Mathematics - 2015-09-10 (page 1 of 2)

12:18 AM

off topic - Is it possible for two 2x2 matrices to be similar over $\mathbb{C}$ but not over $\mathbb{Q}$? I have two matrices and I put them in rational canonical form to deduce that they aren't similar over $\mathbb{Q}$; does this also show that they aren't similar over $\mathbb{C}$?

PVAL

@BalarkaSen @MikeMiller A proof that top homology is zero for non-compact manifolds is in the section of Hatcher ch.3 labeled under "Poincare Duality". It should be a good deal easier than the deformation retraction onto an $n-1$-complex which really requires handle (or morse) cancellation in the top dimension and some care in the case where theres infinitely many handles. I think the fact the top homology is zero is actually used to actually find the $n$ and $n-1$ handles which will cancel.

So you might not want to use that to show such a fact about homology

Mike Miller

1:02 AM

@PVAL: I'm aware of that proof, the discussion stemmed because I claimed there should be an easier method; it seems fundamental. I agree the cancellation is too hard though. Anyway it's not too hard at least to show that $H_n(M;\Bbb Z)$ is torsion - we can show $H^n_{dR} = 0$. For pick an exhaustion by compact manifolds $M_i$ of your manifold.

Then for any $n$-form $\omega$ you can find a form $\eta_i$ defined on $M_i$ minus a couple balls, which you can choose to be near the boundary, such that $d\eta_i = \omega$. You may further assume that except possibly in $M_i - \nu \partial M_i$, $\eta_{i+1} = \eta_i$. In this way you continue to have "problems" but you can take a limit as $i \to \infty$ to get a form $\eta$ such that $d\eta = \omega$.

On the other hand the other way via poincare duality is not really that hard and I am using the de Rham theorem here... maybe I am being too picky.

Rudy the Reindeer

1:27 AM

Anyone any idea why I just got a downvote here?

Semiclassical

1:42 AM

evening @chat

Jose Antonio

Hi. I have a problem in linear differential equation and I don't know how to solve it or how even begin. \begin{pmatrix}{1&1/\epsilon\\}\end{pmatrix}

\begin{pmatrix}{1&1/\epsilon\\ 0&-2}\end{pmatrix}

I know that the system \dot{x]=Ax defined by the above matrix it have all the solution stables

but the problem ask to find a vector solution, with initial condition with unitary magnitude and find two times 0<t_1<t_2 such that for all t between t_1 and t_2

we have that the magnitude of the solution is greater than C//\epsilon for all t \begin{pmatrix}{1&1/\epsilon\\}\end{pmatrix}between t_1 and t_2

but after seen the matrix exponential in wolfram alpha I can't see why we can find this solution or how even begin

I'd appreciate any help Thanks

Semiclassical

chatjax isn't processing that, so you may want to doublecheck what you've got above

$$\begin{pmatrix}1&1/\epsilon \\ 0 & -2 \end{pmatrix}$$

that?

Stan Shunpike

Hello chat

Jose Antonio

ok let me write it correctly in a question.

Stan Shunpike

1:58 AM

Question: suppose I have the elements A,B,C,D and I want to form a set with them. I want to only permit sets where members are unique. For instance, {A,B,C} would be allowed but {A,A,B,C} would not be allowed. What is the name of this property? Is "uniqueness" the right term?

Semiclassical

a set of distinct elements?

Stan Shunpike

Thank you!

I could not call to mind the word

Gracias

Semiclassical

de nada

Jose Antonio

2:24 AM

This is the problem where I have doubts with the method to find the periods of high growth in a stable solution of a linear DE: math.stackexchange.com/questions/1428920/…

Stan Shunpike

3:28 AM

Hello!

Rigor

5:10 AM

Hello!!

anon

@StanShunpike sets automatically don't have repeated membership. {A,B,C} and {A,A,B,B} are the same set.

