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4:02 PM
you could take $f(x)\in I\implies f(x^n)\in I$ as a criterion unto itself, but it'd be nice to have a more familiar way to present that. (there might not be one)
 
Max
@DanielFischer ... because this means that the difference $f(x)-g(x)$, which is equivalent to zero, stays zero when raising $x$ to a higher power?
 
You can say it so.
 
Max
@Semiclassical yeah i was looking for the easiest way to explain it to someone who has not previously given any thought to the problem
 
@RandomVariable Thank you. Well, I don't have other better option, I need to highly believe in myself and work crazy hard, this is the only way to success in my opinion. I wanna created amazing questions, I wanna obtain amazing proofs, I wanna be like Ramanujan, it's one of my dreams.
 
Haven't been able to simplify the criterion yet
 
4:08 PM
@Chris'ssis Good! You are already Ramanujan, sort of!
 
Max
thanks a bunch for everything so far guys
 
@Committingtoachallenge You may have that opinion, but I don't think Chris's Sis is narcissistic.
 
@RandomVariable Have you ever seen this question?
$$\int_0^{\pi/2} \frac{x \cos^2(x) \cot(x)}{3+\cos(4x)} \ dx =\frac{\pi}{128} \log(9232+6528\sqrt{2})$$
I created it last days.
 
i think trying to adjudicate narcissim on an internet forum is a fools errand :/
both the accuser or the accused are far more likely to get frustrated (and frustrate others) than to arrive at anything at all persuasive
 
That's right.
 
4:14 PM
@Chris'ssis I think I saw you post that yesterday. But other than that, I haven't seen it.
 
@Semiclassical I think I need to check the dictionary whenever you write, lol.
 
i'll take that as a compliment :)
 
I have a surprisingly small vocabulary. I usually express all my ideas using very simple words.
I have read almost nothing except math books after high school.
 
i have a fairly wide vocabulary, which also means that i occasionally remember words entirely wrong. harder to keep precise definitions in your head when you have too many of them
 
@RandomVariable OK
 
4:19 PM
suppose it's also a matter of styles. i prefer knowing a lot of things somewhat imprecisely than having a precise mastery of one subset
 
I do not know precisely what I know or what I know precisely, lol.
 
snerk. i know one thing precisely: that i know a great deal of imprecise things :>
 
Hi all
there is something still wrong. What you have is called the cosine integral, it does not have an elementary antiderivative, hence you cannot use FTC on this.
Is this correct? It has $\text{Ci}(e^t)+c$ why can't someone use this and FTC to calculate a definite integral?
 
to clarify, you're wondering whether that comment is valid?
 
Yes
 
4:38 PM
Hi @Anastasiya-Romanova, lol.
 
@JasperLoy Yuhuu...
Need help...
4
Q: Integral Contest

Anastasiya-RomanovaBefore you answer this OP, please read all the terms and conditions below. Thank you... Today I hold an unofficial little contest on brilliant.org. Now, I will hold it here on Math S.E. It's just for fun guys. (>‿◠)✌ Before we start the contest, here are the rules of my little contest that you ...

Anything you're able to do
 
@Anastasiya-Romanova You can ask a question, but you cannot hold a contest here. So I do not disagree with the closure of your post.
 
Anastasiya I'll let you know something that will make you happy; It has no poles, so feynman style is the only style.
Well, dunno what you include in Feynman style, but I don't think it's possible with complex analysis.
 
@JasperLoy There's no explicit rules for restricting to hold a contest here
 
@Anastasiya-Romanova The community are the rules.
 
4:46 PM
@Anastasiya-Romanova The rules on this site are made by the users and they may change from time to time. They need not be written explicitly.
 
That's the point of this site.
 
@UserX Any methods welcome
 
@UserX Exactly. So you may post, but they may also close your post. QED.
 
My best shot would be complex analysis
 
Ohhh, I hate demoCRAZY...
 
4:48 PM
Oh I manipulated the integrand a little too
Let me check if I got it somewhere
 
Majority decision hah!
 
Am I already done?
 
@Chris'ssis You mean you finished your book?
 
@JasperLoy No ...
 
Dominant and tyranny of the majority, poor minority...
 
