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12:03 AM
@TedShifrin what's the question?
12:26 AM
How broadly applicable is an ODE being a sum of a state free solution and an input free solution? Is it only for Linear ODEs? Only for constant coefficients? Etc.?
1:00 AM
Definitely need it to be linear. Otherwise the whole idea of decomposing one solution into other simpler solutions fails
$(x,\phi(x)) \mapsto \big(\frac{1}{\phi(x)},\frac{1}{x}\big)$
Is there a name for this transformation?
compactification mapping?
@LeakyNun Makes sense, since ultrafilter many elements satisfy "there is a unique min/max" (And there's $\mathfrak c$ biinfinite ones then)
The wizard does not choose the wand, the wand chooses the wizard
@Semiclassical Why is it impossible even in principle to decompose a nonlinear solution in any way?
@AlessandroCodenotti wait I'm no longer sure whether there are only two
originally I thought (0,0,0,0,...) and (0,1,2,3,...) were the only degree-1 vertices
then I realized that (0,1,0,3,0,5,0,7,...) should also be a degree-1 vertex
or something like that
wait oh nvm
1:36 AM
@Ultradark is that from Harry Potter :P
It is indeed from Harry Potter. Who was the groundskeeper before Hagrid?
He doesn't say.
The groundskeeper before Hagrid was Ogg
briefly mentioned in the fourth book
I will allow you to perform an essential task for me, one that many of my followers would give their right hands to perform...
2:00 AM
> Many are called, few are chosen.
@LeakyNun Here you go.
5 hours later…
7:30 AM
@LeakyNun Can you tell me the solution of the thing you told me earlier; how to prove that if for all fields $L$, $F$-embeddings $K \hookrightarrow L$ have image $K$ implies $K$ is a splitting field over $F$ without using $L = \overline{K}$?
@BalarkaSen use the normal closure kek
Oh I see. I thought about that briefly but imagined you had some smaller thing in mind
it's like the smallest thing
Normal closure of $K$ is essentially the smallest subfield containing $\sigma(K)$ for all $\sigma \in Aut(\overline{F}/F)$, right?
7:33 AM
Got it. I kept thinking about some composite of $K$ with some splitting field over $F$, etc etc
Alright I'm happy with this
7:48 AM
OK, next exercise. Say $K_1, K_2$ are finite extensions of $F$ contained in some bigger field $K$. If both are splitting fields, then I have to prove $K_1 K_2$ is a splitting field. This is because if $K_1$ is splitting field of $f$, $K_2$ is splitting field of $g$, then the splitting field of $fg$ must be the smallest field where $f$ and $g$ both split - since $f$ splits it must contain $K_1$ and since $g$ splits it must contain $K_2$ which forces it to contain $K_1 K_2$.
In $K_1 K_2$, $fg$ does split, so we have found the splitting field.
Easy enough
Second part says $K_1 \cap K_2$ must also be a splitting field. This follows from our favorite circle of ideas, since if there's an irreducible polynomial in $F$ which has a root in $K_1 \cap K_2 \subseteq K_1, K_2$, it must split in both $K_1$ and $K_2$. Factorization in $K[x]$ is unique, so that's a splitting in $K_1 \cap K_2$.
Now I wonder if $K_1 K_2$ and $K_1 \cap K_2$ are both splitting fields, does that imply something about $K_1$ and $K_2$? What's an example where neither $K_1$ nor $K_2$ are splitting fields?
I can make one of them to not be a splitting field, $K_1 = \Bbb Q(2^{1/3})$, $K_2 = \Bbb Q(\omega)$, easy enough.
is it true that if $f_t'$ is bounded then $f_t$ is Cauchy?
yeah because $f_t$ is Lipschitz lol
8:08 AM
Take $K_1 = \Bbb Q(2^{1/3}, i)$ and $K_2 = \Bbb Q(2^{1/4}, \sqrt{3})$, maybe.
$K_1 K_2 = \Bbb Q(2^{1/12}, \zeta_{12})$, I think, because $\zeta_{12} = (\sqrt{3} + i)/2$
$K_1 \cap K_2$ I of course don't know but I am positively certain the numbers involved are linearly independent, and it is indeed $\Bbb Q$.
