Mathematics - 2017-08-15 (page 2 of 5)

Kirill

11:01

@LeakyNun ok, that is again not the purpose of that text :) sure, you can do things different ways, but if you exercise, you limit yourself on some technique, as if there were not any other technique.

Any ideas about that?

Leaky Nun

@Kirill about what?

Kirill

6 mins ago, by Kirill

I don' understand. The text says, "we can interprete $\frac{1}{1-\gamma_ix}$ as the geometric series $\sum_{n=0}^{\infty}(\gamma_ix)^n$". But we can do that only if $\gamma_ix < 1$, and for $\gamma_i$ being $\frac{1+ \sqrt{5}}{2}$ that is not the case. On the other hand, we are looking at the formal power series . That means, we are not interested in the convergence, but at the same time we took the limit of this series for our calculations. Is that not cheating, at all?

Leaky Nun

3 mins ago, by Leaky Nun

@Kirill that's just not rigorous, lol

Kirill

not rigorous in what sense? I would say it is all wrong what is written there. But, if the text says that is ok, so maybe I should understand why the expression in the form of the geometric series is possible.

I know only the limit $\frac{1}{1-q}$, if $q<1, \sum q^n$

Leaky Nun

@Kirill It is ok within the radius of convergence

we are not interested in the value of $x$

Kirill

11:06

@LeakyNun like, "you can find some $x$ that that is true"?

Leaky Nun

@Kirill yes

Kirill

ok, that's not fair

Leaky Nun

@Kirill no, that's just not rigorous

we are not trying to find $x$.

it works for an interval

Kirill

@LeakyNun it is hard for me to imagine some interval, if we still think about the Fibonacci sequence. Where is an interval in the FIbonacci sequence?

Leaky Nun

2 mins ago, by Leaky Nun

we are not trying to find $x$.

Oh, I understand the partial fraction method now :D

The sequence is not in $x$. The sequence is in the coefficients.

Kirill

11:10

still, "intervall for a sequence" doesn't make sense.

Leaky Nun

@Kirill I mean it works when $x$ is in some interval

Huy

wtf

Leaky Nun

e.g. $-1 < x < 1 \implies \dfrac1{1-x} = 1+x+x^2+\cdots$

Kirill

@LeakyNun I think I've understood the partial thing for $x^2$, but for $x^5$ one should play with powers of $x$ or gammas somehow...

Leaky Nun

@Huy what

@Kirill where do you see $x^5$?

Kirill

11:14

if you want to decompose $\frac{x}{1-x-x^2}$ in $\frac{b_1}{1-c_1x}$ and $\frac{b_2}{1-c_2x}$ that it's ok. But if you want to decompose $\frac{x}{x^5-x^4-x^3-x^2-x+1}$, those fractions are wrong.

Leaky Nun

@Kirill lol

good luck

but other methods would also face the same problem

you essentially need to solve a quintic equation

Huy

easy

jyotishraj thoudam

0

Q: Cartesian Tensor Notation

I have a problem in proving a problem, which is easy to do in basic multivariable calculus but I couldn't do it in tensor notation. Please help $$\nabla.(r^n \vec{r})=(n+3)r^n$$ Here $r=\sqrt{x^2+y^2+z^2}$ is the magnitude of the position vector $\vec{r}=xi+yj+zk$. I tried representing(i don't ...

Kirill

@LeakyNun that was just an illustration for that, that I understand the partial fraction thing for a degree $\le 2$. Do not ask me to decompose it for other degress.

Leaky Nun

@Kirill other methods would also need to solve a similar equation

including the matrix method

so I don't get your complaint

jyotishraj thoudam

11:19

hi

can someone please answer my question?

Kirill

@LeakyNun not a complaint! I just underline that my knowledge of partial decompositions is limited, e.g. there is a constant $C$, such that knowledge $\le C$ for the whole knowledge. And, $C=2$ here.

mick

Hi all. Plz vote to reopen this ...

i closed it myself but now it is edited !

