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19:15
I'll calculate the rest later, but it seems it is working. I can even compare their sizes:
$\textbf{Tier 4: Tetration}$

Or, because of the very cheaty way we define exponetiation to get around both the $\omega^{\epsilon_{\alpha}}=\epsilon_{\alpha}$ and $\alpha^{\beta\gamma}=(\alpha^{\beta})^{\gamma}$, we can move up the tetration just like previously in exponentiation:

$\{\omega,\epsilon_0,{}^{\epsilon_0}\omega,{}^{{}^{\epsilon_0}}{}^{\omega}\omega,...\}$

We can see that only the first two levels coincide with the epsilon number hierarchy.

To get a better impression on where the epsilon numbers lie wrt this Tier it is good to compute a few examples:
19:27
If $\lim_{n \rightarrow \infty } \left( \sum_{i = 0}^n a_i \right)^{\frac{1}{n}} = k > 1$, what does this tell us about $a_i$ asymptotically?
19:42
Sometimes a galton board is used to demonstrate a binominal distribution. Is there a name for one of higher dimensions?
@Semi Oh right, thanks. I was worried that if we had a value $a$, we wouldn’t necessarily have a unique plane that belonged to it (for $a=k_xx_0+k_yy_0+k_zz_0$, and it isn't that obvious to me that $x_0,y_0,z_0$ are necessarily uniquely determined, in the sense that they always belong to one plane).
However, as you say, we can pick the vector $\vec r=a\vec k/k^2$, which satisfies $\vec k\cdot\vec r=a$ - and therefore we can always find a point on the surface. For now, it's convincing enough for me that $\vec k\cdot\vec r=a$ determines a unique plane. I’m guessing I could show it using contradiction; assume there where another plane that is also described by this equation.
Then we can find a vector parallel to $\vec k$ that lies on this plane. However, this vector is necessarily $a\vec k/k^2$ (because any other vector won’t yield the value $a$), so our plane is unique.
Right.
Here's another more geometric way to think about it. If I divide both sides of the vector equation by the length $k$, the equation becomes $\hat{k}\cdot \vec{r}=a/k$.
But the LHS is just $r\cos\theta$ where $\theta$ is the angle between $\vec{r}$ and $\hat{k}$.
So $\hat{k}\cdot \vec{r}$ is just the $\hat{k}$-component of $\vec{r}$.
In this link: youtube.com/… question says that $f$ is continuous at $x=k$. Isn't that a mistake?
Hence the set of vectors $\vec{r}$ such that $\hat{k}\cdot \vec{r}=a/k$ is the set of vectors whose $\hat{k}$-component is $a/k$.
@Semi oh of course, that's just projecting vectors. Right, thinking about it from several perspectives helps a lot.
19:50
Right.
It literally is just the requirement that the scalar projection is fixed.
yea
also hi @dami
This also tells you the following, if I'm visualizing it right: If I change $a$ by $\Delta a$, then the plane shifts by $\Delta a/k$.
yes I think so too
Hey @Sha! And everyone!
in your second geometric interpretation, that's obvious immediately
I think I like that interpretation best so far, though finding the point along $\vec k$ is also nice.
I'm almost done with my "mathematical foundation" of waves (chapter 2 from Hecht). Then I'll be cramming optics this weekend
19:55
I think that last observation also fits well with the plane-wave connection: If we shift the phase of the wave by $\Delta a = 2\pi$, then the plane of constant phase will shift by $2\pi/k$ i.e. one wavelength
oh righttt
I like it :P
@ShaVuklia awesome @ShaVuklia, how is it going?:P Long time no see!
definitely best one so far
Definitely
haha hi @Waiting well, if I survive my vibrations&waves and optics exam this tuesday, I'll be alive and well:P for now, I'm not sure where I'm at!:P How are you?
19:58
Of course, this also fits with the whole plane wave idea in that we don't care so much about individual points on the wave so much as the overall waveform
@ShaVuklia hehe, you'll succeed! :P Just returned from jogging, barely find energy to type. :P Not that bad (in general), I'm developing something interesting in mathematics, a new theorem.