$[K_1 : \Bbb Q] = 6$, $[K_2 : \Bbb Q] = 8$, $[K_1 K_2 : \Bbb Q] = 48$, so $[K_1 \cap K_2 : \Bbb Q] = 1$
Nice man!
3 hours later…
10:50 AM
It's kind of clever why $F^p = F$ is exactly the right condition for every irreducible polynomial over $F$ to be separable, where $\mathrm{char}\,F = p$
11:07 AM
How does one interpret "$RsR$ consisting of only finitely many left $R$-cosets"?
bit confusing, does one just mean that $RsR$ decomposes into a finite union of left $R$-cosets?
I s'pose so
we're defining some weird algebras of double cosets of $\operatorname{GL}_2(\Bbb Q)^+$
because once we have that (the Hecke algebra) then we can start defining operators (Hecke operators)
Oh that stuff
I dunno anything about that
11:13 AM
me neither
but I think they define endomorphisms of the space of entire forms
and then you can move between levels or smth (something that I vaguely read like a month ago)
yeah I've literally just read that there's a ring homomorphism of the Hecke algebra to the endomorphism ring of the space of entire forms that sends a double coset to the endomorphism sending a form to its image under the action of the "weight $k$ Petersson slash operator at the Hecke operator at the double coset" rofl
11:33 AM
So it turns out you can actually do some kind of an invariant subsapce decomposition for bounded operators. Interesting. Then there must be some way to give a jordan canonical form for the operator. I can see some functional calculus being required.
Suppose we have a affine subspace v+S. So S is a subspace. Let b be a basis of S. Is a basis of v+S in the form v+b?
11:53 AM
@loch Is there a way to make sheaf cohomology with respect to a specific sheaf (which is made sense of by pushing it forward, or pulling it back) functorial on $S$-schemes? Meaning $H^i(-,\mathcal{F})$ (and not meaning $H^i(X,-)$)
12:06 PM
I also asked the big boys:
in Homotopy Theory, 2 mins ago, by Ted E
Does a morphism of schemes $f:X\to Y$ induce a group homomorphism on sheaf cohomology? Like $H^i(X,\mathcal{F})\to H^i(Y,f_*\mathcal{F})$ if $\mathcal{F}$ is a sheaf of groups on $X$, or like $H^i(Y,\mathcal{G})\to H^i(X,f^*\mathcal{G})$ if $\mathcal{G}$ is a sheaf of groups on $Y$? If so, is there a way to reframe this, such that $H^i(-,\mathcal{F})$ is a functor on $S$-schemes (fixing the issue where there is really a different sheaf for each $S$-scheme)?
12:28 PM
@LeakyNun In general if $F$ is a field of characteristic $p$ and $f \in F[x]$ is irreducible inseparable, then since $f' = 0$, $f$ has to be in the image of the Frobenius endomorphism on $F[x]$. By iteratively using this fact you'd end up with an irreducible separable $g \in F[x]$ such that $f = g^{p^n}$
12:39 PM
Yikes, I said that wrong. I meant $f(x) = g(x^{p^n})$ as polynomials in $F[x]$.
$f' = 0$ implies $f(x)$ is a polynomial in $x^p$
(This is NOT equivalent to saying it's $p$-th power of some polynomial, because $F \neq F^p$ in general - $F$ need not be perfect. Separability is trivial there)
OK, so say $K = F[x]/(f(x))$ is the inseparable extension defined by $f$, and $E = F[x]/(g(x))$ be the separable extension defined by $g$. If $\alpha$ is a root of $f$, $K = F(\alpha)$. Since $\alpha^{p^n}$ will define a root of $g$, $E = F(\alpha^{p^n})$, so $E$ embeds as a subfield of $K$
$K/E$ I bet is purely inseparable.
Yeah of course, $E$ is the image of the $p^n$-Frobenius endomorphism $F(\alpha) \to F(\alpha)$.
So for simple extensions $K/F$, we can decompose it into $K/E/F$, a purely inseparable and a separable part.
