Given a sequence of continuous functions $f_n$ such that $f_n \to f$ uniformly in a set $E$, iwe have $\lim_{n\to \infty} f_n(x_n)=f(x)$ for each sequence $x_n \to x$ with $x \in E$. I proved this using the continuity of $f_n$, namely estimating $|f_n(x_n)-f(x)| \le |f_n (x_n)-f_N(x)|+|f_N(x)-f(x)|<\epsilon$ where $N$ is the maximum integer such that $f_n \to f$ uniformly and $|x_n-x|<\delta/2$ where $\delta$ is the $\delta$ of the definition of continuity of $f_n$ corresponding to the index $N$
However, the solution uses that in these hypotheses $f$ is continuous, and then uses the estimation $|f_n(x_n)-f(x)| \le |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<\epsilon$. Is my work valid as well or it doesn't work?
I am using that $f_n \to f$ uniformly, hence there exists $n_1 \in \mathbb{N}$ such that $n \ge n_1$ implies $|f_n(x)-f(x)|<\epsilon/2$. Since $x_n \to x$, there exists $n_2 \in \mathbb{N}$ such that $n \ge n_2$ implies $|x_n-x|<\delta/2$. Since $f_n$ is continuous, there $|x_n-x|<\delta/2<\delta$ implies $|f_n(x_n)-f_n(x)|<\epsilon/2$. If $N=\max\{n_1,n_2\}$, $n \ge N $ implies $|f_n(x_n)-f(x)| \le |f_n (x_n)-f_N(x)|+|f_N(x)-f(x)|<\epsilon$.
Is (one of) the issue(s) the $\delta$ depending on $n$?
@TedShifrin I thought that, since $|f_n(x_n)-f_n(x)|<\epsilon/2$ holds for each $n \ge N$, it would hold for $n=N$ in the term $f_n(x)$ as well. But now I am not so sure about this anymore.
Ok, thanks for the suggestion! One last thing about the order: when we say that $f_n$ is continuous at $x_0$, we mean that for each $\epsilon>0$ and for each $n$ there exists $\delta=\delta(\epsilon,n)$ such that for each $x \in E$, $|x-x_0|<\delta \implies |f_n(x)-f_n(x_0)|<\epsilon$?
Uhm, then I have no idea how to define the continuity of $f_n(x)$ at $x=x_0$ rigorously...
You said that the order above made no sense because, when I claim "$n \ge N$ implies ..." I have that my $\delta$ varies due to the fact that I have $\delta(\epsilon,N)$, $\delta(\epsilon,N+1)$, etc.; so, varying the $\delta$, I have no control over the $|x_n-x|<\delta/2$ anymore?
That is, to each different $n$ I have a different $x_n$ and a different $\delta$ and so I am not sure if the inequality $|x_n-x|<\delta/2$ is always true and so I can't deduce the implication $|x_n-x|<\delta/2 \implies |f_n(x_n)-f_n(x)|<\epsilon/2$?
Sorry for the barrage of questions, but as you can see I am clearly confused :(
that example problem is a substantial chunk of what you're supposed to get out of a real analysis class. people being loosey goosey about that stuff as late as 200 years ago is why real analysis exists. definitely not time wasted.
ted: you holding up? we did not see ducks today at the pool.
I hope you get over your covid soon, Ted & leslie (and family). I managed to avoid it by being a hermit, and wearing masks when I had to go out. I currently have a sniffle & sore throat, but I think that's just from a dust allergy...
Life's mostly back to normal here in Sydney, although it's still common to see a few people with masks in shops & on public transport. I live in an area with a lot of Asian students, so it was common to see a few people wearing masks in the streets even before 2019.
Most people over here were pretty good with masking up. But unfortunately we decided a bit prematurely that the worst of it was over & we didn't need masks any more. Oops.
@TedShifrin I guess there are some similarities, but there are big differences too. They both (kind of) build a map of the internet, but GPT's map is highly compressed. However, I only have a rough idea of how GPT works, and I know even less of the details of how Google works.
Ok. Yes, the page rank algorithm has some things in common with what GPT does. Except that GPT doesn't care much about pages, it just looks at sequences of words.
However, the Google page rank algorithm is continuously evolving, and they have to keep a lot of the details secret to make it hard for people to game the system.
