Mathematics

2:08 AM

Like you said, it's a sophisticated joke

Try the Homotopy chat room.

Mike Miller

3:50 AM

this was 2.5 years ago...

...nvm

4:53 AM

Hi there! :-)

I wondered whether removing homework will somehow bring new batch of generalist badges. It seems that algebraic-topology and commutative-algebra have similar numbers of questions and probably sometimes one of them is ahead (and in top 40 tags) and sometimes the other one. Removal of homework tag will put both these tags into top 40 tags.

@MartinSleziak: Good time! Is there any reference in which we can have almost all about Trigonometry ? I couldn't find such this on the site. Thanks

What "almost all" means.

List of trigonometric identities at Wikipedia?

Good book recommendations on trigonometry

@MartinSleziak: I mean such a book , a famous one.

I don't know what you are looking for. (But I am not particularly knowledgeable of trigonometry.)

5:02 AM

Thanks. I got the second link. Thanks.

Hans-Jochen Bartsch?

@MartinSleziak: May I ask you: Do you know Dr. M. Eshagi at Semnan university?

No, I don't.

Maybe you will ask something useful among questions tagged trigonometry+reference-request.

Thanks again.

6:12 AM

Team 900 will make math.se an empty castle! Or, someday, I will visit this site only for the same reasons as I do visit the EJMR.

r9m

6:24 AM

@Sush what is EJMR ?

The Substitute

I am reading notes on analysis that proves (1) Every bounded sequence in the reals has a subsequence that converges to the supremum of the sequence and (2) The Bolzano-Weierstrass Theorem, which says that any bounded sequence has a convergent subsequence. Is there any practical use in treating (1) and (2) as if they are not equivalent?

6:56 AM

@blue On the other hand, nobody appreciated the gif I posted.

eXtremiity

@TheSubstitute.
(1) says that a bounded sequence has a subsequence that converges TO THE supremum of the original sequence. That is, (1) says where the subsequence will converge to.

(2) says that given any bounded sequence, it has a subsequence that will converge. Furthermore, the original sequence can be bounded and DIVERGENT, but will still contain a subsequence that converges. However, the theorem does not indicate where it converges to.

Hold on, I'm having second thoughts. (1) can be a bounded divergent sequence, will still have a supremum, and therefore the subsequence will converge to it. Hmm.

7:20 AM

It's virtually impossible to keep the pace with 900 sit-ups a day in the number of the votes cast. He leads this week, month, quarter and year.

Please let me know why $\sum_{j=1}^n|x_j|\le\sqrt{n\cdot\sum_{j=1}^nx_j^2}$?

@MartinSleziak

@Sush Isn't that just Cauchy-Schwarz, when one of the vectors is (1,1,...1).

@MartinSleziak, ok, let me see!

BTW it seems to be basically this question: Why is this true: $\|x\|_1 \le \sqrt n \cdot \|x\|_2$?

I have to leave. See you later!

Later

7:28 AM

@MartinSleziak, thanks for making my mind walk little forward.

@skullpatrol, hi!

Hi pal :)

@DanielFischer, thank you so much for helping me last night.

You're welcome, @Sush.

Dobry den, @MartinSleziak.

Has the excitement over winning the World Cup settled down yet @Daniel?

@skullpatrol I never won any world cup, as far as I remember.

The excitement is about to start, because next weekend the season is going underway, @skullpatrol.

7:39 AM

What season is that?

@skullpatrol The league, football/soccer.

Well, being world champions makes it that much more special, no?

@DanielFischer, my book has Cauch-Schwartz as $|x\cdot y|\le \|x\|\|y\|$, where $x$ and $y$ are vectors. So, letting $y=1$, we get $|\sum_{j=1}^nx_j|\le\sqrt{n\cdot\sum_{j=1}^nx_j^2}.$ But triangular inequality gives $|\sum_{j=1}^nx_j|\leq\sum_{j=1}^n|x_j|,$ so how is $\sum_{j=1}^n|x_j|\le\sqrt{n\cdot\sum_{j=1}^nx_j^2}$ guaranteed?

blue

do CS with (|x_1|,...,|x_n|)

@Sush Consider $\tilde{x} = (\lvert x_1\rvert, \lvert x_2\rvert,\dotsc,\lvert x_n\rvert)$. Since $x_k^2 = \lvert x_k\rvert^2$ ...

And, @Sush, Schwarz without 't', it's Hermann Amandus, not Laurent.

7:45 AM

@DanielFischer, ok! sorry for mistake!

@Sush No problem. A lot of people mix them up.

@DanielFischer, thank you so much!!! I didn't know that such little things as $x_k^2 = \lvert x_k\rvert^2$ can be so useful!

@Sush The little things are immensely useful. One just doesn't notice it so often because they are so little.

@DanielFischer, I just saw "A Beautiful Mind". Did Nash really stuck with those people of his past? Did he work for Pentagon as code breaker? I have doubt because some sources say that he did not give a speech while receiving Nobel Prize. So, there might be other things just to make movie beautiful, I think.