4:53 PM
@Anastasiya-Romanova The noun is dominance.
 
The adjective is dominant though.
Another example is important and importance.
 
Doesn't help A LOT but splits it into two integrals
 
@UserX Sorry. No hint, lol
 
I'm not interested in your contesr
Contest* Just trying to help you evaluate it
 
4:59 PM
I'm trying to think of under what constraints an MSE contest would make sense
 
Does anyone else feel that we can somehow use Cauchy-Schwarz inequality on math.stackexchange.com/questions/986074/… ?
 
and the best I can come up with is to proceed via the Blog rather than MSE itself
since otherwise it really just doesn't fit with MSE's intention
 
I wonder where Balarka is
 
Hi @alizter, lol.
 
hi @JasperLoy
 
5:06 PM
hmmm
is this question on-topic?
0
Q: Discovering the mathematical nature of Nature - Galileo's inclined plane experiment

GMcPIn 1638 Galileo published Two New Sciences, in which he described his inclined plane experiment. He discovered that the acceleration of gravity was uniform, and could be modeled mathematically by the simple equation < Distance = c * t² >. Question: Was this the first discovery of the fact ...

i added soft-question + math-history tags, but it still seems a stretch
 
@Semiclassical it is a question about the history of mathematical modelling.
Probably belongs on that new site
math history
 
@Anastasiya-Romanova the first thing that struck me when I saw your integral was mathworld.wolfram.com/BinetsLogGammaFormulas.html
 
it seems rather vaguely formulated
 
@Semiclassical if you are unsure it is prob best for other s to deal with it
 
@Chris'ssis Maybe
 
5:11 PM
nod. hence why i'm not taking further action than the tags
 
Ubuntu 14.10 will be out in a few hours.
 
Woo hoo
I don't use Ubuntu though...
 
I will stick to Debian, Ubuntu, and Mint, using the MATE desktop.
 
My brother will be delighted for sure
 
It's nice to have something to install every six months, but some people don't want that.
 
5:14 PM
I am on mobile since my brother took my pc along with his new one because he didn't have the time to spot and copy/reinstall all programs/projects he had on mine
 
I think installing once every two years is fine, so I usually go for the long term support releases.
 
How does a freeware have technical support?
 
@JasperLoy Computers age too fast
 
@UserX Here support means free automatic security updates.
@Alizter Humans age faster.
Perhaps I should watch the movie Defiance on TV later. It's two and a half hours long.
 
@JasperLoy so $\text{Humans}(x)=o(\text{Computers}(x))$?
 
5:17 PM
Six months are fine.
 
@Alizter I have never used the big or small O notation in my entire undergrad, lol.
 
I have never understood big/small o ever
 
@JasperLoy I didn't get this from analysis. It is because I am currently studying number theory :P
 
But I never took programming either, so the algorithm speed approach is useless to me
 
@Alizter I see. I think it happens often with analytic number theory, which is analysis, lol.
@UserX It is just a notation expressing some limits.
 
5:19 PM
@JasperLoy I use $O(.)$ maybe in taylor series in Analysis now that I think about it
@UserX programming =/= computer science
 
Hi @Parth @Anastasiya-Romanova @UserX @Alizter
 
I am scared about my mathoverflow question
It is going to either be taken well or badly
 
@Sawarnik Celebrating?
 
@Alizter Link?
@ParthKohli Crackers?
 
@Sawarnik In any way.
 
@ParthKohli Not really.
Not today.
What about you?
@Alizter Alright then.
 
I don't ever celebrate with the use of crackers.
Probably because I'm scared of the bigger ones and am too bored of the smaller ones.
 
@ParthKohli lol, you know its exactly the same thing wimme :D
:D :D
How do you celebrate otherwise? With friends?
 
With my family, yeah.
 
Like what you do?
 
5:38 PM
downvotes without comments are kind've annouying
(at least when you're on the receiving end of them :P)
 
@Semiclassical What happened?
 
-1
A: Infinite Series -: $\psi(s)=\psi(0)+\psi_1(0)s+\psi_2(0)\frac{s^2}{2!}+\psi_3(0)\frac{s^3}{3!}+.+.+ $.