GPT breaks text up into tokens, which are mostly words, but also parts of words (prefixes, suffixes, root words), numerals, punctuation, and raw bytes (so they can handle Unicode). There are 50400 tokens. Each token is attached to a vector in a space with a lot of dimensions. The idea is that tokens which are likely to occur together in the training data are likely to be neighbours in the vector space.
GPT-2 uses vectors of length 768. It's open source, and small enough to run on a home computer. GPT-3 uses vectors of length >11000. They won't tell us what GPT-3.5 or GPT-4 use. I suspect that it's >768 but <11000
Well, first they train it by feeding it a huge load of text. The training process basically arranges the token vectors into appropriate locations. There's rubbish in the text, but there's also lot of ok stuff, so average good stuff tends to dominate over bad outliers.
Once the initial training is done, then they feed it lots of examples of prompts & suitable responses.
And then there's a phase where it interacts with human trainers to try and get rid of undesirable responses. But I'm jumping ahead of the story...
Once we have our big ball of tokens, created by the initial training, we could generate text by a simple Markov process. Even a process using chains of length 3 produces text that vaguely resembles sane English. However, just to get all the probabilities for chains of length 3 requires 50400^3 tokens, and we don't have that much training data. So we have to get creative...
That's where the Transformer neural networks come in. I'm still not quite clear on how they function, and nobody quite understands why they are as effective as they are. They learn the Markov probabilities in the sequences in the training data. And they can extrapolate from the training data to produce usable probabilities for sequences that they haven't seen.
Not really. But ChatGPT has read most of internet, including all of Wikipedia, Stack Overflow, and GitHub. But it's only looking at token sequence patterns. So its grasp on actual concepts is very tenuous.
When it makes up Markov probabilities it's just doing a simple weighted mean to interpolate between known probabilities. But with enough training you can get weights that work well enough. Mostly. ;)
Stephen Wolfram has written a pretty good article, with nice diagrams: What Is ChatGPT Doing … and Why Does It Work?. And he mostly avoids reminding you of how awesome Stephen Wolfram is. ;)
> If you’d asked me 20 years ago, “how will Wolfram react to one of the great AI developments of the 21st century?” — I would’ve said, “presumably, by figuring out some way to make it about him.” He doesn’t disappoint! 😀
Here's some stuff I posted in the Physics chat a few weeks ago.
Here's a crude analogy. ChatGPT is like a person doing a jigsaw puzzle. You give it a prompt of a few joined pieces and it adds more pieces, one by one. The picture on the jigsaw consists of a bunch of words written in English, and the jigsaw pieces are cut so that when you join them together the result will be syntactically correct English.
However, the guy doing the puzzle doesn't understand English. When we see the result, it looks impressive because we understand English, but ChatGPT just knows how to stick bits of jigsaw together.
ChatGPT is expert at manipulating language patterns, but it doesn't "know" anything about concepts or real world objects, actions, etc, or how words connect to that stuff.
Except that it's not a 2D jigsaw puzzle, it's got a lot more dimensions.
ChatGPT doesn't attempt to say true or logical statements, and doesn't have any mechanism to evaluate the truth or logical validity of its utterances. All it knows is if those utterances are consistent with the probabilities it learned in training. I said more about this stuff here: meta.stackoverflow.com/a/422397/4014959
Can anyone please explain how did they evaluate $J_0(x)^2$ ? I dont seem to have a clue, on squaring infinite sums
Is this is a known fact ? I mean here, $J_0(x)$ is a Bessel function of the first kind and zero order and is $J_0(x)^2$ 's representation in an infinite series, a standard result , which is used directly ?
you can multiply convergent power series like polynomials and compute their product that way. the coefficient of x^k in the product is determined by only finitely many coefficients of the factors.
Can anyone please enlighten me, about the terminologies : "Domain of the xy plane" and "Closed domain of the xy plane"
The general idea, I posses about domains of 2 variable functions say, D is that D is a subset of $\Bbb R\times\Bbb R$
But while stating about any function of n variable, I am familiar with the notion of domain. But I dont know what's the difference between domain of a function and a closed domain. This looks a lot uncanny in my opinion...
I was studying about differential equation. There, I came accross $2$ variable functions. It happened so, that this is the first time, I am working with them explicitly and rigorously. Up until now, I worked only with one-variable functions. Also, all the terminologies related to one-variable fun...
