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15:00
@Ugo it isn't a problem of wikipedia
you should read what you quoted more carefully
the condition is that ([every irreducible polynomial over K that has at least one root in L] splits over L)
so something needs to split over L
the thing being every irreducible polynomial over K that has at least one root in L
Prove that if $y(t)$ is a non-trivial solution to $y''+q(t)y=0$ where $q$ is continuous on all of $\mathbb{R}$ and $q(t)<0$ for all $t$, then $y$ has at most one zero on $\mathbb{R}$
Does anyone have an idea? I don't want full solution, just a hint
draw a picture
15:16
@Lozansky if $y>0$, then $y''>0$
@mercio okay, where I'm assuming $y$ has more than 1 zero I guess?
@Lozansky you don't need to
@LeakyNun Hmm true
I hate proofs by contradiction with a passion (cc @user21820)
hm I think you can proceed by contradiction
lol
15:17
I like contradiction ...
but yeah it shouldn't be necessary
that's because you're a dirty classical logician
@mercio who are you referring to?
lozansky o..o'
right
proof by contradiction is non-constructive
but beware that one kind of "proof by contradiction" is permissible
because $\neg p$ means $p \implies \bot$
so you need to assume $p$ and prove a contradiction
but to prove $p$ you cannot just assume $\neg p$ and then derive a contradiction
@user21820 see my cc :P
But why?
15:21
@user21820 but why what
Why do you hate something that didn't harm you?
Hi, if i have a linear projection of which projection matrix columns are linearly independent can I assume that it is injective?
lol
@user21820 it's just for fun lol
@JDizzle yes
@LeakyNun I know. You and David love to try to get emotional responses from me.
15:23
@JDizzle columns linearly independent <=> invertible <=> bijective <=> injective
he didn't say it was a square matrix
@mercio do I look like I care?
though um
maybe it is because it's a projection
yes
and because of that the ker(P) is always the zero vector?
write down the definitions of "the only element in ker(p) is 0" and "the columns are linearly independent" and look at them a lot
15:25
oh
@JDizzle Did you try expanding the definition of "injective"?
I get it, thanks @mercio @LeakyNun
lol
my proof is too long
@mercio's proof is much better
honestly i think the part where injective <=> ker(p)= 0 is the hardest
Ugo
Ugo
used this for previous question
$Q((-2)^{1/4})/Q$ is not normal since $f(x)=x^2+2$ has root $(-2)^{(1/4)} \in Q((-2)^{1/4}$ but $f$ does not entirely split over $Q((-2)^{1/4}$ since it has a root $(-2)^{(1/4)}*i \not \in Q((-2)^{1/4})$
15:27
@mercio is it?
Ugo
Ugo
Here another [Q:Q]=1?? no yes??
(because if you write the definitions of the two things i quoted you should be writing down the same sentences)
injective := [T(u)=T(v) --> u=v] <=> [T(u-v)=0 --> u-v=0] <=> ker(p)=0
(whereas for P injective <=> ker(p) = {0} there are actual algebraic manipulations to perform !!)
@mercio ...
that's what you meant
15:29
it's mostly a matter of taste i guess
[Q:Q] = 1 yes
@Ugo are you sure $x^2+2$ has root $(-2)^{1/4}$?
Ugo
Ugo
typo $x^4+2$
your whole sentence is correct if you change $x^2+2$ to $x^4+2$
@Ugo if a degree 2 polynomial has a root then it splits ...
@mercio lol
Ugo
Ugo
15:31
thanks
he only needs to be ridiculed once
Ugo
Ugo
LOL
also you have to justify the last statement about i(-2)^(1/4) not being in Q((-2)^(1/4))
Ugo
Ugo
[Q;Q]=1 yes?? No??
indeed
@Ugo yes
2 mins ago, by mercio
[Q:Q] = 1 yes
and stop putting >1 question marks after every question!!!
15:32
Now if somebody wants to know $P(V)$ and P: V->V, then the columns of the projection matrix are the generator of V?
@JDizzle no
take the zero projection matrix that projects everything to zero
the columns generate P(V), not V
what is P(V)?
wait
what did I just see
> V is a linear map -- mercio
15:34
you saw nothing
o..o
*generator of P(V) where P is the projection from V->V
what is P(V)?
P[V]?
the forward image of V by P
i suppose
as in, {P(v) | v in V}?
15:35
who uses that notation
write im P
those who want to confuse students
yeah our prof makes a lot of effort to do that
to confuse students?
amazing
15:36
if I had students I would confuse them so much
about 95% of the student taking his class aborted 1month later
What are functions with identical image and codomain called in English?
@Mr.Xcoder surjective
Thanks, @Leaky :D
no problem
aka onto
15:39
so back to the question was my assumption right that the generator P[V] are the columns of the projection matrix?
the columns of the projection matrix generate P[V] yes
