Mathematics - 2014-01-18 (page 1 of 3)

@anon Nice answer - my only criticism at the moment is the grammar of the sentence "Explicitly, the vertex set is rotation-invariant which implies their vector average is rotation-invariant the only vector of which is zero." :)))

anon

okay, I'll put a comma in there

Jasper Loy

12:10 AM

GN!

N3buchadnezzar

Say a runner came last in the Olympics. Telling him that he is better than 99.9% of the population is probably true, but it does not make him feel any better. Oranges will always compare themselves with oranges, and bananas with bananas =)

Old John

@anon I think I would have said " ... and the only vector for which this is true is ... "

@N3buchadnezzar True - but it never hurts to look on your own achievements from a positive aspect :)

anon

that would have made the last line of the paragraph spill over and end in the first third of the space alloted.

N3buchadnezzar

I do mathematics because I love doing it, and hopefully the same goes for the runner =)

Old John

@N3buchadnezzar that is the only reason I still do maths even though I have retired

Jasper Loy

12:14 AM

I am thinking whether I would be happier trying to become a professor or just doing math on my own working as something else

N3buchadnezzar

@anon wiki.math.ntnu.no/_media/ma3001/2014v/analytisktallteori/… Take a look at 2.14 here =)

Old John

Time I slept - goodnight all

Jasper Loy

Goodnight!

N3buchadnezzar

@OldJohn Thanks for the help =)

anon

Sometimes on MSE a paragraph's last line will simply consist of a single letter (representing a variable) and a period. That really rustles my jimmies.

Old John

12:15 AM

@N3buchadnezzar glad to help

Jasper Loy

@anon Wow, anon has not said so much for a long time.

Old John

@JasperLoy work as something else - and be a maths genius like Ramanujan!

@anon I always try to pad it out with some extra stuff in those cases

Night all!

Balarka Sen

Holy crab, It finally is ready : math.stackexchange.com/questions/621086/…

Jasper Loy

@anon Are you going to be a math professor? Is that your career choice?

N3buchadnezzar

@JasperLoy ballet dancer

Jasper Loy

12:20 AM

@N3buchadnezzar You?

N3buchadnezzar

@JasperLoy Anon

You, banana.

Me butterfly.

Jasper Loy

lol

@N3buchadnezzar Don't forget me when you win the fields medal

Arkamis

One day, I am going to see yet another problem with a title that references something only the OP can know, like "Probability question part b" and I am going to go absolutely insane.

Jasper Loy

I am already very confused between Arkamis and Karl Kronenfield

Arkamis

who?

what?

N3buchadnezzar

12:27 AM

@Arkamis A dream I had about numbers

Jasper Loy

anon refuses to answer me, lol

Arkamis

like this one: "Real Analysis Question 6". Oh, that's useful. Good. Helpful title. There's only been a single Question 6 in all of the real analysis courses, textbooks, and assigmnents ever

As if "Question 6" is supposed to be meaningful

Jasper Loy

@Arkamis I also hate "Hartshorne Chap 3 Ex 6"

Hi @ethan

Ethan

hi

Jasper Loy

How are your applications.

Ethan

12:31 AM

what title should I use to describe the question I give here on MSE

It says "title/position"

under one heading

I don't want it to sound stupid

lol

Jasper Loy

We all know you are the next Ramanujan

Ethan

no one says that lol

Jasper Loy

@ethan Have you decided where to go?

Ethan

Mathematics Question Help Provider?

@JasperLoy you mean where do I want to go lol, I'm still applying

Jasper Loy

@Ethan Good, I like it.

Ethan

12:35 AM

how about mathematics tutor?

or is that

Arkamis

@JasperLoy At least with that there's a solid reference for the question

becko

are there general methodologies to find aproximate, analytical, formulas for integrals that have no exact analytical expression?

Ethan

yes

for example the Euler–Maclaurin formula

becko

@Ethan thanks. Googling it right now. Anything else?

Ethan

what exactly are you trying to approximate, your question was very open

becko

12:50 AM

@Ethan math.stackexchange.com/questions/613373/…

but right now I want to at least learn the names of general methods, whether they apply to that particular integration or not..

Arkamis

@becko your integral is a polynomial.

becko

@Arkamis I'm pretty sure it isn't.

