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17:08
@r9m Also a bit of integration.
r9m
r9m
@Chris'ssis okay ! ^_^
@r9m I'm going to be honest with you: it's a very easy series.
What's a first possible step such that everything flows naturally?
(Good question I think)
r9m
r9m
@Chris'ssis well I really can't see how to evaluate that one ! :o you may write a hint atleast if you wish !
@Chris'ssis Doesn't it look nicer that way for the upcoming new year ? $\displaystyle\prod_{i=2}^{2015}\sum_{k=0}^\infty\dfrac{1}{\binom{i+k}{i}}=2015$ :D
@Hippalectryon lol, that would be crazy nice! :-)
17:17
I'll give that to some friends right after the NY hehe
@Hippalectryon Wait
r9m
r9m
@Hippalectryon well try to find something unique to the number 2015 (if its about the new year) .. not something that is true for all $n$
@Chris'ssis ?
Find all polynomials that satisfy $(x^2-6x+8)P(x)=(x^2+2x)P(x-2)$
@Hippalectryon I was thinking of a modified version , but no, it's perfect that way :-)
17:20
$x^2-6x+8=(x-2)(x-4), x^2+2x=x(x+2)$
When I reach 5 upvotes I give a precious hint to @r9m :-)
3
A: Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$

Chris's sis$$\frac{\pi^4}{1440}-\frac{\pi^2}{3}\log^2(2)+\frac{1}{24}\log^4(2)+\frac{7}{24}\pi^2\log^2(2)+2\log(2) \zeta(3)-\frac{7}{4}\log(2)\zeta(3)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$

r9m
r9m
@Chris'ssis haha !! alright ! I wait till then :D
@r9m :D:D:D
@r9m I was just kidding,
@Hipp!
@r9m $$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}+1/n^2}{2^nn^2}$$ and then write that series as a double integral (that's important!!!). First you get right of the polylogarithm term, and after getting the double integral, and then doing the proper variable change, all, but all flows naturally.
17:22
@BalarkaSen !
r9m
r9m
@Chris'ssis okay ! :P but that was not an unfair demand either :P LOL
The quotient P(x-2)/P(x) will result in -12/(x+2)+4/x+1
r9m
r9m
@Chris'ssis hmm !!! alright! alright! I'm on it ..
@Hippa Cayley graphs are so cool!
you should study some of those
@BalarkaSen Uh I don't really have time :/
17:24
at least have a look
@BalarkaSen send a link
google "geometric group theory" and look at terry tao's blog
Assuming the map $x\mapsto x-2$ won't change the degree of the polynomial we can also assume that the polynomial has degree -2. Which is nonsense. What should I do?
@Chris'ssis today is a busy day irl. I may not be able to be back in any real capacity until tomorrow (UTC)
@robjohn OK. Just asking since I didn't see you around. :-)
17:25
@BalarkaSen It's not on the first page
@r9m 2015 is pretty common :/
@r9m It's close to 27 so It's probably prime :3
It can't even be written as the sum of three squares :/
r9m
r9m
@robjohn is it possible to apply the Euler Series transformation like you did here ... in this case too ? :-)
@BalarkaSen thanks
0
Q: Find all polynomials with real coefficients that satisfy $(x^2-6x+8)P(x)=(x^2+2x)P(x-2)$

UserXFind all polynomials with real coefficients that satisfy $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\forall x\in\Bbb R$$ My work; $$\frac{P(x)}{P(x-2)}=-\frac{4}{x-2}+\frac{12}{x-4}+1\tag{1}$$ $$\frac{P(x-2)}{P(x)}=-\frac{12}{x+2}+\frac{4}{x}+1\tag{2}$$ I also factorised the two known polynomial which di...

Self advertisement
17:45
Is my hint good enought now?
5
A: Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$

Chris's sisHINT: Consider $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{2^nn^2}+\operatorname{Li}_4\left(\frac{1}{2}\right)$ and then express the remaining sum as a double integral. After some work, you get $$\int_0^1 \frac{\displaystyle...

Actually I'd say it's an answer since the rest thing to do is a piece of cake.
r9m
r9m
@Chris'ssis okay !! lemme try ! :-)
@r9m Just let me know if you meet anything hard after that point I mentioned.
r9m
r9m
@Chris'ssis 'kay boss !!
I wanna ask something about probability. Qiaochu Yuan in this answer: http://math.stackexchange.com/a/3540/146687 states:

$\displaystyle\int_0^{1} t^a (1 - t)^b\, dt$ describes the probability of randomly selecting $a+b+1$ real numbers such that the first number is $t$, the next $a$ numbers are in $[0, t]$, and the next $b$ numbers are in $[t, 1]$.

