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1:05 AM
I wonder how many "limits-without-L'H" are homework questions
@SimplyBeautifulArt: The number of them is Θ(t) as t (time) tends to infinity.
@SimplyBeautifulArt Too many
1:25 AM
Is @BrevanEllefsen online?
1:40 AM
@user21820 @Wojowu @Deedlit @amWhy: I'm going to win myself: repl.it/Fomr/2. Don't know why I didn't compress the duplicate strings. I look forward to someone helping me to analyze this. The idea came from my proof of the Hydra game termination, where I used a strong third-order lemma labelled as "Exercise" in my code. I thought it was a waste to only use it as many times as the starting tree was deep, so I allowed downward extension, but it took me a while to find the proof.
R.I.P. everyone else in the competition probably just lost
Haha it's just to make it look bigger now. =)
:( It still scares me that I can make 355 rep in one day without bounties
I should take a break from MSE for the night...
I don't bother making reputation.
In fact the questions I answer very often gets me 0 upvotes.
By the way, there is an election if you guys didn't know
Also, the top 3 users this month all have calculus as one of their tags
1:47 AM
I got various badges for my answering pattern, like Tumbleweed and Unsung Hero.
@SimplyBeautifulArt my phone has a habit of only letting me know about updates 20 minutes after. I am sorta online atm. (on mobile). What's up?
I'm not a member of Worldbuilding SE so I won't vote.
@BrevanEllefsen It's a thing with chat, that's all
A: Slick way to calculate $\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T(\sin(t+1)-1)^4\,dt$ by hand?

Simply Beautiful ArtNote that the integrand is periodic, hence: $$I=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T(\sin(t+1)-1)^4\ dt=\frac1{2\pi}\int_0^{2\pi}(\sin(t+1)-1)^4\ dt\\I=\frac1\pi\int_0^\pi(\cos(t)+1)^4\ dt$$ Binomial expanding and Chebyshev polynomials of the first kind: $$\begin{align}8(c(t)+1)^4&=8c^4(t)+3...