SemiclassicalHere is some progress towards a solution. To review, we have $\psi:\mathbb{R}\to\mathbb{R}^{3\times 3}$ such that $\psi_0=0$. (Notation: I take $\psi_0^{(n)}:=\frac{d^n\psi}{ds^n}\bigr|_0$). We further assume that $\psi_0'$ is a special orthogonal matrix (determinant 1) and $\psi''(s)=(A+Bs)\psi'...

 
:D
 
if there's a reason for the downvote i'd gladly withdraw/modify it, but just sniping it like that is a kick to my ego
 
^it happens :(
 
5:40 PM
@Semiclassical Take your downvotes like a man.
 
yessir
 
don't worry.
 
I am temtped to downvote because I don't like your avatar
 
I use O(.) too in taylor series to denote the error
 
@Semiclassical are you a girl?
 
5:40 PM
pffffft
 
I will downvote because your name sounds silly
 
@Alizter lol
 
I will downvote because I stumped my toe earlier
 
idea
 
5:41 PM
I will downvote because i am totally well jelly of your skills
 
"i will downvote because orion is aligned with the red moon!"
 
But I do it without understanding how the notation is used in other cases
 
@Semiclassical In all seriousness ignore the odd downvote. Or you will go crazy.
 
nod
i should find a better avatar, lol
 
@Semiclassical That was not meant to be taken literally :P
 
5:42 PM
this one's nice enough but not exactly creative
snerk, still
 
@Semiclassical have a half classical equation and half quantum
 
heh. if i was going to go that route, i'd do
$$\oint p(x)\,dx=2\pi \hbar(n+\gamma)$$
 
teh
 
bohr-sommerfeld quantization
LHS is the classical action for an orbit of a 1D particle
 
nods looking like he understands
 
5:44 PM
^
 
RHS is in terms of $\hbar$, and so s essentialy quantum
 
@Semiclassical As a mathematician. The RHS looks like pi times some weird thing then an integer with euler gamma
 
so the RHS tells you to only care about the LHS when it happens to give an integer multiple of Planck's constant
$\gamma$ is what's known as a Maslov index
usually it's something like a half-integer
 
almost euler gamma
 
and $\hbar$ is Planck's constant / 2$\pi$
 
5:46 PM
so is that not just h?
 
right
it's a taste thing :P
i like \hbar better than h, ergo dumb ways of writing things :)
 
I remember studying schrodinger equation a bit ago
you know, for laughs
 
though, i should point out that this just boils down to a statement about asymptotic analysis of schrodingers equation
 
Didn't get very far.
i like quantum more because I don't and won't have to do the experiments
 
setting aside some other constants, Schrodinger just says $-\hbar^2 \psi''(x)+V(x)\psi(x)=E \psi(x)$ for a given real function $V(x)$
 
5:48 PM
engineers do it
 
if you take $\hbar$ to be small (in the units of the problem) then you can carry out various approximations
which amount to the integral requirement i quoted above
 
ah
What is the hamilton operator I forget
 
$-\hbar^2 D_x^2+V(x)$ in this context
i'm taking the mass to be exactly 1/2 :P
 
Huy
@Semiclassical: I looked at it a bit again today, but still didn't manage. :(
 
i read mass as mess :P
 
5:50 PM
what's cute is that, if you take that potential to be something like x^4-a x^2 (which is a nice model)
then the integral $\oint p(x) \, dx$ is nothing more than some complete elliptic integral
which is some function of energy $E$.
 
Huy
@Semiclassical: I have seen the thing on wikipedia before, but the references were rather cumbersome.
 
so inverting that integral requirement amounts to finding $E$ as a function of some integer $n$---in other words, energy quantization!
@huy: nod, i just wanted to point that out
there's also a reference in there to a 1967 paper which may clarify things (if you can access it)
so doing semiclassical physics ends up having a lot to do with classical mathematics (in the sense of elliptic integrals and the like)
 
Huy
@Semiclassical: I could access it at uni, strangely enough it doesn't work at home. Can you tell me which was reference number 23 in this paper? (scitation.aip.org/content/aip/journal/jmp/8/4/10.1063/1.1705306) I think it was something from Snider, but I don't recall the title.
 
