I mean, cubic forms can't have local minima, obviously, unless they're dead 0. So if you end up with a nontrivial cubic, it's a saddle point. When you get to quartic forms, I don't know any theory. @Thorgott
Let $\mathcal{A}$ be some complex algebra (i.e., algebra over the complex numbers), and let $a_1,...,a_n \in \mathcal{A}$ be pairwise commuting elements (i.e., $a_i a_j = a_j a_i$ for all $i,j=1,...,n$). Call a subalgebra $\mathcal{B}$ inversion closed if $x \in \mathcal{B}$ and $x$ is invertible...
@user193319 do it in two steps: first show that the subalgebra generated by commuting elements is commutative, then show that the smallest inversion-closed algebra containing a given commutative subalgebra is commutative
for the second step, you can describe the elements explicitly: if $R$ is a commutative subalgebra, then the "inversion-closure" of $R$ by $\{x^{-1} y \mid x,y \in R, \text{ $x$ is invertible}\}$ (which is just a localization, basically)
for the first step, all elements are polynomials in the generators
Note that if $x$ and $y$ commute and $x$ is invertible, then $xy=yx$ implies $yx^{-1}=x^{-1}y$ by multiplying with $x^{-1}$ from both sides
If I have an operator $S$ on a Banach space $X$ such that $X = R(S) \oplus F$ for some subspace $F \subseteq X$, and $\dim X/R(S) = \beta < \infty$, why does it follow that $\dim F = \beta$?
Rudin makes this claim in his proof of theorem 4.25 of his FA book, but I don't see why it's true.
Then we take the lattice $\mathrm{ro}(\Bbb P)$ of regular open sets of $\Bbb P$ with join the intersection and meet the interior of the closure of the union
(negation is the interior of the complement)
And the map $\Bbb P\to\mathrm{ro}(\Bbb P)\setminus\varnothing$, $p\mapsto\mathrm{int}(\mathrm{cl}(\{q\mid q\leq p\}))$ is a dense embedding of posets
Is it still dense if we take the boolean algebra of open sets instead?
@RyanUnger It is, but it is for an analysis course
The embedding construction I mentioned above is important when reformulating forcing in terms of boolean algebras and boolean valued models from the formulation in terms of posets
@RyanUnger I know the big examples of PDEs, I took an undergrad course in mathematical physics that had a semester on PDEs, but the approach was very classical since nobody had taken functional analysis at that point
Green's functions, maximum principles, stuff on harmonic functions and so on
Operator algebras is where cutting edge fundamental physics is happening. Look up Connes' Embedding Conjecture (CEC) in the theory of von Neumann Algebras and Tsireslon's problem about quantum correlation functions and entanglement. CEC is astonishingly equivalent to Tsirelson's problem.
The theory of von Neumann algebras (vNAs) is basically noncommutative measure, subfactor theory in vNAs is noncommutative galois theory, and $C^*$-algebras is noncommutative topology---really cool stuff!
vNAs also have connections with knot theory---through the Jones' polynomial.
Also, operator algebras turns out to be the right framework in which to do continuous logic--although I am not too familiar with the details of this.
Let $f:\mathbb R^2\mapsto\mathbb{R}$ and $(r,\varphi)$ polar coordinates defined using following relations: $r=\sqrt{x^2+y^2}$ $x=rcos(\varphi)$ $y=rsin(\varphi)$. If $h(r,\varphi)=f(rcos(\varphi),rsin(\varphi))$ then:
Generally , this makes sense to me
But I'm confused with the usage of arguments and functions here
Shouldn't we stick with $r$ and $\varphi$ as arguments only ? If we did, then what would partial derivative of f over x mean ?
Isn't x a function then ?
I would appreciate any help
partial derivative of f over x would mean with respect to x btw
Are equations on the picture mathematically correctly written would be my first question
@Kenkar $f$ starts out as a function of $(x,y)$. By composing with the given functions, you end up with $h(r,\varphi) = f(x(r,\varphi),y(r,\varphi))$. So one must differentiate $f$ with respect to $x$ and $y$, since those are the only variables $f$ recognizes.
This is the same as the standard chain rule in one variable, but people abuse it with Leibniz notation: If you start with $f(x)$ and $x=g(t)$, then you get $h(t) = f(g(t))$ and $h'(t) = f'(g(t))g'(t)$, or $\dfrac{dh}{dt} = \dfrac{df}{dx}(g(t))\cdot \dfrac{dg}{dt}$.