Stan Shunpike

@anon In what context then does repeat membership arise?

anon

multisets

Stan Shunpike

What is the definition of a set? That is, how do you know that sets don't have repeat membership?

anon

they just don't

that's not a thing for them

something is either a member of a set or it isn't

Stan Shunpike

5:19 AM

That's true

user61230

Sets have one containment operator: "is [element] in the set?"

user61230

There is no "how many [element] are in the set?" defined for a set.

Stan Shunpike

I see!

So I must be messing this up then

The distinction is

I can count A,B,C

but there is no distinction of multiple As

that would imply just another type of element D

there can't be more than one A

user61230

That's... not completely true, though it's pretty close.

Stan Shunpike

LOL

how is it wrong?

anon

5:22 AM

also, the only reason we allow ourselves to write down elements multiple times in a set (which is redundant) is because in practice doing so implicitly makes things way easier. for instance I can say the union of {x,y} and {z} is {x,y,z} without having to split into cases depending on if z is also x or y.

user61230

I can create something that describes a set, for instance using set builder notation, that strictly speaking "has" multiple instances of the same element in its description.

Stan Shunpike

@anon nicely phrased. that's cool

user61230

But, when evaluated, that set is equivalent to - in that it literally means exactly the same thing as - the set with only one instance, because that's all equivalence on sets checks.

Stan Shunpike

@Emrakul that is a very subtle distinction. in what context would that come up?

anon

when using set-builder notation

user61230

5:26 AM

It comes up because most sets aren't written out in completion, but are rather written out using a description.

user61230

You can, for example, describe an set with an unbounded number of unique entries in a way that describes an infinite number of '1' entries, but the set itself only has one '1' in it despite whatever the description nominally says.

Stan Shunpike

@Emrakul that's awesome!

anon

Say $U$ is a subset of $X\times Y$, and and $p:X\times Y\to X$ is the projection map $(x,y)\mapsto x$. If you write $p(U)=\{x\in X: \exists y\in Y: (x,y)\in U\}$ then chances are there are many $(x,y)$ with a given first coordinate $x$, so that that given coordinate is "written" infinitely many times in the set-builder notation.

user61230

Yeah! Lots of weird things like that.

Stan Shunpike

@anon That's cool too.

Are these concepts related to equivalence classes? or is that something else...

anon

5:31 AM

not really

Stan Shunpike

ok lol

user61230

tangentially

anon

it's just that it'd be a pain to force ourselves to only "write" elements down once (meaning only specify/refer to them in one instance in whatever notation we use). then we'd be forced to investigate our formulas and membership conditions and revise them until we get them to only spit out every element once, which is pointlessly complicated

user61230

equivalence classes rely on the uniqueness of each element, but beyond that, the two ideas are very different

air

That is cool @anon and @emrakul. But I guess in a sense one could say that sets are just equivalence classes of multisets (partitioned based on their unique elements)?

anon

5:36 AM

ugh

a bad case of "trying too hard"

user61230

I... guess?

air

Sorry :P it just seemed that this is what @StanShunpike was hinting at.

Stan Shunpike

I wasn't hinting. I was probing lol

I had no idea

shot in the dark

air

and I guess I caused an awkward silence by trying too hard :(

user61230

mm, s'okay, things go quiet on their own

air

5:46 AM

ah okay

Rigor

...and shooting in the dark is not an effective way of learning math :P

Agawa001

@Rigor shooting in the dark can kill bugs

3

user61230

shooting in the dark is thus good for learning programming

Agawa001

^

Rigor

^^

Balarka Sen

6:34 AM

@TobiasKildetoft Perhaps you know the answer to this question?

Tobias Kildetoft

@BalarkaSen Hmm, not off the top of my head, no

Balarka Sen

ah, ok.

VermillionAzure

excuse me

question

once again, how can i mathematically represent the distinction between "is a" and "has a" relationships in set theory without including them as a part of the statement?

Or should I start looking into category theory?

The Substitute

7:19 AM

off topic - if a 3x3 matrix has eigenvalues 3,3,5, how do I determine whether I have 2 jordan blocks or 3 jordan blocks?