@G.T.R: ...imitation is the sincerest form of flattery, i guess?
 
5:53 PM
@G.T.R He knows.
@Alizter :D
@Semiclassical Yea.
 
@huy: that's not uncommon, your uni's library has paid for access to that content
 
Huy
@Semiclassical: I expected to being able to log in with my uni account but it isn't listed there.
 
@huy: at my uni I can get access off-campus if I log in through my campus's library website
so you may have to do some similar rigmarole
 
@Semiclassical are you familiar with $\sigma$ algebras ?
 
nope
i know people sometimes talk about quantum in terms of those
but i've never touched them in my physics coursework
 
5:57 PM
@Semiclassical you majored in physics ?
 
physics+math. i've been in grad school for it since then
 
Huy
@Semiclassical: Don't you need them to understand how to integrate operators in a function space?
 
not if you're a lazy physicist :P
the standards of rigor are quite a bit looser
it's the sort of thing we'd presumably recognize quickly if we had to read up on it
 
Huy
At my uni, the standard of rigour of some physics lectures seems inexistent which is why I went for maths instead of physics.
 
but we don't necessarily take the time to learn that language, since the coursework usually doesn't do so
on the other hand, we do talk about hilbert spaces a lot
 
5:59 PM
@DanielFischer With regard to an answer I saw posted on MSE, does $f(z) = z^{z}(1-z)^{1-z}$ not have a branch point at infinity? It certainly appears that $f(1/z)$ doesn't have a branch point at the origin, but I feel like I'm doing something careless.
 
so it's really a cultural thing
 
Huy
@Semiclassical: It depends on the lecturer too I guess. Some have the reputation of doing very mathematical courses whereas others not so much.
 
@huy: reference 23 in the Wilcox paper is to "R. F. Snider, J. Math. Phys. 5, 1586 (1964), Appendix B"
 
Huy
@Semiclassical: Thanks.
 
@Anastasiya-Romanova's question reminded me of a questin I wanna consider my POTD in the following days.
 
Huy
6:02 PM
@Semiclassical: I have some ironing to do first. I'll check it out later. :P
 
$$\int_0^{\pi/2} \log\left(\frac{2 \sin(x)}{\sin(x)+\cos(x)}\right) \log(\cot(x)) \ dx = \frac{\pi}{4}$$
 
nod. i'm looking up the paper now to make sure it's been cited correctly
 
Huy
@Semiclassical: I recall having seen it in the paper, but with a long derivation and I was looking for a simpler way. Not sure though.
 
@RandomVariable For $\lvert z\rvert > 1$, we can write that as $$z\cdot z^{z-1}(1-z)^{1-z} = z\left(1-\frac{1}{z}\right)^{1-z}$$ using the principal branch of the logarithm for $(1-z^{-1})^{1-z} = \exp\left((1-z)\log (1-z^{-1})\right)$ (that is one branch of the function defined on $\lvert z\rvert > 1$), so $\infty$ can be an isolated singularity, hence is not a branch point.
 
yeah, it's there and it's pretty tedious looking
would imagine that a more modern reference would give a better proof
 
Huy
6:04 PM
@Semiclassical: My prof told me it is just a short computation, less than half a page. :D
 
lol
i should point out that while the quantum context is the most obvious source of this
 
@robjohn did you see my integral above (before)?
 
it also comes up in the classical mechanics context of hamiltonian evolution
which is perhaps not so surprising when one considers how many analogies there are between quantum mechanics and the hamiltonian formulation of classical mechanics
 
Huy
@Semiclassical: I didn't pay much attention in classical mechanics, I'm afraid. Do we even have operators there? $H$ is just a function there, no?
 
it is. but you still have time evolution of those quantities
i'm having to stretch my memory, i'll confess
 
Huy
6:07 PM
Yes, I recall propagators in classical mechanics too.
 
but i know you do run into similar things in that context, and perhaps even that same formula
 
Huy
But I understood them as just shifting solutions of differential equations?
 
right. what i have in mind is that you can treat H like a matrix if you've got more than just 1 position and 1 momentum
 
Huy
Oh, yes, of course.
 
very different interpretation than in the quantum case, but i think the formalism might be similar
 
Huy
6:09 PM
@Semiclassical: Well I did think about proving it for finite-dimensional spaces first, i.e. $H$ being a matrix.
 
