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.?
@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 $L = \overline{K}$?
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.
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.
It's kind of clever why $F^p = F$ is exactly the right condition for every irreducible polynomial over $F$ to be separable, where $\mathrm{char}\,F = p$
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
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.
@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 $H^i(X,-)$)
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)?
@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 $f = g^{p^n}$
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 $E_i/F$
@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?
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?)
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 $\Bbb R$.
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.
@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)
@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.
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?
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
@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.
@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
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
@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.
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.
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
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.
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
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.
@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.)
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.
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)
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.
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?
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.?
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))
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