The Substitute

7:30 AM

nevermind, the number of blocks is the sum of the dimensions of the eigenspaces :)

skill patrol

8:06 AM

@VermillionAzure do you mean "is a" as in "equals" and "has a" as in "belongs to"?

What do you mean by "without including them as a part of the statement?" @VermillionAzure

VermillionAzure

8:20 AM

@skillpatrol Yes, equality or element of a set n the first case

"Has a" meaning possession of a feature or ownership

e.g. a human has/owns an eye

so human R eye where R is the relation

what is the symbol for "has"?

skill patrol

a flipped "belongs to"

skill patrol

8:43 AM

@VermillionAzure \ni

or \owns

VermillionAzure

@skillpatrol ???

@skillpatrol I can't find this one

skill patrol

@VermillionAzure click on the link

the blue "\ni"

$\ni$ \ni
or $\owns$ \owns

skill patrol

9:13 AM

Also for $\not\ni$ \not\ni

or $\not\owns$ \not\owns

VermillionAzure

@skillpatrol Yeah but the first answer considers it the same symbol, just "mirrored" in a different order

Like, a set owns a certain object but that's membership

I wanted a "set collectively owns a characteristic feature" like "dogs have fur"

skill patrol

have = owns :-)

Silent

@robjohn, $$\lim_{n\rightarrow \infty}\frac {n!}{n^n}=\lim_{n\rightarrow \infty}\frac {n}{n}\cdot \lim_{n\rightarrow \infty}\frac {n-1}{n}\cdots \lim_{n\rightarrow \infty}\frac {1}{n}=1\cdot1\cdots0=0$$ Is this argument correct?

VermillionAzure

@skillpatrol Yeah but

is \neq has

is \\neq has

wtf why cant i use it

skill patrol

use what?

robjohn

9:22 AM

@Silent That is not quite right... you can't do that with a variable number of limits that grows without bound.

VermillionAzure

latex

skill patrol

@VermillionAzure here

VermillionAzure

\neq

still not working

i have all 3 bookmarks

skill patrol

you need the dollar signs

robjohn

@Silent Using that logic you'd get $$\lim_{n\to\infty}\sum_{k=n}^{2n}\frac1k=0$$

VermillionAzure

9:24 AM

@skillpatrol dollar signs?

$neq

\$neq

skill patrol

two

VermillionAzure

$$neq

\$$neq

skill patrol

on each side

VermillionAzure

$$neq$$

\$$neq\$$

skill patrol

one on each side

VermillionAzure

9:25 AM

$neq$

oh

skill patrol

what about the \

robjohn

$\neq$ $\to\ \neq$

Silent

@robjohn, oh I got what you said! thanks!

VermillionAzure

$\neq$

Ahhhhhhhhhhhhh okay there

skill patrol

:D

VermillionAzure

9:26 AM

Well what I meant was

is $\neq$ has

Silent

@robjohn, so insteed i should apply squeeze theorem, right?

VermillionAzure

because inheritance is different than composition

If I inherit genes, I have genes as a feature. I am not genes. I am, however, partially made up of genes

robjohn

@Silent There are many ways... You can simply note that $$0\le \frac{n!}{n^n} =\frac1n\frac2n\frac3n\cdots\frac nn \le\frac1n$$ and use the squeeze theorem, yes.

VermillionAzure

It's incorrect to say $genes \in human$

It's also incorrect to say
$human \in genes$

Silent

@robjohn, thank you so much

VermillionAzure

9:28 AM

The first answer on the question that was linked to said that
$\ni$
is a re-arrangement of $\in$ viable alternative

But that just means that $dog \in mammal = mammal \ni dog$

skill patrol

yes...

