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6:00 PM
yeah sorry, that was one big careless mistake. Let me fix it...
$$\int \sqrt{\tan x} dx$$
\begin{align}
&\int \sqrt{\tan x} dx\\
& = \int \sqrt{\frac{\sin x}{\cos x}}dx\\
& = \int \sqrt{\frac{(\cos x)'}{-\cos x}}dx\\
& = \int \sqrt{-\frac{f'}{f}}dx\\
& = \int \sqrt{-(\ln(f))'}dx\\
& = \int \sqrt{(\ln(\frac{1}{f}))'}dx\\
\text{Let }
u^2 & =(\ln(\frac{1}{f}))'\\
2udu & =(\ln(\frac{1}{f}))''dx=2uu'dx\implies \frac{du}{u'}=dx\\
& \int \sqrt{(\ln(\frac{1}{f}))'}dx\\
& = \int \frac{\sqrt{u^2}}{u'} du= \int \frac{|u|}{u'} du\\
\end{align}
My suspicion is confirmed: Had $u=\tan x$ and $u'=\sec^2 x$ does not have the trigonometric identity $1+\tan^2x=\sec^2 x$,
 
@Secret $u' = \sec^2 x$
and there is an elementary antiderivative (or else why would i ask you)
 
No, I know there's an elementary derivative, but what i want to say that it exists simply because of the trigonometric identity (I am overthinking something way beyond the challenge you gave me inspired from this integral)
 
and for this purpose you can drop the absolute
 
also typo: $u^2=\tan x$
The reason why this integral has an elementary antiderivative is because the trigonometric identity allows the following equation to hold: $u'=1+u^2$
 
seriously, I don't see why the need of introducing $f$
you never really used $f$ except to confuse yourself
all you used was $(\ln(\dfrac{1}{f}))'$, and that's equal to $\tan x$.
 
6:12 PM
Initially, I want to see if I can use inverse chain rule, before it leads nowhere and I use that u^2 substitution
Sorry typo again: the trig identities allow $u'=\frac{u^4+1}{2u}$, thus making it to have an elementary antiderivative
 
Hey @LeakyNun !
Why is the vertical component of velocity at the highest point of projectile motion zero?
 
Now I am starting to wonder, did Rische algorithm actually have incoporate the trig identities, because via it the antiderivative of this integral will guarenteed to be found. I guess I need to read that section again and check...
 
@Abcd because it isn't moving upward
 
@LeakyNun any other reason?
 
@Abcd why is my reason not enough?
 
6:24 PM
@LeakyNun But it isn't going horizontally too.... so horiz. comp. should also be 0 then
 
@Abcd why can't it be going horizontally?
if it were moving upwards, then it would still be reaching higher, so it would not be the highest point
 
@LeakyNun TY. It is a satisfactory explanation.
 
you may want to consult @Secret on matters of chemistry
 
Oh @Secret are you free to explain?
17
Q: Difference between shells, subshells and orbitals

AmunaWhat are the definitions of these three things and how are they related? I've tried looking online but there is no concrete answer online for this question.

 
Is $\sin(r\pi)$ algebraic for all $r\in\Bbb Q$? If not, what would be a counterexample?
 
6:27 PM
Not really, let me see th question first and then I decide
 
DO orbits and suborbits look like the one in the diagram of the first answer of the second question @LeakyNun @Secret ?
 
@LegionMammal978 I would say they are
@LegionMammal978 it would be sufficient to prove it for $r \in \text{unit fractions}$
 
@LeakyNun Why is that?
 
@LegionMammal978 because e.g. $\sin(2x) = 2\sin x \cos x$
i.e. if $\sin x$ is algebraic then $\sin nx$ is also algebraic where $n \in \Bbb N$.
@Abcd what do you mean by "the second question"?
 
@LeakyNun I linked you to two different questions.
 
6:34 PM
so it seems that suborbit and subshells are the same
 
@LeakyNun :(
@LeakyNun I know that.
I didn't ask that..
@LeakyNun My question meant: Is the diagram factually correct?
 
Quantum mechanically, those chemistry terms are linked as follows:

1. Shells = principle quantum number n, gives the energies of the orbitals (not related to its shape)
2. Subshell = azimuthal quantum number/angular momentum quantum number l, responsible for the spherical, dumbell, clover, flower like shapes of s,p,d,f orbitals. Their orientation in space is controlled by the magnetic quantum number m.
3. Orbitals = probability of finding electrons in a region. This is done by squaring the wavefunctions with different n,l,m. The boundary is defined to be where you will find electrons 90-95
 
@LeakyNun Is this correct?
 
