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02:00 - 18:0018:00 - 00:00

2:45 AM
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
 
yeah, internet convos are peculiar in a lot of ways...
... that certainly is one of those ways.
 
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
 
 
2 hours later…
4:35 AM
$$\frac{\partial }{\partial \alpha} u(\sqrt{\alpha}\eta,\sqrt{\beta}\zeta) = \sum_{i=1}^n u_{\sqrt{\alpha}\eta}(\sqrt{\alpha}\eta,\sqrt{\beta}\zeta) \frac{\eta_i}{2\sqrt{\alpha}} \neq \sum_{i=1}^n u_{\sqrt{\alpha}\eta}(\sqrt{\alpha}\eta,\sqrt{\beta}\zeta) \frac{\eta_i}{\sqrt{\alpha}}$$?
 
5:29 AM
The spherical mean $M_u(y,t) = Q[u]$ has the interesting property that given any linear operator $A$, $M_{Au}=Q[Au]=AM_u$
 
@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 ?
 
 
1 hour later…
6:53 AM
Are all the sets except b uncountable above? @LeakyNun
 
@Abra001 ;)
@Abra001 I got it
 
@Silent right
 
Calculation error
Whatnot
 
thank you!
 
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.
 
7:13 AM
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
 
7:31 AM
\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}
 
8:11 AM
Out of algebra, analysis, geometry, and topology. Is geometry the most versatile field of mathematics?
 
Well, considering how hard it is to define geometry and it found its places in almost every fields of maths, yes
2
 
What is an angle?
 
8:26 AM
Something that is defined by an inner product?
 
Yes I think it is the usual way to define it, using $\cos$ (I don't really know).
 
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
 
Yes, I agree, but it seems complicated to find the good criteria.
 
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
 
8:54 AM
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).
 
9:15 AM
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 $(x,y) \in \Bbb R^2$?
 
a subspace contains zero
 
Oh right. Forgot about that
thx
 
;)
 
@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?
 
9:26 AM
right
 
thanks
 
How do you prove that for any complex matrix $A$, $range(AA^T) = range(A)$ ?
 
This is sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌​ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌​ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌​ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌​ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌​ooooooooooooooooooooooooooooooooooooo coool
 
so, what is it?
 
yup, cool
 
Is $\left(-\varepsilon,\varepsilon\right)^n$ a valid vector space (or subspace of $\Bbb R^n$)?
 
No
 
@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.
 
@Silent: Yes real matrix sorry
Here it is AA^T
it's range
hmmm wait
 
9:40 AM
@Nûr, see this proof.
 
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?
 
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
 
@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 $\mathbb{R}^n$
 
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.
 
An element in $\mathbb{R}^k$ is a k-tuple, which isn't an element in the set of n-tuples
 
9:46 AM
sorry $D_xg$ is not invertible since its not square but it has full rank $k$ supposedly because $D_xf$ has full rank $n$
 
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
 
It is not stable
by scalar multiplication
 
@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?
 
Is it $k$ since $T \mid_{\Bbb R^k}$ has only the first $k$ columns of $T$ which are still linearly independent?
 
I have remarked it is never precised when $(x_1,\ldots,x_k)$ is identified with $(x_1,\ldots,x_k,0,\ldots,0)$
 
9:53 AM
@philmcole yup
 
@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?
 
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
 
Is $T$ supposed to be invertible ?
 
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
 
Alright, thanks!
 
9:55 AM
No problem, see you!
 
Good night then
 
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?
 
no idea except that the matrix is symmetric
gateprep dies not exist
 
https://i.stack.imgur.com/8Fbmm.png
Anyone good at the pumping lemma here? lol
 
10:10 AM

  Logic

This room is meant for discussion about logic, including found...
[Random]
 
@Secret ?
 
Consider the glomerulous atkinis:
$$\int_{S^{S^S}} \frac{1}{(dx)^2}$$
We are interested in distilling the merokos lupin:
 
@ManeeshNarayanan do you ask for real or complex eigenvalues ?
 
$$\int_{S^{S^S}} \frac{1}{(dx)^2} = \int_{S^{S^S}} 1 +\sum_{n=1}^{\infty} (n+1)(-1)^n(1-dx)^n$$
 
they mention this much @Nûr
 
10:18 AM
?
I don't see, so please tell me
 
::operation temporarily on hold::
 
they are asking for non zero distinct eigenvalues. it might be complex. right?
@Nûr
 
Ok thx
Sometimes it is not precised but it s real eigenvalues
 
@ManeeshNarayanan that sentece will became clear over time. Ignore that for now
 
Hmmmm, @ManeeshNarayanan I got two distinct eigenvalues if $ab^t$ is symmetrical
And one if $ab^t$ is skew-symmetrical
 
10:38 AM
if and only if *
for the last
 
Hello everybody
 
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
 
10:59 AM
@Nûr Can we use the fact that $A$ and $A^t$ have same eigenvalues?
 
11:21 AM
It should be useful indeed
When I have spoken about distinct eigenvalues it was not non-zero distinct but just distinct
 
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?
 
11:36 AM
How does one prove if a grammar G is in LR(1)?
 
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?
 
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
 
hmm, it is an exercise i got, do you have a counter-example?
 