So yeah, geometry ftw
@Waiting oh right, well take a good rest then!
@ShaVuklia Yeap, perhaps. :P
Do you have any Fourier transform stuff for your exam?
It seems like you'd have to for a math-optics course, though I know well enough how hairy that can get.
20:05
@Semi well, I've seen a bit of Fourier transform in the vibrations part of the course, and also a bit in quantum. But not sure about the optics part
Well, it's all wave stuff
In optics I chiefly have in mind the intensity profile of a slit experiment as a Fourier transform of the relevant aperture
are you familiar with Hecht?
too bad
i just looked at the table of contents, and they do include a chapter called "Fourier Optics"
20:09
we won't do that though, it's one of the last chapters
Ah, too bad
alright, gotta start! you'll hear from me when I get stuck :P
@Hippalectryon how are you doing?
@Waiting Great and you ?
20:11
@Hippalectryon cool. I went jogging a bit and now I'm pondering over some mathematics I created.
So, a small idea.
I look at the sequence $(a_i)$, the sequence converges to 0.
Also, I know that $\sum_{i=1}^\infty (a_i/2^i) = 0$.
I want to minimize $max(|1-a_1|, sup_{k\ge 2} (a_k))$
I think I can't minimize it to anything less than 1/2, but I find it hard to rigorously explain.
Sorry, obviously that's $sup_{k\ge 2} (|a_k|)$
Also, hi @Ted
So you want to show, for a specific set of sequences, every element differs from the sequence 1,0,0,0... by at most 1/2.
(I have no idea how to solve it, I just like saying it that way)
20:27
Hi can someone take a look to my bounty question?
https://math.stackexchange.com/questions/2292760/uniform-convergence-in-convex-set
Actually, I want to show that I can't bring it to be that way.
That for a specific set of sequences, there's an element differing from the sequence 1,0,0,0... by more than 1/2
I can show that
$|a_1|<sup_{k\ge 2} (|a_k|)$
You can start by showing that all your sequences are of the type $1/2, -1/2, -1/2, -1/2$ or $0,1/2,-1/2,-1/2,-1/2$ or $0,0,1/2,-1/2,-1/2,-1/2$ and so on.
@AaronMartinez I don't get your first question. Where in the equality does it say that $H(x_0) (x-x_0)= f_{n}(x)-f_{n}(x_{0})$ ? Its just a telescoping sum
Forgot the half the first time.
@Jayesh Necessarily? I don't think so.
20:34
I mean, it can be any $a$
There I think I lost you - what do you mean?
Hi @Studentmath, @Hippa, et al
Okay, suppose that $a_0$ is not $1$ and instead is fixed to be some $c$, now show that the sequence with minimal sum is of the form $c, -c, -c, -c$ and so on.
Finally done packing
20:36
wee
Packing whom, Demonark?
Hi, a Balarka
Hi :) I have a question. If $G$ is group and $g,x \in G$ this $|gxg^-1|=|g|$ How to prove it ?
So many typos, my bad.
@Ted He can't pack himself.
Packing Balarka actually
20:38
Balarka, Chat Jax finally quit
But yeah I've got one more paper I'll do when I get back home and then finally will be free to do stuff
@Jayesh Ahah. And then I show that such a sequence is impossible, since $|a_1|<sup |a_k|$, and conclude that it can't be 1/2 (or less)?
Also rip in chatjax @Ted
So you go home and then back again in a month?
I had to make a new bookmark.
20:39
Nope, I'll be here throughout
Ted is here
@Ted I figured how to derive the Gysin sequence from Serre spectral sequence.
discovered it myself accidentally, it seems
@Studentmath Yes, something like that.
@Hippalectryon I doesn't says that, I just deduce that $H(x_0) (x-x_0)= f_{n}(x)-f_{n}(x_{0})$
Thanks @Jayesh, I'll try that out.
