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4:34 AM
Q: If $X \subset \mathbb{R}^n$ is compact then $X \times X \subset \mathbb{R}^n \times \mathbb{R}^n$ is compact - proof verification

D.C. the IIIClaim: If $X \subset \mathbb{R}^n$ is compact, then $X \times X \subset \mathbb{R}^n \times \mathbb{R}^n$ is compact. Key definitions: A set $X$ is closed if every convergence sequence converges to a point in the set $X$. A set $X$ is bounded if there exists $M > 0$ such that for all $x \in X$,...

Well, this puts me in my place.
@D.C.theIII A sequence has just a single index.
I'm guessing my notation is wonky and all over the place. So you saying that doesn't surprise me.
You have $(x_k,y_k)$ convergent. You must now argue that means $x_k$ converges, etc.
No, the logic is backwards.
I wonder if the user questioning your "elitest morality" is the same one of a few days ago but under a new user name.
Who a few days ago? I must have erased my memory. Well, he has a bit of a point, but I wonโ€™t be answering/helping in the future.
Anyhow, do you see the difference between what you started with and what I started with?.
4:49 AM
Not yet. I'm trying to see it
well idea wise I get what you mean. I'm just trying to construct the argument
interesting that a suggestion regarding a lack of effort was characterized as 'moralistic.' it is not clear to me that a lack of effort has moral dimensions.
maybe the OP would like to engage in a discussion of whether we're all sleeping off the hangover of john locke, apologist for capitalism, confusing effort and morals.
or maybe not.
is it moral to be a good student? are OPs students? big can of worms
i'm sensitive to this stuff because my daughter is young and already internalizing stuff she isn't aware of yet. she knows it's bad to be lazy. i don't know why it's bad to be lazy. we don't criticize rich people for being lazy, or at least, not enough.
it's part of the unwanted DNA of capitalism that we inherit that makes us think this stuff.
and when we praise someone for hard work there is often someone else who benefits more from that work, incentivized by that praise, than they do.
This is presumably someone trying to learn from lectures of a physicist, based on earlier post. My exercise gave explicit directions on how to proceed and he did none of that. But he did catch me in a sloppy sign transposition, for which I apologized. Hubris? Overstating it. Maybe a Tromp/Fox supporter.
i'll go back to my communist chats, but, snacks for thought
5:04 AM
Yeah, I knew this wasnโ€™t someone taking a math course. The details of his/her status โ€” no clue.
i think my response to that, which i obviously wouldn't post, is, if the only problem with the attempt was sign errors i don't think anyone would be having this conversation. but please try first.
everyone's going to goof up signs. go ahead and goof a few up in a genuine attempt. it'll go better.
Anyhow, DC โ€ฆ
I think I have to use contradiction to show this don't I?
No, the sign error was in the physics video, and in explaining it I made a small error.
Show that if the sequence of ordered pairs converges, then first each component converges.
i understand that. my hypothetical comment was directed at, if catching other people's minor errors is a badge of honor, show how honorable it is by exerting a minor effort in your own problem. help people help you.
5:08 AM
ohhh.... this goes back to the idea of $|x_i - a_i| < \|\mathbf{x} - \mathbf{a}\|$
that's a useful idea.
the functional analyst in me demands that i remark that it's as useful of an idea if you put a C in front of the right hand side, as long as the C only depends on the space and not the vectors.
all this hard thinking...I should've been an influencer
You were defending my honor, subtly, leslie. It will not go unappreciated.
dc that's mostly what i do now. lesliecoin is a side thing
5:16 AM
well, it funds my lavish lifestyle, but a lot of what i'm doing now is in partnership with brands who want my elegant way of living to somehow enliven the way the public views their products.
so i'll do a poolside tiktok about some cologne, or spray champagne off my balcony, that kind of stuff.
my neck, my back
you can peg lunacoin to lesliecoin with your age ratio
i'm going to assume that's an insult, and let you know that envy gets you nowhere in this business.
5:31 AM
theres people on facebook apparently named leslie coin
scammers. i'll sue them into oblivion.
disturbing fact, armament manufactures stock goes up after a mass shooting.
just thought i would lighten the mood.
there was a segue there, but i have forgotten it already
anyway, here's wonderwall
how about cat pics
scary looking
5:43 AM
sorry no cats, or dare i say it, no i won't, here
So we have: $$\|(\mathbf{x}_k, \mathbf{y}_k) - (\mathbf{a}, \mathbf{b})\| = \|(\mathbf{x}_k - \mathbf{a}, \mathbf{y}_k - \mathbf{b})\| > \|\mathbf{x}_k - \mathbf{a}\|, \|\mathbf{y}_k - \mathbf{b}\|\ \text{(individually)}$$

At the component level taking $|x_k^i - a|$ as an example, I know these individually converge so taking the largest $k$ of them all I can say: $|x_k^i - a| < \frac{\delta}{\sqrt{2n}}$, same for the $y_k^i$ components.