@Sush I know pretty little of Nash. His embedding theorem, of course. And that he was schizophrenic (if that is in fact the correct diagnosis, I'm not sure about that). Beyond that, not so much.

7:56 AM

@DanielFischer, ok, thank you so much!

8:14 AM

@Chris'sis Prove that $$\lim_{n\to\infty} \prod^n_{i=0}\Gamma\left(1+\frac in\right)^{\Gamma\left(1+\frac in\right)/n}=e$$

@robjohn It's late, you should be sleeping.

@Sush The real life is quite different from the movie. Read the book to find out. But the movie is nonetheless inspiring.

8:39 AM

@900sit-upsaday I have deleted my account about ten times, don't worry about it.

8:53 AM

@Chris'ssis above sorry

9:26 AM

Greetings

@Alizter Does it tend to $e$? How is that possible?

@Alizter Question: what is the first thing to note? Don't hurry, think of it.

@Chris'ssis I have a beautiful proof for it :)

@Alizter Teach me.

Start with ze log

OK ... and then ...

It turns into a riemann sum

9:38 AM

@Alizter Yeah, that's clear.

So then the intragral is $\int_1^2 \log{\Gamma(X)}^{\Gamma(x)}dx$

rehband

Hey guys, I have a newbie question here: How do I find the partial derivatives of $$h(x_1,..,x_n)=\int_{0}^{||x||} f(t) dt$$ where $||x||$ is the Euclidean norm of $x=(x_1,...,x_n)$ and $f$ is some continuous function?

@Alizter and then?

Hmm I may have done something wrong here

@Alizter Well, I'm glad you see that. :-)

9:42 AM

wait no

Digamma function

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: It is the first of the polygamma functions. == Relation to harmonic numbers == The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as == Integral representations == If the real part of x is positive then the digamma function has the following integral representation...

@Chris'ssis Yes

I have a mistake ooops

@Alizter This one would have been a nice variant though $$\lim_{n\to\infty} \prod^n_{i=0}\Gamma\left(1+\frac in\right)^{1/n}$$

OR

$$\lim_{n\to\infty} \prod^n_{i=0}\Gamma\left(1+\frac in\right)^{(1+i/n)/n}=e^{\large\frac{3}{4}\log(2\pi)-\log(A)-\frac{5}{4}}$$

I don't know what I was thinking :)

@Alizter try this one - Prove that

$$\sum_{n=1}^{\infty}\left(\psi^{(0)}(n)+\frac{3}{2}\psi^{(1)}(n)+\frac{3}{2}\psi^{(2)}(n)+\frac{1}{2}\psi^{(3)}(n)+\frac{1-n}{n^2}-\log(n)\right)$$
$$=2+\frac{\gamma}{2}-\frac{\log(2\pi)}{2}-2\zeta(2)+3\zeta(3)$$

Actually I'm only interested to tell me the first step to do. Take your time! :-)

This is about the art of solving problems.

I'm sure many would do it with the risk of getting a messy proof. The point is to get a marvellous proof.

Mew

10:04 AM

hi

anyone in?

I am here, but I cannot help you with math.

Mew

aw

pretty simple question tho

how to solve tan(x) = sqrt(2) - 1

i know the answer is pi/8, but what is the easiest way to get it

$$\tan(a/2)=\sin(a)/(1+\cos(a))$$

$$ \sqrt{2} - 1=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}/2}{1+\sqrt{2}/2}=\frac{\sin(\pi/4)}{1+\cos‌(\pi/4)}=\tan(\pi/8)$$

@Chris'ssis Wow, you are a genius!

10:38 AM

@MikeMiller Although I agree with your comment about spamming others, I don't think people will necessarily look at his question.

@Chris'ssis I start with $\sum_{k=n}^\infty\frac1{k^{m+1}}$

Mew

ty Chris's sis

11:47 AM

@Chris'ssis I assume you have one...

@robjohn You are up early, lol.

@robjohn Sure. Welcome back! (btw) :-)

@Chris'ssis I was worried about the $+\frac32\psi'(n)$ term until I noticed the $-\frac1n$ term later

@Chris'ssis What is the distance you run every day?

@JasperLoy approx. 3-4 km

11:51 AM

@Chris'ssis Very impressive.

@JasperLoy It really makes you feel better, especially mentally. :-)

@robjohn You have a solution, right?

@Chris'ssis I just saw the question. I know the asymptotics, so I know whether it converges.

@robjohn That's a very good starting point.

@robjohn Are you familiar with Banach manifolds or integrals of Banach-valued functions?

@Alizter see what @robjohn said above.

11:57 AM

Hello

I ask because Lang uses Banach manifolds in his "Fundamentals of differential geometry" and integrals of Banach-valued functions in his "Real and functional analysis", but it seems they are not mainstream @robjohn.

Is there anyone familiar with Munkres topology?

@MathsLover What do you want to ask?

I wonder If metric spaces(two sections of chapter 2) are required to study chapter 3( about connectedness and compactness)

@JasDer

@JasperLoy

@MathsLover I can't tell you offhand like that, but metric spaces are very important, so you should study them.

