Mathematics

Secret

02:45

I really hate how conversations on the internet often abruptly stopped without warning

It seems as if the other party simply blinked out from existence and no more response to the questions are given

This won't happen in face to face because the moment someone wants to leave the conversation, it becomes obvious

skull

yeah, internet convos are peculiar in a lot of ways...

... that certainly is one of those ways.

Secret

I will figure out how to continue that conversation with Acuriousmind. Unlike JD and vzn, I actually can use equations and I will show him what I mean and see if it is at least self consistent

Currently, I think the best way to gain an intuition of classical dynamics is to first solve for the solution of some equation of motion, and then take time slices and arrange those in chronological order

So in theory, 2 time physics can be understood that way without too many assumptions of our intuition

Secret

04:35

$$\frac{\partial }{\partial \alpha} u(\sqrt{\alpha}\eta,\sqrt{\beta}\zeta) = \sum_{i=1}^n u_{\sqrt{\alpha}\eta}(\sqrt{\alpha}\eta,\sqrt{\beta}\zeta) \frac{\eta_i}{2\sqrt{\alpha}} \neq \sum_{i=1}^n u_{\sqrt{\alpha}\eta}(\sqrt{\alpha}\eta,\sqrt{\beta}\zeta) \frac{\eta_i}{\sqrt{\alpha}}$$?

Secret

05:29

The spherical mean $M_u(y,t) = Q[u]$ has the interesting property that given any linear operator $A$, $M_{Au}=Q[Au]=AM_u$

Abra001

@AvnishKabaj ya that's what you call l'hopital rule, cause it will take alot of mathematicians to hospital.

how did you distribute the log there ?

Silent

06:53

Are all the sets except b uncountable above? @LeakyNun

Avnish Kabaj

@Abra001 ;)

@Abra001 I got it

Leaky Nun

@Silent right

Avnish Kabaj

Calculation error

Whatnot

Silent

thank you!

sockevalley

If I would pick the string $a^p b^p+1 c^p+2. How could I show an contradiction?

If p = 1 and v = a and x = b. This would mean that we get abbccc which >= pumping length of 1.

Hawk

07:13

Can someone expand on the hint on math.stackexchange.com/questions/2150788/… ? It isn't clear to me how to prove the other inequality

how do i delete my own msg here

Secret

07:31

\begin{align}
(\triangle_x - \triangle_{t'}) u & = u_{tt}\\
u (x,t' \leq 0, 0) & = f(x,t')\\
u_t (x,t' \leq 0, 0) & = g(x,t')\\
u_{t'} (x,t' \leq 0,0) & = h(x,t')
\end{align}

DarkVampiric AbstractArtist

08:11

Out of algebra, analysis, geometry, and topology. Is geometry the most versatile field of mathematics?

Secret

Well, considering how hard it is to define geometry and it found its places in almost every fields of maths, yes

2

Nûr

What is an angle?

Secret

08:26

Something that is defined by an inner product?

Nûr

Yes I think it is the usual way to define it, using $\cos$ (I don't really know).

Secret

Meanwhile, I don't know if the notion of angle is well defined in a Banach space, for example

I guess one interesting thing we can ponder is what is the minimal criteria for something to be called an angle

We can build almost any geometry by starting with some topology, so that will be a good place to start in generalising the notion of an angle

Nûr

Yes, I agree, but it seems complicated to find the good criteria.

Secret

Another complication on what exactly is geometry is that when we drew some lines in euclidean space that are perpendicular with each other, it is obvious to us by looking at it. However, algebraically, the condition for perpedicular is actually an equation with n variables for example

Same applies for the notion of direction, which is not obvious when we only looked at the components. The notion of direction seemed to be some kind of relation need to be satisfied between the components, for example

Nûr

08:54

For those who understand french youtube.com/watch?v=ZW9JpZXwGXc

you can translate what is said using this script: archive.org/stream/Allons-nousContinuerLaRechercheScientifique/… (it starts page -10-).