Ethan

It's not a polynomial for example if $p_i$ or $q_i$ is a non integer

Arkamis

Oh sorry, my font didn't display right, it looked for a moment that you had $p_i > 1$, not -1

Ethan

oops

Arkamis

12:57 AM

@ethan that's still not a problem for positive exponents.

Ethan

i deleted it meant to say somthing else

my bad

lol

Arkamis

I'm not entirely convinced it's a problem for $p_1 > -1$ either, though.

becko

You can assume that there are no singularities, as in the comment.

Ethan

yea I don't know

well then that puts constraints on q_i and p_i

becko

no. Assume that there are no singularities inside the interval of integration

It puts constraints on a and b

Ethan

12:59 AM

oh

well you can't have all the pi_'s and q_i's sum to somthing always smaller then k

otherwise you would have singularities

becko

what is k?

ohoh... sorry, missed my own notation

Charlie

Hi @ethan

Arkamis

Nevertheless, en.wikipedia.org/wiki/Binomial_theorem#Generalisations

becko

Yes. I already tried using the Binomial series. It didn't help.

Ethan

alot of series multiplication

@becko take the exponential of the logarithm of the expression

Arkamis

1:01 AM

Series multiplication is trivial

Ethan

approximate the logarithm some how

becko

I kept only the first two terms of the Binomial series for each factor.

Ethan

@Arkamis computationally not really

becko

It didn't go too well.

@Ethan exponential of log, and then what?

Ethan

the product will turn into a sum

Arkamis

1:01 AM

@Ethan vector convolution. Cauchy product

becko

yes.

Ethan

and the sum could be easier to calculate by numerical methods

Arkamis

You have to get a little clever with non-integer exponents

becko

I tried playing exp log a little, but I got nowhere.

Arkamis

but it's not impossible.

Ethan

1:02 AM

@becko I don't know I have to work on my college apps

sorry

Arkamis

becko, typically with numerical integration issues, a method slows down if you have to continuously compute the function at a number of nodes. If you can memoize certain parts of the function, you can avoid much of the computation.

In this case, you can directly cast the product as a summation by using the generalized binomial theorem. This will allow you to memoize the coefficients and as such if you choose to do a numerical quadrature routine, will greatly accelerate the process.

becko

@Arkamis The problem I had with the binomial series was that the series doesn't converge whenever x is close to P_i or Q_i

Arkamis

It's a finite series. It converges.

becko

it is not finite if p_i and q_i are not integers

Arkamis

Right, keep forgetting that part.

Ethan

1:07 AM

well if the interval [a,b] is very large

eventually all you need to really look at is the main x^n term

Arkamis

But your P_i, Q_i are distinct

becko

@Ethan ok. Good luck.

Arkamis

so why not use the series for the cases when it's not close, and then handle the cases where it is independently

Chris's sis

@Ethan $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{1}{i^2(i^2-i j+j^2)}$$

Arkamis

In any case, if x is near P_i for some P_i, then your function is close to zero

for that term

Alternatively

becko

1:09 AM

@Arkamis Yes. Probably using that in conjunction with the binomial series leads somewhere.... that's what I was trying to do now.

Arkamis

if your P_i, Q_i are in the interval [a,b], then you have finitely many such points

Break the domain of integration down into sub-intervals

becko

yes

Arkamis

so [a,P_1],[P_1,P_2], etc. (here I'm just assuming that P_1 < P_2 etc but it need not be the case)

becko

ok

after you split the interval of integration like that, the singularities are guaranteed to be only in the extreme points of integration

I'll work on that idea... see where it leads.

Arkamis

You said you have no singularities

But in that case, yes, that would force them to the endpoints

becko

1:12 AM

Yes. That's what I meant when i said no singularities. That you had split the integration, so that the singularities were at the endpoints. As in the comment by Kyle in the question.

Arkamis

As long as your singularities are non-essential, you should have no problems.

becko

@Arkamis I'll see where it leads. I'll post in the question if it works.

Sush

1:33 AM

How to show that number of $r$-circular permutations out of $n$ objects $Q_r^n=\dfrac{P_r^n}{r}$ is always positive integer?

becko

1:56 AM

@Sush that's the same as proving that n!/(n-r)! is divisible by r

In fact, you can make a stronger claim: n!/(n-r)! is divisible r!