VermillionAzure

@skillpatrol But the problem is that the concept of ownership is different than the concept of "also being"

Owning features is different than inherently having them

The reason I'm looking into this is because I'm currently pondering "inheritance vs. composition" in C++ and other OOP-enabled langauges

Where both are considered equivalent but not equivalent design patterns to use

To objectively reason about them and to compare them using formal logic, I found myself lacking to describe the exact nature of them with my basic knowledge of sets:

$\in \land \lor \implies \iff =$ and the sorts

Basically, an object owning an object is different than inheriting properties of an object

skill patrol

...yes

VermillionAzure

@skillpatrol So it's incorrect to say that $hair belongs to human$ and $hair is human$ at the same time

Because hair is a feature of a human or being human

Once applied to the real world, even the concept of "a human" becomes too abstract to truly use, as humans are often missing one feature or another, like one may be missing and arm and another, an eye. But does that make both mutually exclusive to each other as human? No. Perhaps being human has something to do with cardinality--can that be expressed in set theory?

It's even worse once we try to apply this to code and try to "create" our own logics systems and worlds and confuse features for subsets and members and such so...

What theory currently deals with these sort of issues and what can I look for to help assist me in my exploration of these concepts?

skill patrol

cstheory.stackexchange.com/…

The Artist

9:46 AM

@Chris'ssistheartist @robjohn How to find $$\int_{0}^{\pi/4}\int_{\pi/2}^{\pi}\frac{(\cos x-\sin x)^{y-2}}{(\cos x+\sin x)^{y+2}}\,dy\,dx$$ without using the substitution $t=\frac{\pi}{4}-x$ and then substituting $u=\tan t$. Any better or shorter method? :)

VermillionAzure

@TheArtist Can't you simplify the trig somehow?

The Artist

@VermillionAzure are you referring to the $-2$ and $2$ ?

VermillionAzure

@TheArtist yeah

If I calculate correctly you can simplify that to $1/(cos^2 x - sin^2x) ^ 2$

but that doesn't really do anything

The Artist

@VermillionAzure yes thats the correct. I thought about it, but I think the original form is much cleaner looking.

VermillionAzure

@TheArtist I hate calculus lol

The Artist

9:55 AM

@VermillionAzure say that to @Chris'ssistheartist :P

@VermillionAzure she absolutely loves it :)

skill patrol

@VermillionAzure you may get that comment flagged as "offense" in here pal :P

VermillionAzure

@skillpatrol But seriously what do you do with that integral

skill patrol

::think::

VermillionAzure

@skillpatrol idk man

There's no direct trig identity to make some magic happen

@TheArtist Maybe there's some way to get mileage out of $1/(cos^2x - sin^2x)^2$

it doesn't make anything nicer though

oh hey

$1/(cos(2x) + 1)$

The Artist

@VermillionAzure like i said :) I think the original form is much cleaner :)

@VermillionAzure dont you think the original form looks cleaner? :)

VermillionAzure

10:06 AM

@TheArtist I do too

Wolfram Alpha time~~!!!

@TheArtist Can this even be solved

The Artist

@VermillionAzure oh yes :) I have a solution. I was just asking for an alternative solution.

@VermillionAzure @VermillionAzure I mentioned the method on the original chat message :) substitute $t=\frac{\pi}{4}-x$ and then $u=\tan t$.

@VermillionAzure that should give you $\frac{1}{4}\ln\left(\frac{\pi^2 - 1}{(\pi^2/4) - 1}\right)$ :)

VermillionAzure

@TheArtist How do you even come up with this magic

what is this

The Artist

@VermillionAzure OMG This is not magical at all :O magic is @Chris'ssistheartist work :D . I learn alot from her :) You should too ;)

VermillionAzure

@TheArtist I don't understand Calculus

specifically integraton

the derivation of these formulas is like finding unicorns

It's like there's no order to it, only hands grasping at the wind and suddenly we find Mr. Solution Bigfoot

skill patrol

math is like that sometimes

get over it

:P

The Artist

10:20 AM

@VermillionAzure me too can totally relate to this, I suck suck suck big time. But the more you run to @Chris'ssistheartist with questions, the more you progress. Atleast I feel like I am atleast slowly improving with this strategy.