but don't say "the generators"
ok, great
ever
how to say that in English?
like i said it o..o
15:40
@JDizzle "the columns of the projection matrix generate P[V]"
or "a set of generators of P[V] is ..."
or " a generating set of ..."
because generators are not unique
haha ok i thought it was the wrong term, in Germany we use something like the generating system
is anyone here familiar with the laws about homeschooling children in new york
and is it legal to have them not see anyone ever and not do any standardized test at the end of the year
Hello all. Do you have any tips to drastically increase math level ? I feel like my level is stagnating while I have an extrmely difficult test to pass in a few monts.
15:44
Same ^^, except mine is easy :D
I find that you improve at anything by struggling at it :c
@Lucas what topic?
Mathematics in general (2 years postgraduate)
(for ENS Ulm contest)
I need to be in the 60 first of France :S
15:46
hmm you still have a few months
I guess you should look at all the previous exams for the previous years
you need to be good at abstraction cuz their exercises are very different from the standard eazy computational stuff
I met some people who developed a very strong mental representation of mathematical concepts. It seems that it helped them a lot.
i have no idea what that can mean
@mercio intuition
@LeakyNun @mercio Let $y(t_1)=y(t_2)=0$ where $t_1<t_2$ and $t_1,t_2 \in \mathbb{R}$. Consider the point $(t_1,0)$. It can be proven, using the existence-uniqueness thoerem, that $y(t_1) \neq 0$. Thus, since $y$ is continuous, we know that in a nbhd $t\in(t_1, t_1+\epsilon)$, $y(t)\neq 0$. Since $y(t_2)=0$ we know that there must be a maximum (the proof for minimum is analagous) point $t_0 \in (t_1, t_2)$. Thus $y''(t_0) \leq 0$. But since $y(t_0) > 0$, we must have $y''(t_0)>0$ so contradiction
I said
I
hate
contradiction
15:50
did you just write $y(t_1) = 0$ and $y(t_1) \neq 0$ on the same line
did he?
@LeakyNun :s why?
Oh
I did
it should be the derivative
Having a working representation of linear algebra in 3d spaces is very important
@Mr.Xcoder because it isn't constructive
15:51
Yes especially
(constructive is a precise math term)
But I can't edit, because if I add the apostrophe the msg becomes too long
lol
So it stays
15:52
how do you prove that $y'(t_1) \ne 0$?
As a reminder to the reader to be observant
s/we must have/so/
what's the existence-uniqueness theorem?
@LeakyNun Well it's assumed $y$ is non-trivial
Rolle's theorem
15:52
how does that say anything
The problem is that a purely academic approach can not suffice since the exercises are totally destabilizing and difficult.
$y''+p(t)y=0, y(t_0)= y'(t_0)=0$ has unique solution according to said theorem
they are destabilizing because you have no experience of anything outside eazy stuff
But the trivial solution is a solution but we said $y$ is nontrivial so contradiction
this doesn't say that...
@Lozansky do you really have to use "contradiction" here
15:54
yes
It can be proven, by contradiction, that it's the only thing that works
you're just going to put "contradiction" whenever you can, huh
someone just want to watch the world burn
and you get experience by doing those exams and struggling on difficult exercises
Have already been posed unresolved problems to see how the candidate is going to approach the problem, or theorem demonstrated in 2015.
15:57
that shows what reflexes you have when gauging the relative difficulty of various propositions
@Lozansky that's an interesting theorem
@LeakyNun Existence-uniqueness?
can you prove it?
Yes it was the purpose I think.
@Lozansky yes
15:59
things like "it is enough to prove this in THIS case only cuz I have an eazy proof that it generalizes from this special case"
I have 2 projection matrices from the same projection and the basis B,C. One matrix is B,B the other C,C. What are the steps to get the matrix B,C?
@LeakyNun I haven't actually seen a proof of it
My book said something like "nah too difficult"
@JDizzle adjoin the first column of the first matrix to the first column of the second matrix
@Lozansky what's the condition on $p$?
Continuous
Yes, It is appreciated to start by dealing with special cases (dimension 1 or 2 in algebra) or making drawings.
16:00
interesting
@JDizzle what if the projection is the identity but the basis are different ?
adjoin? Never heard that
@JDizzle it means stick them together with a glue
it's an English word
@LeakyNun It is like this. Consider the initial value problem
$$y'' + p(t)y'+ q(t)y = g(t), y(t_0) = y_0, y'(t_0)=y_0'$$