Let K/k be a finite seprable extension. Let L/k be subfield of K. Show that K/L is separable, and L/k is separable.
How to prove this?
I can show that L/k is separable.
But I’m not sure why K/L is separable.
I can show that K/L is separable if char K=0.
But I’m not sure how to handle the case when char K=p.
Claim 1: L/k is separable. Take any a in L. Look at its minimal polynomial p(x) in k[x]. Since L sits inside K. a is inside K. Hence By sep. of K/k, we get that p(x) is separable.
For K/L, take any a in K. Let q(x) be its min poly. Suppose q is not sep, then derivative q’(x) must be 0. That is, if char K=0, we must have q(x)= constant = 0 because q(a)=0. This contradicts irreducibility of q(x).
Hence if chat K=0, K/L is sep.
So we now have to consider the case chat K=p
Suppose on the contrary that q’(x)=0. It follows that q(x)= g(x^p) for some g(x) in L[x].
Let $L/M/K$ be a tower of fields. The proof that $L/K$ is separable iff $L/M$ and $M/K$ are also separable is contained in a lot of notes and texts I've come across, subject to the assumption that the extension $L/K$ is finite. An equivalent condition to separability, that $L\otimes_K\overline{K}...
is the definition of the limit of a function in terms of neighborhoods invalid anywhere where the usual epsilon delta definition of the limit of a function valid? I feel like the former is much more succinct...
i'm not 100% sure of what you have in mind but the 'topology'/neighborhood/open set flavored notions of limits generalize the 'metric' flavored ones that involve notions of proximity
broadly speaking anyway, you can come up with examples that maybe don't fit that pattern, but you sorta have to work at it
like the epsilon-delta from R or R^n is a metric space notion, and any metric space is also a topological space where you can phrase the limit in terms of neighborhoods/open sets without explicit reference to metric
as i've noted before at least a few times, he even calls chapter 2 'basic topology' without ever defining what a topology is, or even using the word at all outside of the chapter title
Mapping/function can be replaced by "transformation". But analytic/holomorphic is referred to a special kind of transformation. May be we can call "analytic transformation "
I was thinking of a way to make articial life, and i found a very disappointing result, as it's still a result i share it, maybe someone can do smtg with it, don't rly know if it's the place.
Lets take a function of $F: \{0;1\}^n \rightarrow \{0;1\}^n $ bijective.
Let $a <n$
Let $ S = \{s: \{0;1\}^n \rightarrow \{0;1\}^n \}\; | \;\forall c \ne a \; : s(x)_{c} = x_{c} \text{ and } x \rightarrow s(x)_a \text{ is constant} \}$
I'm reading up on Tonelli's theorem for sums here, i.e. if the terms of a double series are non-negative, the order of summation may be interchanged. They write:
> If you would have an uncountable number of non-zero elements $f_{ij}$ then all the sums would be infinite and the result would be trivial.
What do they mean by "uncountable number of non-zero elements $f_{ij}$"? I'd be grateful if you could explain with an example.
If I have to show some uniform convergence of power series $\sum_{n=1}^\infty a_nz^n$ on some domain, then can I just set $f(z):= \sum_{n=1}^\infty a_nz^n$ and $f_k(z) = \sum_{n=1}^k a_nz^n$ and show $|f(z)-f_k(z)|<\epsilon$ uniformly on the given domain for large $k$? Here, I didn't show $f$ converge pointwisely priori.
@schn intialy the sum on a familly is defined as the the sup of the finite sum of differently indzxed element of the family. If uncontable number of indexation lead to non 0 let say all stricly positive, then let consider the partition [1/2^k; 1/2^k+1[ there is automaticly one of these portion that contain an infinite number of indexation leading to this portion, by suming these element we have +00, so the initial sum is +00
@Koro I have no context for your question. However, an embedding is usually a map $\iota : X \to Y$ such that the map $\iota : X \to \iota(X)$ is an isomorphism.
