Mathematics

Associated with Math.SE; for both general discussion & math qu...
Jul 17 11:26
sorry for caps
Jul 17 11:26
WHY ARENT FUNCTIONS DEFINED AS OPTIMIZATION SOLUTIONS ??
Jun 28 10:42
I assume people like to know such things that look similar
Jun 28 10:42
@leslietownes Like I said ; just a comment. Not the main question but a similar looking integral
Jun 28 10:41
@leslietownes ofcourse it gives 2018 vibes. its an old post from 2018 :p
Jun 27 23:50
HELLO
Jun 27 23:42
in the comment i made a conjectured value. is it correct ?
Jun 27 23:42
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Q: $ y = \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx $

mickConsider $$ y = \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx, $$ where $\operatorname{li}(x)$ is the logarithmic integral. Is there a closed form for y ? It appears that a good approximation is $ 10 \cdot \operatorname{Ci}\bigl( \frac{56}{19}\bigr)$, where $\operatorname{Ci}(x)$ ...

Jun 16 07:12
Why does Alessandro Codenotti say no kind of algebraic topology should be allowed ( STARRED TWICE HERE ) ??? Is it too advanced ? For mathoverflow ? Or was he joking ? and so why ?
Jun 5 00:06
PRETTY WILD HMM
Jun 5 00:06
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Q: Connected complicated continued fractions ??

mickConsider $$A = 1+\dfrac{2+\dfrac{3+\dfrac{4+\cdots}{5+\cdots}}{4+\dfrac{5+\cdots}{6+\cdots}}} {3+\dfrac{4+\dfrac{5+\cdots}{6+\cdots}}{5+\dfrac{6+\cdots}{7+\cdots}}} $$ and $$B = 1+\dfrac{3+\dfrac{5+\dfrac{7+\cdots}{6+\cdots}}{4+\dfrac{6+\cdots}{5+\cdots}}} {2+\dfrac{4+\dfrac{6+\cdots}{5+\cdots}}{...

Jun 5 00:05
@copper.hat I HOPE NOT
Jun 5 00:04
who wants to see an unlikely weird conjecture ???
Jun 5 00:04
hi@Jakobian
Jun 4 23:11
guess who's back, back again? micky's back, tell a friend. It gets so lonely without me.
May 19 19:55
Hi guys. I was wondering, suppose an account gets hacked, a moderator goes crazy or a technical error occurs etc … is there a backup for all the stuff here ??
Apr 14 21:49
hi @Thorgott
Apr 14 21:48
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Q: Zebra tours on a 10x10 board?

mickI was thinking about knight tours. https://en.wikipedia.org/wiki/Knight%27s_tour I also wonder about knight tours for bigger boards such as 10x10 or 12x12. Maybe this belong more in a puzzle group or math group, but I take my chances here. Migration is an option as I am also active on mathstack. ...

Apr 14 21:48
i had a question about moving chess pieces of math nature. I posted it on chess exchange but i am uncertain if it belongs there or here or on puzzle stack ?
Apr 14 21:46
guys
Apr 14 21:46
hi huys
Mar 30 23:27
hi
Mar 25 12:08
So yes I got a new question
Mar 25 12:08
@XanderHenderson it is partially resolved. I was able to, but not on my mobile.
Mar 25 08:24
I can’t post questions anymore !!??
 
Jul 1 07:49
@jjagmath yes but I am not convinced. Are you ?
Jul 1 07:49
Why is t transcendental ? I cant even prove $\exp(1/e)$ is transcendental.
 