VermillionAzure

@skillpatrol what

that makes no sense, especially for such a constructed way of logic

where is the elegance?

skill patrol

It is there too.

Nothing comes easy...

...just like your "language" analogy.

Functional Analysis

What is $H_0^1$ space?

Sobolev space

I will check out a PDE book, thanks.

Balarka Sen

11:26 AM

Ugh, no good questions to answer.

skill patrol

ask one

Balarka Sen

Don't have anything interesting to ask for now.

Alec Teal

11:38 AM

So my two daily downvotes just came in, I don't care about stuff that's still positive but math.stackexchange.com/questions/564960/… has been tipped to -1, please take a look.

M.S.E

@Chris'ssistheartist \int_0^1 \frac{\ln(x+\sqrt{2})}{\sqrt{x(1-x)(2-x)}} dx ?

mickep

12:07 PM

Oh no, no more integrals from "evil Chris" today! :) I'm still struggling with math.stackexchange.com/questions/1426826/… !! :)

4

user147690

What are you thinking about @BalarkaSen?

user147690

Still haven't got my marks back for advanced algebra, but I got 100% for functional analysis Assignment 1 and 100% for algebraic physics assignment 2

user147690

So I am doing something right this semester xD

Balarka Sen

Not much, I am reading a non-math book.

user147690

What book?

Balarka Sen

12:17 PM

Excellent, @AlexClark!

user147690

Jul 15 at 13:48, by Balarka Sen

rock climbing is useless. everything excluding math is useless.

user147690

@BalarkaSen Things go downhill from here surely :P

Balarka Sen

@AlexClark not something you'd be familiar with. I read Beckett's Waiting for Godot a few days ago, though

user147690

I'll read the french version ;)

Balarka Sen

"I'll"?

You mean you plan to read it?

user147690

12:19 PM

I-will

user147690

It sounds funny from the wiki page

Balarka Sen

yeah, I know, I was wondering why you put the future tense in there.

it's not at all funny

user147690

' is an absurdist play by Samuel Beckett, in which two characters, Vladimir and Estragon, wait endlessly and in vain for the arrival of someone named Godot.'

user147690

Wait what? What past tense?

Balarka Sen

In what sense does that sound funny?

user147690

12:21 PM

I don't know, I guess it depends on how you mentally fill the scenario

user147690

I imagined two guys waiting for a friend to turn up, while sitting on a park bench, and the entire play progresses with him not turning up

Balarka Sen

I found it a very depressing read, really.

user147690

'tragicomedy' :P

Balarka Sen

I'd recommend you not to read wiki pages before reading the story.

user147690

Oh definitely won't

user147690

12:23 PM

Just the first line and the cover of the book

Balarka Sen

@AlexClark Yes. It's a play with two parts. The first part is quite funny and very confusing.

user147690

It'll be more confusing in French for me

Balarka Sen

But the depressing truth Beckett wanted to convey sinks in when you start reading the second part

Just read it in English

user147690

Also you didn't clarify the past tense reference before? What I wrote definitely isn't in past tense

user147690

I've read the french version is past tense

Balarka Sen

12:25 PM

Look at the message again, I have edited. It was a typo.

user147690

Ohhh okay my bad :P

user147690

Can you explain enveloping algebras to me?

Balarka Sen

Don't even know what it is

Alec Teal

lmgtfy.com/?q=enveloping%20algebras

user147690

It's used in Lie algebras, but also in non-associative algebra which I thought you had looked into

Balarka Sen

12:28 PM

there are no non-associative algebras, @AlexClark. this ain't 1st april.

user147690

:P

Balarka Sen

@AlexClark Out of curiosity, why do we study Lie algebras? I presume you know more about them than last time when I asked you. I have heard that tangent spaces of Lie groups get the structure of a Lie algebra (maybe DanielF told me), but that is still not enough motivation for me (although it's an interesting fact)

user147690

@BalarkaSen The thing you just mentioned is all I know(as of yesterday haha)

user147690

The course is structured so we do theory for the first half

user147690

and then application for the second(which started last lesson)

Balarka Sen

12:38 PM

derp. oh well!