@Liad $F$ is a $F_p$ vector space. So try writing down a basis and count how many elements you can have
 
@Liad $F = \Bbb F_3[X]/(X^2+1)$
 
11:42 AM
@LeakyNun why is it a counter example?
 
Because $|F:\Bbb F_3|=2$
 
@Liad $[F:F_p]=2$
tu...
:c
 
Took me twelve edit to write that degree properly but I'm gonna claim I sniped you anyway :P
 
Hehe ^^
 
triste so
 
11:45 AM
oh i misread
 
i dont see why the index is 2, can one of you explain?
 
@Liad because it is generated by $1$ and $X$ as a $\Bbb F_3$-vector space
 
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
 
12:05 PM
sorry! i had network issues.
@LeakyNun How did you get this matrix? Can you explain please?
 
@ManeeshNarayanan are you familiar with the correspondence between matrices and linear transformations?
 
then just do it, lol
that matrix is what corresponds to the linear transformation $xa+yb \mapsto (x(a \cdot b) + y \|b\|^2)a + (x\|a\|^2+y(a\cdot b))b$
 
12:30 PM
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
 
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
 
Thank you very much @LeakyNun
 
@LeakyNun @AlessandroCodenotti there was a typo in the exercise, the assumption is $[F: F \ ^ p] \lt \infty$ and need to prove that $[F: F \ ^ p ] = p \ ^ n$
 
what's $F^p$?
 
@AlessandroCodenotti image of F under Froebenius homomorphism, I suppose
 
12:41 PM
$\{ x \ ^ p : x \in F \}$
@LeakyNun excatly
anyone see why it's true?
 
1:15 PM
@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 $x-t=0$)
 
2:09 PM
@loch interesting
 
@LeakyNun why does it suffices?
 
@Liad because then you can extend $F^p$ to get $F$ step by step, each step having index $p$
and then tower law
 
yea what i thought but we assuming $F$ is finite?
 
@Liad no we aren't
we're given that $[F:F^p]<\infty$, so this process would stop
 
Right!
 
2:38 PM
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 ?
 
no idea
 
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?
 
@LeakyNun, What does $|x|+1\le |y|$ mean?
 
@Silent It means what it means
 
"$|x|+1\le y$ 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
 
2:46 PM
neither
 
What is the significance of f(x)+f(1/x)=x?
 
@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
 
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...
 
@LeakyNun but $|x|\ge 5$ means $x\ge 5$ or $-x\ge 5$ right
 
@Silent nvm I'm wrong
no
yes
 
2:52 PM
confused
me
@LeakyNun So, $|x|+1\le y$ or$|x|+1\le -y$ is correct?
 
right
 
$|x|-1\le y$ 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?
 
can't be bothered
 
ok
"can't be bothered" means "do not disturb" right?
 
it means, I don't want to think about it
 
3:04 PM
oh, thanks for clarification :)
 
3:25 PM
I want to understand graphically what do we mean when we say the nth derivative of x^n
 
3:40 PM
@RishiKakkar do you understand graphically what it means when we say the 1st derivative of x^n?
 
@anakhronizein Yes... I do... but what does the n-th derivative mean?
 
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.
 
relevant: latest 3b1b
clickbait much
 
@anakhronizein I quite realise that but does it have any geometrical significance?
 
this was amazing video!
 
3:50 PM
I have stopped watching 3B1B
Math is for nerds my dude
 
Is that not enough geometrical significance for you, @tatan ?
 
why?
 
helo chat
 
@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?
 
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
 
3:58 PM
The nth derivative gives you $$\underbrace{\text{the rate of change}, \text{ of the rate of change}, \dotsc, \text{ of the rate of change of } f}_{n\text{ times}}$$
 
@anakhronizein Do you ever actually use it (in analysis or anywhere for that matter)? I am interested in its application part...
 
You use it all the time in analysis.
Look up things about the Taylor expansion of a function.
 
actually it ends up at n!
 
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...
 
which is interesting as per me
graphically I can't represent n!
 
4:03 PM
@anakhronizein Maybe I like that
 
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.
 
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 ??
 
I don't follow, @RishiKakkar
 
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
 
4:24 PM
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!
 
4:54 PM
@AlessandroCodenotti, will you please verify calculation there:
2 hours ago, by Silent
$|x|-1\le y$ or $|x|-1\le -y$
 
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?
 
5:19 PM
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
 
is f continuous?
@Mr.Xcoder hi
 
@LeakyNun Hi
 
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?
 
@LeakyNun If that is necessary... yes it is
 
5:35 PM
@WhatGives Take the example of things1, thing2 both being linear functions of n with positive slope.
 
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

Boson BearI 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...

 
@Semiclassical Sorry I don
 
Pick two such functions, whichever ones you want.
 
@Semiclassical Sorry I don't understand how that question impacts my question =[
Oh ok sure y=x?
 
That's one.
You need two.
 
5:46 PM
y=2x?
 
Okay. Take that as your thing1 and thing2
both of those go to infinity as x->infinity.
 
yes
 
does thing1-thing2 go to zero as x->infinity?
 
aha! but I though I could do like this: lim n-> inf(thing1 - thing2) <=> lim n-> inf thing1 - lim n-> inf thing2
 
that only works if the two limits actually exist as finite quantities.
 
5:49 PM
hmm okay
 
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.
 
Okay I am going to google that stuff and get into it. Thank you!
 
02:00 - 18:0018:00 - 00:00

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