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12:48 AM
Hi
How to show C being a club if C is defined as a collection of limit ordinals $< \kappa$?? I know that definition of club means closed and unbounded so I have to use proof by cases to show that C is closed and C is unbounded. What I got for unbounded is that since $\kappa \in C$ so that implies sup($C \cap \kappa) = \kappa$ but that's just a one liner.
 
1:13 AM
isn't a club where you dance and stuff
 
No, it's what people play golf with.
 
2:12 AM
It's also one of the four suits ... so it's important for bridge, poker, cribbage, etc.
 
Hi there, I'm reading a book in control systems and the author states "The Routh–Hurwitz criterion is a necessary and sufficient criterion for the stability of linear systems. The method was originally developed in terms of determinants".
Could you please state any reference that discusses this issue with determinants. Thanks
 
psa
@TedShifrin I've got an explanation in a multivariable textbook that I don't really understand. Can I share it with you?
 
2:44 AM
Afternoon
 
2:56 AM
@ÍgjøgnumMeg Nacht
 
@Leaky alles klar?
 
im etwas
 
hä was?
 
@ÍgjøgnumMeg you don't know the expression?
 
Ich weiß nicht welchen Ausdruck du meinst
Meinst du "alles roger in Kambodscha?" oder wie?
 
3:05 AM
yeah along those lines lmao
 
3:31 AM
I have a more complex compound interest question involving partial sums. Suppose one opened a savings account with principal P at rate i compounded monthly, then also contributed an amount c every month. What would be the closed form to find the total value of the savings account after k terms?
In other words, (((P + Pi) + c) + [((P + Pi) + c)]*i) + c) + [(((P + Pi) + c) + [((P + Pi) + c)]*i) + c)]*i...
 
 
1 hour later…
4:53 AM
user image
4
 
What a cute dog
 
@AkivaWeinberger nice meme 10/10
 
 
1 hour later…
6:16 AM
So, about the Kabbalah maths...:
New Foundations actually has the exact same axioms
So New Foundations is Kabbalah
 
6:33 AM
@psa Sure. I will probably be around some Tuesday (but I have no idea where in the world you are).
 
@TedShifrin hi
 
psa
7:15 AM
@TedShifrin I'm in Canada (PST). Going to bed now.
 
8:02 AM
So does anyone have a good example of an existence proof for a bijection between two sets? I was just reading that the multiplicative property of the totient function can be proven by such a proof, but it only outlines the proof, as does not include the parts I cant work out myself
 
9:00 AM
@usukidoll $\kappa$ is not in $C$ is $C$ is defined as the limit ordinals strictly less than $\kappa$...
Rather for unboundedness let $\alpha<\kappa$. Can you contstruct a limit $\beta$ with $\alpha<\beta<\kappa$?
 
@AkivaWeinberger but I think that would be a good thing, at least some aspect of our species would survive, as the artificial intelligence that we made, but that's from the perspective of someone that believes the extinction of the human race is inevitable
 
9:48 AM
@AlessandroCodenotti I don't know a lot of number theory at all, but you could read a copy of David Burton's introduction to elementary number theory. Contains this proof in depth.
Oh sorry, wrong tag!
@Adam check out the book
 
 
2 hours later…
12:03 PM
@TedShifrin hi thanks you're answer regarding my question about the semidefinite cone. You wrote: "Assuming you mean symmetric when you say semidefinite, the vector space of symmetric $n\times n$ matrices has dimensions $n(n+1)/2$. The positive semidefinite cone is the closure of an open subset of this." Do mean, that semidefinite cone is a closed set if you write closure? Or does is mean something more specific?
 
 
2 hours later…
1:58 PM
Hi, I have a question: I have a paper that says the following: Let T be a stochastic matrix. It says that "Note also that stochastic matrices preserve the $L_1 norm$, i.e. for every vector u, $L_1(u.T)=L_1(u)$". What does it mean by saying "L_1 norm"?
I don't understand! Is it a function? since he states L_1(u.T) [which vector u and matrix T]
Note that I understand Linear Algebra, basics of abstract algebra, etc.
 