Thus: $$\|(\mathbf{x}_k - \mathbf{a}, \mathbf{y}_k - \mathbf{b})\| = \|(x_k^1 -a_1, \dots, x_k^n - a_n,y_k^{n+1} - b_{n+1}, \dots, y_k^{2n} - b_{2n})\| =
$\ge \max(.,.)$.
are you inferring $\max(|x_k^i-a_i|, \dots, )$?
5:53 AM
the first line should be $\ge$ :-)
ohhhh, not strict. so the $\max$ of those components
just being a pedant
whoah yikes. i thought that was just a conspiracy theory
it's cool. I do need that because I am guilty of dropping less than or equals here and there and that comes back to bite me
max albany grates again
5:55 AM
my machinations are in cipher Leslie
i want a t-shirt that says "where were you on 1/6 ?"
@CalvinKhor If you are not a good student, how are you going to get admission in a prestigious university? Then you won't be able to get a high salary job, you won't have a handsome wife, you won't be able to pay your debt to your family and country, you would be an burden to society.
never should have trusted those extremisers
@PrithuBiswas why not just give up?
what is 'good' student?
the one scoring the highest in every exam?
5:59 AM
@CalvinKhor what extremisers?
Sooo..... is this the sort of argument Ted was referring to? I don't see how this is opposite to what I had written except for "assuming" that $(x_k,y_k)$ converges, which I don't see why we can assume the
@copper.hat Sinn Fein :p
@Koro Idk
@D.C.theIII sorry, i was not really following and an about to hit the sack.
@copper.hat lame excuse =P
when i grew up sinn fein were extreme. since the history is forgotten among the younger generations, they have become respectable
6:02 AM
i still like the 'surely you know someone who can fit a timer' tweet.
even if it is apocryphal
@copper.hat Seems a few movmeents have been able to do that. Marcos in the Phillipines comes to mind.
scary thought
i used to travel with a bunch of chips (the ic sort), wires, etc. no one ever asked
ok, good night folks, i need my ugly sleep
haha. one of my friends, immediately after 9/11, traveling with a hard disk. "uh, can you turn it on?"
no, i can't.
6:05 AM
@D.C.theIII why are you assuming convergent sequence in $X\times X$ as $(x_k,y_j)$?
the only thing i was ever caught up on was food
@copper.hat good night =)
@Koro I changed it to one index so $(x_k,y_k)$ instead of two indices
night joe
@D.C.theIII ok
Good night @copper.hat
WHat you removed was interesting could you post it again
6:08 AM
Is it good to have a midlife crisis at the age of 17?
@Koro Yea, after I posted my solution Ted told me about adjusting it.
I removed it because you had already considered the change.
But the change is not yet made in your post.
@PrithuBiswas you wouldn't know it is a midlife crisis because you're nowhere near mid-life. It is just a "crisis"
@Koro I considered the change, but I haven't fully grasped the idea behind it. I did it somewhat based on the symbol manipulation aspect
@D.C.theIII Oh, then it is a early-life crisis.
no need for an adjective in front of it.....it is just a crisis.........
6:11 AM
@D.C.theIII The deleted message was: A sequence on a set S is defined as a map from N (the set of naturals) to $S$. So 'assuming' a sequence in $S=X\times X$ should be a map from $\mathbb N$ to S.
hence why I needed to have one index instead of two....got it. thanks
yup! Considering the sequence as $(x_i,x_j)$ is like considering a map from $\mathbb N\times \mathbb N$ to ...
I feel very salty today =(
@Koro I'm still a little perplexed as to why we can assume $(x_k, y_k)$ can converge since that is what I need to show converges
6:31 AM
@D.C.theIII By the first definition in your post.
But the definition is written poorly.
A set X is closed if every convergent sequence in X converges to a point in the set X
The set (0,1) is not closed in R because the sequence (1/n) is in (0,1) but converges to 0 which is not in (0,1).
Right I agree with that and I understand it, but I'm trying to show that $X \times X$ is closed. So That means I have to take a convergent sequence and show that it converges to a point in $X \times X$
6:48 AM
Right....now it rings a bell and is clicking again.