You can prove that n!/(n-r)!r! is an integer by induction. See en.wikipedia.org/wiki/Pascal_triangle

Nick

Guys, does anyone know where I can find a proof for the sum to infinity formula

$S_{\infty} = \frac{a}{1-r}$

Ethan

i'm going crazy

becko

@Nick you're talking about the geometric series?

Nick

@gecko: yes, absolutely

becko

@Nick see en.wikipedia.org/wiki/Geometric_series

Nick

2:05 AM

Oh! As n goes to infinity, the absolute value of r must be less than one for the series to converge.

I didn't know it was that simple

becko

yes.

Nick

@becko: Are you good with infinite series?

becko

not really

Nick

oh,well my questions isn't too complex it's just some simple intuition

S = 1 - 2+ 3 - 4 + 5 - 6 + ...

becko

if you have a question that admits a well defined answer, you should ask it in the forum

Nick

2:07 AM

O_o oh yeah, right then. I think I'd like to here what they's say

well thanks anyway.

Bye Bye

hugs for everyone

Astrum

Why does $0=c(T-\lambda _1I)...(T-\lambda _m I)v$ mean that the map $(T-\lambda _i I)$ is not injective?

becko

@Nick look here: en.wikipedia.org/wiki/…

Nick

O_O

What the heck? There's a wikipedia page for that!!?

But mine is different, it's S = 1 - 2+ 3 - 4 + 5 - 6 + ...

becko

you can obtain your series from the one in wikipedia. the one in wikipedia is your S times 2 minus 1

Nick

But if I take the double of my series in a certain way, then I get that

becko

2:15 AM

@Nick exactly

What's the rate of convergence of the binomial series?

Where can I find an estimate of the error in keeping the first n terms?

Nick

Woah!

I just found out my series can be two things if I double it

1 - 1 + 1 - 1 + 1 - 1 + 1...

which is 1/2

and

4 + 8 + 12 + 16 + 20 + 24 + ...

which is infinity!

becko

@Nick that tends to happen when the series diverges

Nick

But the grandi's series. I have the answer to that.

So, can I say that S = 1/4 ?

Maybe.

Mike

oh no.

Alizter

2:35 AM

@Nick Don't believe everything you see on TV.

Nick

@Alizter: hehehe

@Alizter: So, tell me. What is the right answer? 1/4 or infinity?

Mike

numberphile is the literal devil

in the standard sense of summation (limit of partial sums) your series goes to infinity

Nick

I take it that you've all seen the the whole sum of natural number is -1/12 thing

Mike

of you go ahead and define summation differently you can obtain different values for it

Nick

Well, we're not defining summation differently.

We have an infinite number of terms

If I had an infinite number of rooms and they were all full and I had to make space for you, I'd tell everyone to shift one room and then there'd be a room for you.

Mike

2:42 AM

yes you are

you obtain -1/12 by doing something called zeta function regularization

Nick

Yeah, I watched that proof as well

Mike

in the standard sense (limits of partial sums) the sum over the naturals is very much infinity

Nick

But So, 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ... = 4 + 8 + 12 + 16 + 20 + 24 + ...right?

Mike

not in the sense you're probably thinking

Nick

mhh...

eh, equation systems were made intended for use with finite things. So yeah, it is wrong to equate two infinite series

I'll maybe ask this on main later

till then, bye

I gotta get to school

Mike

2:58 AM

your question will get closed on main

the issue isn't setting two series equal, it's that they are divergent series

anon

4:26 AM

"In order to give it a rigorous meaning, it must be divided by another functional determinant, making the spurious constants cancel." - en.wikipedia.org/wiki/Functional_determinant - What is the rigorous definition of the ratio of two functional integrals like this?

hmm, actually I shall read the next section

anon

4:38 AM

next section doesn't say

Sush

5:54 AM

$$\dfrac{(2n)!}{2^n\times n!}=(2n-1)\cdot(2n-3)\cdots5\cdot3\cdot1$$ how to show it?

@becko, many many thanks!

Sanchez

@Sush, rewrite your denominator as (2n) * (2n-2) * ... *2, by splitting up your 2^n into n 2's, and multiply each 2 to each thing in the product of n!