$L_1(u)$ is the sum of the absolute value of the coefficients
$u.T$ is just another row vector
 
@Astyx Thank you Astyx! It is really helpful
 
Glad to help
 
2:29 PM
Hi
Why is the "Note" after the first definition true?
^ That is from Milnor&Stasheff btw.
I mean what's stopping me from choosing a $\mu_y$ such that its image under $\rho_x\circ \rho_y^{-1}$ is $-\mu_x$?
The definition of local orientation itself doesn't prevent that right?
 
$M$ is a manifold ?
 
@Astyx yes.
 
What's $H_n(M, M-x, \Bbb Z)$ ?
 
relative homology of the pair $(M, M-x)$
with $\mathbb{Z}$ coefficients
 
2:44 PM
I guess you want it to be continuous in terms of $y$ if that makes any sense
 
3:03 PM
Meaning that if you make a choice for $x$ you have only one "sensible" choice for $y$
 
39 mins ago, by feynhat
I mean what's stopping me from choosing a $\mu_y$ such that its image under $\rho_x\circ \rho_y^{-1}$ is $-\mu_x$?
@Astyx But the definition itself doesn't prevent this from happening, does it?
 
Hey there, everyone! Does anyone know about Walton-Meek Gregory patch? I am having trouble to compute its normal and to solve its zero singularity. Does anyone know to correctly calculate and compute these? Thanks!
 
3:22 PM
@feynhat I think so yes, it's just not "compatible" with the choice you make for $x$
 
So, just to make sure, you're saying that the definition doesn't require $\mu_y$ to map to $\mu_x$ under that composition.
@Astyx
 
What the note means is that if you make a choice for $\mu_x$ then there is only one sensible choice for $\mu_y$
 
Ahh... I see. So, its saying that if we choose $\mu_x$ then this induces a $\mu_y$ for all $y$ in a neighborhood of $x$.
 
Yes
 
3:39 PM
hmm... I thought that it meant, if we have already chosen $\mu_x$ for all $x$, then, these $\mu_x$'s would satisfy that compatibility relation.
^ if this were true, then every manifold would be orientable, right?
 
Yes
Ah in some sense, every function $M\to \{-1,1\}$ would be continuous
 
Thanks.
 
@FelixCrazzolara Yes, that's right.
 
3:58 PM
@Ted do you know why people use A for the 2nd fund form
or why manfredo uses B
 
4:43 PM
@ÉricoMeloSilva Nope, no idea. B for bilinear form? :)
 
4:58 PM
hey
soo i got a question
 
that’s my guess for B
 
while solving homogeneous differential eqns, we assume that "y=vx"
 
but that kind of sucks cause there a billion bilinear forms you might work w lol
 
but even if the variables y and x are related again, it doesnt mean they would be related that way right?
 
yeah, I mean, you need notation. I just hate the EFG efg that Manfredo uses. Ridiculous when teaching ... I opted for lmn.
 
5:00 PM
so how can we assume that?
 
@MartianCactus: I don't understand what you're saying.
 
@TedShifrin that’s just historical though i guess
 
You're making a substitution $v=y/x$ and seeing what you can learn about $v$.
@Eric: I dunno. Who used those?
 
Ah, well, such a crummy source.
Pedagogically, it still sucks.
Or we could use $g_{ij}$ and $h_{ij}$ even for surfaces, as Chern did in his famous Berkeley notes.
 
5:02 PM
i was about to say that’s better
 
I don't find it better for middle to weaker students.
And using $x^1,x^2$ for coordinates is absurd when you're writing lots of equations with $x$ and $y$, especially polynomials.
Some things are better for abstract application than for concrete application.
So, learning lots?
 
im just tired from staying up reading things
a lot of papers to read not enough time
 
All papers, no problem sets?
hi DogAteMy
 
no problem sets but i’m still doing problems to prep for generals
 
5:10 PM
So, DogAteMy, turning into a linguistics major yet? :)
 
oh i get it it was just a misunderstanding sorry
i overlooked the fact that v is a variable
 
No problem, @Martian.
 