Thanks for the jolt
6 hours later…
1:06 PM
@CalvinKhor Hello Calvin, I apologize for my weird reply to you, koro and copper. I guess I was too salty to be ironic.
@PrithuBiswas no worries man its just chat, be cool
also i'm not so sure i understood you ๐Ÿ˜‚
@CalvinKhor My physics teacher said that to us when we didn't do our studies.
ahhh i get you
@porridgemathematics weak convergence means... $\mu_n(f)\to \mu(f)$ for all $f$ in.....? continuous functions?
yeah, since everything is compact continuous functions
(more generally on compactly supported functions)
1:12 PM
@porridgemathematics the assumption is that $\mu_n(f)\to \mu(f)$ is true for all indicators $f$, and hence by linearity for all simple $f$
im guessing you got this far?
how did you get it for all indicators
you mean for indicators on measurable sets? or just on $\mathcal{B}$
isnt that the same as mathcal B
ugh, let me change that notation
it will potentially confuse other readers
sorry :)
no worries, its worth fixing that
1:15 PM
I guess when I am upset over what my teachers say to me,
I like to hide the pain in the shadows of comedy.
well open sets is enough
im guessing your thought is its enough because each of the $\mu_n, \mu$ are outer regular, even with respect to some random basis, but my concern is that the convergence of sums of indicators on this basis to $\mathbf{1}_{B}$ where $B$ is just measurable is going to be different for each $\mu_n$
and here im talking about convergence in $L^p$
but were talking about $L^p(\mu_n)$ for different $n$
ok i need to think more lol
do you see what im trying to say lol?
it came out kind of incoherent, but its hard to really explain without typing out a massive paragraph
basically what im saying is, the argument you have in mind seems to depend heavily on a single measure, and the open sets we use to outer approximate borel sets will also necessarily be different for different measures
well i sort of got wind of an issue when i read "for different n"
1:20 PM
of course, its possible the condition on these basis sets might turn out to make your argument doable
what's the $\mu$ in the condition for the basis?
i just edited, the $\mu$ is really some prefixed measure
I should have defined it before efven defining $\mathcal{H}$
yeah ok
so the $\mu$ is some positive finite measure on $\overline{\mathbb{C}}$, and the $\mathcal{H}$ are opens with $\mu$-null boundary
that form a basis
1:41 PM
whelp, +1'd and will think about later when i have paper
thanks for the upvote
2:14 PM
i think i was heavily overthinking this, i think this should work for any basis as long as the space is compact: there is a finite subcover of $\overline{\mathbb{C}}$, $H_1,..H_k$, and $\mu(X) \leq \sum_{j} \mu(H_j) = M$, then for large $n$, $\mu_n(X) \leq M + 1$, and so for all $n$, $\mu_n(X)$ is bounded. If $\phi_{M} \rightarrow f$ uniformly, then $\int \phi_{M} d\mu_{n} \rightarrow \int f d\mu_{n}$ as $M \rightarrow \infty$ uniformly in $n$
now assume $f \geq 0$, and WLOG say $f \leq 1$, write $\phi_{M} = \frac{1}{2^n} \sum_{j=1}^{2^{n}-1} \mathbf{1}_{f^{-1}(\frac{j}{2^n},+\infty)}$, write each $f^{-1}(\frac{j}{2^n},+\infty)$ as a countable union of basis sets. So each $\phi_{M} = \sum_{p \geq 1} c_p^{M} \mathbf{1}_{U^{M}_{p}}$, $U^{M}_{p} \in \mathcal{H}$ and $c_p^{M}$, converges uniformly to $f$
choose $M$ large so that $\int \phi_{M} d\mu, \int \phi_{M} d\mu_{n}$ are all very close to $\int f d \mu, \int f d \mu_{n}$, then let $n \rightarrow \infty$ so that $\int \phi_{M} d\mu, \int \phi_{M} d \mu_{n}$ become equal for this large $M$, but then $\int f d\mu_{n}$ and $\int f d \mu$ are very close
so formalizing this a little, this should be enough to show $\int f d\mu_{n} \rightarrow \int f d\mu$
Are numerical sequences functions with domain as subset of natural nnumber?