0

Q: The sequence A126022 : $a_{n+1} = a_n + \lfloor a^{-1}(n+1) \rfloor $

Consider the sequence A126022 at OEIS http://oeis.org/A126022 $$ 1,2,4,7,10,13,17,21,25,30, ... $$ We start with $a_1 = a(1) = 1$. $a_n = a(n)$ and $^{-1}$ means functional inverse. ( check the link If you are confused , Leroy explains it more clearly ) If this self-reference reminds you of...

recurrence-relations asymptotics closed-form

Huy

@Kirill: no. you can also solve cubics. and if you're lucky, sometimes also higher order degree.

mick

And answer too :)

Leaky Nun

@Kirill are you asking a question or what

Kirill

11:23

lets try

mick

Who downvoted ? It is a Nice question imho

Leaky Nun

@mick it's still unclear

jyotishraj thoudam

I upvoted yours @mick

Leaky Nun

and the description in the OEIS is much better than yours

Kirill

@LeakyNun does the Schurs' complement of $D$ in a block matrix $\begin{pmatrix}A & B \\ C & D\end{pmatrix}$ exists every time, if $D$ invertable?

mick

11:26

Thanks @jyotishrajthoudam

Yes ar the Oeis it is better , like I Said and therefore the link

jyotishraj thoudam

How do I turn this $x_mx_m(x_ix_i)^{\frac{n}{2}-1}$ into $(x_ix_i)^{\frac{n}{2}}$

Using tensor properties

Tobias Kildetoft

@jyotishrajthoudam First, you need to tell us what the symbols mean

jyotishraj thoudam

-1

Q: Cartesian Tensor Notation

I have a problem in proving a problem, which is easy to do in basic multivariable calculus but I couldn't do it in tensor notation. Please help $$\nabla.(r^n \vec{r})=(n+3)r^n$$ Here $r=\sqrt{x^2+y^2+z^2}$ is the magnitude of the position vector $\vec{r}=xi+yj+zk$. I tried representing(i don't ...

multivariable-calculus tensors index-notation

@TobiasKildetoft

Kirill

(don't know Schur, but I like the formula $\det(X) = \det(D)\cdot \det(A-BD^{-1}C)$

jyotishraj thoudam

@TobiasKildetoft can you please resolve my issue on the problem i posted

Tobias Kildetoft

11:31

@jyotishrajthoudam None of that notation looks familiar to me. I keep forgetting that "tensor" is often used in a way that is practically unrecognizable from the term as I know it (i.e. in tensor products).

jyotishraj thoudam

@TobiasKildetoft ok thanks for your effort

Huy

@jyotishrajthoudam rename $m = i$ and you're done

Kirill

@Huy :)

jyotishraj thoudam

@Huy But that would violate the rule

Since I cannot have more than 2 repeated index

Huy

yes, that's why they are distinct

but they mean the same thing

$x_i x_i$ = $x_m x_m$

Balarka Sen

11:35

@Huy

Huy

what does it matter whether you write $r^2 = x_i x_i$ or $r^2 = x_m x_m$

Balarka Sen

i is farther from m in the keyboard

Huy

I think you mean from x

Balarka Sen

nah, farther from my fingers not from the TAB key lol

Huy

wat

just place your hand properly on the keyboard and win

jyotishraj thoudam

11:37

i know its essentially the same

Tobias Kildetoft

@BalarkaSen something can't just be farther from something else. You need a frame of reference

Balarka Sen

ok, that was good.

can't top that one

jyotishraj thoudam

but put m=1 and i=2

Huy

????????????????????????????

Balarka Sen

@TobiasKildetoft yes, the point is, the frame of reference were my fingers placed in the standard way

Huy

11:38

@jyotishrajthoudam THAT would "violate the rule"

Tobias Kildetoft

@BalarkaSen So i is farther from m than your fingers?

Balarka Sen

mhm

d(my fingers, i) > d(my fingers, m)

Tobias Kildetoft

Or did you mean that i is further away than m?

jyotishraj thoudam

then according to summation convention if we expand the term $x_mx_m(x_ix_i)^{\frac{n}{2}-1}$ it would be diiferent from $(x_ix_i)^{\frac{n}{2}}$

Huy

m and i are indices, not variables

Balarka Sen

11:39

@TobiasKildetoft Oh, the latter. Didn't notice your parsing there.