12:00 PM

I've already studied them but quickly, Not in depth though.

@MathsLover OK, then I guess you can move on to the next chapter first, though it's always good to study books from cover to cover.

Why is that?

@JasperLoy

I mean, why is it better to study books from cover to cover?

@MathsLover To get the most information out of it?

Ah ok, but I've heared that munkres treatment of algebraic topology is not that good. it's better to find another text (anyway, I don't intend to study algebraic topology currently)

@JasperLoy

@robjohn, If you're familiar woth munkres topology, Does the third chapter on connectedness and compactness require solid knowledge of metric spaces( two sections on chapter 2) ?

with*

@MathsLover I've never read it

12:05 PM

Ah ok thanx :))

@MathsLover I recommend you Bredon's Topology and Geometry. It has good treatments of point set and algebraic topology, and treats differential topology too.

Do you know anyone in the site who know the book so I can ask him\her ?@robjohn

@MathsLover Even if he reads the whole book, he cannot answer your question. It is kind of a non-question.

@JasperLoy, I will give it a look(maybe after studying general topology from munkres?! )

@MathsLover Bredon is the only book that treats point set, differential and algebraic topology in some detail.

12:09 PM

Ah cool, @JasperLoy , I will look for it soon :)

Ok @jasper , I will move on to the next chapter. If I found I needs more on metric spaces I can go back to it and then go forward again , isn't a good idea ?

12:32 PM

@MathsLover I imagine there are quite a few, but I don't know of anyone in particular.

Balarka Sen

12:42 PM

@blue of course it was I.

Hi people! Is it reasonable (or just not-crazy) to define a symmetric positive definite kernel on $\mathcal{X}\times\mathbb{S}_{++}^n times \mathcal{X}\times\mathbb{S}_{++}^n$, where $\mathcal{X}$ denotes a non-empty set, and $\mathbb{S}_{++}^n$ the space of symmetric positive definite $n\times n$ matrices? Thanks a lot! cc @robjohn, @DanielFischer ( P.s.: Is it ok to mention you, or it's just as annoying it seems to be? )

@MathsLover Yes, that is a good idea.

salbeira

To show that there is no local max or min in a multi-dimensional function at any point, I need to show that the hesse matrix at that point is indefinite, right?

1:03 PM

@nullgeppetto It is not-crazy. Whether it is reasonable (in the sense that it helps you come nearer to your goal) depends.

Thanks @DanielFischer! First it's good that it may make sense and it's not crazy..

@DanielFischer Have you thought of showing another picture, other than the one you are coughing?

@DanielFischer, actually I have just asked a question about, but I'm afraid that posting it here may be considered as spamming.. Should I post it?

@JasperLoy I have thought about that, but haven't searched through everything to see whether I have one where I don't look too stupid.

@DanielFischer I would like to ask you if you have learnt about Banach manifolds or integrals of Banach-valued functions. They don't seem mainstream stuff.

1:09 PM

@nullgeppetto I have looked at it. I don't think I can answer it :(

Thanks @DanielFischer! No problem at all, thanks for taking a look!

@JasperLoy A bit here and there.

@DanielFischer I hesitate to read Lang's Fundamentals of Differential Geometry and his Real and Functional Analysis because he uses them in those books.

@JasperLoy What about just dipping in your toe and seeing whether it is to your liking?

@DanielFischer OK, after I have read the more mainstream books.

1:24 PM

@robjohn I'll show you my proof later on, I didn't put it on paper yet (it won't take too long anyway).

(everything flows naturally)

Alyosha

In the definition of a normal extension, '$K$ is a splitting field for some family of polynomials in $F[X]$', why not just say `$K$ is a splitting field for at least one polynomial in $F[X]$', since a single polynomial still forms a family of polynomials (with {1} being the index set)?

Am I right in saying the two are equivalent?

This is a sanity check.

@Alyosha If $K$ is the splitting field of an infinite family of polynomials, it need not be the splitting field of a single polynomial.

If $K$ is the splitting field of $p$, then $[K:F] < \infty$.

Alyosha

@DanielFischer Thank you.

You are welcome.

barznjy

1:40 PM

Hi,
I have the following integral;

$$y=\int_{0}^{\infty}\frac{e^{x f}}{m+x}\gamma\left(a,h x\right) dx$$
where $f,m, and h \in reals$
f,m, and h are $>$ 0

$a \in integers$, a=1,2,3,...
$\gamma\left(a,h x\right)$ is the lower incomplete gamma function

Can anyone help me how to solve it?
Thank you very much

1:52 PM

Currently there are 12 accounts on math that say "please delete me" Can a mod delete them, or do the users need to have emailed the SE staff? @robjohn

@JasperLoy One needs to email the staff.

(And how did you count them?)

@DanielFischer I just searched the usernames for "please delete me" Sorry, I was referring to the usernames, not the profiles.

2 of the 12 are mine. 1 I requested the SE staff to delete, and the other I did not.

I see.