The region inside is where the electron has a 95% probability of appearing
 
6:36 PM
@LeakyNun Is the elliptical path labelled as shell correct? Does it exist?
 
@Abcd no, it is clearly an abstraction
in "reality", the four graphs are juxtaposed
the four orbitals coexist and are concentric
 
@Secret Is the energy of 2s = 2p?
@LeakyNun graphs?
 
@Abcd only for hydrogen atoms with one electron.
 
@Abcd Only when it is an hydrogen atom that they are degenerate
 
@Abcd can you see three sets of x-y-z axes?
(degenerate means same energy)
 
6:38 PM
@LeakyNun I know that ...
 
@Danu I dont know what else there is to say. Holonomy groups act on (a great many) vector spaces $E_b$, where $E$ is somehow associated to the tangent bundle. Then there is a parallel extension of $v \in E_b$ if and only if the holonomy group fixes $v$. So it's possible/plausible that the restricted holonomy group could fix $v$ but not the extended one.
 
anyway, I have to sleep now. It's 4:38 here
 
@Secret What about other atoms? But 2s and 2p are in same energy level right? then why not same energy? @LeakyNun
@Secret Ok, sorry, Bye :) TY.
 
@Abcd in other atoms, generally, 2p > 2s.
 
@LeakyNun But then they are in the same energy level = 2 = L, aren't they?
 
6:40 PM
@Abcd I just said 2p > 2s
 
@LeakyNun Then why are they in the same energy level
 
@Abcd because of some godforsaken quantum effect that I don't remember know
 
2 denotes energy level...
@LeakyNun Oh,
 
I find that people here are really hard with others.. If our question is a little unclear we gets tons of dislike.
And at first people don't try to answer but they try to edit to earn rep.. pretty childish
 
@LeakyNun Is energy of 2s + energy of 2p = energy of energy level 2
@RomainB. True that! Though there are exceptions ...And I do not agree to your second point.."to earn rep"
 
6:43 PM
@RomainB. you're asking us questions. We are not supposed to go through much work to decipher your message.
and how do you know that they edit solely to earn rep?
@Abcd I don't think so.
 
loosely speaking, I don't think it's really well-defined to say "energy of energy level 2" as the different sublevels of 2, like 2s and 2p, have different energy. Unless it's an h-like atom
 
because most of edits on my posts are change of tags or bullshit things like edit LaTex and it changes nothing
 
@RomainB. seriously, good latex looks much better
2
 
@LeakyNun Agreed
 
I always assumed these minor latex changes people make to my posts are for easier cataloging or some such reason
 
6:44 PM
Yep but the people that changes mine just change like (a,b) into text(a,b) or things like that
 
-1
Q: Area of a sphere

Romain B.We know that we can't flatten a sphere to a plane so how can we mesure the area of a sphere ? Here is a gif to illustrate what I mean : https://m.imgur.com/gallery/5RE0Twe

 
How do you expect people to be clear what you are asking?
 
Yeah it's a bit unclear but c'mon you understood
 
@RomainB. no, I don't.
 
6:46 PM
when the sphere rolls on the gif, it's impossible in real life
 
yes, I get the first part, then what
 
Why is time taken in two halves of projectile motion equal?
 
because we can't flatten a sphere without deformation
(problem of world map)
 
yes, continue
 
so I wanted to know if this gif was correct or just a help
because I know we can use triple integrals and things like that
I should re-write it :)
 
6:48 PM
see, you can elaborate.
@RomainB. so your focus is on the gif
that doesn't sound like the focus in the current wording
 
yep
I have difficulty to translate my thoughts in text :/
 
then don't complain about the community for being unclear what you're asking.
 
Why is time taken in two halves of projectile motion equal? @LeakyNun ... sorry for tagging :(
 
They could say "We do not have fully understand can you rewrite it or precise more ?"
 
