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12:00 AM
he's gotta be fucking with me
 
I think this is not even close to the worst passage in that article
 
Look man I get how hard it is to make good notation, I've made plenty of bad and overloaded notstion
But cmon
 
@Jeff What you are saying is all wrong. If you want a rigorous argument for what I suggested, remember that $$\lim(f(x)-15) = \lim\left(\frac{f(x)-15}{x-4}\cdot (x-4)\right) = \lim\frac{f(x)-15}{x-4}\cdot\lim(x-4).$$
 
@MikeMiller ah yes, straightforward mochizuki style. but he probably got those 3m for a reason
I'm taking SPDE next semester and have not a single clue how anything he talks about is even slightly connected to that, after like 4 consecutive lectures on probability
 
12:18 AM
@TedShifrin Where do I go from here than? I guess I don't understand what you said last night.
 
@Jeff: Did you work out what I typed there? What did it tell you?
 
I see it. I'm not sure what it's telling me. We know that the left limit = 3 and the lim(x-4)=0
@TedShifrin
 
So what's the product of those limits?
 
zero
 
So what does that say $\lim (f(x)-15)$ is ?
 
12:20 AM
zero?
 
Why are you putting the ?
 
not sure. why. but lim(x-4)/lim(x-4) =1, so lim(f(x)-15) must be zero
 
that above screenshot unironically looks like something out of the random math paper generator
 
The first equality in what I typed used the fact that $\dfrac{x-4}{x-4}=1$ whenever $x\ne 4$, yes.
No limits there.
 
@Ted TY
 
12:23 AM
So, what I said last night was that you should "know" that if $a/b$ has a limit and $b\to 0$, then $a$ must go to $0$ as well. This algebra is the rigorous proof.
 
@user2103480 I imagine it's clear why he wants to do this to experts in SPDE who will understand the failed attempts at doing what he succeeds at doing
Unfortunately math papers tend to not be the best place to learn math
 
LOL, yes, that's true. Mostly journals try to shorten articles and therefore minimize exposition and explanation. But I fooled a few :P
 
 
4 hours later…
4:36 AM
@AlessandroCodenotti Hahaha lmao
I thought somehow you were doing C^* coalgebras
 
Similar to how one can write an integral in terms of a power series, is there a standard way to convert a power series to an integral?
Think of, say, the Fresnel integral, but backwards
 
4:57 AM
@xcodeking What do you mean by integral in terms of power series specifically? That is say if you're calculating $\int f dx$, and there is some open neighborhood around a point in the domain of $f$ such that $f = \sum_n a_nx^n$, then you have $\int f dx = \int \sum_n a_nx^n$
Writing this as a power series would mean you have to interchange the integral and the summation, so you have to have (I think) some bound on the partial sums and then you can interchange by DCT. Is this what you have in mind?
 
 
2 hours later…
7:06 AM
@TedShifrin Yay ! :D Well I've been quite busy for the last few years. Partly doing maths away from this chat ;)
In fact I'm more into computer science and electronics nowadays
What about you, anything new ?
 
 
2 hours later…
8:57 AM
@BalarkaSen oh god no
 
9:19 AM
If N = 0, then what does it mean to have a free rank of 0? I thought N = 0 means that it ISN'T free.
 
9:32 AM
zero module
 
It has no basis, so what does free here mean?
 
the basis is the empty set
"the empty set is a basis" is different from "it has no basis"
 
 
2 hours later…
11:11 AM
 
12:10 PM
 
12:26 PM
These are different definitions of limit at infinity, first from Hunter's intro real analysis book, second from Bartle's intro real analysis book, and third from baby Rudin.

Is there a standard definition of limit at infinity? Which definition should I adopt?
 
Second definition is a special case of the first definition and first definition is a special case of the third definition
They're all the standard definition, just on different levels of generality
 
12:46 PM
@Thorgott Does the third definition (from Baby Rudin) coincides with Definition 4.1 when A and x are real? (Rudin claimed that it does in his book).

(Definition 4.1) $\quad$ Let $X$ and $Y$ be metric space; suppose $E \subset X$, $f$ maps $E$ into $Y$, and $p$ is a limit point of $E$. We write $f(x)\to q$ as $x\to p$, or $\lim_{x\to p}f(x)=q$ if there is a point $q \in Y$ with the following property: For every $\epsilon > 0$ there exists a $\delta > 0$ such that $d_Y(f(x),q)<\epsilon$ for all points $x\in E$ for which $0<d_X(x,p)<\delta$.
 
I haven't even read the definitions here but I see no reason to doubt Rudin on this --- try proving the equivalence
 
Yes, it does
 
@Thorgott I thought it doesn't coincide with definition 4.1. I came up with an example, where did I go wrong?

Let E = {0,1}. f is defined on E with f(0) = f(1) = 2. Then according to definition 4.33, f(t) -> 2 as t -> 0, since for every neighbourhood of 2, I can take the deleted neighbourhood of 0 with radius 2 (which is {1}).
But 0 is not a limit point of E.
 
1:02 PM
@yh05 You need an actual neighborhood of the point according to that definition. The only neighborhoods of $0$ are $\{0\}$ and $E$.
 
ah
true, the definitions only coincide when $x$ is a limit point, because the other definition isn't concerned with that case
 
@skullpatrol I've found that gas often helps with social distancing
 
at second look, there seems to be an error in Rudins definition
$V$ should be a deleted neighborhood of $x$
 
So I need to add the assumption that x is a limit point to definition 4.33 when x is real?
@TobiasKildetoft I chose the neighborhood (-2,2) in my example.
 
@yh05 as which one?
 
