fact that it is not definite

I can always make it a definite

Ted Shifrin

I don't see why all this meddling into identities is needed, @Jasper.

user174558

@TedShifrin OK. It's just so that at least we know who is who.

Chris's sis the artist

@Michael it seems you don't like that much what I call the real stuff.

morphic

@JasperLoy likes to be a detective

Ted Shifrin

We certainly have no idea who chris'ssis is. Leave it alone.

user174558

12:45 AM

Hey @Sajindia if you want to talk, let me know.

Michael

@Chris'ssistheartist Haha I guess. No but can you seriously check this equation imgur.com/CZLc35C . I have an assigment due tomorrow that requires me to use somethign similar

Chris's sis the artist

@Michael I also have an assignment, amongts many other things, not to pay attention to these challenges and see of my work. You said you know nothing about integration, and then you put questions that come from a different level and try to laugh.

user174558

@Michael Does not mean anything when you have not specified your variables...

Michael

find the length between 45 and 1200

its an indefinite

user174558

@Michael You seem to jump from question to question. That is not good.

Chris's sis the artist

12:47 AM

My mistake I entered this conversation.

Michael

@JasperLoy Ok this is actually dead serious. This is something my level.

user174558

@Chris'ssistheartist You can leave it anytime.

user174558

@Michael But you cannot ask the expert such an elementary question. If you are serious, post on the main site. The expert is tired.

Michael

@JasperLoy but the expert likes helping bad people right, mrs expert?

@Chris'ssistheartist Please say yes

user174558

@Michael No, you seem like a troll now.

Michael

12:49 AM

@JasperLoy I SWEAR TO GOD THIS IS AN EQUATION THAT I DERIVED. I need to make sure it can calculate the length of a quadratic in the form of y=ax^2+bx+c

user174558

@Michael OK, post the question properly written on the main site if you are so serious.

Michael

I did

user174558

Then wait, LOL. End of story.

Michael

0

Q: Need help with a certain integral

Sorry, this is my first post ever and formatting is bad. I appreciate all assistance Given the equation y= $ 3.8x^2-4.4x+1444 $ Find the definite arc length integral between 0 and 1200. I am not sure on how to format $(dy/dx)^2$ ,so I left it alone. $$\int_0^{1200} \sqrt{1+ (\frac{dy}{dx}}...

I got an answer, I just need to make sure I understand it and it is correct

I don't know why everyone is being mean. I am asking for help in understanding

Nvm. I'll go fail

user174558

@morphic Not as much of a detective as @alexclark.

Chris's sis the artist

12:54 AM

@Michael I think you only try (unsuccessfully) to make fun of people since some minutes ago you seemed very in trouble with simple differentiations. In the meantime you finished some other classes.

Michael

@Chris'ssistheartist I wasn't making fun of anyone. I was sarcastically saying I had a phd...

@Chris'ssistheartist I thought we were going on with the sarcasm.

@Chris'ssistheartist but I can't force you to help... I'll pretend it is right and just go with it.

anon

@TedShifrin I put it over here

1

Q: Finitely-presented torsionfree group whose abelianization has torsion

In chat someone asked Does anyone know of a torsion-free group (finitely presentable) whose abelianisation has torsion?

abstract-algebra group-theory

spose I should ping @DanRust too

L33ter

Lagange :D

hi @anon

anon

hello

Chris's sis the artist

@Michael How would you calculate $$\int_0^1 x^{n-1} \log^3(1-x) \ dx$$? In case you're interested ... (I think this one shouldn't be a difficulty for you)

Michael

12:59 AM

I can't see it, sorry

L33ter

hi @TedShifrin

Chris's sis the artist

@Michael It's what I call to be a light integral. You can finish it even mentally, seriously while jogging, doing some shopping and other activities.

Michael

@Chris'ssistheartist Uhmm, sec let me try to get this chat thing to work.

L33ter

@TedShifrin do you know about phase space and configuration space ?

Michael

@Chris'ssistheartist Wait so how far does your knowledge of integrals go?

user174558

1:01 AM

@Michael I did not call her world expert for nothing.

Chris's sis the artist

I'm out to take some sleep.

user174558

@Michael I think you should not waste her time anymore.

user174558

@Chris'ssistheartist Goodnight.

Chris's sis the artist

@JasperLoy Good night.

user174558

Happy Thanksgiving Day to America! Good night!

Sajindia

1:09 AM

Is every covering space $p : E \to X$ induced by a properly discontinuous action on $E$ of a group $G$ so that $X=E/G$?

anon

those are the regular/normal/Galois covers aren't they?

Sajindia

I don't know

so not every covering space has that property?