@BrevanEllefsen I thought you'd be interested in that
And I'm interested in other answers as well
@user21820 Ok
I kinda like worldbuilding
it's like when you were a kid with lego blocks
Some of the answers are very interesting, especially when it starts like "Actually it already exists on earth..."
But I never felt a need to ask or answer a question.
1:51 AM
@user21820 I felt the need to upvote things far too many times
Ah I see.
Well usually I get there by the Hot Network Questions List, by which time the good answers already have tons of votes so no need to add another.
But on Math SE I upvote answers that are good that I see.
I just can't help it XD
As for that integral, isn't it actually easier to just expand?
1,4,6,4,1 isn't hard.
1:55 AM
The OP probably was thinking about changing sine into exponential form
and apply binomial theorem again
Your answer honestly looks like more work.
Lol, compared to what?
just directly integrate then find limit.
But that's tedious
how are you going to deal with all the $\sin^n(t+1)$?
Hmm I guess you're right.
I thought maybe it wouldn't be so bad since we do know those integrals.
But not sounding enticing anymore.
I guess the only trick is the shift due to periodicity, which you could justify a bit more.
2:02 AM
The integral on [-T,T] differs from that on [-T-1,T-1] by an amount that is bounded since the integrand is bounded.
Oh wait, easier to just go to the multiples of π.
Rounding the integration limits off to the nearest number that is one less than a multiple of π would change the integral by a uniformly bounded amount.
So that discrepancy vanishes in the limit.
I just like rigour hahaha.
I'm trying to take a derivative with multivariable chain rule
And I keep getting the wrong answer
Oh, nvm
I hope I didn't overkill with my answer
I mean my newest answer
@user21820 So you know those x^x^x type problems where people take the derivative?
I just love doing it with the multi-variable chain rule
y=u^v, u=x, v=a^b, a=x, b=x
$$\frac{dy}{dx}=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}$$
And then expand $\frac{dv}{dx}$ in the same manner
I'm probably the only person who enjoys doing that :-/
2:23 AM
It probably looks like magic to some people
Is there a situation where it actually makes things easier?
The last time I used the multi-variate chain rule was in undergraduate introductory multi-variable calculus.
So long ago.
2:25 AM
Well, maybe it doesn't make it easier, but it does avoid a lot of rewriting you would usually do
Also lets me fit possibly long derivatives in a few lines
I see.
Well I always rewrite exponentiations anyway.
For asymptotic expansion, that's always the way to go.
So I've never even realized the multi-variate chain rule would avoid it.
I wonder how a professor would feel if his first year calculus students brought him multivariable chain rule in some derivatives
2:28 AM
I think it's fine if the student knows what he's doing. Specifically multi-variable calculus is often taught quite sloppily so there are pitfalls that are hard to see.
I finally overcame the bounty I placed since 7 o'clock
XD no longer ranked with negative rep
@user21820 you know, when you place a bounty you lose rep
And the week just restarted
2:31 AM
Yea I know. Didn't realize what you meant at first.
My bounties on my own questions didn't work on average so I gave up using them.
I use them mostly for other people questions
Doesn't really matter how it works out
@user21820 if rep is not for bounties and privileges, it's practically useless
Well it's not totally useless; with enough rep it allows me to vote to delete troll/crank posts.
But that's if it already has a negative score; the terrible part is that I can't even vote to delete if it has non-negative score.
:-/ but you don't actually use up rep for that
Well so I need the rep to get that ability first. And I need 1 point to downvote before I can vote to delete.
But yes that's roughly why I'm not farming rep since I don't use it much.
I also severely dislike giving answers to questions with no context or no effort.
Which is the bulk nowadays.
Well, I gotta head to bed, so I'll see you tomorrow
[so sleepy, I literally made about 5 typos just now (>.<)]
2:43 AM
Sure good night!
7 hours later…
9:58 AM
@user21820 While I might be totally wrong on that because my understanding of notations as strong as that is very minute, from what I have heard (but AFAIK which has not been confirmed by any sort of a formal proof) the matrix system which Deedlit has coded has strength beyond the second order arithmetic, which would make his program give a larger number than yours
Also note that I'm not just trying to deny your victory (I try to emphasize, as much as I can, that I don't know if I'm right)
Also, would you mind explaining what it means, in your program, that "[the label?] is duplicated in a downward chain of height (c)"?
10:27 AM
@Wojowu: On first glance the lemma I used (the decreasing sequences from any well-ordering S are lexicographically well-ordered) seems third-order. By the way, what exactly do you mean by "goes beyond"? If a function cannot be proven total in a system S, does it say anything about its growth rate? I was wondering about that but could not see a direct relation. I only know that S cannot prove totality of well-orderings beyond its proof-theoretic ordinal.
@Wojowu Here I meant "[the node] is duplicated..." Namely, if the leftmost leaf x has a non-blank label y, encoded as x = [y] in the program, then you first reduce the label before replacing the resulting node by a chain of copies of height c. If c = 2 and y is reduced to y', then x is replaced by the downward chain ( y' -> y' -> y' ), encoded as [[[y'],y'],y'] in the program, since the label is the last list item.
10:55 AM
Indeed, just because a function is not provably total doesn't say much about its growth rate, but with "goes beyond" I meant precisely that it is faster than all functions provably total in SOA
As for third-orderness, I think it can be framed in second-order terms, since all decreasing sequences are finite and hence can be encoded (with some sort of encoding) as numbers
Ah, now I understand what you mean with the reduction. This makes it quite unlike the Buchholz hydra, but I don't know how they compare
Yea it's just trying to pump the lemma as far as I can.
On one hand, Buchholz hydra duplicates a larger part of the tree, not just one node, but it has smaller set of labels
I think the key difference is that it has non-monotonic labels.
I'd lean towards guessing Buchholz hydra is stronger, but I might be wrong
I'll probably be reading about it soon anyway.