1:13 PM
I mean, you don't have to assume $x$ is a limit point for the definition to make sense, though I don't think there's ever any reason to consider the other cases
But I think you do have to assume $V$ to be a deleted neighborhood of $x$ to get anything sensible
Otherwise limits at isolated points are non-unique by that definition and that's generally undesirable
 
Ahh, indeed the conclusion that the limit at 0 is 2 is correct
It is counter-intuitive, but it makes more sense if you think of extending the function to be equal to 2 everywhere but at 0
And then it also becomes clear that it should indeed be a deleted neighborhood
 
my bad for the blind defense of Rudin!
 
I think Rudin mean deleted neighborhood for the case that x is real in the definition. At the end of the definition it reads for all $t\in V\cap E, t\ne x$
 
Ahh, that does make it all fit together then
Rudin should just have ordered the quantifiers correctly and it would have been much clearer
 
Yeah, the important bit is that you want $V\cap E\setminus\{x\}$ to be non-empty and not just $V\cap E$ to be non-empty, cause if $V\cap E=\{x\}$, then the latter condition is vacuously satisfied and limits aren't unique anymore
 
1:20 PM
True
clearly we should write a replacement for Rudin, because I for one feel up to doing it better :)
 
My understanding so far, definitions 4.1 and 4.33 do not coincide when A and x are real. If I pre-suppose that x is a limit point of E, then the definitions coincide. But it is necessary for x to be a limit point of E since limit is not well defined if x is an isolation point of E. So in the end, definition 4.33 is fine?
 
Definition $4.33$ should be changed as to demand $V$ to be a deleted neighborhood of $x$. It might very well be the case that changing this definition does not affect the correctness of anything else in the book, but it is definitely the more mathematically sensible definition.
Everything is well-defined in any case, you just don't get uniqueness with the original definition 4.33 in case $x$ is isolated. It's not necessary for $x$ to be a limit point of $E$ for definition 4.33 to make sense, but I'm like 99% sure you will never need to consider a limit at a point that's not a limit point.
Perhaps the better way to put this is that definition 4.33 subsumes definition 4.1
 
anyone know how to make an image in the question body smaller?
 
I will change neighborhood V of x to deleted neighborhood V of x in my book. Thank you for your help.
 
1:40 PM
This is supposed to be something basic. But I can't figure out the mistake in the reasoning. Let $(X,\le)$ be a poset such that every chain has an upper bound.

Let $\Sigma$ be a (non-empty) chain of $X$ and consider the set of all its upper bounds. Let's denote it by $\mathsf{UpBound}(\Sigma)$. Consider any chain of $\mathsf{UpBound}(\Sigma)$. Since by hypothesis, every chain has an upper bound, this chain much have an upper bound as well, say $x$. Now we claim that $x$ is the maximal element of $X$. To show this observe that if $y\in X$ be such that $x<y$ then $y\in \mathsf{UpBound}(\Sigm
Question: Where do we implicitly use Axiom of Choice in the argument?
 
$x$ is not an upper bound of $\operatorname{UpBound}(\Sigma)$, it's an upper bound of some non-specified chain in $\operatorname{UpBound}(\Sigma)$
 
Oh man!
Thanks @Thorgott.
 
np
 
2:00 PM
I need a hint on how to rotate $z^2-y^2-x^2=1$ so that the forward sheet is in the first octant $(+,+,+)$
 
One question, which definition of limit at infinity (the ones I posted above) is more commonly used?
 
They all agree wherever they apply
an author will just choose whichever formulation/level of generality they feel appropriate for what they're trying to achieve
 
 
2 hours later…
3:55 PM
Show that for $N$ large enough, there exist $a,b \in \mathbb{N}$ such that $N<a^2+b^2<N+N^{2/3}$. Any hints on this one would be appreciated.
 
 
3 hours later…
6:30 PM
Stupid question, if $y$ is really a function of $x$, but we want to find $\frac{dx}{dy}$, is it sufficient to solve for $x$ and follow differentiation rules as though $x$ were a function and $y$ an independent variable, or do you have to do something in reverse when applying, i.e., the chain rule?
My understanding is that, at least when the relation between $x$ and $y$ is both right-unique and left-unique, the roles of dependent and independent variable are defined by what derivative is being taken, rather than being a property of the expression or equation itself, so that $\frac{dx}{dy}$ defines $y$ as the independent variable for that derivative, but I wanted to make sure this is correct.
 
 
1 hour later…
7:40 PM
@user10478 It's a theorem (Inverse Function Theorem) that if $\frac{dy}{dx}$ evaluated at the point $(x_0,y_0)$ is nonzero, then locally you can solve for $x=x(y)$ near that same point (even if you cannot do so explicitly), and then $\frac{dx}{dy}$ at that point is the reciprocal. So, yes, locally you can make $y$ the independent variable. There may be lots of different $x$ values for that $y$, but we're staying near the point $(x_0,y_0)$.
 
 
2 hours later…
9:10 PM
Is there a name for functions that can be recovered by taking their derivative repeatedly? sine and cosine functions for example
 
In mathematics, an analytic function is a function that is locally given by a convergent power series. There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others. Functions of each type are infinitely differentiable, but complex analytic functions exhibit properties that do not hold generally for real analytic functions. A function is analytic if and only if its Taylor series about x0 converges to the function in some neighborhood for every x0 in its domain. == Definitions == Formally, a function...
 
Ah I mean as in $f^{(n)}(x)=f(x)$ for instance
for some n
 
then I don't know if there's a name for that
 
I know no name.
If you fix any $n$, you can write down the power series for any such function.
 
They're linear combinations of $e^{\omega x}$ where $\omega$ is an $n$th root of unity
 
9:25 PM
aha, because it's a linear differential equation
 
9:59 PM
Hello
This looks new
Hmm ....
 

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