Can I interpret his answer as follows:

Say I randomly pick ten real numbers where the first number is $\frac{1}{2}$, four numbers between $0$ and $\frac{1}{2}$, and five numbers between $\frac{1}{2}$ and $1$, then its probability is $\displays
How to link a word or sentence like in the main page?
17:59
@Venus That answer troubles me too :/ as the integral is independant of t
If anyone answers you, ping me !
@Hippalectryon OK. I tried to chat @Did but somehow I can't tag his name.
@Venus No. After you have got the first number, $t$, the probability that the next $a$ numbers are $\in [0,t]$ is $t^a$, and the probability that the following $b$ numbers are in $[t,1]$ is $(1-t)^b$. If $t$ is fixed, that's it, and the probability is just $t^a\cdot (1-t)^b$. For $t = \frac{1}{2}$, we get a boring $1/2^{a+b}$. The integral comes in because the first number, $t$, is randomly drawn uniformly distributed in $[0,1]$, just like the following $a+b$ numbers. So to get the ...
the chat's '@ list' only lists the users that were here ... hours ago
... probability of the total event of the choice of $1+a+b$ numbers, you have to integrate with respect to the conditional probabilities given the first number (times the probability density, but that is $1$ here).
@Hippalectryon Ping ^^
@DanielFischer But how can QYuan say that B(a+1,b+1) describes 'the probability of randomly selecting a+b+1 real numbers such that the first number is t ....' as B does not depend on t ?
18:05
@DanielFischer Oh I forgot that you're good at every math subject. OK, it means to get the probability as the beta function, we can't set a value for $t$?
@Hippalectryon We call the result of the first selection $t$.
@Venus We must integrate over all possible values of $t$. It's probably easier to see in the case of only finitely many possible outcomes, where you get sums instead of integrals.
@DanielFischer How about like this:

I randomly pick ten real numbers where the first number is $t$ where $t\in[0,1]$, four numbers between $0$ and $t$, and five numbers between $t$ and $1$, then its probability is $\displaystyle\int_0^{1} t^4 (1 - t)^5\, dt$. Is that correct interpretation?
@Venus The order of the picks, that you first pick the four numbers $\leqslant t$ ($< t$ is probably better, since the probability that you pick $t$ again is $0$) first, and then the five numbers $\geqslant t$ ($> t$), matters. Without taking the order into consideration, you must multiply with the number of ways to distribute the $< t$ and $> t$ events, which is $\binom{a+b}{a}$.
(Which Qiaochu does next.)
@DanielFischer If so, then what is the interpretation of $\displaystyle\int_0^{1} t^4 (1 - t)^5\, dt$?
@Venus Urk. Got it mixed up above, sorry.
The above was for the integrand. For the integral, what you said is right.
18:20
OK, thank you so much @DanielFischer. Next time if I have other questions, I'll summon you ^_^
@r9m is this giving you trouble? $$\int_0^{1/2} \frac{\log^2(1-x)\log(x)}{x} \ dx$$? Just use your imagination a bit. How can we cleverly make use of beta function here?
r9m
r9m
@Chris'ssis I'm weak with beta ! .. I need more time please !
@r9m I'll consider this integral my POTE (evening). I'm sure lots of the students I talk to will be in trouble (not you, of course, you're in the high league)
@Chris'ssis pote ?
@Hippalectryon problem of the evening
18:26
xD
r9m
r9m
@Chris'ssis I'm somewhere along the lowest rungs of hell !! a.k.a. The Low League (TLL)
@r9m Actually, using my trick here you also get the value of that series you talked to in the post.
It's very important to know how to properly treat this integral!
r9m
r9m
TLL sounds better than r9m .. I might change my username someday to TLL :D :P
@r9m lol :-)
@r9m Maybe you can use substitution $x=1-2t$
18:29
@Venus My way is straightforward. I doubt, really, one can find something better.
@r9m and @Chris'ssis I mean $x=\dfrac{1-t}{2}$
@r9m beta function or Euler's generalized harmonic sum (as you wish).
Let's summarize: well, I know that $$1) \int_0^{1/2} \frac{\log^2(1-x)\log(x)}{x} \ dx$$
What does that mean?
It also means that I know to compute $$2) \sum_{n=1}^{\infty} \frac{H_n}{2^n n^3}$$
And what does that also mean?
Well ... It also means I immediately know to compute $$3) \sum_{n=1}^{\infty}(-1)^{n+1} \frac{H_n}{n^3}$$
That's all.
@r9m thanks! I had so fun this evening with your question! :-) And ^^^
People, apparently I might have not succeded in proving the existence of my paper on Robin's inequality. But then I thought: how about submitting to vixra, in order to have a certification of existence, and then give my paper to my endorser and submit to arxiv?