That is, it is a map from $X$ to $Y$ which preserves the structure of $X$. In essence, it puts a copy of $X$ into $Y$.
if $S=\{a,b,c,d,e\}$ and $T$ is the 10-element set containing all two element subsets of $S$. I want to sho that there exists no injective function $f:S\rightarrow \{0,1,2,...,10\}$ s.t. $g:T\rightarrow \{1,2,..,10\}$, $g(\{i,j\})=|f(i)-f(j)|$ is bijective.
My idea was assuming there exists an injective $f$. So I know that $f(i)=f(j)$ implies $i=j$. Since $g$ is bijective for all $l\in \{1,...10\}$ there exists $i,j\in S$ s.t. $|f(i)-f(j)|=l$ but I don't see how to continue. @XanderHenderson can you maybe help me?
in this example, the 'sine( ) marking' is just the number that's being fed into the logarithm
it happens to be the sine of something in whatever was interest to napier, but that's somewhat conceptually different from whatever his version of the slide rule was designed to do
as best as i can understand it, napier tabulated things as the sine of something simply because it was common at the time to consider numbers as trig functions of other numbers in tables
"most practitioners who had laborious computations generally did them in the context of trigonometry. Therefore, as well as developing the logarithmic relation, Napier set it in a trigonometric context so it would be even more relevant" - this doesn't tell me much, but that's what i get out of it
Suppose $B$ is the bilinear form corresponding to symmetric uniform elliptic operator $Lu=-\sum (a_{ij}u_{x_i})_{x_j}+cu$. Then there exists $beta>0$ such that $\beta||u||_{H^1}^2\leq B[u,u]$.
This looks like Garding's inequality $\beta||u||_{H^1}^2\leq B[u,u]\leq \gamma||u||_{L^2}^2$ but here gamma depends on $c$ too. I don't get why $\gamma=0$.
@robjohn thanks, that helped. Here's a nice proof of your assertion. Quick question; why in the proof are we assuming $x_\alpha\in [0,\infty]$ and not $x_\alpha\in [0,\infty)$. What is the point of allowing $x_\alpha$ to be possibly infinite?
@SouravGhosh But Strange, I dont find Picard's theorem in SL Ross. My version of Picards theorem is : The differential equation $\frac {dy}{dx}$, $f(x,y)$ , $y(x_0)=y_0$ has a unique solution on $|x-x_0|\leq h$ $(h>0)$ if both $f$ and $\frac{\Delta y}{\Delta y}$ are continuous on the domain $D=\{(x,y)| (x-x_0)\leq h, |y-y_0|\leq k, h,k>0\}$. Is this what your version of Picard's theorem as well ?
@robjohn yes. I'm currently working on how this statement implies that an uncountable sum of positive non-zero elements equals infinity. I guess it is simply the contrapositive of the claim in the above link, although I'm unsure how you negate "then $x_\alpha=0$ for all but at most countably many $\alpha\in A$".
the negation of "S is at most countable" is "S is uncountable," even if S is {alpha: x_alpha isn't 0}, although i guess what that means more concretely might depend on your set theory
@leslietownes I guess the conclusion "then $x_\alpha=0$ for all but at most countably many $\alpha\in A$" is the same as saying "$\{\alpha\in A: x_\alpha \neq 0\}$ is at most countable". So then the negation would be "$\{\alpha\in A: x_\alpha \neq 0\}$ is uncountable".
as with anything, lotsa logically equivalent ways to write it out, but thats definitely how i'd think of it internally, rather than some condition holding or not almost all of the time, some set being at most countable or not
A diagram chases is just using properties of arrows (epimorphism, monomorphism, isomorphism, etc) taking elements, and "chasing their images" around the diagram.
Until you reach the proof goal
For example existence of a certain map would be a proof goal in a diagram chase. As well as proving the map is well-defined should their be quotient groups involved.
So I have that screenshot. That's fresh. But it's no farther along than the last time I tried this project idea out. I'm carefully moving forward. Maybe in a month it will do some useful feature.
I rewrote essentially everything so that I could use Bootstrap Studio (kind of like adobe dreamweaver or webflow) to design the GUI. The backend is Django (Python) / Neo4j
@SouravGhosh Ohh..Actually there's a whole chapter dedicated to it. I can't find it 😫. I don't know if I should ask you this, but can you please point out the portion you are talking about in the book. I have a pdf version of the book i.e ia801500.us.archive.org/33/items/… .