@RomainB. that's your responsibility
 
6:49 PM
but no, people see that, they see that it hard to understand DUE to MY fault and they just skip and -1 me
 
you're the one asking the question
we have no responsibility to answer it
we're all volunteers
 
ofc
but you could be polite volunteers
just -1 really unclear question or homework !
But I've learned that I need to be more precise :)
 
don't take the downvotes personally
take half a minute to read the text of the downvote
 
Haha I fully do ..
Or there is ?
oh*
I didn't know
Thanks I will see that
 
<a class="vote-down-off" title="This question does not show any research effort; it is unclear or not useful">down vote</a>
 
6:52 PM
Arf..
I will re-write it asap
 
this is all what the downvote means.
 
Ah I just though that people who downvote could write a little sentence to help me correct it
but no..
 
@Abcd set t=0 to be the time when the particle is at the highest point
 
I have a completely different question, can we put in our CV that we volunteer on this website ?
 
notice that the journey when t<0 is basically the same as the journey when t>0
 
6:54 PM
@MikeMiller I don't know this
 
@RomainB. that's up to you to decide
 
So why isn't it important to classify full rather than restricted holonomy groups? @MikeMiller
 
@LeakyNun ok
 
And what if a group occurs only as full holonomy group, not as restricted (i.e. does not occur for simply connected manifolds)? Can this not happen?
 
7:02 PM
Can a closed question be re-opened if wee re-wrote correctly and clear the question ?
 
@Danu Why do you care about holonomy?
 
@MikeMiller I care about quaternionic Kähler manifolds
And I want to properly motivate them
 
Why do you care about quaternionic Kahler manifolds?
 
For my thesis.
 
Why are you writing your thesis?
 
7:13 PM
lol
 
@MikeMiller Because, perhaps unlike you, I have to in order to be able to continue doing mathematics next academic year.
 
7:27 PM
@RomainB. yes, certainly.
 
Hi, I am looking at the statement that if we have undirected multigraph then it has euler cycle <=> every nodes are with even degree ,however suppose we have euler path and all nodes have degree 2 ,if we create new edge between 2 nodes again we will have euler path but there will be two nodes with degree 3 meaning they are not with even degree?
 
Zee
7:56 PM
@LeakyNun I figured the problem, the notion of convergence used was the topological one
 
8:53 PM
Why MBST is not necessarily a MST?
 
9:16 PM
@MikeMiller you're just being pedantic at this point.
 
10:03 PM
nobody home
 
10:22 PM
crickets
 
crickets some more
 
cicadas
 
kek
 
10:52 PM
and the silken, sad, uncertain rustling of each purple curtain / thrilled me, filled me with fantastic terrors never felt before
dammit i can't sleep
 
rip in sleep
 
dammit damminark
2
 
What?
 
@BalarkaSen just listen to Remain in Light by Talking Heads if you're already gonna be up, it's a spectacular album
it has nothing to do with insomnia but I recommend it to anyone
 
@Daminark I thought you'd recognize a pun when you see it
 
10:56 PM
Oh well now that there's an edit...
 
@Fargle Noted. Sounds like a job for a Sunday :)
I have work to do in the morning so I wouldn't be able to listen to it right now. But consider it listened tomorrow
i will stay here for 2 more minutes
and i'm all outta minutes
 
rip
 
press f to pay respects
 
I was about to give him the fair warning that David Byrne is among the weirdest and most unhinged vocalists pop has ever seen.
 
Well, do you want to talk about higher homotopy groups?
Specifically, the fact that they are abelian groups? This came up but I haven't yet seen a satisfactory proof
 
11:02 PM
I know nothing of homotopy groups besides the good old fundamental group.
 
Oh, well this is perfect, I can explain and see if I get it lmao
 
You should do that in our room.
That way I'll be able to go back and see it more easily.
 
So, remember that you can identify $S^n = I^n/\partial I^n$
Oh alright, sure
 
Hi everyone :) I have a set : $(\bigcup_{n \ge 1} \mathbb{R} \times \left\{\frac{1}{n} \right\}) \cup (\bigcup_{n \ge 1} \mathbb{R} \times \left\{\frac{-1}{n} \right\}) \cup ( \left\{ 0\right\} \times \mathbb{R}) $.
 
@PawełKusz Oh wow, this looks like a double-sided version of the topologist's comb.
 
11:45 PM
Is the discrete space is homeomorpic with euclidean space?
 
@PawełKusz No--discrete space is totally disconnected, but Euclidean space is connected.
 
@Fargle Thank you :)
 
The easier way to think about it (though I'm only doing this to undermine Fargle) is that the preimage of a point (which is open) in a discrete space is a point (which is not open)
 

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