PVAL

2:32 AM

@anon Didn't this get answered in chat as well...

Well I upgraded my comment in chat to an answer there..

Michael

3:06 AM

hello

Can you have an n'th integral?

Michael

3:25 AM

nvm

I'm mostly just an idiot

5:46 AM

Hi friends

How are we all?

Hello @TobiasKildetoft. Are you a student?

Don't know if people usually ask that sort of question here

Tobias Kildetoft

@I'mmostlyjustanidiot I am a postdoc

I'm mostly just an idiot

Oh wow, and it's meant to be really hard to get a postdoc position I have heard, something like 1% of people get one or something

Tobias Kildetoft

@I'mmostlyjustanidiot They are certainly competitive, but I think it is way more than 1%, at least if you only count those who actually apply for them

I'm mostly just an idiot

Oh, well that's very good to hear then. I am not a very good student because I'm not very smart, so I am trying to work extra hard to succeed

I'm taking a gap year now, so I have an extra year of preparation for my next year of university

Question: Why do we bother to name idempotent elements?

Tobias Kildetoft

6:05 AM

@I'mmostlyjustanidiot Because it is an important property

they are for example related to the decomposition of rings into direct products

I'm mostly just an idiot

So the book lets $x\in A$ have $x^2=x$, then says $x'=1-x$ which clearly is also idempotent, since $(x')^2=(1-x)^2=1-2x+x^2=1-2x+x=1-x=x'$. We obtain $x+x'=1$ and $x-x^2=0$.

They then state that $x$ and $x'$ are 'complementary orthogonal idempotents', does that literally just mean that $x+x'=1$ and $xx'=0$ where $x,x'$ are idempotents?

Then they say that with this, we can take $a=ax+ax'$ for any $a\in A$ and we have that $A$ is a direct sum of rings $A=A_1\oplus A_2$ where $A_1=Ax$ and $A_2=Ax'$

Tobias Kildetoft

@I'mmostlyjustanidiot Yes, that is precisely what complementary means here

I'm mostly just an idiot

That's a beautiful construction

I'm mostly just an idiot

7:32 AM

Why does complex Lie algebra $g$ being solvable imply that $g'$ is nilpotent?

I can show the converse easily

Paradox 101

7:53 AM

@robjohn can you explain how $$\binom{-\frac12}{k}=\left(-\frac14\right)^k\binom{2k}{k}$$? Is this a result of a formula?

Tobias Kildetoft

8:27 AM

@I'mmostlyjustanidiot Yeah, the revere direction is a lot more tricky

It is because a solvable Lie algebra is isomorphic to a subalgebra of the algebra of upper triangular matrices

I'm mostly just an idiot

@TobiasKildetoft In the finite case I got an awesome answer from someone

Tobias Kildetoft

(usually formulated in terms of stabilizing a full flag)

I'm mostly just an idiot

@TobiasKildetoft Yep that's exactly what he went for: math.stackexchange.com/a/1547042/284674

Tobias Kildetoft

@I'mmostlyjustanidiot Yep, that's the result (and I agree that it probably fails for infinite dimensional Lie algebras)

though it might still work if one assumes it to be locally finite or something like that

I'm mostly just an idiot

What does locally finite mean, if it isn't too complicated

Tobias Kildetoft

8:30 AM

@I'mmostlyjustanidiot It means that any finitely generated subalgebra is finite dimensional

I'm mostly just an idiot

Thanks :)

Riggs

8:57 AM

is there some inequality relating (x - y)^2 to x^2 + y^2? Or is there nothing we can say about the relationship between these two expressions?

Tobias Kildetoft

@Riggs Their difference is precisely $-2xy$ (or $2xy$ depending on the order)

so it is determined by the signs of $x$ and $y$

Paradox 101

9:13 AM

Can anyone have a look at this? I'm not sure if I'm doing this correctly and am not sure how to go on from here

What will the interval of convergence be in this case?

Vrouvrou

9:49 AM

@robjohn are you there ?

Antonio Vargas

11:01 AM

@Paradox101 Why not try the ratio test?

user174558

11:13 AM

I am now an orange.

Chantry Cargill

o_o

I'm mostly just an idiot

11:57 AM

@TobiasKildetoft Your suspicion that it fails in the infinite case is verified here: math.stackexchange.com/a/1306293/284674

@JasperLoy That's a really nice color

user174558

@robjohn We are now of the same colour.

user174558

@I'mmostlyjustanidiot Thanks no need to call yourself idiot.

I'm mostly just an idiot

@JasperLoy Only mostly an idiot, not fully

robjohn

@JasperLoy what happened to your blue?