Alec Teal

There are two ways to approach it. Affine geometry (I can get you the exact name) which is more traditional. Or you can drag abstract algebra into it with lie groups

Both let you pass data about the 'manifold' around. Like a unit length between charts say.

Balarka Sen

I mean, can you use Lie algebra structures on tangent space of Lie groups to get information about the Lie group? If so, what kind of?

Can you give me an example?

lol! good question.

Alec Teal

You might get lost in the footwork. I recommend Introduction to Smooth Manifolds by John M. Lee for this, it's thick and filled with examples (only catch is it considers only metrisisable manfiolds)

Balarka Sen

Not every manifold is a Lie group, so I am not sure if I can learn about Lie algebras from there. Besides, I am not trying to read textbooks on it. Just look for having a talk with someone who can tell me interesting things about them.

But thanks!

Alec Teal

I'm against such 'touring' on principle, but manifolds are well worth the study.

If you've done topology to any great level you get manifolds 'free' (almost)

user147690

12:43 PM

So the Lie algebra of a matrix group G uniquely determines the structure of G in a neighbourhood of the identity and then we start getting into physics pretty much, but until a few weeks from now I won't have any fun things to say I would guess

Balarka Sen

I don't know much about the smooth category. I am learning some multivariable calculus right now, so that I can understand that stuff.

user147690

We showed in class that the tangent space to the identity of a matrix Lie group constitutes a Lie algebra

Alec Teal

You don't need to go much beyond substituion in iterated integrals. It's not as much calculus as you'd think

user147690

So nothing much in the way of structure

Balarka Sen

can you elaborate on "uniquely determines the structure of G in an nbhd of the identity", @AlexClark? what do you mean by structure and uniquely determined there?

I'm interested.

@AlecTeal Fair enough, but I need to know the implicit and inverse function theorem to get going.

Alec Teal

12:46 PM

I'll be honest, I'm not even sure I know those theorems :P

user147690

Hmmm the best explanation I can give is pretty bad, but it's because $\forall x\in L_G$ where $L_G$ is the Lie algebra of the Lie group $G$, $e^x\in G$, and we choose a tangent curve specifically so it passes through the identity of $G$.

The result ends up being that the Lie algebra for the Lie group is precisely the tangent space to the identity of $G$

user147690

I.e. $A\in L_g \implies e^A \in G$

user147690

And sufficiently close to the identity we can(haven't done this yet) get a converse

Balarka Sen

OK, thanks.

user147690

With the converse we should get unique information about the structure

user147690

12:54 PM

(in a neighbourhood of the identity)

user147690

I shall sleep now, I'll try to formulate something better to say xD

user147690

Hopefully Tobias doesn't come in here and say that's all wrong!

Chris's sis the artist

1:34 PM

wow, I received a lot of messages in chat. :-)

Oh, this one? $$ \int_0^1 \frac{\ln(x+\sqrt{2})}{\sqrt{x(1-x)(2-x)}} dx $$

Kind of question I need to think of for a while. I have to admit the first thing that crossed my mind when I saw it was the Eisenstein Series.

Mambo

Hi

skill patrol

Hi

Chris's sis the artist

1:51 PM

Significant progress is already made ...

Mambo

@skillpatrol Are you a grad student?

evinda

Hi!!!

Agawa001

hi

evinda

How are you? @Agawa001

Agawa001

@Chris'ssistheartist how was the game scored yesterday

@evinda nickel

evinda

2:04 PM

nickel? @Agawa001

Chris's sis the artist

@Agawa001 2-1 (sets) for Halep :D

Mambo

is there any phd student online now?

or sophomore or final year

Agawa001

@evinda the fault of my french :p, i m perfectly fine

@Chris'ssistheartist the romanian did it again

evinda

@Agawa001 Where are you from?