@Chris'ssis This one?
 
@robjohn Yeah
 
@Chris'ssis It is the sum/difference of 4 integrals of products of logs. It should be doable by methods we've used before.
 
@huy: looking at the Wilcox paper, the method of proof is to show that both sides are solutions to some appropriate inhomogeneous ODE with initial condition
which isn't a bad approach at all; i like that kind of method when doing BCH calculations
 
Huy
@Semiclassical: BCH = ?
 
6:15 PM
Baker-Campbell-Hausdorf
$e^X A e^{-X}$ type identities
 
Huy
Oh, I see.
 
@DanielFischer Was it for some reason not proper to look at $f(1/z)$ and see that it doesn't have a branch point at the origin?
 
interestingly, Wilcox also mentions the following identity which he ascribes to feynman
 
When I was younger and at university my professor told me how he got only top grades throughout his studies. I replied to that and said "Impressive" without really acknowledging that I could not have done the same.

It was very foolish of me to cut the discussion short like that. I should have listened to my professor more. Being overly positive in a such situation is actually wrong. It is a kind of a arrogance or narcissism to do that. I can't really express this thought clearly. But I think what I am trying to say is that if your peer tells you that he got only topgrades like he did, and
 
$$\left[\dfrac{d}{d\epsilon}e^{\alpha+\epsilon\beta}\right]_{\epsilon=0}=\int_0^‌​{1} e^{(1-s)\alpha}\beta e^{s\alpha}\,ds$$
 
6:19 PM
@robjohn I think there might be some issues due to $\log(\sin(x)+\cos(x))$
 
Huy
@Semiclassical: I think I have seen that identity at some point before.
 
ugh, should be $\int_0^1$
 
Huy
@Semiclassical: Or something very similar.
@MatsGranvik: A lot of young people are arrogant at first. At least some learn over the course of time.
 
he moreover asserts that the identity we're interested in can be then obtained by taylor expanding $H(\lambda+\epsilon)$ in powers of $\epsilon$ and identifying the first-order term
which is a neat idea, but not one i can verify at a glance
 
@RandomVariable No, that's fine too. I just found the above more obvious at that moment in time.
 
6:28 PM
@robjohn Mathematica has a very hard time with this one Integrate[Log[Sin[x] + Cos[x]] Log[Sin[x]], {x, 0, Pi/2}]
Again, I'm so happy for Simona Halep, she suffered a lot when she lost some important games last months. Even some of our papers talked against her. Anyway, who works hard will finally win and break all the limits!
By the way, the previous question can lead us to another very nice integral ...
$$\int_0^{\pi/2} \log\left(\sin\left(x+\frac{\pi}{4}\right)\right) \log(\cot(x)) \ dx$$
 
6:44 PM
here's something for anybody who knows how to use fractional operators operators (I don't)
i know the following is valid for $\alpha$ a nonnegative integer
$$(1-D_x)^{-\alpha}g(x)=\int_0^\infty ds\, e^{-s}\dfrac{s^{\alpha-1}}{\Gamma(\alpha)}g(s+x)$$
can it be extended to real $\alpha$?
 
Max
Is the condition $f(x) \in I \implies f(x^n) \in I$ equivalent to

Let $I=(g_1(x),..,g_m(x))$, then condition is: $(g_1(x^n),..,g_m(x^n)) \subseteq I$?
why doesn't it show who i replied to? ;o
 
it does show the message, if you click on the little $\leftarrow$ on the upper-left corner of your message text
 
Max
ah i see
 
7:05 PM
comments containing personal attackish usually get a load of stars.
 
@BalarkaSen hi
 
hello
 
@BalarkaSen I am really interested in PNT
@BalarkaSen do you want to chat in numT?
 
well, read up then
@Alizter OK
 
@BalarkaSen and you also like to bring your star contributions to it, don't you? :-)
 
7:20 PM
@Chris'ssis sure thing :P i actually did star that one.
 