Huy

and you said it: you use them for a sum

jyotishraj thoudam

what the fuck is going on. I'm asking a perfectly valid question & making fun of me nice

Tobias Kildetoft

ok, so the latter (btw, I am uncertain of whether the word "farther" is really correct, since I recall Tolkien going on about the fact that he used it whatever people said)

jyotishraj thoudam

Yes indices following the levi-Cevita summation convention

Huy

have you ever seen Levi-Civita set $i = 2$ but $m = 1$ in the same equation?

jyotishraj thoudam

11:41

In tensor Notation, $x_m$ is a vector

And $x_mx_i$ is a second order tensor

Huy

so?

if $x_i x_i = r^2$, then $x_m x_m = r^2$ too

it's a number

forget tensors

jyotishraj thoudam

Yes

mick

Ok I made the final edit now.

Plz vote reopen

Also upvote and answer :)

Huy

Plz send money :)

3

Tobias Kildetoft

@mick Also provide us with some idea of what you are talking about.

mick

11:44

In the year 3000

0

Q: The sequence A126022 : $a_{n+1} = a_n + \lfloor a^{-1}(n+1) \rfloor $

Consider the sequence A126022 at OEIS http://oeis.org/A126022 $$ 1,2,4,7,10,13,17,21,25,30, ... $$ We start with $a_1 = a(1) = 1$. $a_n = a(n)$ and $^{-1}$ means functional inverse. By $\lfloor a^{-1}(n+1) \rfloor$ we mean $\max K $ such that $a_K = a(K) $ and $a_K = n+1 $ or $a_K < n+1$. (...

recurrence-relations asymptotics closed-form

Balarka Sen

Man now I realize how much I miss these Huy jokes

Huy

<3

mick

That is What I talk about

Kirill

How does 1 in $R[[x]]$ look like?

Huy

@Kirill: 1

@BalarkaSen: come to high school in Switzerland, then you have to hear my jokes every day.

Kirill

11:46

nope, thats not the notation for $\sum_{n=0}^{\infty}a_nx^n$, @Huy

Huy

@Kirill: yes, it is, if $a_0 = 1$ and all other $a_j = 0$

Tobias Kildetoft

@Kirill how is $1$ not such an element?

Balarka Sen

@Huy I wish I could.

On the positive note I'll get kicked out of school next year

Huy

I wish I could kick pupils out of school

Kirill

@TobiasKildetoft I mean: 1 in $\mathbb{Z}$ is 1, 1 in $\mathbb{Z}^{2 \times 2}$ is $\begin{pmatrix} 1&0\\0&1\end{pmatrix}$. That's formally not the same.

Huy

11:49

@Kirill: but this time it is formally the same

Mike Miller

you can, it's just a much more physical process than Balarka is thinking of

Tobias Kildetoft

@Kirill Notationally, $1$ means the unit everywhere

Huy

welcome back Mike

Tobias Kildetoft

@MikeMiller BalarkaSen Said you might know Hodge theory.

Leaky Nun

@Kirill 1 in $\Bbb R$ is also 1. Your point?

Kirill

11:51

think of eigenvalues: you cannot write $(A-\lambda)v=0$. Correct would be $(A-I \lambda)v=0$.

Mike Miller

Depends what that means

Huy

no

that is still wrong

0 is a number, not a vector

Kirill

@Huy no, that's ok.

Huy

NO

Balarka Sen

lol

Kirill

11:53

@Huy I like that. Why not?

Tobias Kildetoft

@MikeMiller I was hoping you could tell me that part. I would like to try to understand what the "original" Hodge theory is, since some people have developed something they call an algebraic analogue and used that for some pretty amazingthings.

Huy

if you want to be pedantic, be pedantic

Tobias Kildetoft

I think it involves something called Hard Lefschetz

Balarka Sen

@Kirill Huy is making a point here. If you don't want to say 1 is a power series, don't say 0 is a vector.