(It is a conference by Grothendieck in which he develops his ideas on society).

philmcole

09:15

Hi, is $\Bbb R^2$ as a subset of $\Bbb R^3$ simply all $(x,y,0)$ for $(x,y) \in \Bbb R^2$? or could it be also all $(x,y,17)$ for $(x,y) \in \Bbb R^2$?

Nûr

a subspace contains zero

philmcole

Oh right. Forgot about that

thx

Nûr

;)

Secret

Silent

@mercio, we can't assign non-zero real values to $x_0, x_1$ so that the recurrence relation $x_{n+2}=2x_n$ converges, right?

mercio

09:26

right

Secret

Silent

thanks

Nûr

How do you prove that for any complex matrix $A$, $range(AA^T) = range(A)$ ?

Secret

This is sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌ooooooooooooooooooooooooooooooooooooo coool

skull

so, what is it?

Secret

09:31

Introducing Paint 3D

►

skull

yup, cool

philmcole

Is $\left(-\varepsilon,\varepsilon\right)^n$ a valid vector space (or subspace of $\Bbb R^n$)?

Nûr

No

Silent

@Nûr I think your question is wrong on two points: a. It should be rank, not range: see this, and b. It is not true for complex matrices, see this.

Nûr

@Silent: Yes real matrix sorry

Here it is AA^T

it's range

hmmm wait

Silent

09:40

@Nûr, see this proof.

philmcole

I can't find the reason why a function $f: \Bbb R^n \to \Bbb R^n$ restricted to $\Bbb R^k$ for some $k \lt n$ is of the form $g: \Bbb R^k \to \Bbb R^n$ with $g(x_1,\ldots,x_k)=f(x_1,\ldots,x_k,0,\ldots,0)$. Why are the other variables necessarily set to zero?

Nûr

Yes the hypothesis are what I've said in the last messages

Ok, I had already solve that, I just wanted to know if it was not a very easy question

Daminark

@philmcole what do you mean exactly when you say "restricted to $\mathbb{R}^k$"?

The problem is, $\mathbb{R}^k$ isn't technically a subspace of $\mathbb{R}^n$

philmcole

It is part of a proof and $f$ is a diffeomorphism and its derivative $D_xf$ is invertible. They argue that $D_xg$ has rank $k$ too since $g$ is a restriction of $f$ to $\Bbb R^k$. I'm trying to understand this argument.

Daminark

An element in $\mathbb{R}^k$ is a k-tuple, which isn't an element in the set of n-tuples

philmcole

09:46

sorry $D_xg$ is not invertible since its not square but it has full rank $k$ supposedly because $D_xf$ has full rank $n$

Daminark

The point is that there's a clean way to define $\mathbb{R}^k\subset\mathbb{R}^n$ as having the last n-k coordinates be zero

Nûr

It is not stable

by scalar multiplication

Daminark

@philmcole so think linear algebraically

Let's say you have a linear map $T:\mathbb{R}^n\to\mathbb{R}^n$

If you restrict $T$ to $\mathbb{R}^k$ (where we identify $(x_1,\ldots,x_k)$ with $(x_1,\ldots,x_k,0,\ldots,0)$)

Then what is the rank?

philmcole

Is it $k$ since $T \mid_{\Bbb R^k}$ has only the first $k$ columns of $T$ which are still linearly independent?

Nûr

I have remarked it is never precised when $(x_1,\ldots,x_k)$ is identified with $(x_1,\ldots,x_k,0,\ldots,0)$

Daminark

09:53

@philmcole yup

philmcole

@Daminark Okay :) And this does also generalize to $T:(-\varepsilon,\varepsilon)^n \to \Bbb R^n$ even thought $(-\varepsilon,\varepsilon)^n$ is not a vector space anymore?

Daminark

I'm about to go to bed but the exercise now is to convince yourself that it works when you talk about derivatives as well

Nûr

Is $T$ supposed to be invertible ?

Daminark

I'm not really going for that here so much as, if you now have a smooth function from $\mathbb{R}^n$ to itself, and you restrict to the first k coordinates, find out what happens to the derivative

philmcole

Alright, thanks!