20:42
I hate moving @Daminark, I had to move my sisters countless times when they moved around. Then my girlfriends brother asked for my help to move a couple months ago, and I said yes. was a full day of moving, and then they asked my girlfriend if I'd like beer as a payment for helping. Then, they decided that I "owed" them for some reason or another, I received less than a thank you for helping them.
@Hippalectryon is it a telescoping sum?
Which is the worst thing in the world.
Eek @Dodsy, that's really annoying
@AaronMartinez It's just saying $a-d=a+b-b+c-c+d$. Notice that the expression is on two lines
@Pawel you mean order with that sign, right?
20:43
Turns out, I "owed" them because I borrowed his truck to move years ago, though he said if the truck broke down, I was liable and I had to pay for the gas myself. He was also given a better car to drive during the meantime...
So, moral of the story, I no longer help people move or ask people for help when I move!
I am never going to move from the rathole below the floor. I'll live there till I die.
@Studentmath What a sign ?
@Hippalectryon I know, but also the meaning of the H(x_0) is the derivative?
|g| is the order of g, i.e. |g|=n means $g^n=1$ (the least $n$ that holds that)
@BalarkaSen Good idea
20:45
Correct?
O lawd @BalarkaSen
@AaronMartinez Yes, $H(x_0)$ is the limit derivative of $f_n$ at $x_0$. What about it ?
If so, consider $(xgx^{-1})^n=xg^nx^{-1}$
or why can we state the equality $H(x_0) =\frac{ f_{n}(x)-f_{n}(x_{0})}{x-x_0}$, NOTE it doesn't have the limit @Hippalectryon
If the limit were include, then the expression would be the derivative @Hippalectryon
@AaronMartinez That equality is false. It's not written anywhere.
20:49
of course is written, that's my first question, (just divide by {x-x_0})
@Hippalectryon
@AaronMartinez That's not what I mean. You talk about that equality, but nowhere in the proof do I see it. Why do you think that equality holds ?
because I analyze the I+II+III expression and then I conclude that weird equality @Hippalectryon
@AaronMartinez I don't see how we can deduce it from I+II+III. It's just a telescoping sum.
I agree with you, its a telescoping sum, but if you delete some elements from it, you'll see that you'll get this $H(x_0) =\frac{ f_{n}(x)-f_{n}(x_{0})}{x-x_0}$ @Hippalectryon
@Jayesh actually, I can mark the sup with T. then I can show that $|a_1|<T$. So it's immediate that $max(|1-a_1|,T)>1/2$
20:54
@AaronMartinez Can you write down in detail how you get there ?
@Studentmath Thank you :)
@Pawel welcome :)
okay @Hippalectryon just hold on
@Hippalectryon forget About it haha. I've just see it, I was wrong with my assumption
yep
Alright on to question 2 :-)
@MikeMiller Morning.
21:04
thanks for making me see it.
@AaronMartinez Also, how it's the form of zz, is it a simple vector in X, or has the form of z0z0?? what does that refer to ?
@Hippalectryon because z_0 has the form z_{0}=t_{0}x+(1-t_{0})x_{0}
@AaronMartinez Ah, it's referring to the pink z after ** 2 ** ?
sigh
the windows calculator sucks
and my nephew put sticky stuff all over my calculator...
yes @Hippalectryon
21:10
@AaronMartinez I'm not sure I get what your question is this time. Can you clarify it ? $z$ is just an element of $R^n$.
this is long division of polynomials, right?
oh weird
what they did was take x + 1 and make it f(-1) = 0
then input -1 into the equation for x
G'night, Mike and Nate.
goodnight
@Hippalectryon I get confused about the form of z because I could took an element z_0 of the form z_{0}=t_{0}x+(1-t_{0})x_{0} and I worked with it with the gradients. So , my question is 'is z an R^n element on the whole proof?' or 'z has this form z=t_{0}x+(1-t_{0})x_{0}?
Nate hasn't learned yet ;)
21:15
Hi chat
:C
@Dodsy for a polynomial, $f(-1)$ is the alternating sum of the coefficients.
@AaronMartinez z represents any element of X. The sentence goes like this in the proof: "... for all z in X". It's not linked to z_0.