A general algebraic extension is a direct limit of simple extensions so this decomposition pushes though I think: If $K_1 \subset K_2 \subset \cdots$ and $E_1 \subset E_2 \subset \cdots$ are directed systems such that $K_i/E_i$ are purely inseparable, then $K/E$ is also purely inseparable, where $K$ and $E$ are the respective limits.
Almost by definition. The same is true of the separability of $E/F$ from those of $E_i/F$
1 hour later…
2:16 PM
Question: Has anyone ever considered a set of axioms where $\bigcup\limits_{i=1}^\infty\emptyset$ is axiomatically defined to be $\{\emptyset\}$?
1 hour later…
3:28 PM
Can one say that a number is nothing but a 1x1 matrix? Can one add and subtract zero lines and columns to a matrix as he wishes?
@MadSpaceMemer Well, what is meant by a number? Does it belong to a specific set? Does a $1\times1$ matrix belong to a specific set? Do you want the set these belong to to possess any specific structure? Is there an isomorphism in whatever sense is appropriate?
@BalarkaSen yeah that's where you get the decomposition from
@BalarkaSen Have you listened to 16[485] by Agrypnie? Did you like it?
let $k$ be from the reals for an example can then every $k \in \mathbb{R}$ be expressed as a matrix of 1x1. As for your other qustions. I am not really sure its just an idea I had since a victor is a $K^n$ Matrix then I thought (ah okay so a number could be a 1x1?)
(*When the divisor and the constant c are the same then is the middle \
column sequence nonperiodic?*)
Clear[t, n, k];
nn = 180;
divisor = 2;
c = divisor;
t[1, k_] := t[1, k] = If[k == nn, 1, 0];
t[n_, k_] :=
t[n, k] =
Mod[t[-1 + n, -1 + k] + t[-1 + n, k] +
c* t[-1 + n, -1 + k] t[-1 + n,
k] + (+1 + c* t[-1 + n, -1 + k]) (+1 + t[-1 + n, k]) t[-1 + n,
1 + k], divisor]
Table[t[n, nn], {n, 1, nn}]
ListLinePlot[Table[t[n, nn], {n, 1, nn}]]
ArrayPlot[Table[Table[t[n, k], {k, 1, 2*nn}], {n, 1, nn}]]
3:40 PM
Well here are some candidate answers:

Well, what is meant by a number?
It sounds like by number you mean an element of $\Bbb R$ for now. Note that $\Bbb R$ is also a (one-dimensional) $\Bbb R$-vector space.

Does a $1\times1$ matrix belong to a specific set?
Here we mean a $1\times1$ matrix with real coefficients, so $M_1(\Bbb R)$, which is an $\Bbb R$-algebra (hence a fortiori a $1$-dimensional $\Bbb R$-vector space)

Do you want the set these belong to to possess any specific structure?
Well they both have vector space structure over $\Bbb R$.
3:55 PM
That is too much fire power for my taste. I will just ignore this question till I have gathered more information about what you are saying. it is not that important anyway. thanks tho.
4:07 PM
Turkish Sign Language manual alphabet is much closer to the written version
Well that should make sense.
@AkivaWeinberger nice
(As compared to, say, ASL)
@MadSpaceMemer There was no fire power, I was just saying that the pseudo-proposition "Can one say that a number is nothing but a 1x1 matrix" isn't really precise enough to assign a true or false value to it. But after being more precise, it upgrades to a proposition, and you can assign to it a true or false value. (And I answered the question that you probably intend to ask in the affirmative)
(at the cost of being two-handed)
4:12 PM
My hands won't let me do P
@TedE Yes that I have understood surely. It is however this "upgrade" in the form of the questions you have placed was for my level of knowledge too much of a "fire power" since I did not understand/read/had most of those terms you have used.
I am just a newb Ted :) :D
Okay memer, carry on :-).
@ÍgjøgnumMeg Imagine scratching your middle finger with your second finger
@Akiva my middle finger just bends back on itself when I try to do that and my tendons ache hahaha
I guess I'll never be able to perform the alphabet for a deaf Turk :(
4:17 PM
youtube.com/watch?v=X_G-FBSf1UI This is actually incredible
that's really funny :d
@ÍgjøgnumMeg That is great hahaha
I accedantly played it too loud with my headsets now the people sitting behind are looking weird to me ( library )
As long as you're not in an elite UK university you won't be lynched for listening to it
4:24 PM
@ÍgjøgnumMeg cool
That's some interesting notation
my favourite youtube version of Debussy's arabesque uses a similar graphical score
it's great
4:47 PM
I wish there wouldn't be a mathematics chat but a chatroom per mathematics discipline... Algebra chat, probability theory & statistics chat, etc.
