Hello, some fellow mathletes often say that geometry can be solved using a so-called "Jeslava" lemma. I'm very confused, but non of them can give a full explanation. So I want to ask you, does this lemma even exist? What is it about? Or is it just their joke?
not: no, you don't, but you (not the book) appeared to bring up the question of whether the book's hypothesis implies that the variables might be identically distributed, and it does
as copper noted there might be a missing or unstated independence assumption in (a) [the theorem that (a) is a 'corollary' of has this assumption]
@leslietownes I mean they didn't mention that x_1,x_2,...,x_n are same in antecedent and the conclusion doesn't seem to be right with that kind of antecedent
so should mention that
@leslietownes yes that also Made me confused
it it made me confused in thinking does any kind of random variables implies that conclusion of corollary (a)
youth: not that it means anything ('mathlete' math is almost its own genre of math with its own famous things) but i have never heard of it
independence could be worked into the book's definition of "random sample"
if it has one.
then again, a lot of people who write these kinds of books don't pay close attention to definitions, and it might be a mistake to pay more attention to it than the author intends you to
PDE question: I know the general solution to $u_t=xu_x+xu$ is $u(x,t)=e^{-x}f(xe^t)$ for arbitrary functions $f$, but what's the general method by which you would find this?
Okay, I've finally done it. Half-proved twin primes. Check it out here: https://math.stackexchange.com/questions/4643410/elementary-attack-on-the-twin-prime-conjecture-using-an-ideal-generated-by-i2
Extremely simple concept using an ideal
You need a value $x \in I$ such that $\gcd(x^2-1, p_1 \cdots p_n) = 1$. So we simply prove that that's always possible. But the proof still needs completion. I don't think the lemma is that hard seeming.
Hi! I understand that if I blow-up a 4-manifold at a point, topologically it corresponds to taking the connected-sum with -CP(2). But imagine I have an immersed surface with, say, only one singularity which is a transverse self-intersection (i.e. locally two transverse discs). Then I take the blow-up at that singular point. So the ambient manifold gets modified by adding -CP(2), and what about the surface itself? What does it become? You remove the two intersecting discs, and what do you replace them with?
This is wild conjecture on my end, but I think you want to take the strict transform? This should effectively correspond to looking at $D^2\times\{0\}\cup\{0\}\times D^2$ in $\mathbb{C}^2$, removing the origin, taking preimage in the affine blow-up and then taking the closure again, which leaves you with one $D^2$ lying over $[1\colon0]$ and one $D^2$ lying over $[0\colon 1]$ in $\mathbb{CP}^1$, resolving the intersection. Not sure if this is accurate, though.
Yes this seems to correspond with what I have in mind! Is it easy to see how the normal bundle changes? At least regarding a relation between the self-intersection numbers of the surface before and after the transform?
(I don't seem to fully grasp it...)
Thing is: I'm not at all familiar with algebraic geometry and complex geometry; whenever I see sheaves, varieties, divisors etc., I'm immediately lost by the jargon x')
Suppose that X is a topological space. Suppose that $f,g,h: I\to X$ are given paths such that their concatenation is defined. Then I want to prove that $([f][g])([h])=[f]([g][h]),$ where [.] denotes the path homotopy class of .
I drew an image here and using that I arrived at $F:I\times I\to X$, ...
They applied L Hospital rule, but then how did sinx/(ln(1+x)) come ? It was there in that former huge expression, I know, but then how did they directly write this from there?
@Koro What's the plug and chug ?!
How did it simplify to that? I have no clue abou this!🥲🥲🥲
Now using limit rules, forget about them. $\lim_{x\to 0} \cos x=1=\lim_{x\to 0}\frac 1{x+1}$ so forget about them too using the limit rules. Then, you have the third equality.
Why can we do it? To get the answer: Prove that if $\lim_{x\to a} g(x)=1$, then 1) the limit $\lim_{x\to a}f(x)$ exists if and only if $\lim_{x\to a} f(x) g(x)$ exists. 2) If $\lim_{x\to a}f(x)$ exists, then $\lim_{x\to a}f(x)=\lim_{x\to a}f(x)g(x)$
@Franklin So the point of that is to have an example where $\lim_{x\to\infty}\frac{f(x)}{g(x)}$ does not exist but $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$ does exist?