Wouldn't that be fine? I guess submitting to vixra leaves indeed a trace.
(I do not know my endorser, that's why I need to prove the existence of the article before giving it to me)
r9m
r9m
18:49
@Chris'ssis okay !! :-)
r9m
r9m
@Chris'ssis I need to try that ... still no ideas in my tiny little head !
@r9m To try what? OK, whatever is, take your time.
r9m
r9m
@Chris'ssis evaluating the integral using your hint !
@Chris'ssis , @r9m Please, could you consider my question? Unless you don't know viXra
18:52
@r9m Are you serious?
r9m
r9m
@VincenzoOliva I don't know .. sorry :(
@Chris'ssis yes !! -_- why is it so hard to believe ? :O
@r9m I see, thanks anyway :D
@r9m Let's start with the integration by parts, that is $$\int_0^{1/2} \log^2(1-x)(\log^2(x)/2)' \ dx$$
@r9m what do we get then?
$$\int_{0}^{1/2} \frac{\log(1-x) \log^2(x)}{x}\ dx+\frac{1}{2}\log^4(2)$$
@r9m how can we write that integral?
$$\left(\int_{0}^{1}-\int_{1/2}^{1}\right) \frac{\log(1-x) \log^2(x)}{x}\ dx+\frac{1}{2}\log^4(2)$$
@r9m lastly we see that $$\int_{1/2}^{1} \frac{\log(1-x) \log^2(x)}{x} \ dx=\int_{0}^{1/2} \frac{\log^2(1-x) \log(x)}{x} \ dx$$ by variable change
@r9m using the symmetry we're done.
Q.E.D. (please do not post my work anywhere)
r9m
r9m
@Chris'ssis okay !! I guess I must thank you for spoon feeding me !!! :D Happy Children's day in advance !! :D
@Chris'ssis no posts shall be made even with remote similarity to this context !! Thank you :)
19:06
@Chris'ssis I'm wondering, how do you get $$\int_{0}^{1/2} \frac{\log(1-x) \log^2(x)}{x}\ dx+\frac{1}{2}\log^4(2)$$from $$\int_0^{1/2} \log^2(1-x)(\log^2(x)/2)' \ dx\,?$$
r9m
r9m
@Chris'ssis accepted answer !! thanks !
I know you used by parts, but what are $u$ and $dv$ here?
@r9m some typos there in denominator :D (I also talk by phone)
For what it's worth I got $$\int_{0}^{1/2} \frac{\log(1-x) \log^2(x)}{1-x}\ dx$$after by parts.
@r9m more typos there!!!
r9m
r9m
19:11
@Chris'ssis where ?
@r9m I'll rewrite it again.
19:27
@r9m We have that

$$I=\int_0^{1/2} \frac{\log^2(1-x)\log(x)}{x} \ dx$$

$$=\int_0^{1/2} \log^2(1-x)\left(\frac{\log^2(x)}{2}\right)' \ dx$$
$$=\int_0^{1/2} \frac{\log(1-x)\log^2(x)}{1-x} \ dx+\frac{1}{2}\log^4(2)$$
$$=\int_0^{1} \frac{\log(1-x)\log^2(x)}{1-x} \ dx-\int_{1/2}^{1} \frac{\log(1-x)\log^2(x)}{1-x} \ dx+\frac{1}{2}\log^4(2)$$
$$=\int_0^{1} \frac{\log(1-x)\log^2(x)}{1-x} \ dx-I+\frac{1}{2}\log^4(2)$$