There are, they just aren't as active
Hi, could anyone tell me how to attack this problem from this pdf (math.berkeley.edu/~stankova/MathCircle/Multiplicative.pdf) $$\sum _{d|n} \mu (d) \phi (d)$$ I would have solved it, however, notice that it is not a Dirichlet convolution so I cannot really reduce it. Even if I replace $\mu$ or $\phi$ with their Dirichlet expressions, the other term would remain a multiplicand with a nested summation and I am out of ideas on how to process that. Could anyone help me please?
5:07 PM
At a first glance: I think you can just sum over primes dividing $n$ exactly once (and one). There's an easy expression for $\varphi(p)$ and probably some counting argument to get a closed for
$1 + \sum_{p || n} \mu(p)\varphi(p)$
that might be rubbish though
5:27 PM
hi @ÍgjøgnumMeg
Hi @Ultradark
Hi everybody
Hey @Rithaniel
hi @Rithaniel
hi @Rithaniel, @ÍgjøgnumMeg, @Ultra
5:37 PM
Yo @TedShifrin
hi @TedShifrin
Heya Ted
@MadSpaceMemer What if you want to multiply an $m\times n$ matrix by a scalar? You certainly can't multiply it by a $1\times 1$ matrix. So, depends on context.
heya, demonic @Alessandro
Henlo @Alessandro
5:38 PM
Hi @ÍgjøgnumMeg
So many greetings
Soon we'll have vacuous greetings.
Can we identify an ideal which is maximal with respect to containing a vacuous greeting?
The answer is clearly "That is a gibberish question, Rithaniel"
I wonder if this is Rithaniel's lurking friend.
5:43 PM
Nah, he just likes to read the chat over my shoulder when he should be listening to the professor
Well, you shouldn't be chatting during lectures, anyhow!
Shouldn't you be listening to?
Technically speaking, I shouldn't have the chat open during lecture, but I do still pay attention
Who listens in lectures? Pff
Uh huh. Sure. This reminds me of tantrums I used to throw in class sometimes.
No phones, laptops, etc.
5:44 PM
Can we invent a new functional operator?
Today is only for dysfunctional things.
If those are the rules, I'd respect them
If the professor doesn't mind, then I usually find myself looking up topics related to the lecture during the lecture
@TedShifrin I wanted to be an ass and make some kind of Latin plural for tantrum instead of "tantrums", but apparently the etymology is unknown!
Which is cool
One of my lecturers during my undergrad used to walk up to the person scrolling through their phone and just quickly run his finger up and down the screen rofl
I once made a student let me answer his phone when it rudely rang. That stopped that.
Possibly from Tamil!
Yeah! An unexpected twist
There's a youtube video where a teacher had a rule that he would answer a phone on speaker phone if it went off in class, so the students conspired to prank him and had a friend ring during class pretending to be a caller from planned parenthood
5:49 PM
Yeah, if a phone rings, that'd be a problem
The teacher went bright red rofl
If you think you might need to take a phone call, set it to silent and leave the room before answering if it does go off
I'm pretty much 100% laptop, though.
I need a new one, at that
If the lecturer has a set of notes, I print them and make notes in the margin as the lecturer lecturificates
Otherwise I frantically scribble down every single thing the lecturer says for fear of missing smth, which I think is bad technique lol
5:54 PM
I've actually found that I struggle to understand something if I focus too much on taking notes during lecture.
Like, my attention is all on making sure I transcribe things accurately, instead of making sure I understand what is being said.
Yeah I have the same problem, but if I don't make the notes then I have nothing to refer to
bit of a conundrum
@ÍgjøgnumMeg: Yes, copying everything verbatim is what most of us do, but it's not the best note-taking. Often you miss important verbal cues ... although I always did try to put important remarks on the board.