Well, yeah, I was disturbed by that paper I received that was suggested to be related to my proof of Au-Yeung series, that one changed my mood and that's why I said many things after that. The next time I'll have another achievement I won't make it known here (that's sure - to avoid any other such a discussion).
@BalarkaSen OK ;)
 
Why is it so quiet?
 
@UserX Because I work on ... $$\int_0^{\pi/2} \log\left(\sin\left(x+\frac{\pi}{4}\right)\right) \log(\cot(x)) \ dx$$ (this is why I'm quiet --- well, not that quiet)
 
@Chris'ssis Have you ever seen $$\int_{0}^{\infty} \frac{x \cos (2 \cos^{2} x) \sinh (\sin 2x)}{x^{2}+p^{2}} \ dx = \frac{\pi}{2} \left[ \sin \left(1+ e^{-2p} \right) - \sin(1) \right]$$

or

$$ \int_{0}^{\infty} \frac{\cos(2 \cos^{2} x) \sinh(\sin 2x)}{(x^{2}+p^{2}) \sin x} \ dx = \frac{\pi}{2 p \sinh p} \left[\sin(2) - \sin (1+ e^{-2p} ) \right]?$$

These are specific cases of a couple of simple formulas I derived without using contour integration. Mathematica has difficulty approximating both integrals for specific values of $p>0$, especially the first one which only converges condition
 
@RandomVariable No, but I can say they look awesome ...
@RandomVariable Have you ever considered to write a book of integrals, series and limits you discovered? I strongly believe it would be very appreciated and I'd like to have it.
 
7:33 PM
@Chris'ssis No, not really.
 
If there is a thing that shock me to a certain extent is that I cannot manage to find such integrals in some books on market (with some exceptions --- well, the integrals in the books I read were not that complex)
Like this one
 
@Chris'ssis That is just the integral you've stated a few lines above
 
@Max Yes; for write an arbitrary element of $I$ as $f(x) = \sum a_i(x)g_i(x)$. Then $f(x^n) = \sum a_i(x^n)g_i(x^n) \in \langle g_1(x^n), \dots, g_m(x^n)\rangle \subset I$.
 
@Chris'ssis I think the expansion of $\log(1-(1-a^2)\sin^2(x))$ might be useful there
 
@robjohn I'm not sure I get that point. Are they the same? hmmm, let me check again ...
@robjohn I saw that expansion in one answer of yours.
@RandomVariable Well, maybe it's time to think of it. People really need such books. It's a pity not to have your own book, since as far as I know you also have a lot of stuff.
@RandomVariable I said to @robjohn the same thing some time ago. He should definitely have a book too.
I believe this: a hand of people can change the face of the world by their actions (in a good sense). Sometimes we miss the proper books to see the beauty of things around us, the books that make us to fall in love with things.
 
7:47 PM
@Chris'ssis That integral is $0$
Note that the integrand is odd around $\frac\pi4$
 
@RandomVariable I recommend you "Limits, Series, and Fractional Part Integrals" by Ovidiu Furdui, from cover to cover, you'll have a lot of fun, there are very nice questions with very nice solutions.
@robjohn Yeah, true.
 
@Chris'ssis If I ever get to a point where I'm more confident in my mathematical abilities, I might consider it. But right now I feel like there is way too much that I don't know. I still ask silly questions at times.
 
@RandomVariable everyone asks silly questions, if you subscribe to the idea that any questions are silly.
 
@RandomVariable I think everybody asks silly questions, they never end depending on the level we are. This shouldn't be an obstacle at all.
 
@Chris'ssis Noting that, I think this integral is fairly simple using the Fourier expansion of $\log(2\sin(x))$
 
7:52 PM
@robjohn Indeed. :-) (but noting that integral is $0$)
 
@Chris'ssis Yes, but you need to multiply it by $\log(2\cos(x))-\log(2\sin(x))$
One of which is simply the alternating version of the other
 
@robjohn Or we can simply make use of the beta function.
 
@Chris'ssis I am not so familiar with the formulations of the Beta function using sin and cos, but I imagine you are correct
 
@Chris'ssis I've worked them out before, but I use them so rarely that I don't remember them
 

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