Can someone please explain me the first 3 points
2:37 PM
@Belucat I can't quite see the picture in your message. Maybe only my computer is bad?
the ifrst three sentences are definitions. they aren't used in the example 1.6.1
Yeah I don't understand the definition
Can you please explain me the definition I am learning calculus by my own please help
it's a tricky definition. you can spend a months-long class on it. it helps to work with examples.
yeah an example would do
note that the quoted section doesn't define what it means for $lim \x_n = \infty$ there, although it does make a definition in terms of that. if it doesn't define $\lim x_n = \infty$ somewhere else before, that's a logical hole and the concept has not been defined yet.
2:40 PM
what is the use of if for any e>0
it's just the definition. one intuition is that a sequence converges to $L$ if no matter how strictly you interpret $\approx$, the equation $x_n \approx L$ holds for all sufficiently large n.
then there's "sufficiently large" which is itself a thing.
will check
a lot of intuition that falls short of the full definition (or something close to it) will leave something out. e.g. "x_n goes to L if x_n gets closer and closer to L as n gets larger" has a lot of tiny issues with it. for example 1/n gets closer to 0 as n increases, but it also gets closer to every negative number as n increases. so the thing in quotes doesn't single out 0 as the thing the sequence is moving toward.
this quoted "intuition" also arguably makes it sound like each x_n needs to be strictly closer to L than the last in order for the sequence to converge to L, which is not the case.
How can 1/n get close to negative numbers?
if you plot those points on a number line, they move to the left as n gets larger.
2:46 PM
yeah They stop at zero
moving to the left means getting closer to the negative numbers.
well that's part of the reason why the quoted language doesn't capture the full idea. it doesn't get that "stops at zero" part of the example. it only requires "closer."
lots of intuitive 'translations' of the definition are like that. they will have something that conveys something slightly short of the full idea.
which doesn't mean they aren't useful. but it does, maybe, explain why the definition is as complicated as it is.
2:47 PM
so you can prove that 1/n doesn't converge to a negative number using the definition, and that's where epsilon will come in, and it will become significant that the definition requires something to hold for every positive epsilon.
for example 1/n doesn't converge to -1/3 because the defining condition fails for epsilon = 1/3. you can't find any N for which |-1/3 - 1/n| < 1/3 holds for all n > N. in fact, it doesn't hold for any single n ever.
what's N(e) here
and N(epsilon)
well, there isn't one. that's the point.
2:51 PM
How do u define them
you can prove that a sequence fails to converge to a number by exhibiting a positive epsilon for which no "N(epsilon)" can be found that would make that inequality in the definition true.
which is also, in general, tricky stuff! i bet somewhere on that list of problems (the link is just to a MSE search) there will be some like that.
Can we say that $\sum_{n=0}^\infty (1+(-1)^n)$ is divergent because on the even we are summing 2 and on the odds we are summing 0 or is this incorrect because we must be careful because for non absolutely convergent series there are troubles in arranging the terms in different orders?
Like splitting evens and odds, for instance
2:54 PM
za: the standard way of doing that would be to use the divergence test (i.e., lim 1 + (-1)^n does not exist, whereas it would be 0 if there were convergence). it complicates things to split it up as a sum of 1 and a sum of (-1)^n's. note that the series as written is a series of nonnegative numbers (which will converge absolutely if it converges at all), but once you begin looking at the pieces, you do get into handling -1's and where they go.
i.e. the numbers are "arranged" already so that the sequence is nonnegative. it complicates things to begin fiddling with that.
Can't a limit tend to neagtive ?
I mean never?
cause A sequence {xn} is said to be small if limxn=0
sure, the definition just gives sequences going to 0 a special name. absent context, i wonder why. you do see that in some constructions of the real numbers from the rationals.
oh ok
Please check if I am right Epsilon is any small number which we take to check our limit ryt?
and shouldn't that only be possible for convergent limits?
3:11 PM
if a limit exists, then there's some L where no matter what epsilon is, you can find N(epsilon) as in that definition up there. it might be that for particular values of epsilon and L you can also find N(epsilon) as in the definition up there, but it won't be possible to find "N(epsilon)" as in the definition for every epsilon unless the sequence is convergent and L is the thing that it converges to.
human language that avoids the definition, as i tried to do so much just there, is no substitute for working directly with the definition, and examples. x_n = 1/n is a good example.
Yeah am I right? for any epsilon at some time the inequality holds good
well these translations aren't capturing all pieces of the definition, but yes.
like let epsilon be 1/1000 for 1/n sequence after 1/1000 we can give terms where the inequality will get satisfied
What am I missing
the inequality with L = 0 plugged into it. yes. and not just giving "terms" where the inequality is satisfied, but a point (N(1/1000)) past which every term of the sequence satisfies the inequality.
another good example would be the sequence x_n given by x_n = 0 if n is even and 1 if n is odd. note that no matter what epsilon is, you can always find infinitely many n so that |x_n| < epsilon. but the sequence does not converge to 0.