where $p, q$, and $g$ are continuous on an open interval $I$ that contains the point $t0$.
Then there is exactly one solution $y = \phi(t)$ of this problem, and the solution exists throughout the interval $I$
16:03
Do you have something similar to the ENS contest abroad (extremely selective contest) or is it specific to France ?
I'm afraid I don't know
Let B be the standartbasis of R3, C = ( (1,2,0)(1,0,2)(0,2,2) ). Then the columns of the matrix B,B are the vectors bi, the same for C with ci. The matrix B,C multiplied with (1,2,0) should then be equal to (1,0,0)
@LeakyNun should I multiply the columns? And do that with every column?
16:22
@JDizzle what the hell is "the matrix B,C"?
WE were told if you talk about a projection matrix you need to note the 2 basis on which they operate
$M^B_C(P)$
thats our notation
Input is on basis B output on basis C
I have never ever ever seen this notation in my life
yeah he "invented" it for class
then you should have "stated" it
yeah, next time I will
16:26
Could you ask me questions, not long to answer but subtle enough to quickly assess my level (2 years postgraduate) ? :-)
@Lucas prove the intermediate value theorem :P
the image of connexe is connexe :p
by continuous function
(I can give examples if you do not see what kind of questions I mean)
What to say about a linear application which leaves stable any hyperplan?
If a function is integrable, does it tend to 0? (lol) What assumptions are sufficient? (e.g. UC, or decrease)
Ugo
Ugo
At end want to find Emb_f(L,L) where $L=Q((-2)^(1/4),e^(i*pi/2))$ . I thnk that but might be wrong that [L,L]=1 so the only mapping is the identity. appreciate hints
@Lucas French spotted :P
why ? ^^
16:40
"connexe"
connected
lol
16:55
Example of a non-continuous linear function? non-compact unit ball?
Show that, in finite dimission, two diagonalisable endomorphisms are in a common basis.
What are the open-closed sets of a normed vector space?
What are the adhesion values ​​of sinus in R? Demonstrate the rank theorem. Find a non-constant sequence that does not admit a smaller period (:p).
> application
function :P
lol
For the last one, of course, the sequence is assumed to be periodic :p
find a dense subset of Q whose complement is also dense
decimals
17:01
decimal numbers, that does not exist in English ?
I don't get what you mean
ah ok
yes it was not clear
You can aslo answer some of my questions if you want :D
hi @BalarkaSen do you have any hints?
For my covering space exercise?
17:06
I don't know how to do it myself. I can see the covering space topologically but it's not clear to me why it should be holomorphic.
That's why I asked you
I can't even see it topologically
Ah, alright, that I can tell you
So the doubly punctured complex plane deformation retracts to a wedge of two circles
Consider this picture
Call that space $X$. There is a map $p : X \to S^1 \vee S^1$ to the wedge of two circles, such that $p$ wraps $b$ around the first circle and $a$ around the second
You can verify that this is a covering map.
The idea should be that $\Bbb C \setminus \{-1, 1\}$ is then covered by a "thickened" $X$. That thickening is in fact a disk
so complicated :o
Yeah it kind of is
But this by no means is a proof; I only constructed a topological covering map $D^2 \to \Bbb C \setminus \{-1, 1\}$ modulo all details. I want a holomorphic covering map.
I see the words "theta functions" flying around in places where it's done
Lapsus in my last question : of course it is not sequence but function
17:13
3 mins ago, by Balarka Sen
Call that space $X$. There is a map $p : X \to S^1 \vee S^1$ to the wedge of two circles, such that $p$ wraps $b$ around the first circle and $a$ around the second
mind = blown
@BalarkaSen I don't see how you can thicken X, because b must touch a...
b must touch a? You mean p(b) = p(a)?
Why would thickening break that?
(As an aside, the thickening should be done carefully; as you proceed away from the origin of the tree you must decrease the thickening factor exponentially)
I mean p(b) would touch p(a) non-trivially
Eh? p(b) = p(a).
oops
I meant the thickening
I see a way to avoid that would be to thicken the origin to something that touches the boundary
I am sorry, I don't at all see what you're saying
17:20
never mind
@BalarkaSen I would have never thought of that covering though
Would the preimage of e, a, b look something like this? @BalarkaSen
under the thickening, I mean
by "preimage" I mean the preimage of the deformation retraction D^2 --> X
Ah, yes, I was confused for a while.
That's right.
18:03
I am looking at the following:
Compute the function $\displaystyle{F_f}$ for $\displaystyle{e^{-a|x|}, x\in \mathbb{R}}$ and then $\displaystyle{\overline{F}(F_f)(0)}$.