So you don't even need to do long division to solve for k?
@Hippalectryon Oh ok. My intuition said that, but I wasn't sure
21:18
Hey everyone
Second, the gm,ngm,n function how is defined? I think it's domain is [0,1][0,1] but I don't know which is it'd codomain. As said in the proof, it's defined by $g_{n,m}(t)=f_{m}(tx+(1-t)x_{0})-f_{n}(tx+(1-t)x_{0}),\quad t\in\lbrack0,1]$ which means that its domain is [0,1] and its codomain is R^m
@Dodsy Nope. $x+1$ is a factor if and only if $-1$ is a root of $f$. So you just have to solve the equation $f(-1) = -1 - k - k - 7 = 0$ for $k$.
Right. So how can you tell the difference between when you should do division or use this method?
is it because of the uknown "k"?
is it because we don't need to factor?
I don't get it.
In the commutative diagram above (taken from Page 5 of Milnor's Topology from the Differentiable Viewpoint) shouldn't it be $M$ instead of $\mathbb{R}^k$?
21:21
@Hippalectryon I see
@Dodsy Experience and intuition, I guess.
Okay
Factoring something with an unknown in it seems rather difficult, though
I'll have to study that part, no worries.
21:23
The above image gives some conext to my question
in structure theorem of modules, en.wikipedia.org/wiki/…
@Perturbative Does it matter?
invariant factor decomposition, can i permute the invariant facotre
@AaronMartinez Third, why do we take m→∞? could have been n? and why taking m→∞ implies the next inequality?? $m$ and $n$ are indeed equivalent here. We take $m\to\infty$ so that $f_n\to f$.
@AaronMartinez The inequality comes from combining $\left\vert \frac{f_{m}(x)-f_{m}(x_{0})-[f_{n}(x)-f_{n}(x_{0})]}{\Vert x-x_{0\Vert}}\right\vert \le2\epsilon$ and the convergence
Hey all--anyone care to explain why the flags? Did someone want their own messages deleted, or was someone worried about an IP-infringement on that page-snippet above?
21:28
weird internet
@nitsua60 Wait, you mean someone's been flagging again ?
@nitsua60 Look at the starred messages :( not much we can do
G'night again, Mike
@SteamyRoot Three messages (incl. the page snippet) were regular-flagged.
It looks like they were dismissed across the network, so I'm not sure it'll be easy to figure out which the other two were.
@Hippalectryon These were not messages with profanity. I believe it was those last two of perturbative's, and I can't recall the third.
@SteamyRoot Well $M \subseteq \mathbb{R}^k$, but to me that diagram implies either $g$ or $h$ could map points of $U_1$ and $V_1$ to points in $\mathbb{R}^k$ outside the manifold $M$
21:31
@Hippalectryon oh ok
@AaronMartinez So far so good ?
@Perturbative Their images are already defined, though. The only thing that really matters in that diagram is that it commutes, I'd say.
(still had it open elsewhere)
@SteamyRoot Ah okay, I guess I'm being a bit pedantic
Those flags don't make any sense >.>
21:33
@Hippalectryon Yeah, but I knew there had been flag-drama here before, so thought I'd see what's going on.
@Hippalectryon yes I think My confusion was because of the m, We dont have convergence for f_m, but if m=n then we have the convergence
@Hippalectryon One thing I'd recommend you-all do is poke your mods about assigning some room owners who are actual regulars. It seems like your four room owners aren't actually active chatizens?
3
That's a good idea
@Hippalectryon Don't spam stuff at me like that.
@t
@TheGreatDuck Sorry >.>
21:36
What was that anyway?
Tern is an owner and is often here (under a different name). Admittedly, robjohn isn't around too much.
@nitsua60 I own a room and I'm only not here as I was kindly asked to stay away from chat by the mods here.
and on that note I'm out of here
@TheGreatDuck I typed fast and mistook you for another user
oooh
kk
@nitsua60 who flagged them if I'm allowed to ask that?