Oh ok
so every terms after it should also satisfy the inequality
Also How are divergent sequences defined?
3:16 PM
it varies depending on the resource or textbook. usually one just says a sequence is divergent if is not convergent. i.e. if there is no L that will fit into the above definition.
No finite limit?
sometimes people will augment this by giving special treatment to some sequences that diverge in specific ways. the page you scanned above maybe is going there, with sequences that have limit "infty."
books that have defined that term aren't always 100% consistent on how they handle that, but 'no finite limit' is probably a good default.
If the limit tends to infinity how can epsilon be defined?
i mean Can an epsilon exists where the inequality will be satisfied thereafter?
well, the page you mentioned above doesn't explain that at all. which is weird.
I am getting doubts into my mind
3:19 PM
it defines what it means for x_n to have limit L. then it says, okay, if x_n has a "finite" limit... well, it didn't define anything else.
there's a fairly standard definition that they could have snuck in right after that. maybe it comes later?
how is that possible?
unless n tends to infinity no such epsilon exist
yeah, you need another definition.
So the one defined is for only convergent?
3:22 PM
usually: $\lim x_n = \infty$ if for any $M > 0$ there is a positive integer $N(M)$ such that $x_n > M$ holds for all $n > N(M)$.
Thank you soo much
this stuff has a well deserved reputation for being tricky. good luck with it.
4:01 PM
I am starting to suspect that my post might not have any answers.
like , somewhat interesting , but absolutely unanswerable.
i'm surprised it hasn't seen more engagement, but maybe a lot of people had my reaction. "wow, looks interesting and also tricky. ๐Ÿ‘€"
@leslietownes I guess the only thing I can do personally is keep a list of all of the constructable sets I come across and hope to see a pattern.
4:43 PM
feels like there ought to be an answer, particularly as it is inspired by a popular textbook problem. might be one for mathoverflow, actually.
intuitively it feels like the answer should be 'huh, nobody knows' or 'here is an obscure paper from 1973 on this exact question.'
If $k$ is in your set, for what $\ell$ is $k+\ell$ in there?
5:12 PM
I believe that's the general chat room of Mathematics Stack Exchange. I'm asking for advice if some person can to explain to me what is the product of Word Press
I should to buy, when I'm interested in a blog similar than the blog of a famous mathematician: I want a blog of Word Press but I don't know what is their product from which I can to contract a blog in which I could to edit mathematical formulas. Can you add the reference (how is called the service) of Word Press for such solution? Many thanks-
there is a "mathjax" plugin for wordpress. that might be what people are using. mathjax is also used here in the chat.
@leslietownes many thanks for your reply. Then I understand that I can to contract any blog of Word Press that mathajax works in it. Is it?
that's my understanding, too, although i do not have a wordpress blog in which i can personally test it.
I hope more feedback of other users, but many thanks for your help @leslietownes
5:27 PM
One can always enable MathJax on a webpage by adding a few lines to the header.
I do the same thing with a header on local pages.
5:43 PM
Howdy @robjohn
Good morning
Just visited the Vampire Room (Quest).
Are you feeling ghoulish?
2 hours later…
7:52 PM
@TedShifrin The sun seemed a bit too bright on the way home...
8:11 PM
You are indeed a shady character.
8:21 PM
I take umbrage at that comment
The conversation dims to opaque.
I would mirror that sentiment, if I had one.
What is the appropriate way to use the term "likelihood"? I have always used "the probability that I bring an umbrella given that rain is forecasted" and "the likelihood that I bring an umbrella given that rain is forecasted" interchangeably, but now that I am learning about maximum likelihood estimation, I wonder if "the probability that I bring an umbrella given that rain is forecasted" and "the likelihood that rain is forecasted given that I bring an umbrella" are synonymous instead?
Or is this a case of an unfortunate divergence between technical meaning and colloquial meaning?
8:41 PM
You should reflect on that, robjohn.
โ€œLikelihoodโ€ does not seem quantitative to me. But you should ask statisticians, not us.
8:58 PM
@user10478 It is a divergence between technical and colloquial. In technical stats terms the likelihood function is the probability of the observed data as a function of a given parameter. Whereas when you use "probability" to describe the rain you are talking about the probability of the data "it is going to rain" occurring as a function of a given parameter
So when you speak of "likelihood of rain" or "probability of rain" what you are doing implicitly is using a probability density function with a given parameter to "calculate" what the chances of rain are.
did someone mention goulash?