Daminark

09:55

No problem, see you!

philmcole

Good night then

Maneesh Narayanan

in CSIR-TIFR-ISI-NBHM, 5 mins ago, by Maneesh Narayanan

Consider two non-zero $p-$dimensional column vectors $ a$ and $b, p ≥ 2$. How many non-zero distinct eigenvalues does the $p×p$ matrix $ab^t + ba^t $have?

Do you have any idea?@Secret

Can you check whether my idea correct or not?

Secret

no idea except that the matrix is symmetric

gateprep dies not exist

sockevalley

https://i.sstatic.net/8Fbmm.png
Anyone good at the pumping lemma here? lol

Secret

10:10

♦ $Mathematics$ Logic

This room is meant for discussion about logic, including found...

1

[Random]

Maneesh Narayanan

@Secret ?

Secret

Consider the glomerulous atkinis:

$$\int_{S^{S^S}} \frac{1}{(dx)^2}$$

We are interested in distilling the merokos lupin:

Nûr

@ManeeshNarayanan do you ask for real or complex eigenvalues ?

Secret

$$\int_{S^{S^S}} \frac{1}{(dx)^2} = \int_{S^{S^S}} 1 +\sum_{n=1}^{\infty} (n+1)(-1)^n(1-dx)^n$$

Maneesh Narayanan

they mention this much @Nûr

Nûr

10:18

?

I don't see, so please tell me

Secret

::operation temporarily on hold::

Maneesh Narayanan

they are asking for non zero distinct eigenvalues. it might be complex. right?

@Nûr

Nûr

Ok thx

Sometimes it is not precised but it s real eigenvalues

Secret

@ManeeshNarayanan that sentece will became clear over time. Ignore that for now

Nûr

Hmmmm, @ManeeshNarayanan I got two distinct eigenvalues if $ab^t$ is symmetrical

And one if $ab^t$ is skew-symmetrical

Nûr

10:38

if and only if *

for the last

Rosa

Hello everybody

Nûr

and if it is not symmetrical and not ati-symmetrical, its kernel is of dimension $n-2$, so there is 1 or 2 or 3 distinct ones ...

@ManeeshNarayanan

If you suppose all coefficients are real. it is not 1 or 2 or 3 but 2 or 3.

and the if and only if I mentionned is also true for real coefficients. If not it is 1 or 2 for the case $ab^t$ is skew-symmetrical

Silent

10:59

@Nûr Can we use the fact that $A$ and $A^t$ have same eigenvalues?

Nûr

11:21

It should be useful indeed

When I have spoken about distinct eigenvalues it was not non-zero distinct but just distinct

Leaky Nun

in CSIR-TIFR-ISI-NBHM, 2 hours ago, by Maneesh Narayanan

Consider two non-zero $p-$dimensional column vectors $ a$ and $b, p ≥ 2$. How many non-zero distinct eigenvalues does the $p×p$ matrix $ab^t + ba^t $have?

sockevalley

11:36

How does one prove if a grammar G is in LR(1)?

Liad

hi i got a question:
$F$ is a field with $char(F) = p \gt 0$. $[F:F_p] \lt \infty$, i need to show that $[F:F_p] = p \ ^ n$ for some $n \ge 0$, someone can help?

Leaky Nun

Let $C = ab^\top + ba^\top$. Observe that $Cv = ab^\top v + ba^\top v = a(b \cdot v) + b(a \cdot v) \in \langle a,b \rangle$. Therefore, any eigenvector with non-zero eigenvalue must be in the form of $v = xa+yb$. Then, $Cv = (x(a \cdot b) + y \|b\|^2)a + (x\|a\|^2+y(a\cdot b))b$. This can be represented by the matrix $\begin{bmatrix} a \cdot b & \|b\|^2 \\ \|a\|^2 & a \cdot b \end{bmatrix}$, whose characteristic polynomial is $t^2-2(a \cdot b)t+(a \cdot b)^2-\|a\|^2\|b\|^2$.

