So you look at $C^\infty$ maps from $[0,1]$ to $\Bbb R^2$ with the two conditions $f(0)=(0,0)$ and $f(1)=(1,1)$. Then there's an issue with reparametrizing the same set. But basically it's that infinite-dimensional Frechet space.
Oh, that restricts you to some subset, sorta yucky 'cuz of $\le$ and $\ge$ not meaning open.
It's most likely going to be infinite-dimensional, regardless. This is where the study of calculus of variations comes in. Trying to do max/min problems in infinite dimensions.
So, you basically want a foliation of the square (minus the two points), with the property that all leaves go through the two points. This comes from a $1$-form with the property that it vanishes at the two points and is non-zero elsewhere.
no, all the functions $P,Q$ with the properties I've specified will give you answers to your question. So you have a more handle-able parameter space: smooth functions $P(x,y)$, $Q(x,y)$ with the boundary conditions.
Let $u \in H^s(\mathbb{R}^n)$.Given that $s \in \mathbb{N}$ and $s>\frac{n}{2}$.
I need to prove that $$u\in L^{\infty}(\mathbb{R}^n)$$ and $$\|u\|_{L^{\infty}(\mathbb{R}^n)} \le C\|u\|_{H^s(\mathbb{R}^n)}$$ $C$ depending only on $s$ and $n$.
Appreciate your help in getting some hints on provi...
@PrashinJeevaganth I don't really know how to explain it, but if you know the meaning of the symbols, then you just have to find the logical statements. It asks for which is equivalent to $p$ if and only if $q$, so the only ones that work are I and II. They read: 1. (Not $p$ or $q$) and ($p$ or Not $q$) 2. (Not $p$ and Not $q$) or ($p$ and $q$). If you think of it like that, the logical equivalency becomes more obvious.
In particular, when $n$ is even you'll also have $-1$ as a solution. Hence when $n/2$ is even---as n=80000 is---you'll also have the possibility that 1-x=-x^2
This doesn't have a real solution, but it does have complex solutions
@JasperLoy Just wondering: What about the book makes it so difficult. I'm only on the first chapter, so of course, I haven't had trouble so far. (I'm asking out of genuine curiosity, not cockiness.)
@CaptainAmerica16 Well, many of the concepts may not make sense to you if you have not been exposed to some undergraduate math already, so you may think you understand the material when you really don't. But technically, other than informal logic, there are no formal prerequisites.
Some calculus: The function $x^n$, when $n$ is a positive integer, is an increasing function for any $0<x<1$, and as such $(1-x)^n$ is a decreasing function on $0<x<1$
@JasperLoy I might be thinking of the wrong movie then. If I get sick of Marvel, I watch DC. If I get sick of DC, I watch Marvel. I never get sick of either though, so that's basically all I watch. Well, that and Monk.
@JasperLoy He made some really good films. The Informant is probably the least well known, but it's also really good. I finished the film without realising it was him lol.
(I suspect that DC was perceived as pandering, where-as Marvel is perceived as earnest. Though, that is entirely speculation on my part, and colored by my own knowledge of the companies outside the movies themselves)
Also, yeah, I was a kid when the Harry Potter books were coming out, so I got into those. What Warner Bros did with the movies is part of why I distrust them as a company.
I agree. The original Ironman has a lot of poor plot points, but excelled in that was a very good origin story. As an extension, it could serve as an origin story for the entire Marvel universe.
I don't know if any of you have seen the movie "Good Will Hunting" but there is a particular mathematics problem in the movie that is of interest to be. One of the problems used in the movie is
"Draw all homeomorphically irreducible trees of size $n = 10$."
I've been able to draw all ten valid ...
That's crazy! Maths in movies is weird; there was a recent one where a genius kid calculates the Gaussian integral and then solves Birch and Swinnerton-Dyer conjecture next or something.
Proof, A Beautiful Mind, Pi, An Invisible Sign, Gifted... all have characters with a mental illness. I believe A Beautiful Mind is based on a true story, but the rest aren't.
@EricSilva I read that in the Turing film they had to throw extra doses of social anxiety and awkwardness at the character (whereas in reality Turing was a very social guy).
he also wasn’t mega into crosswords like the movie would have you believe
i mean it’s really i think a problem w just how biopics kinda are, you have to fictionalize a lot to get a good movie, peoples lives just aren’t good film plots, and it’s hard to strike the balance between honest and good filmmaking
especially w math nerds cause unfortunately the things that would excite us are a snoozefest to look at
I know that normed linear space is banach space iff iff every absolutely convergent series is convergent in E. So basically we know norm so we take summation norm of (x1,x2)_n and then summation (x1,x2) converges. where norm is given by defining the operation .
@LeakyNun I'm trying to power through my last calculus assignment. My teacher uses a million different representations for function composition. (fg)(x) just means f(g(x)), right?
A proper map between topological spaces is a continuous map such that the preimage of a compact set is compact. Does one mean quasi-compact, or quasi-compact Hausdorff?
isometry is distance preserving map or norm preserving map. X is real banach space so it has standard euclidian metric and for first we have define complexification norm.
@TedShifrin, @mercio: I'm sorry for taking up so much of your time and effort yesterday. I was very stressed and started asking questions without thinking first. (Referring to the part where I was to show the results for $E=\ell^p$ and $E=c_0$.) I'll try to refrain from doing that in the future, and if having a prolonged conversation preferably starting a separate chat room for that chat as not to take up the entire chat here. I'm however of course very thankful for your help.
Without twisting the surrounding environment a lot, a $\mathfrak{c}+1$ level zoom will really only give you that given number surrounded by a sea of infinitesimals
There cannot be any other real numbers nearby because any of them will be at least a real number apart from the number in the center of the window, which is not infinitestimal
So yes, you can do a continuum level zoom, its just not interesting until you mark out certain family of infinitesimals (which will follow more or less the same pattern due to the way the surreals are constructed)
Therefore, for any zoom level $m < \mathfrak{c}$ you will get something like the first picture
once you hit $\mathfrak{c}$, the number at the center is the only one you will see
Now to figure a projective geometry that can project the two infinitely far ends in the 3rd picture to points at infinity, so that the relation between the rationals and the irrationals can be shown by clearly
Let $u \in H^s(\mathbb{R}^n)$.Given that $s \in \mathbb{N}$ and $s>\frac{n}{2}$.
I need to prove that $$u\in L^{\infty}(\mathbb{R}^n)$$ and $$\|u\|_{L^{\infty}(\mathbb{R}^n)} \le C\|u\|_{H^s(\mathbb{R}^n)}$$ $C$ depending only on $s$ and $n$.
Appreciate your help in getting some hints on provi...