and hence

$$I=\frac{1}{2}\int_0^{1} \frac{\log(1-x)\log^2(x)}{1-x} \ dx+\frac{1}{4}\log^4(2)$$
I'm really having trouble doing some category theory problems. If I show the specific problem(s), could someone give me some general hints (and I'll perhaps need hints in general about how to tackle category theory problems).
@r9m sorry, in some days I'm very tired and my attention is dropped, but the very good ideas are always there. :-)
19:44
The interesting fact is that if I want to draw a picture, say, and I miss a line, that one is there in my mind as if I look at the picture in mind without using my eyes at all. After a while I might realize that the line is not actually there.
@Alyosha I can give it a shot buddy
I'm no category theorist but I done a class or two
Specifically questions $3-5$ here, but I think if I get the point about $3$ I can do $4,5$. here is the referenced question sheet in question $3$ of the first-linked sheet.
Thanks!
@Alyosha alright, so, where are you stuck? just keeping it all in your head?
Pretty much.
I understand what it wants you to do but I'm not sure how exactly to help. You may have to definition chase here.
19:54
Okay. To be precise, which part of question $5$ on example sheet $1$ do you think it wants me to use?
Yes, I think I am fairly inexperienced in cat definition chasing.
@Alyosha either (b)(i) or (b)(ii), which both use essentially the same concept
Maybe first try to identify where the cone should go
Where it goes in what sense?
r9m
r9m
@Chris'ssis thank you !! :-) I got that !! :)
like, which part of the metaphor with the example.
@r9m Isn't it amazing that one can establish such awesome relations? :-) Welcome.
r9m
r9m
19:58
@Chris'ssis :D
The cone should probably be the $H(-,A)$, given that in $1.5.b$ this functor has a representation.
@r9m Wait, it's even more than that ... you can look up for other integrals of this type. This is just the very beginning.
Exploiting the symmetry for higher powers.
r9m
r9m
@Chris'ssis I tried to work out a binomial identity like robjohn used for $H_n/n^22^n$, but I couldn't :( ...
Okay, I'll give it another bash. I may repost here if for some reason I'm still having trouble. Thanks, @AlexanderGruber!
20:07
@r9m I think I have a very brilliant way for $H_n/(n^22^n)$
This one is also interesting ...
10
Q: Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$

Chris's sisI remember that some time ago I was asking this question Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$ , and now, while I was making a review, I asked myself if we can get the closed form of $$\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$$ by using the similar tools as in that proof. The probl...

@Chris'ssis I'm gonna need sunglasses :/
@Hippalectryon Why? :-)))))
Because your answers are too brilliant :DDD
2
@Hippalectryon lolllllllllll :D
@r9m Now I'm computing $$\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(3)}}{2^nn^2}$$
Goddddddddnnnnnnnnneeeeeeeeeessssss!!!!!!
What an awesome problem I just came across!!!!
@Hippalectryon
20:37
@Chris'ssis What one ?
@Hippalectryon $$\int_0^{1/2} \frac{\log(x)\log(1-x) \operatorname{Li}_2(x)}{x}\ dx$$
@BalarkaSen not exactly; there will be infinitely many nodes of vertex degree 3 as well as 2. try drawing the graph with vertices representing elements of word length <=3 and you should pick up on the pattern.
@AlexanderGruber Okay, I'm having trouble with the showing that $\text{Cone}_{\mathscr{JC}}(-,DA)$ is representable.
@Hippalectryon I'm done with it. There are many things on the paper to collect and put together for the final answer. I'm looking for a shorter way.
20:47
@Chris'ssis Hacks :c
@MikeMiller Is it correct that the functors $X,Y$ are isomorphic if there exists a transformation with an inverse (natural transformation with an inverse, dropping the naturality condition)?
That is, that the condition is pretty weak?
@Hippalectryon this version is slightly easier $$\int_0^{1} \frac{\log(x)\log(1-x) \operatorname{Li}_2(x)}{x}\ dx$$
I seem to remember that being true; if it is, it's in MacLane's book
Is it usual that a representation of a presheaf is not required to be natural?
I'm not sure if our lecturer left naturality out on purpose.
21:12
Hello Professor @TedShifrin
hi @skull et al
Why French -__-
Latin, @Hippa
et alia
21:15
I thought you meant et @Alyosha
pas moi ...
@TedShifrin Hello.
@Hippalectryon Suppose you break something in a shop in France What is the first thing you say to the sop owner?
@PedroTamaroff How much, yeah
@PedroTamaroff And sorry, I guess :)
@Hippalectryon I mean in French.
21:24
@PedroTamaroff Je suis vraiment désolé, ça coûtait combien ?
@Hippalectryon Right, that was what I heard.
:-)
Where ?
One says "Je suis tres désolé."
From my father.
@PedroTamaroff Très before désolé sounds weird, but we understand :)
@Hippalectryon Why does it sound weird?
21:26
Désolé is already pretty strong
No need to add très before
Vraiment is better
Vraiment is "truly", right?
Indeed
@PedroTamaroff Très désolé would be like very sorry, doesn't it sound weird ?
0
Q: Find the closed form of $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{H_n}{kn (k+n)^3}$

Chris's sisHere is a challenging double series question I wanna share with you $$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{H_n}{kn (k+n)^3}$$ together with the question: what tools would you like employ for computing it?