Using Cauchy-Schwarz, one can show that it has two distinct roots

and those roots are non-zero, so there is two distinct non-zero eigenvalues @ManeeshNarayanan @Silent @Nûr

@Liad that's a lie

Liad

hmm, it is an exercise i got, do you have a counter-example?

loch

@Liad $F$ is a $F_p$ vector space. So try writing down a basis and count how many elements you can have

Leaky Nun

@Liad $F = \Bbb F_3[X]/(X^2+1)$

Liad

11:42

@LeakyNun why is it a counter example?

Alessandro Codenotti

Because $|F:\Bbb F_3|=2$

Leaky Nun

@Liad $[F:F_p]=2$

tu...

:c

Alessandro Codenotti

Took me twelve edit to write that degree properly but I'm gonna claim I sniped you anyway :P

Liad

Hehe ^^

Leaky Nun

triste so

loch

11:45

oh i misread

Liad

i dont see why the index is 2, can one of you explain?

Leaky Nun

@Liad because it is generated by $1$ and $X$ as a $\Bbb F_3$-vector space

Liad

ok. does the following question is true ? ^^
i need to show that if $a \ ^ {1/p} \notin F$ then $x \ ^ {p \ ^ n} - a $ is irreducible

nvm i got to go.

thanks ! @LeakyNun @AlessandroCodenotti

Maneesh Narayanan

12:05

sorry! i had network issues.

@LeakyNun How did you get this matrix? Can you explain please?

Leaky Nun

@ManeeshNarayanan are you familiar with the correspondence between matrices and linear transformations?

Maneesh Narayanan

yes

Leaky Nun

then just do it, lol

that matrix is what corresponds to the linear transformation $xa+yb \mapsto (x(a \cdot b) + y \|b\|^2)a + (x\|a\|^2+y(a\cdot b))b$

Maneesh Narayanan

12:30

How can we guarantee that $a$ and $b$ are linearly independent?

I got matrix by setting $x=1,y=0$ and $x=0,y=1$.

@LeakyNun

Leaky Nun

2 mins ago, by Maneesh Narayanan

How can we guarantee that $a$ and $b$ are linearly independent?

we can't. but if they are linearly dependent, the question becomes trivial

Maneesh Narayanan

Thank you very much @LeakyNun

Liad

@LeakyNun @AlessandroCodenotti there was a typo in the exercise, the assumption is $[F: F \ ^ p] \lt \infty$ and need to prove that $[F: F \ ^ p ] = p \ ^ n$

Alessandro Codenotti

what's $F^p$?

Leaky Nun

@AlessandroCodenotti image of F under Froebenius homomorphism, I suppose

Liad

12:41

$\{ x \ ^ p : x \in F \}$

@LeakyNun excatly

anyone see why it's true?

loch

13:15

@Liad It suffices to prove that the min poly of any element $t\in F$ over $F^p$ is given by $x^p-t^p=0$ (or $x-t=0$)

Leaky Nun

14:09

@loch interesting

Liad

@LeakyNun why does it suffices?

Leaky Nun

@Liad because then you can extend $F^p$ to get $F$ step by step, each step having index $p$

and then tower law

Liad

yea what i thought but we assuming $F$ is finite?

Leaky Nun

@Liad no we aren't

we're given that $[F:F^p]<\infty$, so this process would stop

Liad

Right!

Liad

14:38

i proved that $x \ ^ p - a \ ^ p$ is irreducible for each $a$ in the basis over $F \ ^ p$ this way :
this polynomial is equal to $(x-a) \ ^ p$ so $a$ is inseparable so we can write it as $g(x \ ^ p)$ so the degree of the min poly must be at least $p$ so it is exactly $p$ , what do you think @LeakyNun ?

Leaky Nun

no idea

tatan

If for a function f, for all x in dom (f) there exists 1/x in the dom (f) and f(x)+f(1/x)=x... then what is the set of possible values of x?

Silent

@LeakyNun, What does $|x|+1\le |y|$ mean?