Tobias Kildetoft

@I'mmostlyjustanidiot Neat

user174558

11:58 AM

@robjohn Well, I thought I would have a change, at least for a while.

user174558

It's troublesome that there are so many definitions of the Fourier transform.

robjohn

@JasperLoy IMO, the one that Fourier Analysts should use is $$\int_{-\infty}^\infty f(x)e^{-2\pi ix\xi}\mathrm{d}x$$

user174558

@robjohn I see. Rudin uses another one though. Does it mean he is not a mathematician?

robjohn

@JasperLoy Does he use the one with $e^{-ix\xi}$?

user174558

@robjohn Yes, same as Michael Taylor and Lawrence Evans.

robjohn

12:03 PM

@JasperLoy That is the one that physicists usually use since it is easier to write. However, the inverse transform is different and other properties are not as nice.

Aldon

It's weird that the chat still does not support Mathjax

user174558

@robjohn That is the one Stein uses, and also Gerald Folland.

robjohn

@JasperLoy Folland was a Stein student, if I am not mistaken

user174558

@robjohn I see. I think I will use what you suggested then. That also seems to be the one most Fourier analysts use.

user174558

@robjohn Stein must be your student then!

robjohn

12:04 PM

@JasperLoy Perhaps I should modify my claim, then...

Paradox 101

@AntonioVargas how will a simple ratio test help? I have to use uniform convergence by the M test as well. Although for the M series I used the ratio test for values of $r$

user174558

@robjohn I am trying to like the Stein books but they just don't seem to contain enough material. The 4 Lectures in Analysis books.

Paradox 101

Is that what you meant or should I apply the ratio test from the start? @AntonioVargas

robjohn

@JasperLoy Those were written much too late. I never read them.

user174558

@robjohn Of course, the other 3 main ones are great. Anyway, most top algebraists seem to define ring to have 1, so I will also follow that convention.

I'm mostly just an idiot

12:09 PM

@Aldon It does, you just have to get the link off of the thing on the right

user174558

Wow, this orange is really beautiful. Maybe I should start eating oranges now and not bananas.

robjohn

@Paradox101 $$\begin{align}\binom{-\frac12}{n} &=\frac{\left(-\frac12\right) \left(-\frac32\right) \left(-\frac52\right)\cdots \left(-\frac{2n-1}2\right)}{n!}\\ &=\frac{(-1)^n}{2^nn!}(2n-1)!!\\ &=\frac{(-1)^n}{2^nn!}\frac{(2n)!}{2^nn!}\\ &=\left(-\frac14\right)^n\binom{2n}{n}\end{align}$$

I'm mostly just an idiot

That orange makes me happy

user174558

@robjohn Today is Thanksgiving Day. I am thankful for my mum and my books, but nothing else.

user174558

@I'mmostlyjustanidiot The sun rises just before it becomes darkest.

I'm mostly just an idiot

12:11 PM

@JasperLoy How?

user174558

@I'mmostlyjustanidiot Just a saying.

I'm mostly just an idiot

I don't like it

user174558

@I'mmostlyjustanidiot It means something like don't give up when you feel like.

user174558

Did you know that Book Depository was started by a former Amazon employee and is now acquired by Amazon? Interesting.

I'm mostly just an idiot

Jasper, are you a math student?

Would you like to discuss Noetherian rings?

user174558

12:14 PM

I am a mental patient trying to get well, not a math student, sorry.

I'm mostly just an idiot

I am a math student trying to get mental

user174558

You can discuss math with the experts here, I am only an orange.

Chantry Cargill

@JasperLoy But you're the apple of my eye.

I'm mostly just an idiot

user174558

@ChantryCargill Hehe. Your name is interesting. It sounds Indian.

Chantry Cargill

12:16 PM

@JasperLoy It's from a cowboy book I believe.

user174558

@ChantryCargill So it is not your real name? I am using mine here.

Chantry Cargill

@JasperLoy It is my real name.

It was based off of a series by Louis L'amour. I think it was borden chantry.

user174558

@ChantryCargill Ah? Your parents like cowboys then.

user174558

Interesting that cowboys ride on horses and not cows. They should be called horseboys.

Chantry Cargill

@JasperLoy Interesting observation.

user174558

12:18 PM

@ChantryCargill I read that you want to go to grad school and do math. What are you interested in?

user174558

Oh my mum is back, time for dinner, bye.

Chantry Cargill

lol

He must be hungry.

Paradox 101

@robjohn ok I got i. Thanks :)

Vrouvrou

hello @robjohn

Paradox 101

@robjohn can I ask you a question relate to analysis?