Indeed, I think I'm gonna need to develop my note taking a bit over the next 2 years lol
So, as time has gone on, I have slowly begun taking fewer and fewer notes and just listening to the speaker. I actually have gone a little too far in some cases. Like, certain formulas are impossible to memorize at a glance, and you need to copy them down for reference
I think studies have shown you learn/retain more when you take some notes. Also, recopying notes after class to make them complete/correct improves comprehension.
6:00 PM
Makes sense
Hey, that's probably a good idea
It's kind of like working problems, but without the headache
Ergh, I've been sitting staring at a problem for about an hour and I have no idea how to even start it lol
so I'm procrastinating here
I've been working on side projects all weekend.
It's indubitably number theory, so I won't be of any help, @ÍgjøgnumMeg.
uhh indirectly, but it's more algebra/linear algebra I think
6:03 PM
Well I'm trying to show that $(\operatorname{GL}_2(\Bbb Z), \operatorname{GL}_2(\Bbb R))$ isn't a Hecke pair, which just means showing that there are $\operatorname{GL}_2(\Bbb Z)$-double cosets that don't decompose into finitely many left cosets
but I don#t really know how to work with these objects fluently yet so it's a bit weird
So we're talking about row and column operations on real matrices over $\Bbb Z$.
Left action is row operations; right action is column operations.
Did I tell you something you already grokked?
No, I've just been staring at the problem tbh
6:09 PM
So the left coset of $A$ is all the matrices you obtain by (integer) row operations on $A$.
So you should play around with some simple examples.
Okay I will do so :) Thanks
The double coset is all the matrices you obtain by (integer) row and column operations.
hi all
Let $p: C \rightarrow X$ be a covering map. Is $\deg p$ only defined if $X$ is connected?
Long time no see, @JoeShmo.
yeah, hey Ted!
6:17 PM
Different components of $X$ might be covered different numbers of times.
great, thanks.
but all these paragraphs are talking about $\deg p$ without mentioning that $X$ is connected. So I just want to make sure that's implied.
What book?
um, Cappell lectures out of his head. So the accompanying texts I'm using are mostly Hatcher and Chicago usually has very good notes about everything
So what pages of Hatcher?
the ambiguity came from one of the homework questions, and a few paragraphs I saw online. Everyone repeatedly omits the fact that $X$ is connected so it looked suspicious.
6:22 PM
I'm suspicious about your claims. That's why I wanted to track it down.
Hi chat
so I had to miss class last week, but for example here are two questions straight out of the notes that someone sent me. Let $f_1: Y_1 \rightarrow X$, $f_2: Y_2 \rightarrow X$ be covering maps. Let $f = f_1 \cup f_2$. Prove $f$ is a covering map and give a forma for it's degree
Hi @Astyx
Salut, @Astyx.
no mention of $X$ connected. He probably said that in class, but it wasn't mentioned in the notes
6:29 PM
But did Sylvain declare that he's always assuming $X$ path-connected or connected when discussing covering spaces? A lot of people would make that agreement.
I believe Hatcher does that in the text.
he probably did. but i wasn't in class last week unfortunately so I'm going based off of someone else's notes. its broken telephone.
That convention might have even occurred earlier. I dunno.
probably. just making sure for my sanity
Hatcher on page 56 briefly discusses this towards the middle of the page
but not quite, he says that the "number" (cardinality of $p^{-1}(x)$) is constant if $X$ is connected
6:34 PM
See p. 63.
from which it follows of course that the degree is well defined..
In fact, Hatcher puzzlingly doesn't force covering maps to be surjective, so in the disconnected case, a whole component could not be covered at all.
@user10478 If the DE is linear and has solutions y1,y2 then every linear combination ay1+by2 is also a solution. (That's what it means to be linear.) If it's a nonlinear DE, then y1+y2 will not in general be a solution even if y1,y2 are. (I guess it's possible that -some- solutions of a nonlinear DE can be decomposed into others, but not arbitrary linear combinations.)
Howdy, @Semiclassic.