Leaky Nun

@Silent It means what it means

Silent

"$|x|+1\le y$ and $|x|+1\le -y$"

OR

"$|x|+1\le y$ or $|x|+1\le -y$"

i was copying and pasting :)

but you are so quick

Leaky Nun

14:46

neither

tatan

What is the significance of f(x)+f(1/x)=x?

Leaky Nun

@tatan note that the left hand side of your equation is symmetric under taking reciprocal

(what a fancy way to convey a simple idea)

therefore so must the right hand side

tatan

So, x must be =1/x and the we get

@LeakyNun Thanks!! You are real quick man!

@LeakyNun I really liked your statement - "what a fancy way to convey a simple idea"... I think that's the only think done in every exam I give...

Silent

@LeakyNun but $|x|\ge 5$ means $x\ge 5$ or $-x\ge 5$ right

Leaky Nun

@Silent nvm I'm wrong

no

yes

Silent

14:52

confused

me

@LeakyNun So, $|x|+1\le y$ or$|x|+1\le -y$ is correct?

Leaky Nun

right

Silent

$|x|-1\le y$ or $|x|-1\le -y$

implies

$|x|-1\le y$ or $-|x|+1\ge y$

implies

$|x|-1\le 1-|x|$

implies

$|x|\le 1$

@LeakyNun, is above calculation correct? I have doubt about $|x|-1\le 1-|x|$ step, i mean, does 'or' lead us there?

Leaky Nun

can't be bothered

Silent

ok

"can't be bothered" means "do not disturb" right?

Leaky Nun

it means, I don't want to think about it

google.co.uk/…

Silent

15:04

oh, thanks for clarification :)

Rishi Kakkar

15:25

I want to understand graphically what do we mean when we say the nth derivative of x^n

anakhronizein

15:40

@RishiKakkar do you understand graphically what it means when we say the 1st derivative of x^n?

tatan

@anakhronizein Yes... I do... but what does the n-th derivative mean?

anakhronizein

Well the 2nd derivative is the same thing as the 1st derivative of the 1st derivative.

And the 3rd derivative is the same thing as the 1st derivative of the 2nd derivative.

...

So the nth derivative is the same thing as the 1st derivative of the (n-1)th derivative.

Leaky Nun

relevant: latest 3b1b

What they won't teach you in calculus

►

clickbait much

tatan

@anakhronizein I quite realise that but does it have any geometrical significance?

Silent

this was amazing video!

Balarka Sen

15:50

I have stopped watching 3B1B

Math is for nerds my dude

anakhronizein

Is that not enough geometrical significance for you, @tatan ?

Silent

why?

Tuki

helo chat

tatan

@anakhronizein No. I realise that that first derivative has some real meaning. It is the slope of the tangent of a function. It gives you a definite information of a function. So the usefulness or significance of anything for me lies in its utility. How does n-th derivative give me some definite information about a function?

Harry Evans

I am studying for an exam in functional analysis and I need help on this item here:

Let $\{x_n \}_{n=1}^{\infty}$ be a sequence in a Hilbert space $\mathcal{H}$ such that $x_n \rightharpoonup 0.$ Recall that this means that $\langle x_n, u \rangle \rightarrow 0$ for all $u \in \mathcal{H}.$

1. By induction, prove that there exists a sub-sequence, $\{x_{n_k}\}_{k = 1}^{\infty}$ such that $\left| \langle x_{n_k}, x_{n_j} \rangle \right| \le \frac{1}{k}$ whenever $k > j.$

2. For $N \in \mathbb{N}$, define

anakhronizein

15:58

The nth derivative gives you $$\underbrace{\text{the rate of change}, \text{ of the rate of change}, \dotsc, \text{ of the rate of change of } f}_{n\text{ times}}$$

tatan

@anakhronizein Do you ever actually use it (in analysis or anywhere for that matter)? I am interested in its application part...

anakhronizein

You use it all the time in analysis.

Look up things about the Taylor expansion of a function.

Rishi Kakkar

actually it ends up at n!

anakhronizein

Taylor series

In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. The concept of a Taylor series was formulated by the Scottish mathematician James Gregory and formally introduced by the English mathematician Brook Taylor in 1715. If the Taylor series is centered at zero, then that series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century. A function can be approximated...