Evening all
 
ponders whether to speak to rude @ÍgjøgnumMeg
 
5:13 PM
@Ted Hey! You managed to type my name!
heheh
Reading more about modular forms, finally getting to some number theory :) Unfortunately I'm currently hanging on a lovely "obviously this is equivalent to" statement
 
Luckily for you, I know essentially nothing about modular forms.
 
I'd like to say I know something about modular forms..
 
I've coauthored a paper with modular forms in it, but ...
 
Hmm, I haven't seen @Mathein around for a while and he hasn't been on WhatsApp, maybe he's burying himself in his bachelor's dissertation
 
I hope he's OK. I haven't seen him a while, either.
 
5:19 PM
Yeah I hope so too, I usually see him around in uni but I haven't been in for a couple of days
 
So this makes sense, right?
$${\{a_j}\}_{j=1..n} \subset \mathbb N$$
$$n\,\varphi\Bigl(\sum^n_{j=1}a_j\Bigr) \leq \gcd(a_1,a_2,...,a_n)$$
 
@TedShifrin are you here?
 
No, my evil twin is here.
 
I have a stupid question
Let $A=\{a_1, \cdots, a_n\}$ be a basis of $H := \{ (x_0, \cdots, x_n) \mid \sum x_i = 0 \} \subset \Bbb R^{n+1}$
Let $\Lambda$ be the lattice in $H$ generated by $A$
I have two seemingly sensible ways of finding the covolume of $\Lambda$ that gives different answers
first way: the vector $\frac1{\sqrt{n+1}}(1,1,1,\cdots, 1) =: u$ is perpendicular to $A$ and has length $1$
so take the absolute of the determinant of $[a_1 ~ a_2 ~ \cdots ~ a_n ~ u]$
 
It does?
OK, it does.
 
5:29 PM
second way: decompose $\Bbb R^{n+1}$ into $H \times \Bbb R$ by sending $v$ to $(v - \frac1{n+1} \sum v_i, \sum v_i)$
 
What does the first component mean?
 
then there is a Haar measure on $H$, called $\mathrm d\mu$, such that $\text{Lebesgue measure on $\Bbb R^{n+1}$} = d\mu \times \text{Lebesgue measure on $\Bbb R$}$
$H$ was defined to be $\{ (x_0, \cdots, x_n) \mid \sum x_i = 0 \}$
btw the inverse map $H \times \Bbb R \to \Bbb R^{n+1}$ sends $(x,t)$ to $(x_i + \frac1{n+1} t)_i$
 
I don't understand how you subtract a scalar from a vector.
 
oh woops
 
The fancy Lebesgue measure discussion is still doing what you did in your first approach.
 
5:32 PM
send $v$ to $(\left(v_i - \frac1{n+1} \sum_j v_j\right)_i, \sum v_i)$
 
Huh?
 
so subtract the scalar pointwise
just to project it to $H$
 
Oh, I see what your notation means. OK. Yeah, multiply by $(1,\dots,1)^\top$.
Anyhow, if you do it correctly, it should be identical to the first.
 
the problem is that it isn't
 
You haven't done it correctly.
 
5:33 PM
no it really isn't
 
You have to scale the $\Bbb R$ factor carefully.
 
yeah so my question is why isn't this $\Bbb R$ factor scaled correctly
why doesn't this projection scale it correctly
 
Because you want an isometric isomorphism.
Well, with all the typos you've made, it can't tell if you're doing it correctly. You need to take a unit vector in the normal direction, or else it won't work.
 
how do I determine the measure on $H$ to make sure it's isometric?
 
It's just $\iota_{\vec n} dx_0\wedge\dots\wedge dx_n$, where $\vec n$ is the unit normal.
 
5:36 PM
yeah I think here $(0 \in H, 1 \in \Bbb R)$ is sent to $\frac1{n+1}(1,\cdots,1)$ which is not a unit vector
 
So this is $\frac1{\sqrt{n+1}} \sum (-1)^i dx_0\wedge\dots\wedge\widehat{dx_i}\wedge\dots\wedge dx_n$.
 
but algebraically my projection looks nice lol
 
so none of these symbols are to be interpreted as being of their conventional meaning here, right?
 