Is $\displaystyle{F_f}$ the Fourier transformation of $f$ ? But what exactly is $\displaystyle{\overline{F}(F_f)(0)}$ ?
I have never see that notation for the Fourier transform, but maybe
What else could it be? And about the second notation? @KevinDriscoll
Honestly, I have no idea what else it would be. If it is the Fourier Transform, then bar F might be the inverse Fourier Transform
Ah ok! Thank you! :-)
19:02
@MaryStar It would be helpful to know where that exercise and notation came from
I've never seen $F_f$ for the Fourier transform, though I have seen $\mathscr{F}(f)$ (and maybe $\mathscr{F}_f$?), so $F_f$ doesn't seem unreasonable. $\bar{F}$, however, does not seem like natural notation for the inverse Fourier transform.
When looking at determining the stability at a critical point for a first order autonomous system, when can we be sure linearization will work?
Hello! Could somebody clarify Harish Chandra Rajpoot's last part of the answer where $=\lim_{x\to \infty}\frac{\ln2+\frac{1}{x}\ln(1+2^{-x})}{\ln3+\frac{1}{x}\ln(1+3^{-x})}$ transforms into $=\frac{\ln2+0}{\ln3+0}=\color{red}{\frac{\ln2}{\ln3}}$ on the post math.stackexchange.com/questions/1445294/… I don't quite get how 0's came about.
$\lim_{x\to\infty} \frac{\ln(1+2^{-x})}{x} = 0$
Roughly speaking, observe that $\ln(1 + 2^{-x})$ decreases to zero as $x \to \infty$ (since $2^{-x}$ decreases to 0, and $\ln(1) = 0$)
and $\frac{1}{x} \to 0$ as $x \to \infty$
Then use the fact that the limit of a product is the product of the limits (supposing that all of the limits involved exist)
oh... uh... ping @Tug'Tegin
@XanderHenderson Thank you for your explicit explanation!!!
19:46
Hey all, I've observed that if you look at squares of natural numbers, and define the sequence x(n) by taking the second digit of binary representations of those squares, we can then define a_1 and a_0 by counting consecutive 1s and 0s in sequence x and adding their number as terms of a_1 and a_0. Then, a_1(n)/a_1(n-1) converges to sqrt(2). Same for a_0(n)/a_0(n-1). Why is this true?
20:08
who can I calculate the integral $\int \frac{(2x+3)^2}{\sqrtx}$ without expanding the polynomial
20:20
Hi, I have a projection P: R3 -> R3 which projects orthogonal onto a plane U with basis ( (-1,0,1), (0,-1,1) ) and normal vector (1,1,1). Then the projection matrix to this basis would be ( (1,0,0)^T,(0,1,0)^T
(0,0,0)^T )
is this correct?
@Trey I'm not sure that there is an obvious way of evaluating the integral without expanding the binomial, but if you make the substitution $u = \sqrt{x}$, you might be able to save yourself a little time.
And it is only a binomial---it isn't that hard to deal with
Yeah I was trying to generalize it
in what way?
and for what purpose?
I don't have a specific purpose, math.stackexchange.com/questions/2570909/…
I'm having a hard time with integrals, actually
20:36
i've thought of some more constraints on the problem i posted here last night
19 hours ago, by heather
let's say i have some vector $\vec{x}$ and some finite set of matrices $A$ in $\mathbb{C}^n$. How can I calculate the average number of steps to get from $\vec{x}$ to any other vector in $\mathbb{C}^n$?
1. all matrices in $A$ are unitary and of finite dimensions.
2. you can also multiply the matrices, as in $A_1A_2\vec{x}$
3. the randomly picked vector $\vec{y}\in\mathbb{C}^n$ that I wish to get to would be a unit vector.
4. $\vec{y}$ can also be rewritten as $a_1\hat{i} + a_2 \hat{j} + a_3 \hat{k} + ...$; that is, it is a linear superposition of the basis states. Also, $|a_1|^2 + |a_2|^2 + |a_3|^2 + ... = 1$, and $a_1, a_2, a_3, ...$ can all be complex numbers.
Does that make the problem solvable?
is the euler integral taught in SingleVariable calculus/ calculus 2?
21:10
Anyone good at ODE's here atm?
21:31
@Trey I've never seen it in any course like that.
Is $f(y) = -y^2, y\in \mathbb{R}$ negative semidefinite or negative definite?
@XanderHenderson Could you please again have a look in this my problem: in the post $math.stackexchange.com/questions/1446831/…$ Rajpoot writes that $\lim_{t\to\infty}\frac{\ln(1+3^{-t})}{3^{-t}}=1$, isn't it 0 in fact because, as you can see, x approaches infinity so the fractional parts should be 0? It's 1 when x approaches 0.
21:48
0
Q: Need help setting up and solving dual problem