Then I can try to see for myself haha..
11:01 AM
Good luck, I guess ;)
The paper I'm referring to is "A Relationship among Gentzen’s Proof-Reduction,
Kirby-Paris) Hydra Game, and Buchholz’s Hydra Game
(Preliminary Report)*"
kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0912-7.pdf.. I've not read it yet; hopefully I know enough to understand it.
I'll definitely take a look at that
It's hard to find actual papers about these stuff haha..
There seems to be quite a bit of technical detail there
Which is always good :D
And yeah, I agree
Yea I only have some basic proof theory background..
11:10 AM
Hm, it seems to me that this paper only talks about a very weak ("one dimenstional") variant of the Buchholz hydra, which is what I know under the name of Beklemishev worm
Yea, I'll have to hunt in the references...
Here is Buchholz's original paper if you want epub.ub.uni-muenchen.de/3842/1/buchholz_wilfried_3842.pdf
It is somewhat terse though
Ah okay thanks I'll take a look.
Terse indeed.
Where's the big picture lol.
What big picture?
I mean, the paper has mostly symbols and no big picture (overall intuitive sketch)
11:22 AM
There isn't much place for intuition in proof theory :P
And even then, I'd guess most mathematicians wouldn't bother with that and instead just go straight to the proof
(as Buchholz does in this paper)
Heheh.. If I ever get through it (or prove it myself) and have an intuition for it I'll tell you.. I actually think logicians do like intuition and it's only in papers that they often just give the proofs.
Might be so
This is probably the case in most of mathematics
Right now I feel like taking a nap haha..
See you!
Cya :)
12:05 PM
@user21820 I believe this answer on MO speaks of the same operation on well-orderings as you use (passing from an ordering to its decreasing sequences), just defined a bit differently (but equivalently): mathoverflow.net/a/139015/30186
It shows that PA (or, equivalently, ACA_0) can prove, for any well-order X, that this new ordering (denoted $\omega^X$) is also a well-ordering
So I feel like it might be possible for your function to be proven total in ACA (without the 0)
1 hour later…
1:22 PM
@Wojowu I don't see how it is equivalent. Their statement is not about decreasing sequences (with unbounded length), but about finitely nonzero sequences with tail-first ordering, so cutting off at the first sequence of a descending chain implies that all subsequent sequences in the chain have nonzero coordinates at most that of the first sequence's last nonzero coordinate. Then it is a straightforward induction on the coordinates.
In contrast, the lemma I used does not share that feature; if you cut off at the first sequence of a descending chain, you still have no guarantee on anything, and my proof is not completely constructive in the sense that I need to recursively define a sequence that lower bounds the chain (using infinitary pigeonhole, namely a union of finite sets is finite) and then show that that lower bound is eventually constant and hence must have been in the original chain.
So if there is some equivalence over PA I am not seeing it; feels very different to me.
1:50 PM
@user21820 Indeed, I was wrong about them being equivalent, but they "almost" are; consider instead that the maps are from X to {0,1}, not to $\omega$, then almost-everywhere-zero sequences from X correspond bijectively to (necessarily finite) decreasing sequences of elements from X.
Under this identification, the "tail-first ordering", as you call it, of these sequences corresponds to lexicographic ordering of decreasing sequences from X - in essence, in the latter you first compare the largest elements, while in the former, you first compare the position of the first position of a nonzero element
You can prove that the ordering of sequences is a well-ordering by a simple induction as well - but in that case, you do the induction along the largest element of X appearing in the decreasing sequence (which corresponds to the largest index of a 1 in the other ordering)
So your ordering doesn't take X to $\omega^X$, but rather to $2^X$.
@Wojowu: In my lemma, I stated (non-strict decreasing sequence) and it is crucial to allow my function to grow fast. However, I see that you could use ( X -> ω ) since each number is the count of how many times the item appears.
Ah, so with "decreasing" you mean "non-strict", that makes sense
But I still have to think about whether it is equivalent enough, because as stated in the MO post one explicitly quantifies over finitely nonzero sequences on X, which does catch all decreasing sequences from X, but that's external to the system.
Well technically we should use X -> omega+1, since the last element might repeat infinitely many times
Haha no it's just ω still, I only need finite decreasing sequences.
2:00 PM
Alright, so omega it is :)
I'm not sure what you mean with "that's external to the system"
My lemma states "for every well-ordering S, the set of all decreasing sequences from S ...". "the set" here can externally (in ZF say) be seen to be equivalent to the form you stated above.
But it may make a big difference internally.
Because two statements that are equivalent in a stronger system may not be provably equivalent in a weaker system.
I do not know enough to judge this case, but I feel that there is some little bit of non-constructiveness in the equivalence proof (in ZF).
But a decreasing sequence can be encoded using a number, and thus the set of decreasing sequences can be seen as the set of natural numbers
And I disagree, the equivalence proof is about as constructive as it can get
I provide an explicit order-preserving bijection
Wait let me think a bit.
Take all the time you need to think :P
@user21820 If you don't have enough time, just make a code to think for you, since computers are allowed infinite time in this contest.
2:11 PM
Ok I agree here. I kept forgetting that the well-ordering I use in my super-tree proof is always computable.
@Wojowu But let me come back to this comment you made earlier.
@user21820 Okay I see the error is with my comment here.
Previously zero coordinates can become nonzero later in the decreasing chain, so it is not a straightforward induction on coordinates.
@user21820 Did you just reply to yourself?
I've no choice if I want to link to my own comment haha..
I like full documentation. =)
If we know sequences with all nonzero indices <alpha are well-ordered
Yes that's right.
Then consider a set of sequenes with indices <=alpha
2:15 PM
Q: Asymptotics to the sequence leading up to an infinitely nested radical