@AaronMartinez On to the next one then. Fourth, why do we take the n=nϵ?? I really don't see this step. We take $n=n_\epsilon$ so that the inequality on I still holds
21:39
@TheGreatDuck [rummages...]
Another flag?
great
it wasn't me (i hope)
@Hippalectryon so its just like notation, just to empathizing n depends on \epsilon ?
So many flags you'd think it's the UN
Anyway
21:42
badum tss
@AaronMartinez Well, since the inequality $I\le2\epsilon$ holds for any $n\ge n_\epsilon$, we could have taken any $n\ge n_\epsilon$. But for the sake of simplicity we take $n_\epsilon$.
@TedShifrin the thing is anon is an owner, tern isn't :P
@Daminark that joke is horrible. Go to the punitentiary.
21:45
maybe tern should make himself the owner instead
@nitsua60 I gotta go. When you find something, ping me please.
@TheGreatDuck Turns out I can't find the flagger. CM can, though.
@TheGreatDuck Cya!
a CM (shog, in fact) did investigate the issue a few days ago and identified the flaggers and warned them
@Hippalectryon oh ok n_\epsilon then
not sure if the same person is involved
if yes i vote for a good long suspension
21:49
Gentlemen, I would like to learn many theorems by heart. Do you have any idea about how this should be done?
I'm currently in contact with other mods and CMs about it.
@AaronMartinez Fifth, We do we need to repeat the proof?? ins't x0 arbitrary?? x0 is not arbitrary, because of hypothesis 2)
@Kirill Every time you have to look one up, write it down on a card. Then throw away the card. (That's what I have my students do.)
@Hippalectryon right,so the proof will need to be repeated?
@AaronMartinez Yes. If you repeat it for all $x\in X$, then you're done.
21:52
Flagger is chat-banned for a few days.
@nitsua60 that is a nice one! Could you also advise me about the categorization? I have 17 theorems in one lecture, 31 in another. I think it would be pretty untalented to learn them by numbers. I tried to do some graph networks between definitions and the theorems where these definitions are used, but the whole graph is too difficult then and cannot be reproduced on the one paper.
I can't promise I'll be here daily, but I'll try to pop in often enough that I'm pingable. Please don't be shy.
@Hippa Comment on fait pour y aller à Notre Dames des Champs en rer ?
@Astyx Tu pars de quelle gare ? Luxembourg ?
@Kirill I would, but my SO's yelling at me to make dinner. (And they're right--I should have stepped away to do this ten minutes ago!)
21:54
@Hippalectryon La même que toi a priori, si t'es bien à Palaiseau
@nitsua60 Thanks
@Astyx Non je suis rentré sur Paris. Je vais te dire ça.
@nitsua60 Nice, thanks.
@nitsua60 could you write me a message then when you have time? Is that ok?
Thanks @nitsua
@Hippalectryon so would I have to use x instead of x_0 in the whole proof ?
21:57
@Astyx Lozère -> Denfert Rocheraux puis prend la 4 direction Montrouge et descend à Vavin (2 arrêts)
@Hippalectryon J'imagine que je peux m'arrêter à Port Royal
Et marcher un petit km
@AaronMartinez Yes, but it doesn't actually require you to rewrite the whole proof. Basically it goes as following: during the proof, we've shown that $f_n$ converges anywhere. Therefore any $x$ follows hypothesis 2), and if it works for $x_0$ it works for any $x$.
@Astyx oui, ça fonctionne
Nice, merci
À demain du coup, moi je suis encore en préparation pour les oraux !
@Astyx ok, bonne chance :-)
@Hippalectryon oh I see
22:02
@AaronMartinez And the last question, how can the proof be formal?, i.e. at the beginning it must be always the ϵ>0ϵ>0, and maybe the x,x0∈Xx,x0∈X must be given too? or they must be given in the middle of the proof? or where should they be? or how? I'm not sure I see what you mean
@Hippalectryon I just thought or had the impression that the proof wasn't written in a formal way, but maybe its just my nonsense idea
@AaronMartinez It looks OK to me. Did you have a specific point in mind ?