Krijn

12:27 PM

@JasperLoy The explanation is easy: A cowboy is the boy who herds the cows.

Agawa001

12:43 PM

@JasperLoy is that a closed-eye mean square that u had in avatar, i mistook u for rob from far

Chantry Cargill

@Agawa001 It's actually the back of his head.

Agawa001

lol :D

Antonio Vargas

1:05 PM

@Paradox101 I missed the part where you have to use the M-test.

Agawa001

@evinda try this v = symsum(4/(k^2*pi^2)*(sin(k*pi/2)*sin(k*pi*x))'*exp(-k^2*pi^2*t), k, 1, Inf)

Paradox 101

@AntonioVargas then is what I've done correc? If so how do I poced from thee?

Antonio Vargas

1:20 PM

@Paradox101 I'm not sure what you have done. It wasn't clear from the image.

Paradox 101

1:41 PM

@AntonioVargas this is what I've done: $ \sum \frac{1}{\sqrt n}(\frac{x}{1+x})^n \le \sum \frac{1}{\sqrt n}x^n=\sum \frac{1}{\sqrt n}r^n= M_n$ from ratio test $r<1$ $\implies$ $-r<(\frac{x}{1+x})^n<r$

Antonio Vargas

@Paradox101 Probably the first step is not correct (unless you also assume $x>0$?). You'd be better off setting $x/(1+x) = y$ and finding the radius of convergence for $y$ in $\sum n^{-1/2} y^n$ first, then substitute back in $y = x/(1+x)$ to find the interval of admissible $x$'s.

Chantry Cargill

How does achille swap out n for x d/dx here? math.stackexchange.com/a/1547255/109879

Huy

does Heine Borel imply locally compact?

Paradox 101

1:56 PM

@AntonioVargas if I do that, then wouldn't it end in the same manner? $ \sum \frac{1}{\sqrt n}(\frac{x}{1+x})^n \le \sum \frac{1}{\sqrt n}r^n=\sum \frac{1}{\sqrt n}r^n= M_n$ from ratio test $r<1$ $\implies$ $-r<(\frac{x}{1+x})^n<r$

Antonio Vargas

@Paradox101 Sure, except now you've taken a valid path there :)

Mats Granvik

paulcooijmans.com/ethics/abbrev.html

Paradox 101

@AntonioVargas so then in the final answer we'll simply replace $r$ with $1$?

robjohn

2:20 PM

@Paradox101 sorry, I was away.

Paradox 101

@robjohn no problem

robjohn

@JasperLoy No, I was a Stein student.

@Paradox101 did you get your question answered?

Paradox 101

@robjohn yes almost. Given the question regarding the M-test in the picture is the last point in this solution $ \sum \frac{1}{\sqrt n}(\frac{x}{1+x})^n \le \sum \frac{1}{\sqrt n}^n=\sum \frac{1}{\sqrt n}r^n= M_n$ from ratio test $r<1$ $\implies$ $-r<(\frac{x}{1+x})^n<r \implies |(\frac{x}{1+x})^n|< 1$ <$ the answer?

robjohn

2:36 PM

what picture?

Paradox 101

@robjohn this one

robjohn

@Paradox101 You need $\left|\frac{x}{1+x}\right|\lt1$

Paradox 101

@robjohn oh ok but after we remove $n$ from there that will simply be the answer we need?

robjohn

@Paradox101 Not sure what you mean by "remove $n$ from there"

@Paradox101 To get uniform convergence, you need $\left|\frac{x}{1+x}\right|\le r\lt1$

Have to go for a while.

jukka.aalto

2:51 PM

How do I sketch the support for a probability function p_{X,Y}: supp p_{X,Y}={(x,y)|x=-1,0,1,y=1,2,...} ?

I can't imagine how it looks!

If anyone could lead me on the way or draw a crude sketch, it would be immensely helpful

Paradox 101

3:06 PM

@robjohn I mean that as we've assigned $r$ the value $(\frac{x}{1+x})$ then there will be no $n$ here. So will this simply be the answer we were looking for?

ok

anon

3:41 PM

@PVAL I looked through some of the transcript to see if it got answered but apparently not enough

jukka.aalto

I've reformulated my troubles with discrete, stochastic vectors in a example question: math.stackexchange.com/questions/1547402/…

Feel free to help

I have enumerated the things I don't know how to work out

This is for my exam prep

AlpArslan

3:58 PM

Hey guys, just a question.

So, I have that, if X is a complex inner product space and T a linear endomorphism of X, then, if <Tx,x> = 0 for all x in X, we must have T = 0.