Ah I remember the good ol' days during my bachelor, when asking a question on main would immediately get someone subtly calling you a moron and answering your question
hehe
 
So your decomposition is not isometric.
 
5:38 PM
@TedShifrin I don't like this
it looks nice algebraically
lol
 
Looking nice is not sufficient for correctness?
Anyhow, I gave you the correct volume form for $H$.
 
thanks
 
You can see immediately that this is identical to your original determinant formulation.
 
aha
 
@ÍgjøgnumMeg: Are you suggesting I'm calling Leaky a moron?
 
5:39 PM
@Ted nah, I asked a question on the main site and usually I would get a very quick comment/answer hehe
 
So, Leaky, are you convinced?
 
yeah
but I'm required to deal with the two measures at the same time lol
poor me
 
Why so?
 
it's related to Dedekind's zeta function
 
Nice
Haar measure and some weird measure on Minkowski space?
 
5:42 PM
yeah
they used a non-isometric decomposition, as Ted would say
 
omg that is the creepiest cat sound from outside
 
@ÍgjøgnumMeg so the covolume of $|\mathcal O_K|^\times$ is supposed to be $\sqrt{r_1+r_2} R$ under the "correct" measure
 
Oh, I see what you mean, @Leaky.
 
now they decomposed it differently and the measure becomes $R$ instead
 
5:46 PM
poppy-cock
 
 
1 hour later…
6:54 PM
Why is it that one talks about "holomorphic functions" and not about "holomorphisms"?
 
I don't know. Probably comes down to someone just favoring that particular naming convention when they were building the literature, and it stuck
 
Looking for help with this question if anyone has an idea: math.stackexchange.com/questions/3442479/…
 
"Holomorphism" is easier to say, but it also might be too close to homomorphism in pronounciation, too
 
homeomorphism isn't that far away either
though it at least has a syllable more, I guess
 
What is the definition of the Lie algebra $\mathfrak{so}(4,\Bbb{C})$? Does it have a nice basis?
 
6:59 PM
does anybody know anything about the group generated by row and column swaps on a matrix?
is anything know about it?
 
Do you mean the group of permutation matrices?
 
Well, my instinct would be to look at the direct product of symmetric groups
 
I don't believe so
 
One symmetric group for the columns and one for the rows
 
oh you're right
yes thank you
 
7:02 PM
I don't know if that ends up overcounting in some cases, though
It also depends on if certain columns and rows are identical
 
@Rithaniel: But you're looking at it as acting on the space of all matrices.
 
a row swap always commutes with a column swap
so yes it's just a direct product
 
Fair enough, I do tend to overthink things at first glance
 
Right, @JackM. You can think of one in terms of multiplying by an elementary matrix on the left and the other in terms of multiplying by an elementary matrix on the right.
Yikes, it's crazy how people can post totally wrong answers and think they're correct.
2
 
Cool plot I made.
 
7:22 PM
@TedShifrin Anything in particular, Ted?
 
A youngster who wrote a detailed answer explaining how $f'(x^2)$ was the derivative of $f$ with respect to $x^2$. He removed it when I said it was totally wrong.
But makes me wonder how many just-plain-wrong answers are out there.
Ha ha. And someone downvoted my correct answer.
 
yeah, that's pretty bad
 
Oh, my correct answer had a typo. Rather than downvoting, it would be good if people just point out the typo.
 
Hello, I have a pde u_x(x,y)=0 with initial condition u(x,x^3)=h(x). I found the characteristics (t+C,D) with C,D constants. How do I continue from here to find u? Thanks
 
I usually do that and then downvote after a day when the OP ignores me or fights with me. :P
 
7:32 PM
(First time with pde's...)
 
And then a French mathematician comes along and writes $(f(x^2))'$, which I totally discourage students from doing, because it leads to all sorts of crap. Differentiate functions, not formulas.
 
Hi @TedShifrin, I am a newbie here, its my first time. I am a computer science undergrad and now pursuing an open university BS Mathematics. I am taking Vector Calculus this year, and I found your YouTube videos on multivariable calculus to be pretty terrific resource. Thanks for giving back to the community.
 