Formally what the likelihood function is doing is finding all the possible values that you can use in you probability density function to "calculate" the most accurate probability.......I'm trying to not be completely technical in my response so the stats PhD's in the chat don't come at my neck for not using the right technical terms
you know the chat is gone down hill when I'm actually providing answers instead of looking for them...........
I forget some basic facts about Cauchy's integral formula. Given a continuous function $g\colon\partial\mathbb D\to\mathbb C$, we can define a function $f\colon\mathbb D\to\mathbb C$ given by $f(\zeta)=\frac1{2i\pi}\int_{\partial\mathbb D}\frac{f(z)}{z-\zeta}dz$. When do $f$ and $g$ glue into a continuous function on the closed disk?
what had $g$ got to do with it?
In the integral formula, it should have been $g(z)$, not $f(z)$. Typo.
9:16 PM
How does induction behaves when distinguish n even or n odd? For instance, if I want to show that a property holds for any n in N and it is simpler to split cases evens and odds, for n even I must show that it is true for n=0 and then assume it true for n even and show it is true for n+2? Similary, for n odd I must show that it is true for n=1, assume it true for n odd and show it is true for n+2? I am asking this because if I use the "standard" induction I get wrong results.
I mean that for n even n+1 is odd, and so I get stuck because I am not proving anymore something for even numbers in the step for n+1 in induction. If I put n=2m, for m+1 it works because 2(m+1)=2m+2 is even, but I should consider n+1 from induction and not m+1. Intuitively, I get that I must consider the next element of the set I am proving (next even for evens, next odd for odds), but I don't see why I can just use n+1 in the standard form of induction by substituting n+1 in place of n.
Can someone help me understand this, please?
9:31 PM
@Gwyn Yes, you could do that.
Or, if you prefer, you can prove that your result holds for all natural numbers of the form $2n$ (where $n$ is natural), and all natural numbers of the form $2n+1$ (where $n$ is natural).
But you really shouldn't get caught up in the idea that the induction step goes "assume case $k$, then prove the case $k+1$."
Rather, you "assume some case, then prove the next case", whatever that 'next' case is.
Indeed I was guided by that idea of "next case" instead of "next number", thank you for the confirmation Xander:)
10:28 PM
@Yai0Phah As a partial hint, what do you get when you take $g(z)=\bar z =1/z$ on $\partial \Bbb D$?
10:40 PM
I could see why you needed four lectures to discuss quadratic forms and the second derivative test.......this ain't your one variable second derivative test....things get complex really quickly....
Well, there was interesting linear algebra there, not just calculus.
Yea that is where the complexity is coming from at least for me
I'm about to watch the lectures. I just finished reading that's why i commented
Is there a higher degree hessian type of matrix if we wanted to examine a function at a higher degree?
And then it ties into eigenvalues at the very end of the book โ€” see exercise on Sylvesterโ€™s law of inertia.
Yea I"m actually looking forward to getting to the eignevalue stuff because in Insel it was just dry with no motivation behind it
So up to now I've never actually played with eignevalues beyond just doing the usual finding the bases of eignevectors, cayley hamilton in an abstract way
but that is a long ways away for the moment
Sometimes applications are good for understanding.
10:53 PM
can anyone tell me why mathematica is giving me the input back as output?
It's MiniMaxApproximation[Exp[x], {x, {-1, 1}, 5, 0}]
I figured it out, nevermind
@Derivative It does that when the syntax makes no sense to it.
11:19 PM
I had forgotten to import the package
A neat thing
Let f.g refer to $f\circ g$ and f;g refer to $g\circ f$. In other words, f.g is "do g then do f", and f;g is "do f then do g"
Suppose f,g,h are transformations of the plane. Specifically, let f be a rotation about a point A (of some angle), g be a rotation about a point B, and h a rotation about a point C
f.g.h refers to rotating about A, then about *where B moved to*, then about *where C moved to*.
f;g;h refers to rotating about A, then about *where B originally was*, then about *where C originally was*.
I don't see that f.g.h starts with rotating about A
It also refers to rotating about C, then about where B originally was, then about where A originally was
The insight is that these are equal
when you say "where C moved to", by what combination of functions?
The previous two steps
(I suppose "refers to" sounds more intensional then necessary, and "equals" is more neutral)

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