Rishi Kakkar

which is interesting as per me

graphically I can't represent n!

tatan

16:03

@anakhronizein Maybe I like that

anakhronizein

A rough summary is that the derivatives up to some fixed n contain enough information about the function to approximate it within a certain neighbourhood to a certain precision.

Rishi Kakkar

I actually thought is the way of cutting the sections for derivatives in more finer and precise location rather than a bunch or set ,Am I thinking right ??

anakhronizein

I don't follow, @RishiKakkar

Rishi Kakkar

i mean as per derivates we are precising the change in a certain quantity with respect to other in as set but as per nth derivate that set is more finer and precise

anakhronizein

16:24

Well I think it's better said the same thing about a new object.

That is, you are looking at the rate of change in the (n-1)th derivative. The geometry that you are discussing lies in the new object, not the old one.

But there is probably a nicer way of thinking about it.

I just don't have any good ideas right this instant, sorry!

Silent

16:54

@AlessandroCodenotti, will you please verify calculation there:

2 hours ago, by Silent

$|x|-1\le y$ or $|x|-1\le -y$

Abcd

Can someone tell me/ teach me what is Newton-leibinz formula for calculation of infinite series using integration.

For instance if I have a series $1^4 + 2^4 + 3^4 +...+ n^4 $, how do I use N-L's formula here?

tatan

17:19

Can you help me a bit with this? if f(f(x))(1+f(x))=-f(x). FInd the value of f(3)...

where f is defined on R

Leaky Nun

is f continuous?

@Mr.Xcoder hi

Mr. Xcoder

@LeakyNun Hi

What Gives

Hi! (inf=infinity)

If i take the limit of an expression like this lim n->inf (thing1 - thing2) and both thing1 and thing2 turns out to be inf. Do they cancel each other out and I can actually say that the expression converges to 0?

tatan

@LeakyNun If that is necessary... yes it is

Semiclassical

17:35

@WhatGives Take the example of things1, thing2 both being linear functions of n with positive slope.

Boson Bear

Hi I have a question on solving nonlinear ODE numerically. Could anyone take a look?

0

Q: Stiff BVP of nonlinear ODE, shooting method and beyond

I have been trying to solve this coupled ODE set. \begin{align} ( \frac{ \mu^2}{B} +1 ) \Phi^2 + \frac{1}{A} {\Phi^{\prime 2}} + \frac{1}{2}\lambda \Phi^4 - \frac{A'}{r A^2} + \frac{1}{r^2 A} - \frac{1}{r^2} &= 0, \cr ( \frac{\mu ^2 }{B} - 1 )\Phi^2 + \frac{1}{A} \Phi...

What Gives

@Semiclassical Sorry I don

Semiclassical

Pick two such functions, whichever ones you want.

What Gives

@Semiclassical Sorry I don't understand how that question impacts my question =[

Oh ok sure y=x?

Semiclassical

That's one.

You need two.

What Gives

17:46

y=2x?

Semiclassical

Okay. Take that as your thing1 and thing2

both of those go to infinity as x->infinity.

What Gives

yes

Semiclassical

does thing1-thing2 go to zero as x->infinity?

What Gives

aha! but I though I could do like this: lim n-> inf(thing1 - thing2) <=> lim n-> inf thing1 - lim n-> inf thing2

Semiclassical

that only works if the two limits actually exist as finite quantities.

What Gives

17:49

hmm okay

Semiclassical

When we say limit = infinity, that's really a statement about the limit failing to exist in a specific way

Additionally, if you take something like thing1 = n+1 and thing2 = n, then you get thing1-thing2 = 1

which remains =1 for all n and therefore the limit would be =1 as well in that case.

So you really can't conclude anything the limit of thing1-thing2 if the individual limits don't exist as finite quantities.

What Gives

Okay I am going to google that stuff and get into it. Thank you!

♦ $Mathematics$ Logic

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$Mathematics$ Mathematics

♦ Logic

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♦ $Mathematics$ Logic

$Mathematics$ Mathematics