Oh, @Quasar, I'm glad they help. It was my students' idea, actually. I tried to get them to record my differential geometry lectures my last semester, too, but they were too tired.
@Eran: So the function depends only on $y$, right? The way they gave you the initial values is confusing, since they use $x$. Can you rewrite that in terms of $y$?
 
I would also find it strange to calling that an initial condition.
It's a condition, to be sure, but "initial" seems like the wrong adjective
 
Well, you fix the values along a curve (that hopefully is transverse to the family of characteristics).
It's not boundary conditions, because we're not on a bounded region.
 
7:36 PM
Ah, yeah.
So initial conditions for the characteristics
 
@Eran: Actually, are you sure you have the characteristics right? Shouldn't they be lines $x=\text{constant}$?
Let's review the definition.
 
I think so. dx/dt=1 --> x=t+C, and dy/dt=0 --> y=D. Is this not correct?
 
characteristics seem unnecessary here, though that's not to say they don't work
 
Where did $dx/dt = 1$ come from?
Yeah, the solution I was suggesting pays no attention to characteristics.
 
u_x=0
 
7:39 PM
But it should be the same :P
 
u_x=0 => au_x+bu_y=c with a=1,b=0,c=0
 
Ok I would love to hear this solution, because I have tried just with characteristics until now.
 
Now what, @Semiclassic?
 
@Semiclassical Thanks, how does it help me?
 
so characteristics gives dx/dt = 1, dy/dt=0, du/dt=0
 
7:41 PM
yes, right
 
So we have to figure out what the function is on a line where $y$ is constant.
 
yeah. note that this gives the characteristics as u,y=const
 
This amounts to what I was saying, I guess. I said $u_x=0$ means $u(x,y)=g(y)$. What I said was, knowing $u(x,x^3)=h(x)$, can you figure out what $g(y)$ is?
 
yes, geometrically u is constant on all curves where y = const
 
Yes, that's correct.
So you know the function on the graph of $y=x^3$. That determines it everywhere, because you start at any point $(x,y)$ and go along the characteristic until it hits the curve $y=x^3$, and there it has the value that was given to you.
 
7:45 PM
a more immediate way to state the characteristics here is that the value of u can only depend on the value of y, so u(x,y)=g(y) for some function g
 
Which is immediate from $u_x=0$ ... right ... independent of $x$.
 
yep. hence why characteristics is overkill here
 
So, my hint, @Eran, was to rewrite $h(x,x^3)$ as $h(?,y)$.
Note that every real number is of the form $x^3$ for some $x$.
 
@TedShifrin So I go along until I hit the curve y=x^3, and there it has the value. But we said the value is constant for all y-axis, so this is the value I'm looking for in the solution of u?
 
that's not what we said..
 
7:48 PM
@TedShifrin Okay I'll try to do this with change of variables
 
it's constant as you move parallel to the x-axis.
 
right, yes.
 
I guess one way to ask this: Where does the line $y=y_0$ intersect the line $y=x^3$?
 
when y_0=x^3?
 
No.
oh. yes
bleh. my brain
so therefore x=y_0^(1/3)
 
7:52 PM
Which means the value of the function has to be what there?
 
right, yes
 
Hush, @Semiclassic :P
 
yeah, just wanted to make clear I was being dumb
 
@TedShifrin in the intersection?
 
to be clear, what point (x,y) is the intersection?
in terms of y_0
 
7:54 PM
(y_0^(1/3),y_0)
 
right. so the method of characteristics tells you that u(x,y_0) = u(y_0^(1/3),y_0)
so now we just need to know what $u(y_0^{1/3},y_0)$ is.
(and yes, we do have enough info for that)
 
right. Just last question about the method of characteristics, so you wouldn't solve it like that: dx/dt=1--> x=t+C, dy/dt=0-->y=D? You would draw/think about it geometrically?
 
generally, yeah.
that said, you can't always get away with that
sometimes the curves are just not so easy to visualize
 
Ok thanks! I'm with you for the next part :)
 
Proof Verification: If A is a commutative ring with identity, then existence of surjective homomorphism $A^m\to A^n$ means that $m\geq n$.
 