Sush

6:10 AM

@Sanchez, Many many thanks, Sir!

Pristine Kavalostka

7:31 AM

Balarka Sen

@Nick : That's usual convergence issue with conditionally convergent series. See Riemann rearrangement theorem. Similar can be stated when a series diverges. But nevertheless, they all make sense.

1 + 1 + 1 + 1 + ... = -1/2

1 - 1 + 1 - 1 + 1 - 1 + ... = 1/2

1 + 2 + 3 + 4 + 5 + ... = -1/12

etc. etc.

@Mike : Zeta function regularization isn't generally the formal way to state their 'sense'. See summability methods.

Oh, and 1 - 2 + 3 - 4 + .... = 1/4

Balarka Sen

1-1+1-1+... = 1/2 can be summed through Cesaro summation

1-2+3-4+5-.... = 1/4 can be summed through Abel summation

1+2+3+4+... = -1/12 is a little bit stronger (see definition in summability methods) that the other two, it can be done by Ramanujan C(0).

Finally and lastly, 1 + 1 + 1 + ... This can be done by C(0) too.

Balarka Sen

So none of them are actually much 'tough' to do. The real strong one is to sum 1 + 1/2 + 1/3 + ..., which doesn't even have a regularized value of the zeta function(*).

In 2003, Berndt proposed Ramanujan C(1) summation, which in a precise sense involving difference equations is unique. This nevertheless sums the harmonic series too.... EM constant!

*EM = Euler-Mascheroni constant.

Balarka Sen

7:48 AM

PS : I just wrote an article on this! (I think I am not alone in MSE). Tag : @Mike and @Nick

Bye!

I consider this one of my best answers. I'll put it in the shelf for now. =D.

TheLittleNaruto

10:10 AM

Hey Everyone :)

@Jasper , :) (Here comes your Papaya Friend ;))

I hope you dint forget me ^_^

Balarka Sen

Hullo.

Did Naruto ever shrunk that much in Shippuden?

I didn't know @Chris'ssis could change his/her identicon. Can anyone do that?

TheLittleNaruto

Hahahaha Lolz. No He dint shrink . This is his chibi mode ;)

Balarka, I need a simple help

this one is for finding area of any polygon depending upon their number of coordinates

I want to find the above series SUM formula ?

Just like we do have SUM formula for Arithmetic Progression

I am sorry if I am being completely rely . :'(

Balarka Sen

10:41 AM

You want to write that in sum notation, is that right?

Otherwise I am not sure what you mean by 'sum formula'.

TheLittleNaruto

Yes Right

Balarka Sen

$1/2 \left | \sum_{i=1}^{n} ( x_i y_{i+1} - x_{i+1} y_i ) \right |$ with $(x_{n+1}, y_{n+1}) = (x_1, y_1)$

TheLittleNaruto

10:57 AM

Thanks Balarka :)

TheLittleNaruto

11:26 AM

see I have implemented this in code

Balarka Sen

11:48 AM

Nice work!

Balarka Sen

12:00 PM

(Dang that inversion theorem)

TheLittleNaruto

Cya Balarka ^_^

Balarka Sen

Byes.

marcelolpjunior

1:35 PM

How is the Bernoulli numbers? For example, as against $B_2$?

Paul Epstein

1:48 PM

Can anyone help me out with a good reference explaining about the Gregorzczyk hierarchy.

I've been looking at the seminal paper that introduces it and I'm losing patience with all the errors and the ambiguities

The foundational paper is here if anyone wants to look at it with me: matwbn.icm.edu.pl/ksiazki/rm/rm04/rm0401.pdf

don't be scared -- no prerequisites

@Mike

Huy

@DanielFischer: Are you busy just now? I have a small problem with the proof of Baire's theorem.

Paul Epstein

@Huy I have looked at this if you mean Baire's Category Theorem. I can help too perhaps.

Daniel Fischer

@Huy What problem?

Huy

Let $(M,d)$ be a complete metric space. A part of the theorem states that for $A \subset M$ with $\operatorname{Kat}(A) = 1$ we have $\operatorname{Kat}(A^c)=2$ and $A^c$ is dense in $M$ (is that the correct English notion?).

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$Mathematics$ Mathematics