Kirill

@Huy I am just as pedantic, as pedantic I should be for the exam.

Huy

11:54

when is your exam and what is it about

Mike Miller

@TobiasKildetoft How long of a conversation should this be and can you point me to the "some people"?

Tobias Kildetoft

@Kirill In that case, $k[[x]]$ does not consist of anything written as sums, but of sequences of elements of $k$

@MikeMiller Unfortunately I don't have time right now anyway. But the people are Elias and Williamson, in the paper The Hodge theory of Soergel bimodules which they published in Annals 3 years ago

Secret

re: seen mick's post: This is the first time I saw OP vtc own question, and also the first time that it is done to do major fix the question

Mike Miller

Ok, another time then. I'd like to have an understanding of the structure they build so I can bring up the relevant aspects of the older theory

Kirill

@BalarkaSen well, it is not about saying, it is about what formal notation will be correct and what not, when I write an exam.

Huy

11:56

write or take?

Kirill

@Huy linear algebra, tomorrow

Mike Miller

You can browse through Voisin's book (or intro) to great value I suspect

Huy

@Kirill: where do you study?

Kirill

in the university, @Huy

Huy

which one?

Kirill

11:57

@Huy in a ceratin one, do not think that makes a point

Huy

just wondering because there are not many universities that would have exams right now

Tobias Kildetoft

@MikeMiller The one called Hodge theory and complex algebraic geometry?

Leaky Nun

could someone construct a field of cardinality $2^\mathfrak c$?

Kirill

@Huy so, for Germany it's ok

Secret

@LeakyNun that is one very cool way to find generating functions. The approach I have encountered so far involve solving for coefficients after multiplying a series by some power series.

Huy

11:58

ok, so not ETH :(

would be a very reasonable time for a linear algebra exam here too

jyotishraj thoudam

anyone knows tensor notation & summation convention. please help

Huy

@jyotishrajthoudam I did already

Kirill

@Huy you mean, Zurich?

Huy

yes

Tobias Kildetoft

@LeakyNun Take the field of rational functions over the polynomial ring that number of variables

jyotishraj thoudam

11:59

@Huy if I replace m as i I would have 4 repeated indices which is not permissible

Secret

Now that I am made away I can simply bubble out powers of x like these and try to left behind the original sum, it might help in figuring out for me on how infinite series works and its sensitivity to details

Leaky Nun

@TobiasKildetoft nice

Kirill

@Huy it is not to far from Munich to Zurich. The same $ich$ at the end.

Huy

@jyotishrajthoudam then stop asking if you don't accept it

foxtrot9

Sorry for interruption. Right now, I am reading Michael A. Nielsen & Isaac L. Chuang for Quantum Computation and Quantum Information. And in chapter 2, 3 equal signs are used. Can someone explain why is that?

Huy

12:00

@Kirill: I see

jyotishraj thoudam

Clearly you do not know

Huy

:(

foxtrot9

3 equal signs meaning one extra - on equal sign =. Sorry for confusion.

Kirill

@foxtrot9 $\equiv$?

Huy

possibly modular arithmetic

foxtrot9

12:02

yes

I got my answer here: math.stackexchange.com/questions/29143/…

wow, simple google search worked.

Kirill

@foxtrot9 en.wikipedia.org/wiki/Triple_bar

foxtrot9

Got it @Kirill , ty.

Kirill

@foxtrot9 cannot say more particular, I use it for congruent triangles and modulos

Huy

in that book, it appears to be misused as a "defined by" sign

Kirill

and the TeX name is $\texttt{\equiv}$...

Tobias Kildetoft

12:07

@Huy I have seen others use it like that. I would not call it a misuse

Huy

I would call it misuse.

Secret

Need to check again whether I can linearly order the set of all indicator functions on $\Bbb{R}$ and if so construct a concrete example

Let's see...

foxtrot9

@Huv, text from book: An
important linear operator on any vector space V is the identity operator, IV , defined by
the equation IV |v> ≡ |v> for all vectors |v>.