The chapter 10 deals with "Existence and Uniqueness Theory" which starts at PDF pg no-469. I am sorry, I disturbed you a lot today, but I am just enervated for all the travails...
so here's a reading-comprehension question. This probability problem showed up on the main site recently:
Choose two random numbers from [0,1] and let them be the endpoints of a random interval. Repeat this n times. What is the probability that there is an interval which intersects all others?
Let $\{I_k\}_{k=1}^n$ be such a set of intervals. What I interpreted the problem as: Is $\cap_{k=1}^n I_k$ non-empty? But what it's intended as is apparently: Does there exist $j$ such that $I_j\cap I_k$ is non-empty for all $k$?
i think as i was reading it my mind may have thought "is this maybe just that the intersection is nonempty?" for a moment, and then "but wait that's a property that couldn't be unique to 'an interval,' it would be shared by all of the intervals or none of them"
i generally agree that using meta rules like "they wouldn't have written it like that if they meant X, so it means Y" is bad, but it feels slightly more like clumsy prose than ambiguous prose to me
that's my ruling. what's next up on the docket in prose court
right. saying that the 'common interval' "intersects all others" doesn't really make sense, because it's not one of the 'others'
that said, I do manage to get a result---the probability of the intersection being non-empty seems to be $2^n/\binom{2n}{n}$
that works out to be 2/3 when n=2, same as the other interpretation. but for n=3 it gives 2/5 whereas the other gives 2/3. (10 of the 15 shown above have one interval which overlaps with the rest, but only 6 of those 10 have a common interval.)
@leslietownes what may have biased me was that it was very simple to test "success" using Mathematica's interval arithmetic: just compute the intersection of a random set of intervals and see if it's empty.
Given $(M,g)$ where $g=\frac{dudv}{uv}-\frac{dr^2}{r^2}-\frac{dw^2}{w^2} \quad \forall u,v,w,r \in (0,1)$ and integrating w.r.t. $g'=dudv-dw^2-dr^2,$ i.e. using the induced measure from $g'$ by means of the volume form to integrate within $M.$ Does the measure actually tell you any information about $M$ itself?
oh, interesting. i guess it makes sense that the result (for your problem) would go to 0 quickly with n. in that there's pretty good chance for every single interval you add to mess everything up.
feels like worse than tossing a coin and hoping to get heads every time.
I guess I'm asking - when I integrate in $M$ with the measure induced by $g'$ can that tell me something about $M.$ $(M,g)$ is a Lorentzian manifold in null coordinates. But using the measure induced by $g'$ erases the Lorentzian nature. I.e. $(M,g')$ is not a Lorentzian manifold. So when I integrate or do an integral transform in $(M,g')$ what exactly does the image space even tell me about? I presume that it cannot tell me about $(M,g)$ nor $(M,g')$ but maybe it can?
I know that integral transforms are linear operators and that they are used to transport equations to different function spaces where the solution can be solved and transported back
please don't make me open that screenshot and read the argument. does the argument have anything to do with extending a map to an automorphism of a larger thing?
Another confusion that I have is: suppose K and L are extensions of F. Then what is the difference between $F$-homomorphism $K\to L$ and $F-$ isomorphism $K\to L$.
@leslietownes that's the part where I stopped understanding the proof. I don't understand what they mean by 'Automorphisms of N'.
do you know that a normal extension is a splitting field of something? sometimes that's a definition and not a theorem, but it's not as if these are apples and oranges.
a field automorphism of N. do you mean, is it a k-automrophism or a K-automorphism or what? i don't know.
maybe they're asking you to grab an idea from the proof of 3.13, more than to directly apply the theorem. i don't like when authors use "see theorem _" for that kind of thing, but it's common enough.
mm, the idea of the "..." is that the numbers in the array are constant down 'diagonals' and the nth entry of the first column is 1/2^n (at least for n = 1, 2, 3, ignoring the 0th row), wouldn't the nth row sum be -1/2^n?
Another confusion that I have is: suppose K and L are extensions of F. Then what is the difference between $F$-homomorphism $K\to L$ and $F-$ isomorphism $K\to L$.
@leslietownes Ok, so if we ignore the first row, we do have $-1/2^n$ for the nth row. So the sum of row sums is $-1+\sum_{k=1}^\infty \frac{-1}{2^n}$. Is this correct?