ALannisterI need to solve the quadratic programming problem $$ \text{minimize}\,\, \sum_{j=1}^{n}(x_{j})^{2} \\ \text{subject to}\,\,\, \sum_{j=1}^{n}x_{j}=1,\\ 0 \leq x_{j}\leq u_{j}, \, \, j=1,\cdots , n $$ I know that the first thing I need to do is form the Lagrangian. Now, for a problem in standar...

I meant when t approaches 0
22:09
So, now I think I have my Lagrangian set up okay. What I am having trouble with now is moving forward...
 
1 hour later…
23:37
We have that $X\sim U[0,2]$ and $Y=\min (X,1)$.

I want to find the distribution of $Y$.

We have that $F_Y(y)=P(Y\leq y)=P(\min (X,1)\leq y)=P([X\leq y] \text{ or } \ [1\leq y])=P(X\leq y)\cup P(1\leq y)$

Is everything correct so far?

Since $X\sim U[0,2]$ do we have that $P(X\leq y)=y$ ? What does it hold for $P(1\leq y)$ ?
23:51
1
Q: Showing that replacing functions with values preserves concavity

pafnutiSuppose I have the following function $H$: $$H=\frac{f(a)}{1+f(a)+f(b)}-g(x),$$ where $f, g:\mathbb{R_+}\rightarrow \mathbb{R_+}$ are strictly increasing, continuous functions. For some arbitrary set of $f$, $H$ is concave over $a$ (including $f=id$, where $id$ is the identity function). Denoti...

Hey guys, would appreciate some person with a better understanding of analysis and (maybe proofs) than me, to have a look at this? Thanks.

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