Simply Beautiful ArtInspired by one of my previous questions, I have found that if $y_0=0$ and $y_{n+1}=\sqrt{2+y_n}$, then $$y_n=2\cos(\pi/2^n)$$ From there, it's easy to see from Taylor expansions that $$y_n=2-\frac{\pi^2}{4^n}+\mathcal O(16^{-n})$$ Does it hold true that if $y_0=0$ and $y_{n+1}=\sqrt{a+y_n...

Then choose the smallest index appearng at alpha
@S.C.B. Hi
*the smallest number which appears at index alpha
Interesting question you found
2:17 PM
And consider the elements S' which have this number at index alpha
Then we know there is the least element of the set S', which we find by "ignoring" the value at index alpha
And the least element of S' is also the least element of S
@S.C.B. Hasn't been busy in my tags
So basically we need induction along the well-ordering S. We can go by induction over the count of that largest element and when the count stabilizes we use the induction along S to finish.
What are the other two users in this room talking about?
The linked post?
2:21 PM
We are talking about well-orderings
@S.C.B. The contest
In the case of the MO post yeap.
In relation to this entry repl.it/Fomr/2 and its proof of termination
(made by user21820)
@SimplyBeautifulArt your question is really about the rate of convergence
@Starfall Of course
2:22 PM
which drops out from something like the banach fixed point theorem + the mean value theorem
@user21820 Do you mean induction along X?
@Wojowu: Lol I accidentally edited my comment up there in response to you. This no-deleting-after-5-min in chat is silly.
haha indeed it's easy to edit by accident
2:24 PM
Because pressing the Up arrow key allows one to (too) easily edit previous comments.
Hey I just got an interesting idea; we could stage a backward conversation.
Sounds like a horrible idea
That people have to read from bottom-up.. you know the kind that makes people hate to read email from mailing lists.
But yeah, the well-ordering proof is about as simple as that
Yea so in the case of the Hydra game every depth-n hydra can be proven to terminate in PA.
But not arbitrary depth because the proof only shows that there is a proof for finite n.
Well, PA doesn't even prove that every depth-0 hydra terminates
(since it's equivalent to Kirby-Paris hydra)
2:29 PM
I meant the original hydra.
Yes my 0-tree is equivalent to the original hydra.
Ah, confused k-tree with depth-k
Yes, PA proves termination of bounded heihgt hydras
So I guess the question is how far does the inability to get from finite height to arbitrary height internally push my super-tree function's growth rate up.
I suppose you think ACA is enough, and I guess I won't be surprised.
Since ACA has full induction I feel like it should be the case
But I might be wrong
Ok I'll wait for Deedlit's feedback as well. =)
I'm going to go off soon. See you all next time!
2:34 PM
Bye \o
2:52 PM
Hello, @Ramanujan
Hello @S.C.B.
Still having difficulty with the ABC problem?
@SimplyBeautifulArt Why did you send me that?
I got $A:B:C=7/120:1/24:1/40$?
@S.C.B. Perhaps it would interest you
@SimplyBeautifulArt There are way to many possible answers. For example, $$\int_{0}^{1} \frac{x^3}{3x^2-3x+1}$$ or $$\int_{0}^{1}\frac{x^5}{5x^4-10x^3+10x^2-5x+1}$$ are both good examples of how this property can be used, IMO.
2:54 PM
@S.C.B., now what's next?
You are too fast
@Ramanujan We distribute the money as so.
Then @S.C.B.
, how to calculate the amount received by each
@S.C.B. May I borrow said examples?
2:57 PM
No, I'm writing an answer, sorry.
Ok then
Ramanujan $A:B:C$ each contributed $$7:5:3$$
@S.C.B., how is that $7:5:3$? Could not understand.
Well, I have to go now
2:59 PM
Multiply $120$ to each thing.
@S.C.B. (:P Told you you might like that question)
Then only multiplying in the RHS?@S.C.B.
Yes. $a:b:c=ka:kb:kc$.
3:08 PM
Why multiplying by 120 only on RHS not disturb the equality,? @S.C.B.
Because $a:b:c=ka:kb:kc$.
What is $K$? Where does it come from? @S.C.B.
Any real.
Anything is possible for $k$.
@S.C.B.: Nope. False for $k = 0$.
OK, a non-zero real $k$.
3:11 PM
Yup. =)
@S.C.B., $A:B:C=a:b:c =ka:kb:kc$?
The probability that a boy can get the scholarship is 0.85 and the probability that a girl can get the scholarship is 0.65. What is the probability that at least one of them will get the scholarship?
So have you resolved any remaining qualms?
Yeah @ S.C.B
3:20 PM
Could you accept my answer then?
I'm not forcing you though.
@Ok @S.C.B., I will surely accept yours.Don't worry
3:38 PM
@SimplyBeautifulArt I've written the answer.
A: Two interesting results in integration $\int_{0}^{a}f(a-x) \ \mathrm{d}x= \int_{0}^{a}f(x)\ \mathrm{d}x$ and differentiation of powers of functions

S.C.B.There are a lot of possible answers. For example, $$\int_{0}^{1} \frac{x^3}{3x^2-3x+1} \mathrm{d} x=\frac{1}{2}$$ or $$\int_{0}^{1}\frac{x^5}{5x^4-10x^3+10x^2-5x+1}\mathrm{d}x=\frac{1}{2}$$ are both good examples of how this property can be used. We can use this property to calculate these compli...

4:29 PM
4:49 PM
Who are you @Stardust?

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