Chris's sis the artist

@Agawa001 Yeah. This night there will be another amazing game.

Agawa001

2:07 PM

@evinda north africa

Chris's sis the artist

@Agawa001 Simona Halep - Flavia Pennetta

Agawa001

@Chris'ssistheartist with who ?

ah

evinda

So is french your native language? @Agawa001

Agawa001

@evinda lol no, but second

evinda

Which is your native language then? @Agawa001

Agawa001

2:11 PM

arabic/berber

and binary :)

have you got some non-sidy questions about maths for today @evinda

evinda

non-sidy? @Agawa001

Agawa001

yes ?

evinda

@Agawa001 I don't know what the word means :/

Agawa001

anyways

evinda

2:26 PM

$$\lim_{n \to +\infty} \frac{1^{2011}+2^{2011}+ \dots+ n^{2011}}{n^{2012}}= \lim_{n \to +\infty} \frac{ \sum_{i=1}^n i^{2011}}{n^{2012}}$$

How can we continue in order to find the limit?

@DanielFischer Do you maybe have an idea?

Daniel Fischer

@evinda One way is to think Riemann sum.

Agawa001

oh i linked to wrong page

evinda

@DanielFischer You mean like that?

$$\lim_{n \to +\infty} \frac{1^{2011}+2^{2011}+ \dots+ n^{2011}}{n^{2012}}= \lim_{n \to +\infty} \frac{n^{2011} \left( \frac{1}{n^{2011}}+ \frac{2^{2011}}{n^{2011}}+ \dots + 1\right)}{n^{2012}}= \lim_{n \to +\infty} \frac{ \frac{1}{n^{2011}}+ \frac{2^{2011}}{n^{2011}}+ \dots + 1}{n}$$

Daniel Fischer

Probably better to use $\sum$.

Agawa001

Faulhaber's formula

In mathematics, Faulhaber's formula, named after Johann Faulhaber, expresses the sum of the p-th powers of the first n positive integers as a (p + 1)th-degree polynomial function of n, the coefficients involving Bernoulli numbers Bj. The formula says Faulhaber himself did not know the formula in this form, but only computed the first seventeen polynomials; the general form was established with the discovery of the Bernoulli numbers (see History section below). The derivation of Faulhaber's formula is available in The Book of Numbers by John Horton Conway and Richard K. Guy. There is also a similar...

evinda

2:39 PM

$$\lim_{n \to +\infty} \frac{1^{2011}+2^{2011}+ \dots+ n^{2011}}{n^{2012}}=
\lim_{n \to +\infty} \frac{ \sum_{i=1}^n i^{2011}}{n^{2012}}= \lim_{n \to +\infty} \frac{n^{2011} \sum_{i=1}^n \frac{1}{n^{2012-i}}}{n^{2012}} \\ = \lim_{n \to +\infty} \frac{ \sum_{i=1}^n \frac{1}{n^{2012-i}}}{n}$$ @DanielFischer Like that?

Agawa001

theres no n

Daniel Fischer

@evinda No, that's incorrect.

Agawa001

or may you use Faulhaber s_formula

Mike Miller

Morning.

Martin Sleziak

@evinda You might have a look at this question and other posts linked there.

The methods which I recall are: Riemann sum, Stolz-Cesaro, comparison with an integral.

You can probably find some other methods in those posts.

Daniel Fischer

2:48 PM

Morning @Mike, dobry den, @Martin.

Martin Sleziak

Guten tag, Daniel!

Or is another Fischer mod from Austria?

Daniel Fischer

@MartinSleziak Neither of us Fischer mods is from Austria. Arthur lives there (currently), however.

Martin Sleziak

Yes, he works in Wien doesn't he?

I don't have an idea where he is originally from.

Daniel Fischer

Yes, he's currently in Wien. Originally, he's from Canada.

Martin Sleziak

@evinda How did you rewrite $\sum\limits_{i=1}^n i^{2011}$ to $n^{2011} \sum\limits_{i=1}^n \frac1{n^{2012-i}}$?