I wonder if there are examples of L with $L(y_1+y_2)=Ly_1=Ly_2=0$ but not $L(ay_1+by_2)=0$
6:40 PM
If you have different degrees of homogeneity in different terms, you'll mess up scalars.
i forget how n-soliton solutions work (e.g. in the KdV equation) but that might work
to give some examples
looks like no: one has an approximate decomposition of a two-soliton solution for large times (i.e. when the two peaks are far apart) but not an exact decomposition
I would still think there are -some- examples possible, but that they're by far the exception.
6:55 PM
Anyone knows anything interresting about supersymmetry ?
@Astyx in physics or in math?
I know a bit about the math.
Is is the maths of the physics ?
It's the mathematics behind supersymmetry, yes.
So superalgebras and supermanifolds.
7:03 PM
questions like this make me sorta wish we only taught definite integration: math.stackexchange.com/questions/3468520/…
That's not a good idea, @Semiclassic.
yeah, I know
But a lot of high school teachers misteach integrals for sure.
The cure is worse than the disease, but the disease is bloody tiresome
@anakhro So what are those ?
7:06 PM
A superalgebra is a Z/2Z graded algebra.
A supermanifold is a little bit more complicated to explain, but it's got a commutative and anticommutative aspect to it.
We never even got around to integrals in my high school
They don't teach integrals in high school here.
What age range is high-school ? 16 to 18 ?
High school is like 14 to 18, yeah.
Some places they roll middle school in with high school and you get 12-18
So, I'm wanting to model some surfaces in blender. I was thinking of trying to generate a seifert surface
7:12 PM
Automatically generate, or just model in blender?
Because it's easy enough to draw in many cases.
At first, just model. But maybe later write some code that will generate a seifert surface from a given knot
@Rithaniel: You're making a milkshake out of surfaces?
@TedShifrin it's a software for 3D modeling.
Uhm, if I can, then yes, a milkshake sounds awesome
Blender is a free, open source 3d modeling software, yeah
I wonder if it makes good smoothies, too.
7:15 PM
I've been working with it in my sparetime and I'm feeling confident enough with it to start learning how to model mathematical structures (for visualization)
Do you hook it up to a 3D-printer, @Rithaniel?
Dangit, smoothie sounds good, too. I suppose I should get some fluids
Nah, I just do modeling for virtual things
I bet there are some addons that allow you to 3d print stuff with it, though
I actually want to eventually develop a graphics engine that will allow for 3d graphics in a space with extreme positive or negative curvature
But that is a ways away
You can always export to a file format that is compatible with 3D printer software.
Well, TIL.
Bye :)
7:22 PM
Hi all! Hi @Ted! Today I have no math problem in mind ...
Hi, @Rudi. That's cool :P
Hi @Rudi
:-) Saves me from grossly loosing
Servus @ÍgjøgnumMeg
Ich drück die Heneraungn
für den 6.2.!
(TIL = Today I Learned)
@TedShifrin Still unsure if you stooped our last conv. because i
7:24 PM
@Rudi was heißt Heneraungn?! :D
I wrote too much bullshit @ÍgjøgnumMeg Chicken-eyes
Ich weiss ned so art Warzen am Fuß
Ahhh hühneraugen hahaha
Everyone cross your fingers for the 12th of December too
what happens then?
unless you're from the UK, in which case don't just cross your fingers, go vote Labour
(Sorry Ted, I forget that abbreviations can cause misunderstanding)
7:26 PM
Oh sure
FIN got a new prime minister today - holy shit I cant think about politics when I see her ...
Oh yeah?
(but I am not completely sure if I got the news right, and if she is already designated PM or if there is some more votes to go)
7:45 PM
Hi guys, i'm wondering if i have a finite division Ring D, can i have $a^{|D|} = a$ for every a in D?
8:17 PM
@AlekMurt isn't a finite division ring necessarily a field?
@Alesandro Ya, do we have that property in a field?
8:28 PM
In finite fields of prime order, that's Fermat's Little Theorem. In other finite fields, hits holds by construction.
8:40 PM
Nvm, that would be kind of backwards. You can just show directly that the automorphism group of a finite field with $p^r$ elements, $p$ prime, is a cyclic group of order $r$ generated by the $p$-Frobenius.
Rule 30
Spreadheet formula in Microsoft Office Excel.