Thing is, consider the space spanned by all finite linear combinations of {e^(imx), m \in \mathbb{N}}

That's a complex inner product space given by the inner product, <f,g> = f \bar{g}.

anon

@AlpArslan why does <Tx,x>=0 for all x imply T=0?

also $f(x)\overline{g(x)}$ is a function, not a scalar. you mean $\int_{-\pi}^\pi f(x)\overline{g(x)}\,{\rm d}x$

AlpArslan

My apologies, I left out the integral.

Yes, that's right.

Ach, no, I found the error in the example.

I was about to say, let T: e^(imx) \to e^(i(m+1)x)

But, yes, yes, I see why it doesn't work.

Hmm, besides just screwing with the algebra, is there a neater way of proving that <Tx,x> = 0 implies T = 0?

anon

well, it's not true for real inner product spaces

nor for bilinear forms on complex spaces (the complex inner product is sesquilinear not bilinear)

AlpArslan

Yes, that's right.

Hmm.

So, the key lies in the sesquilinearity, ofcourse.

But, ach, I'm getting lost in circular algebra

Jake1234

4:20 PM

Guys, recently I've been thinking that my problem solving skills seem to be quite poor - do you think there would be some use in solving things like math olympiads etc, despite the fact that I'm studying at a university?

I think sort of missed that part in my life - while other good students were solving problems of that kind, I was busy playing video games xD

evinda

Hi!!!

Is $\langle x, \langle x,y \rangle y \rangle $ equal to $\langle x,y \rangle^2$ ?

anon

if $\lambda$ is a scalar, $\langle x,\lambda y\rangle$ is equal to $\overline{\lambda}\langle x,y\rangle$ if you're a mathematician and equal to $\lambda\langle x,y\rangle$ if you're a physicist

evinda

I study applied mathematics... and we have seen the property $\langle x,\lambda y\rangle=\lambda\langle x,y\rangle$. @anon

anon

are you using real numbers or complex numbers?

evinda

Real numbers... aa in this case $\overline{\lambda}=\lambda$, right? @anon

anon

4:31 PM

yes

evinda

A ok... Let $X$ be a Hilbert space ( with inner product $\langle, \rangle $).
The following inequality holds:

$$(\star) |\langle x,y \rangle| \leq ||x|| \cdot ||y|| , \forall x,y \in X$$

@anon

@anon I have also an other question

The inequality $(\star)$ is equivalently written as $\langle x,y \rangle^2 \leq \langle x,x \rangle \langle y,y \rangle , x,y \in X (\star \star)$.

If $y=0$ then $(\star \star)$ holds since $\langle x,0 \rangle =0$ and $\langle 0,0 \rangle=0$.

It suffices to show that $(\star \star)$ holds for $x \in X$ and $y \neq 0$ such that $\langle y,y \rangle=1$.

(if $y \in X \setminus{ \{0\} }$ and $||y|| \neq 1$ then we pick as $y$ the quantity $\frac{y}{||y||}$)

Let $y \in X \setminus{ \{ 0 \} }$ with $\langle y,y \rangle=1$ and $x \in X$.

anon

sure you do

if you scale $y$ in $(\star\star)$ you still get a true equality

Claim I: `(**)` holds for all x,y
Claim II: `(**)` holds for all x and unit-norm y

Can you prove Claim I implies Claim II? Can you prove Claim II implies Claim I?

evinda

4:51 PM

@anon We suppose that it holds that $\langle x,y \rangle ^2 \leq \langle x,x \rangle \langle y,y \rangle$.

If $||y||=1 \Rightarrow ||y||^2=1 \Rightarrow \langle y,y \rangle=1$.
So we get that $\langle x,y \rangle ^2 \leq \langle x,x \rangle$.

Now, let's suppose that $\langle x,y \rangle^2 \leq \langle x,x \rangle$ for $y \in X$ with $||y||=1$.
We pick as y , $\frac{y}{||y||}$ and we get: $ \langle x, \frac{y}{||y||} \rangle^2 \leq \langle x,x \rangle \Rightarrow \frac{1}{||y||^2} \langle x,y \rangle^2 \leq \langle x,x \rangle \Rightarrow \langle x,y \rangle^2 \leq \langle x,x \rangle \lan…

@anon Is it right like that?

anon

more or less

evinda

Could I improve something?@anon

anon

your first sentence should say "for all y." you don't need to write any equations down to explain why Claim I implies Claim II. It actually doesn't make sense to say "we pick as y, y/||y||," you have to find a better way of saying that. In fact just your equation is fine for proving Claim II implies Claim I.

evinda

Ok.. Thank you!!! @anon

Transcript for

$Mathematics$ Mathematics