8:00 PM
If you wanted to be more algebraic about it, I'd probably proceed as such: characteristics tell you that $x=x_0+t$, $y=y_0$. That intersects the curve $y=x^3$ when $y_0 = (x_0+t)^3\implies t=y_0^{1/3}-x_0$, i.e. the intersection is at $(x,y)=(x_0+(y_0^{1/3}-x_0),y_0) = (y_0^{1/3},y_0)$
At which point the question is once again: $u(y_0^{1/3},y_0)=?$
 
Amazing. Thank you!
@Semiclassical I'm actually not sure how to solve this.
 
If $A^m\to A^n$ is surjective while $m<n$, take another homomorphism $A^n\to A^m$ by mapping the first $m$ coordinates identically to $A^m$. The composition of the two arrows gives a surjective endomorphism $A^m\to A^m$, which must be an isomorphism.
But clearly this composition cannot be an isomorphism or $A$ would be trivial.
 
Well, $(y_0^{1/3},y_0)$ is the intersection between two curves: the characteristic $y=y_0$ and the curve $y=x^3$. You know that $u$ is constant on the former curve. What do you know about $u$ on the latter curve?
 
Why must it be an isomorphism? I keep wanting to reduce to linear algebra, but there's no field of fractions.
 
That it equals h(x)?
on the curve y=x^3
 
8:04 PM
Right. So $u(x,x^3)=h(x)$. What does that tell you about $u(y_0^{1/3},y_0)$?
 
@TedShifrin Just to confirm are you commenting on my question?
 
Well, of course ;)
 
This is a standard fact which follows from the generalized version of Cayley-Hamilton’s theorem.
 
Another way to say this: If $(x,y)$ is on the curve $y=x^3$, then $u(x,y)$ is just equal to h(x-coordinate)
 
@Semiclassical u(u_0^{1/3},y_0)=h(y_0^{1/3},y_0)?
 
8:06 PM
For any commutative ring A with identity and an A-module M, given any endomorphism \phi: M\to M we can define M as an A[x]-module.
 
is h a function of two variables?
 
Oh no. right, wait I'll fix this.
 
The action is defined by $f(x)\cdot m=f(\phi)(m)$.
 
u(u_0^{1/3},y_0)=h(y_0^{1/3})?
 
Alllmost
take a look at the left-hand side of the equation
 
8:07 PM
Like you said, just equal to h(x-coordinate)
 
and look for the one letter which is wrong :P
 
Typo haha, u(y_0^{1/3},y_0)=h(y_0^{1/3})?
 
Sure, @William, I get that.
 
yeah. So the method of characteristics tells you that $u(x_0,y_0) = u(y_0^{1/3},y_0)=h(y_0^{1/3})$
and since this works regardless of the choice of $(x_0,y_0)$ we just have $u(x,y)=?$
 
Then apply nakayama’s lemma(a special case of Cayley-Hamilton) which states that $IM=M$ for some ideal $I\subset A[x]$, then for some i\in I, $i\cdot m=m$ for all $m\in M$.
 
8:09 PM
u(x,y)=y^{1/3}?
 
where'd h go?
 
Now just observe that with the action I defined previously surjectivity is just $xA[x]M=M$.
 
u(x,y)=h(y^{1/3})
 
there you go.
 
Set $I=xA[x]$ we then get an inverse.
 
8:10 PM
Yes! Thank you so much @Semiclassical
 
To present a more direct solution: $u_x(x,y)=0$, so $u(x,y)=g(y)$ for some function $g$. But $h(x)=u(x,x^3) = g(x^3)$, so $g(y)=h(y^{1/3})$.
 
OK, so Nakayama is the tool. I've forgotten all this stuff that I knew 40 years ago, @William, sorry.
 
So characteristics is really not necessary here. Method of characteristics is more appropriate when the characteristics you get are more involved.
As an example, suppose you had $u_x+u_y=0$. One way to do that is to make a smart change of variables.
 