@Hippalectryon oh then I think its fine :) . No I didnt
@AaronMartinez Great :-)
@Hippalectryon yeah thank you for helping me :D
22:08
No problem. A tip for next time though: I think the main reason you didn't have any answer despite the bounty is because your questions weren't very clear, you might want to improve that next time :-) @AaronMartinez
@Hippalectryon HAHA yes I think you're right. I'll take it to account for the next time
22:33
@BalarkaSen Well, I was supposedly the other flagger. Dodsy was flagging a bit as well. I know it was definitely not me this time. I cannot conceive it occurring besides someone literally hacking the site.
@nitsua60 why did you chat ban someone for flagging?
but who am I talking? I'm chat banned for spamming last week.
:[
Questions about dice and probability: If I have a number of dice, N, what are the odds of getting at least 2 sixes?
@GhotiandChips What's the odds of not getting at least two sixes ? ;-)
@Hippalectryon that's not helpful.
@TheGreatDuck ? It seems easier to me. That way you don't have to sum from 2 to N
@GhotiandChips find the number of permutations where 2 are 6
@Hippalectryon no cause that method requires you to find the probability with more than 2.
@Hippalectryon wat
just find the prob for n = 2
isn't that the question?
22:42
N is arbitrary
im referring to n = number of dice with a 6 on it
find the probability when exactly 2 out of dice have 6 on it.
@TheGreatDuck Ah, well that gives you the prob of having exactly two 6. Not at least two 6
isn't that what they asked for?
"what are the odds of getting at least 2 sixes?"
22:44
:P
I have an easy time finding the probability when resolving just 2 dice – there is only one instance in the sample space (36), so it's 1/36 (1/6 * 1/6), but /at least/ 2 sixes gets difficult for me to understand once you have more than 2 dice
@GhotiandChips What's the probability of rolling n dice and not getting any 6 ?
@GhotiandChips yeah, it's basically the digit problem as I like to call it
@Hippalectryon there's a more intuitive way
@GhotiandChips how many different 3 digit numbers are there?
10*10*10, right?
so the total number of combinations are 6*6*6*6... n times
@GhotiandChips look up how to choose 2 from N
then, that is the number of individual circumstances where it is 2 6's or greater
the rest are the remaining "digits"
22:52
And what if you were looking for 2 successes, where the result 5 and 6 is a success
hm?
i suck at probability
:p
@TheGreatDuck My way works for this one :-)
@GhotiandChips What's the probability of not getting any 5 or 6 when rolling n dice ?
@GhotiandChips is it 2 5's and then 2 6's?
@TheGreatDuck Where did you get the idea that I did?
5,6 or 6,5 is fine
22:54
@nitsua60 this post
At least one 5 or 6, with a total of 2 successes
@GhotiandChips that's a bit vague
@TheGreatDuck Ah, sorry for the unclarity. I was reporting that the flagger had been banned, but not that I'd done it.
@TheGreatDuck A dice is valid if it's on 5 or 6. What's the proba of having at least 2 valid dice out of N ? That's what is meant I think
(I didn't--a CM stepped in when a fourth inane flag popped up while I was still looking into the previous three.)
22:55
Apologies. It's hard to explain. Hippa explained it yes.
@nitsua60 may I ask who it was? We've been dealing with this for a while and i think knowing who it was might be good for everyone. I've been getting blamed for quite some time and i know I've made a couple flags here and there in the past but nothing real recent afaik.
@TheGreatDuck I don't actually know, myself.
granted, the lack of an (official) chat ban does clear that up I suppose
@Dodsy the flagger got chat banned. Will you now please back off?
(the irony if it were dodsy)
but i doubt it
@nitsua60 i'd star those, but it would be a bit silly to do so.
In case anyone's curious, 1, 2, 3, and 4 were the thrown flags that occasioned the chat-ban.
weird
who banned the person?
22:59
A Communiy Manager (SE employee)
i think flag is simply a new way to troll nowadays
what user are they?
the new trend
that's absurd

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