You do not need $\epsilon$-$\delta$ to reason that $e^{\sin x}$ has no limit as $x\to\infty$. It's clear that it wobbles back and forth between $e^1$ and $e^{-1}$.
@SineoftheTime So you trying to mean if $f(x)$ is convergent to a finite L' as x\to a and g(x) is not convergent at a, then $f(x)g(x)$ is not convergent at a?
@TedShifrin So the important fact applied in that image (I mean the problem mentioned) is :if $f(x)$ is convergent to a finite L'(\neq 0) as x\to a and g(x) is not convergent at a, then $f(x)g(x)$ is not convergent at a. Due to this, they could conclude that as e^{-\sin x} wobbles back and forth and the thing written in breckets converges to a finite limit (which is not equal to 0) so they concluded, that ratio to be non-convergent. Is my understanding correct?
And remember my standard complaint that people abuse L'Hôpital altogether. In particular, do not use L'Hôpital to compute the definition of the derivative, as that is totally circular reasoning.
@Anthony Perhaps I'm misunderstanding the situation, but the point of the blow-up is to remove the intersection, is it not? The two transversely intersecting disks turn into two disjoint disks in the blow-up. Intuitively, I think, blowing up means replacing the point with all its tangent lines and the fact the two disks intersect transversely in this point means that the blow-up replaces this with the two separate tangents, making the two disks disjoint.
@Anthony The term I think you're missing is proper transform. It's a little easier to warm up with a curve in $\Bbb A^2$ with a node at the origin. Try the nodal cubic $y^2=x^2(x+1)$. When you blow up the origin, you get two separate points on the proper transform of the curve.
While teaching differentiability for multivariate functions, we see derivative as a linear map. I want to teach it to my friend. He knows one variable derivative. I want to motivate why it's better to see it as a linear map.
Generally, in standard books they argue that normal definition doesn't make sense in $\mathbb R^n\to\mathbb R^m$. So they show that you can write normal definition in a way that you see $f'(x)h$ as a linear map. And then use this definition for multivariable case.
You don't need to start so fancy. The important step is to consider for what slope $m$ you will have the error estimate $f(a+h)-f(a)-mh = o(h)$, i.e., $\lim_{h\to 0} \frac{f(a+h)-f(a)-mh}h = 0$.
I always emphasized the best linear approximation to my Calc I students, regardless.
Then for multivariable it is reasonable to approximate $f(a+h,b+k)-f(a,b)$ by $mh + nk$ with a similar error requirement. No need to be fancy talking about linear maps just yet.
@TedShifrin Yes, also you get directional derivatives if you put h in some direction. Then you can see that it connects directional derivatives and derivatives
That's how I tried to explain.
@CottonHeadedNinnymuggins I explained it to him. But wanted to know how you people explain.
@TedShifrin I had a little query: Although while using L'Hospital's rule, I stay careful. I check just two things: say, while calculating $\lim_{x\to a}\frac{ f(x)}{g(x)}$ (1) The thing, I check first, is that whether $\lim _{x\to a}=\lim_{x\to a} =0$ or $\lim _{x\to a}=\lim_{x\to a} =\infty$ . If this is satisfied, I proceed to the 2nd step,
, (2) Whether both $f'(x)$ and $g'(x)$ is differentiable in a particular neighborhood of $a$ or not. Is this enough, to go and apply L'Hospital's rule after these two validations?
You're checking the hypotheses, yes, although your (2) sentence is messed up. But my point is that if you have something like $$\lim_{x\to 0}\frac{\tan x}x,$$ the hypotheses do hold, but this is the actual definition of the derivative, so using L'Hôpital is totally circular. Even things like $$\lim_{x\to 0}\frac{\sin 2x}{\sin x}$$ really shouldn't be applications of L'Hôpital.