Huy

yes, as I said

they also define Pauli matrices with the same notation

and introduce Dirac notation

(the qubits at the start of chap2)

foxtrot9

you said it is misused. Can you explain?

Kirill

12:11

wow, I am just happy with $\mathrm{id}_V$ then

Huy

in my opinion, that sign should not be used as a "defined by" sign, but there are obviously people who have a different opinion on this

@Kirill: did you already find the formula for Fibonacci with diagonalization?

Kirill

@Huy no, I have one with geometric and power series, the idea of my text is "we ignore the other way"

Huy

but it is a linear algebra course?

Kirill

@Huy the diagonalisation - for sure! We are just not expected to use matrices for that point. I should shorly think about how to rewrite the sequence as matrix.

Huy

ok

hi Jasper. have you started studying this year?

Balarka Sen

12:18

@Huy indeed, the defined by sign is most standardly $:=$.

Jasper Loy

@Huy I have not. I am still very sick.

Huy

:(

Jasper Loy

John Nash could not do any work for 20 years.

I take comfort in the fact that I am not alone in my suffering.

Huy

I hope you can start working in less than 20 years.

Leaky Nun

@Kirill to write $1$?

Balarka Sen

12:22

@JasperLoy you enjoy operas, right?

might be interesting

Kirill

@LeakyNun ?

Leaky Nun

@Kirill how would you write 1 in R[x]?

Kirill

@LeakyNun as $\sum_{n=0}^{\infty}a_nx^n$ with $a_0=1, a_n=0$ for $n\ge1$. I know, it is not handy and even wrong for some.

user21820

Hello everyone! Does anyone here take a fancy to cleaning Math SE from junk? =)

Kirill

as polynomials usually have $n$ at the top, not the $\infty$.

Leaky Nun

12:31

@Kirill and real numbers continue infinitrly to the right, so you should write a_0 = 1.000....

Kirill

but since polynomial is sequence I think I can do that

that look slike trolling @LeakyNun . I do not see any reason for the decimal formalism. It is all about how you define the elements of your rings. If you define your elements as numbers that you cannot write on the paper, you would be right.

Leaky Nun

I'm making a point.

Kirill

@LeakyNun and the point is definitions. If the definition says that elements of $R[[x]]$ have the form $\sum_{n=0}^{\infty}x^n$ then I just bring them in the form, although it is not necessary.

it is the same reason why I write $ax^2 +bx^1 + cx^0$ for the second degree polynomial, as it is just intuitevly clear for me.

Huy

@Kirill: but then you cannot plug in x=0 but always have to take a limit

Kirill

where in, @Huy?

Huy

12:40

in your second order polynomial

Kirill

why not?

Huy

try it

Kirill

=0

Huy

that's wrong

Leaky Nun

@Huy $x^0$ is undefined for $x=0$

Kirill

12:41

oops. =c

Huy

exactly. you need to take a limit.

Kirill

do not understand you @Huy

$0^0$ is defined as 1 in my script, I just follow the convention.

Huy

oh dear lord

who wrote those lecture notes

Vrouvrou

Hello, someone has an idea please : math.stackexchange.com/questions/2394138/…

Kirill

do you think the convention $0!=1$ is better?

Huy

12:43

that is a definition that doesn't violate any rules

Kirill

@Huy 0 is the power of 0 hasn't violated any rule we had, too

Huy

it does, because
$$\lim_{x \to 0} x^0 \neq \lim_{x \to 0} 0^x$$

Leaky Nun

@Huy actually it doesn't

nobody said that $x \mapsto 0^x$ must be continuous

Huy

neither did I

Leaky Nun

so it doesn't violate any rule

Huy

12:46

yes it does

Leaky Nun

which?

Kirill

but you have just said: these sequences have different limits. And? @Huy

Huy

so it doesn't make sense to "define" $0^0 = 1$

Leaky Nun

@Huy it does make sense in this context

i.e. in the context of power series

Kirill

it doesn't make sense to define $0!$ as 1 as well, but it is handy.

Leaky Nun

12:47

@Kirill it does.

the number of ways to permute 0 items is exactly 1.