Hello I am new to network stats and I would like to compare network graphs based on their densities. I normalized the densities using min max. The group size differ significantly (eg. range between 2 and 800). Is this comparison meaningful What would you suggest?
I just simplified Rule 30 to the Excel formula above.
*Spreadsheet formula...
9:06 PM
Hello, assuming I have a pde, I solved it with the characteristics method and received two solutions. (plus/minus the same solution). I'm worried it's an error because of the existence and unique theorem.?
You can run into issues of characteristics crossing or diverging
in which case uniqueness and existence breaks down
What do you mean breaks down?
I have the equation u_y=xuu_x with initial condition u(x,0)=x. I found the characteristics are x(s,t)=+-se^(integral(z(t)dt);y(s,t)=-t,z(s,t)=s. So the solution is u(x,y)=+-x/(e^(integral(z(-y)dy))
well, for instance, there's the characteristics for Burger's equation (the inviscid one)
if they cross, you get a shock wave and therefore there's no classical solution
yeah, that's a nonlinear PDE
and nonlinear PDEs aren't subject to the same existence/uniqueness theorems as linear PDEs
Thanks!! So is it ok I received 2 solutions (plus/minus) ?
actually, quasilinear is the proper term in this case I think. c.f. math.stackexchange.com/questions/2425413/…
well, let's check it first. what are your characteristic equations?
9:19 PM
Thank you!
I mean, what ODEs do you have
yeah, that looks right
Yay :)
And what initial conditions do you have?
9:21 PM
for those ODEs
yeah. so y(t)=-t and z(t)=s check out. that leaves x'(t)=s x(t)
But I think your answer for x(t) should be simpler
You are right, if I substitute z(t)=s it will make it simpler
Ok. Let me know when you've gotten z(t) as simple as possibly
err, x(t)
9:27 PM
integrate both sides:
We know x(0)=s, then:
that gives x(0) = e^s not s
right, fixing it.
integrate both sides:
We know x(0)=s, then:
That's...the same answer as before.
Right. How can I simplify it more?
I think you have a typo now
should just be pm s e^(s t)
Also, though, if x(t)=+- s e^(s t), then x(0) = +- s
so only +s makes sense here
9:31 PM
Great! So here is the one and only solution and not 2..
So you've got x(t)=s e^(st), y(t)=-t, z(t)=s
Yes, Thank you!
Which may seem all well and good. But there's some fearsome complications under the surface here.
Typically, you'd want to solve for s in terms of x at this point
Are you restricting to x>0 here?
Not restricted, but I'm asked to find an implicit solution.
gotcha. Work on that, then.
9:36 PM
It wouldn't work to solve s in terms of x? I'll try and find what's not working.
Well, try plotting x=s e^(s t) when t=1 (including s<0)
and see what problem arises when trying to solve for x :)
Yes it's a problem...
That said, I'm not totally certain it's actually an issue. I think the actual issue is whether you only want y>0
hmm, maybe not
I think I want all x,y (it doesn't say anything else)?
then life is going to get complicated
Try taking the case of s=1 and plotting (x(t),y(t))
9:39 PM
oh, I'm being silly. I did mean plural: s=1 and s=2
making sure to plot both t>0 and t<0
I'm actually not sure how to plot and why is this needed/helpful for the implicit solution?
because those are the characteristics which pass through the points (1,0) and (2,0)
this isn't so important for the implicit solution as such, but it makes clear what technical issues lie beneath
Oh okay!
I'll try
For s=1 It's a straight line.
For s=2 also.
(e^t,-t) is certainly not a straight line
9:49 PM
neither is (2e^(2t),-t)
My bad.
Thank you so much for all your help :)
Unfortunately I need to go but THANK YOU VERY VERY MUCH :)
10:44 PM
Hello guys!! I need help with graph theory
I need to find (the simplests if possible) a graph which has euler path but not hamiltonian path, and one that has hamiltonian path but not euler path
For Euler and not Hamilton I tried this:
The only path of this graph is $P=(v_1,a_1,v_1)$, so since it repeats vertex but it does not repeat edges we found a graph which has euler path but not hamiltonian path
1) Is that correct? 2) How can we find a graph with hamiltonian path but not euler path?
I think with that graph I found a euler cycle not a euler path, but I don't know
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