yeah ok, then it's the full subcategory thereof with central image, meh
anyway, encoding properties such as unicity or associativity diagrammatically is a useful thing in general. it gets rid of element-dependence, so generalizes how we think of these properties to categories that aren't concrete. this is, in a way, the starting point of the theory of algebraic theories or universal algebra from a modern standpoint.
but if this is in a book whose context is solely linear algebra, it's a rather silly definition lol
@skullpatrol We miss you in "whatever quid" and in the Cafe and Tavern, which I linked. quid and I even started a round of echo in your honor, where I linked!
Yeah I wanted to learn English properly for research. So I don't do grammatical mistakes. when I looked at rules it wasn't very axiomatic. Some rules were bent in weird way.
@Karim, have you understood your misunderstanding about mixed partials versus covariant derivatives? A good exercise is that $d\omega(X,Y) = \nabla\omega(X,Y)-\nabla\omega(Y,X)$ when $\nabla$ is torsion-free ($\omega$ a $1$-form).
Let $\alpha:I\to\Bbb R^3$ be a curve parametrized by arc length $s$, with nonzero curvature and torsion. Consider the parametrized surface $$\textbf{x}(s,v) = \alpha(s)+vb(s),\ s\in I,-\epsilon<v<\epsilon,\epsilon>0$$
where $b$ is the binormal vector of $\alpha$. Prove that if $\epsilon$ is smal...
@TedShifrin $\alpha '' = kn$ where $k$ is curvature and $n$ is normal vector. $x_s = t-v\tau n$ $t$ is tangent vector, $tau$ is torsion. $x_v = b$ $b$ is binormal vector. So on $I\times{0}$, $\alpha$ is geodesic. I realize it's $I\times {0}$ not $I$ while writing.
When considering $\mathbb{Z}_p \times \mathbb{Z}_p$, where $p$ is a prime, {(0,1),(1,0)} is a generating set
Also {(0,1), (p-1,p-1)} is a generating set
Likewise there are many pairs of elements which can be regarded as generating sets.
When we consider $(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_q$, a Cayley graph can be generated by a pair of elements, say $s,t$, where $|s|=q, |t|=p$. I'm thinking of cases where $p,q>3$ and $p,q$ are distinct primes. Since for $\mathbb{Z}_p \times \mathbb{Z}_p$, many pairs of elements of order $p$ can be generating sets, likewise can we havve many possible candidates for $s$ and $t$?
For $\phi:\mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$, based on different $\phi$ different semidirect products will be generated. Then whenever it is possible to have a Cayley graph with respect to a generating set with $s,t$ as mentioned above, will $s,t$ be different for different demidirect products? Can we think about possible candidates for $s,t$?
E/F is radical extension if there is $F\subset F(\alpha_1)\subset F(\alpha_1,\alpha_2)\subset\cdots\subset F(\alpha_1,...,\alpha_n) = E$ such that $\alpha_1^{n_1}\in F$ and $\alpha_i^{n_i}\in F(\alpha_1,...,\alpha_{i-1})$ for $i\geq 2$
Suppose $K\subseteq L\subseteq M$ is a tower of fields.
If $M$ is a radical extension of $K$, is $L$ a radical extension of $K$?
My attempt (resulting in a paradox):
I don't know much about radical extensions other than the definition. I tried cooking up this counter-example:
$K=\mathbb{Q}...
@robjohn I am not stating their notation is poor, hence I said "to me it looks like..."/ "*IMO..."
everybody has their own specialities, maths is not what I do on a daily basis
I m trying to understand it and make sense of it. I therefor find it quite unpleasant to notice that "mathematicians" arrogantly shut off other people and tell them to go study a bit more
@traducerad a neighborhood is any open set containing the point of interest. so it would be any open interval containing the extremum you are looking at.
an open neighborhood has no defined size. It can be as small or large as needed.