Jun 8 11:38
@ThomasAndrews but you make it seem like it is not legit. And you are changing the goal or adding questions. Anyway the conjectures were false so ... I can live with it...
Jun 8 11:38
@ThomasAndrews the ellipsis clearly shows how the truncation is done (best). Proving convergeance formally might be a good exercise but it does converge for sure. Might add it later.
Jun 8 11:38
@ThomasAndrews no not just an assertion. It is true. Since the further branches reach unit fractions fast this is a fact. Henry gave a way to compute it formally.
Jun 8 11:38
@Empy2 what is that $n$ ? the level of truncation ? no that cant be. $A$ and $B$ are not functions and not asymptotic to $n$ but constants so I am unsure what you are doing there.
Jun 8 11:38
@ThomasAndrews all of them converge
Jun 8 11:38
@Henry are you sure about those digits for $B$ ?
Jun 8 11:38
Both A and B converge just fine.
Jun 8 11:38
@ThomasAndrews the simple pattern is now added to the OP.
 
Oct 16, 2024 10:43
@GTonyJacobs thanks ! very intresting. I will work on it more some time later. My argument was for certain starting values , not sure how it affected the average.
Oct 15, 2024 15:58
@GTonyJacobs It makes divisions by 5 much much less likely.
Oct 15, 2024 15:57
@GTonyJacobs the conjecture where i say both methods give both a factor above 1 or both below 1
Oct 15, 2024 11:12
@GTonyJacobs so my conjecture is wrong you say ? Can you give an example ?
Oct 15, 2024 11:12
@GTonyJacobs i mentioned that in the OP ? division by 5 was impossible for some starting values or so.
Oct 14, 2024 19:31
Im not sure what to answer. The estimate was really intended to show it goes up and diverges or goes down to 1. And in fact I dare to conjecture that both ways of estimating are either both below 1 or both above 1. For instance a factor by 7/15.071 and a factor by 7/15.077 both suggest convergeance to 1. Notice i made an exception for 11n + 1 in the comments. So these extreme cases not considered in my conjecture. Im not sure if case 13n+1 is a similar case as my 11n+1 exception.

Furthermore I am confused by the question if i consider the probability of one iteration or ideally the full i
Oct 7, 2024 22:15
your method gave 15.071 whereas mine gave 15.077. It .is only a small correction ! You worked with mod 30. Maybe if you work with mod 30^5 you get closer to my 15.077 ?
Or do you think you will get 15.06 ??
Oct 7, 2024 12:58
I wonder if we can make a collatz variant where the estimated change is multiplied by 1 rather than 3/4 or the above ...
Oct 7, 2024 12:58
@G Tony yes i agree with that
Oct 7, 2024 12:58
Btw what happens if you use mod 27000 ? You probably get different numbers then.
Oct 7, 2024 12:58
@G Tony : I am not saying you are making huge mistakes or anything like that. In fact I , as only one btw , upvoted you. So enjoy my upvote rather than feel sad and say this sucks lol. I need to think more about it. I will come back to it. Thank you for your post.
Oct 7, 2024 12:58
Basically we both use different "random" and "nonrandom" assumptions or at least modelling , and I feel these are conjectures on their own. Maybe your method works better for integers below say $10^{100}$. But for very high imput values and many collatz steps im having doubts if one is superior. Despite appreciating your effort nevertheless !
Oct 7, 2024 12:58
The divisors of $30 n + a$ do not only depend on $a$ but also alot on $n$, which btw is nonrandom after a few iterates of the collatz type iteratios. For instance if after 53 steps of this collatz type we get some $30 n + 1$ it is not easy to say how many times the next step ( $30 * 7 n + 8$ ) is divisible by $2$. We get $105 n + 4$ ofcourse but beyond that it is difficult to tell, it depends on the nonrandom $n$ here. I feel like your "n " is not changing when you go from say $a + 30 n $to $b + 30n$ in say 13 steps, but in fact that $n$ is no more the same. Therefore my slight reservations.
Oct 7, 2024 12:58
For collatz on average we multiply by 3 and divide by 4 , so we get a factor 3/4. Just to be clear.
Oct 7, 2024 12:58
@G Tony : so what is the best estimate for the original collatz ? You considered mod 30 but higher mod give different results. IS there a CLOSED FORM converging estimate ? And is your improved estimate not equivalent to testing many numbers ?