No problem. I am also glad to type it down as a review for myself ;)
 
Another is characteristics, which gives $x=x_0+t,y=y_0+t$. That means that the function is unchanged along the line $x-y=x_0-y_0$
which really means that $u$ must be a function of $x-y$ alone, i.e. $u(x,y)=f(x-y)$. in other words, characteristics gives you a way of deducing the smart change of variables
 
8:16 PM
Okay, yes, thanks you so much! You really helped me a lot!!!
 
(and now imagine cases where the characteristics aren't just straight lines, lol)
glad to help
 
If you would like I have another question about pde that I thought I solved good but when I check it in the end it doesn't work (but also straight lines lol)
 
sure.
 
Thank you!!
 
It's lunchtime for this bozo. See y'all later.
 
8:21 PM
u_x+u_y=u^2. Inital condition: u(x,0)=h(x). I defined g(t)=u(x_0+t,t) and then differentiated and found g'(t)=g^2(t). I solved the ODE and received g(t)=-1/(t+C), and then returned to u with u(x,y)=-1/(y+C). I used the inital condition and got C=-1/h(x) --> u(x,y)=h(x)/(1-yh(x)). When I check this solution with differentiating and calculating u_x+u_y=u^2, I don't get the right answer.
If something is not explained good I'll explain again :)
 
so, to be clear, what method are you using here?
 
I think geometrically maybe the characteristics but not formally.
 
yeah, characteristics is what I'd have in mind as well
so what characteristics did you get? you won't have u being constant on the characteristics in this case.
 
dx/dt=1,dy/dt=1/du/dt=u^2
 
okay, i do seem to get another answer.
so what do you have for x(t), y(t)?
 
8:27 PM
x(t)=x_0+t;y(t)=t
 
So what you seem to have in mind is to start on the curve $y=0$ and then follow the characteristic to another point.
 
yes
 
I think the easier way is to start at a generic point (x0,y0) and figure out where it intersects y=0
So x(t) = x0+t, y(t) = y0+t
and u(t) = ?
 
u(t)=-1/(t+C)
 
right. if you evaluate at t=0, that's u(0) = -1/C
 
8:30 PM
yes.
 
so u(t) = -1/(t-1/u0) = u0/(1-t u0)
 
right, yes
 
So what characteristics tells you is that $u(x_0+t,y_0+t) = \dfrac{u_0}{1-t u_0}$ where $u_0=u(x_0,y_0)$
For what value of $t$ does $(x_0+t,y_0+t)$ lie on the curve $y=0$?
 
do you mean y_0+t=0? t=-y_0
 
well, y(t) = y_0+t=0. but yeah
So therefore $u(x_0-y_0,0)=\dfrac{u_0}{1+y_0 u_0}=\dfrac{u(x_0,y_0)}{1+y_0 u(x_0,y_0)}$
now we're on the line $y=0$, so what can we say about $u(x_0-y_0,0)$?
 
8:37 PM
Sorry I lost you a little bit, just a minute, reading it again.
How did you get this \dfrac{u_0}{1+y_0 u_0}?
 
characteristics tell you $u(x_0+t,y_0+t)=\dfrac{u_0}{1-t u_0}$ should be valid for all $t$
so in particular it should be true for $t=-y_0$
So plug that into both sides and see what you get.
 
Ok I'll do that, Thank you!
 
(just to confirm one point: it's not wrong to start from (x0,0) and see where you go from there. it's just not what sprang to mind for me.)
 
Is it true that $$\operatorname*{argmin}_{\lambda \in \mathbb{R}^n : p_j = \sum_{i \in [n]} \lambda_i p_i} \|\lambda\|_1 = \mathrm{e}_j$$ for any $j \in [n]$?
 
Thank you very very much ! :)
I also found my mistake when starting from (x0,0) - Thanks!
 
8:46 PM
cool
 
Do you want another question? :P
 
not right now
 
Okay thanks :)
 
That is, if we express a vector as an affine combination of itself and a bunch of other vectors, does the trivial combination (assigning 1 to that vector and 0 to the other vectors) minimize the size of the coefficients (in the $\ell_1$ sense)?
This is not true for the $\ell_2$ norm.
Forgot to add the constraint $\lambda \cdot \vec{1} = 1$.
 
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