SteamyRoot

$0! = 1$ makes perfect sense everywhere

$0^0$ depends on context

Kirill

@LeakyNun the number of ways to permute nothing is already an esoteric issue I would say.

Leaky Nun

@Kirill we also have the recurrence equation $n! = n(n-1)!$

when $n=1$, you get $0!=1$.

Secret

I wonder how will one handle the following sum: $$\sum_{k=0}^n k^k (x+1)^k$$? The $0^0 = 1$ convention will allow the binomial expansion to be carried out, however, here there's an explicit $k^k$ term which cannot be $1$ (Or perhaps I have constructed a wrong example)

Leaky Nun

@Kirill the number of ways to permute $n$ items is the number of bijections from $n$ items to itself.

Kirill

12:50

it is like a thesis "all aliens in this chat are brown". True, as there is no aliens here.

Leaky Nun

@Kirill no, $0!=1$ isn't a vacuous truth.

SteamyRoot

^

Leaky Nun

@Secret what on earth is that sum?

SteamyRoot

$n!$ is the number of permutations (or bijections) from a set with $n$ elements to itself.

Leaky Nun

and there is exactly one function from the empty set to itself

SteamyRoot

12:52

You can check that the empty relation from the empty set to itself is a bijection, and that it is the only bijection

Leaky Nun

And the formal construction of a relation is as an element of the power set of the cartesian product of the set with itself

SteamyRoot

$0! = 1$ by definition of the factorial. $0^0$ is, in general, undefined; though we often assign it a value depending on the context.

Leaky Nun

$\mathcal P (\varnothing \times \varnothing) = \mathcal P (\varnothing) = \{\varnothing\}$

Secret

@LeakyNun It's an attempt to ask how to pick conventions for the value of $0^0$ if an expression contains both a binomial expansion and also requiring $x\mapsto x^x$ to be continuous, i.e. if an expression contains $0^0$ that is supposed to be used in conflicting ways when alone, how to pick the correct convention to evaluate such expression?

Leaky Nun

@Secret You meant $x \mapsto x^x$ to be continuous

Secret

12:54

ah yes, sorry

Leaky Nun

It might be a surprise to you that $\displaystyle \lim_{x\to0^+} x^x = 1$

Kirill

formally that would be $n!:=\prod_{k=1}^{n}k:= 1$ if $n<k$. So, if $0!=1$ is wrong, then the definition of product is wrong. Is that?

Leaky Nun

@Kirill no, the empty product is $1$.

SteamyRoot

^

Leaky Nun

Look up the definition of product.

Kirill

12:55

corrected

SteamyRoot

The empty sum is $0$ because any sum can be written as $0 + $ that sum

Secret

@LeakyNun O? I thought the limit is path dependent as it approaches zero, at least when evaluated under the context of $\lim_{(x,y)\to (0,0)}x^y$

SteamyRoot

just like the empty product is $1$ because you can always multiply it by $1*

Leaky Nun

"$0! = 1$ is wrong" is wrong because $0!=1$ makes sense.

neither $0!=1$ nor the fact that the empty product is $1$ is wrong

SteamyRoot

@Secret Yes, depending on how you take that limit, you get a different value.

on the line $y = x$ you $1$

Leaky Nun

12:56

@Secret you did say $x^x$.

Secret

ah I see, makes sense

Leaky Nun

People need to stop viewing exponents as iterated multiplication.

SteamyRoot

@Secret This limit is exactly why it's difficult to assign a value to $0^0$ in general.

Secret

hmm, so I guess it is pretty hard to encounter an expression that requires both the $0^0=1$ convention and $\lim_{(x,y)\to(0,0)}x^y$

and I think should that happened, the limit takes precedence

SteamyRoot

What do you mean, requires that limit?

That limit doesn't exist.

Ever.

Leaky Nun

12:59

@Secret here is how you make the limit any value from $0$ to $1$.

Secret

Well, consider $(1+x)^5 + \lim_{(x,y)\to (0,0)}x^y$, I guess the value is "Does not exist" because the limit does not exist

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$Mathematics$ Mathematics