I would not have expected to ever go down that far the rabbit hole when dealing with mathematical finance
especially optimization
funny IMO: some of the mathematical financial people perform portfolio optimization using lagrange multipliers. I was considering to look into gradient descent to optimize portfolio
but when reading up on these algorithms inner working a whole new world opened up for me
apparently gradient descent looks for local extrema, while lagrange multipliers looks for saddle points
meaning both algorithms are not interchangeable at all
@traducerad both methods find places where the local change vanishes at a point. both will give saddle points and both will give local extrema. I don't know why they would separate them like that.
@traducerad Lagrange multipliers do not only give saddle points. However, they can be used for much more complicated situations such as finding minimizing functions, where gradient descent would not be easy or perhaps possible.
@robjohn maybe, but I was considering to try the other way around. To try to see whether gradient descent instead of lagrange. But yhea it does indeed look like gradient descent is just not the best way to go here
@Snapdragon-X it can be applied in places where other methods can also be used, but trying to find a function that minimizes an integral, the best method is usually Euler-Lagrange
to then discover they don t even expain how they solve their optimization, which is what I was looking for. Ie another way to optimize a portfolio: one which is not based on lagrange multipliers, yet is understanable
@robjohn No, I had to enter the envelope function. r=asec(theta-a). You can see the link. I have supplied it too. desmos.com/calculator/enqcdnvy5c I assume given some assistance, desmos may compute envelope on its own. I will have to see the implementation in detail or perhaps you can try because you understand it much better than I do. Desmos is quite capable
Caution: Do not open the link on a phone. The phone may crash/not open the graph as expected.
in a specific situation (that afaik was the historical motivation for the terminology), there is a relationship between "pullback" in the sense of precomposition and "pullback" in the sense of limit, but it can't really be phrased in terms of the contravariant hom functor well
I believe this is the best you can get: if $f\colon X\rightarrow Y$ is a continuous map between top. spaces and $\pi\colon E\rightarrow X$ is a fiber bundle, these together form a pullback diagram. The pullback exists and is known as the pullback bundle $f^{\ast}E\rightarrow X$. Then, there is a natural map $\operatorname{Hom}(Y,E)\rightarrow\operatorname{Hom}(X,f^{\ast}E)$ ($\operatorname{Hom}$ is taken in the category of vector bundles, i.e. these are the global sections) that is induced by pulling back (in the sense of precomposition) sections along $f$.
I have some confusion. If $R$ is PID and $pR$ is maximal for a prime $p$. Let the $K = R/pR$, and let $M$ be a finitely generated p-primary R-module. Then $M/pM$ is a $R/pR$-module
After that definition, professor said a filteration (he just used this terminology without definition) $M\supset pM\supset p^2M\supset\cdots\supset p^nM = 0$ for some $n$
@love_sodam Assuming that by $M_p$ you mean the set you wrote up, then sure. Then we also need to use that it is finitely generated to see that there is a largest $n$ that works for all elements in the module.
if an $R$-module is annihalated by an ideal $I$, then isn't $R$ automatically an $R/I$ module? Won't that immediately apply to your problem @love_sodam , 'p primary' or 'finitely generated' aside?
in your case $M/pM$ is the quotient of $M$ by the submodule generated by the ideal $(p)$, which is certainly annihalated by $(p)$, and so is naturally an $R/(p)$ module
Fun fact of the day: If, in a category with finite products and coproducts, there exists a family of isomorphisms $X\times Y+X\times Z\rightarrow X\times(Y+Z)$ that is natural in $X,Y,Z$, then the category is distributive.
@porridgemathematics The step I can't follow is why $M$ being a generating set for $I$ that does not contain anything from $H$ implies that none of the $P_i$ could contain something from $H$
right, Here it is evening and I am chilling with a nice glass of red (the leftovers of the bottle my wife and I shared yesterday as a replacement for the Christmas lunch I was supposed to have been at but that got cancelled)
Suppose $K\subseteq L\subseteq M$ is a tower of fields.
If $M$ is a radical extension of $K$, is $L$ a radical extension of $K$?
My attempt (resulting in a paradox):
I don't know much about radical extensions other than the definition. I tried cooking up this counter-example:
$K=\mathbb{Q}...