@TedShifrin in one of your vector lectures you asked the students if they remember doing "proofs" where they'd draw a table and writing stuff in it, and I physically cringed when I heard that. I remember doing it in 6th grade while we learned geometry, I hated it
Its a miracle I love geometry despite that torture
@TedShifrin In 11th and 12th grade, we did quite a lot of euclidean geometry proofs in Japan. It was completely different to how we learned in the US, much more fun
Take a function like $f(x,y) =\begin{cases} \frac{x^2y}{x^4+y^2}, & (x,y)\ne (0,0) \\ 0, & (x,y)=(0,0)\end{cases}$. Of course, it's not even continuous, so certainly not differentiable. But all directional derivatives exist. The directional derivative in the direction $\vec v$ is $\frac{v_1^2}{v_2}$, certainly not linear.
The directional derivative is linear in $\vec v$ typically only if the function is differentiable. Of course, you can make up a function all of whose directional derivatives (at a point) are $0$ even though the function isn't even continuous.
@PNDas I don't know enough to translate between different notations. (I have to get all this stuff straightened out in my head to get ready for differential geometry...)
I don't know what I love. I never knew about my preference. I just cross the options which I can't do. For example, I am planning to do my PhD in PDE. Because I'm bad at everything else.
I'm not a mathematician, and this might be simplistic advice, but exposure to different types of math (and interviews with some mathematicians) made me realize that I like things that are visual and have some sense of movement and curvature to them. This leads me to topology, manifolds, differential geometry, physics, and stochastic processes
(I do almost everything myself, so I don't have guidance from anyone)
@TedShifrin Yes, I had a mini-revelation when I realized that group actions allow me to think of almost any group (at least, the ones I know of) as rotating, flipping, translating, or shuffling some object
That was Galois's idea ... introducing groups in the very first case as a group action by permutations. ... But I used to draw colored pictures for cosets and the quotient $G/H$ too.
@TedShifrin I mean, if you gave me a book on world history then yeah, I'd get distracted. Give me a textbook on, say, astronomy/astrophysics and I'd be busy reading all day
The attention span though is a problem. That's mostly because of the structure of modern internet and how its built for content that is bad at retaining attention to the point that longer form content is seen as strange
You aren't gonna get many people to sit through your multivariable calculus lecture video. But those same people will absolutely watch a 10-second video teaching a calculus "hack" for exams
I blame, everyone and everything because I'm too dumb to figure out what's actually causing this
@Seiya I don't disagree with your point. My course/videos were never intended for disinterested students just trying to learn the easiest tricks to pass exams.
And out of the remaining ones, some won't even understand the material properly. That leaves a very small number of students that actually make good use of that lecture
@TedShifrin at least here students are separated by their abilities. Academically capable students attend different classes to those that were either not very capable or just disinterested.
That's why there's not many students in our Calculus 3 classes
The Russian/soviet regime has been subject to or part of so many wars within the last century or so that I'm starting to think that whole continent is cursed
but i was very impressed at its ability to detect subtle emotions in text. it was able to tell "Get in already." suggests impatience, unlike "Welcome, I've been waiting for you."
that's insane
i am fully behind the chatgpt hipe train. succumb to despair as 20 years of education become obsolete due to a text collator
man could never recognize its place in the universe, always aggrandizing itself to immeasurable heights
> HAL's crisis was caused by a programming contradiction: he was constructed for "the accurate processing of information without distortion or concealment"
another bit of fiddle with this is that once you choose the domain, you don't get to choose the codomain arbitrarily, if you want the function to be given by a specific formula
and then you begin worrying about norms and see if it does so in a norm-preserving way
but those two questions may be intertwined somewhat
e..g if you have a function on X you can 'regard it' as a function on any subset Y of X by restricting it, and if Y is nice enough this might even give rise to a "restriction" map on function spaces like the L^2 spaces, but it won't preserve the norm in general unless the thing you're feeding into it is concentrated on Y in the first place
and "preserves norms of some functions but not others" is not going to give you an isometry of hilbert spaces