This would be a true statement:

$$\sum\limits_{i=1}^n i^{2011} = n^{2011}\sum\limits_{i=1}^n \left(\frac in\right)^{2011}$$

evinda

2:52 PM

$$\lim_{n \to +\infty} \frac{1^{2011}+2^{2011}+ \dots+ n^{2011}}{n^{2012}}=
\lim_{n \to +\infty} \frac{ \sum_{i=1}^n i^{2011}}{n^{2012}}= \lim_{n \to +\infty} \frac{n^{2011} \sum_{i=1}^n \frac{i^{2011}}{n^{2011}}}{n^{2012}}= \lim_{n \to +\infty} \frac{\sum_{i=1}^n \frac{i^{2011}}{n^{2011}} }{n}= \lim_{n \to +\infty} \frac{1}{n} \sum_{i=1}^n \frac{i^{2011}}{n^{2011}}= \int_0^1 x^{2011} dx= \frac{1}{2012}$$

@MartinSleziak Is it like that?

Martin Sleziak

@evinda This looks fine!

Balarka Sen

Hello @MartinSleziak, Morning @MikeMiller

Martin Sleziak

Hi!

Balarka Sen

@MikeMiller Lie groups are orientable, right? You start by choosing a local orientation, and then push it around by multiplication on your manifold.

(this is not a rigorous proof. I think of local orientation as a small sphere rotating in some particular direction around that point)

Mike Miller

Yes.

evinda

3:03 PM

@MartinSleziak And do I have to justify this:

$$\lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n \frac{i^k}{n^k} = \int_0^1 dx \, x^k = \frac{1}{k+1}$$

?

Balarka Sen

How can I prove this rigorously with the homology definition?

Martin Sleziak

@evinda Well, you probably have learned some theorem saying that Riemann sums converge to the integral.

That would be an argument why the equality holds there.

Mike Miller

By doing what you said.

evinda

@MartinSleziak As a Riemann sum do we consider this: $\frac{1}{n} \sum_{i=1}^n \frac{i^k}{n^k} $ ?

Martin Sleziak

Yes.

It is Riemann sum if you divide the interval $[0,1]$ using the points of the form $i/k$.

evinda

3:09 PM

A ok.. So can we just state that it is a Riemann sum, stating that we divide the interval $[0,1]$ using the points of the form $i/k$ ? @MartinSleziak

Martin Sleziak

This answer gives quite a detailed explanation.

Re: can we just state. If this is a question whether it will be sufficient for somebody grading your solution, I cannot answer that - I do not know how much details are you expected to give there.

I'd consider more important question whether you understand that it is a Riemann sum.

evinda

@MartinSleziak A student asked me why it holds.
Can I say the following?

It is known that

$$\lim_{n \to \infty} \frac{K}{n}\sum_{i=1}^n f\left(\frac{Ki}{n}\right)=\int_0^K f(x) \, \mathrm{d}x.
$$

where $\frac{K}{n}\sum_{i=1}^n f\left(\frac{Ki}{n}\right)$ is a Riemann sum.

Balarka Sen

@MikeMiller OK, yeah, sure, I can formalize it. Pick a generator of $H_n(G, G- \{1\})$. And then multiply by $g$ to get the map $H_n(G, G - \{1\}) \to H_n(G, G - \{g\})$. Pick the orientation to be image of the generator.

Martin Sleziak

Yes. Or you can use this for $K=1$.

However, I would also mention explicitly the partition which is used.

It would be $0<\frac1n<\dots<\frac in<\dots<\frac{n-1}n<1$ in the case $K=1$.

And correspondingly scaled if you want to use arbitrary K.

Mike Miller

@Balarka: You need to show local coherence, though.

Balarka Sen

3:20 PM

Hmm, good point. I am not sure why multiplying $1$ by $g_1$ and $g_2$ gives the same orientation if $g_1$ and $g_2$ are close enough in the group.

Let me ponder on this.

evinda