@robjohn Please see that Fourier question and show me what you meant by expanding it by Fourier Series, I got nowhere, just give it a try, you're much more experienced!
@Hippalectryon Americans use that all the time.
@Chris'ssis You'll never learn to [altext](link); will you?
21:29
@PedroTamaroff :c
@FreeMind Do you know what a Fourier series is?
@PedroTamaroff :(
@PedroTamaroff Sounds dumb!
@PedroTamaroff Have you even seen my question?!
@FreeMind I couldn't find it.
21:31
@PedroTamaroff The problem is not Fourier Series, the problem is what happens after expanding, I get nowhere.
1
Q: Approximating an integral equation using Fourier or Taylor series

FreeMind$$F(\theta)=\sin \theta \int_{-l}^{l}e^{-ikz\cos \theta}f(z)\,dz$$ is it possible to approximate both $f(z)$ and $F(\theta)$ with Fourier Series or Taylor series, then find its coefficient and check out whether it is a good approximation or not? We don't know what the $f(z)$ is, but we know $|z|\...

@PedroTamaroff The annoying thing is, we don't have the function rule, how the hell am I supposed it crack this question then?! I did various dumb things, it didn't work.
@PedroTamaroff Kid, I'm not playing with your games. Talk like that with your mom, pap, and the professors you want to.
No really, it is part of the guidelines no to spam with questions, and if you do (like many of use do) use [text](link).
2
@Chris'ssis I am not decoding GFY. I think someone else just did.
relax people
skull is here!
@skullpatrol Let's agree that Chris popped.
I told him to stick to the usual rules we live by.
you need to lighten up
be nice
21:36
I'm pretty relaxed.
but not nice
How was I not nice?
@Chris'ssis Please be nice. This is not the first incident of rudeness from you, further ones will lead to a much harsher suspension than the one I just placed on you.
asking "will you never learn"
@skullpatrol It is just facetious.
Like "bro do you even [text](link)"?
21:38
8 mins ago, by Pedro Tamaroff
@Chris'ssis You'll never learn to [altext](link); will you?
@PedroTamaroff You should be nice too. Your history isn't as bad, but be careful in the future.
I know what I said, @skullpatrol.
You'll never learn to
Your gravatar is pink know. That is more unsettling than this. What hapened?
@ManishEarth Sorry $\mu$aster.
You'll never learn
don't come in here preaching rules to us pal
be nice
21:41
@skullpatrol They are not my rules.
They are our rules.
You know them.
I know them.
They knows them.
We got dis down.
@PedroTamaroff You need not be abrasive when talking about rules
Does not matter what you're doing, be nice
@ManishEarth OK, that's harsh.
Explaining rules? Does not get you a pass to be rude
3
21:42
^
Sorry I annoyed you with my intention to pass rules.
Without the rules we're monkeys.
2 mins ago, by skullpatrol
don't come in here preaching rules to us pal
preacher?
Je sius vraiment désolé.
Seppukku.
21:43
@PedroTamaroff That misses the point entirely
@ManishEarth I have a question you might be able to answer. Chemically, why does $H_2O_2$ burn less than alcohol?
U MAD?
@skullpatrol I ain't mad.
I don't get mad at simple conversation.
@PedroTamaroff does it?
@ManishEarth It seems to be. It also bubbles.
21:45
@PedroTamaroff Bubbling is the release of oxygen
@ManishEarth OK. I thought $H_2O_2$ didn't burn, but today I used it and was not amused.
heh
it burns a lot
but that's the sensation, the actual reducing/oxidising power varies
@ManishEarth I also have this question.
Why is it called bicarbonate if the formula is $CO_3^-$?
I don't see no two there.
that's not bicarbonate
@Manish how long 'til Chemistry graduates? I know it's not long now
21:50
$\ce{HCO3-}$ is bicarbonate
$HCO_3^-$?
@MikeMiller shrug they didn't give any timeline
OK, that.
21:51
The prefix "bi" in "bicarbonate" comes from an outdated naming system and is based on the observation that there is two times as much carbonate (CO3) in sodium bicarbonate (NaHCO3) and other bicarbonates as in sodium carbonate (Na2CO3) and other carbonates.
chemistry is harsh
@ManishEarth I'm amused that your profile on chemistry says you'll mainly be seen asking, when you've got thrice the answers as questions
@MikeMiller that's an old old profile :p
I am not great with chemistry
physics student, with a passion for programming. Chem is a side side hobby :p
nice hobby :-)
22:02
@PedroTamaroff Ssup?
Oh a chemistry lover. Time to lay down some questions
@ManishEarth can you identify a model?
probably not
don't know what exactly you mean
I don't know the name of the atomic model I get learnt at school so I can't read more online
@FreeMind I warned him already. Seems like enough.
@UserX Bohr's? (shells)
Historically it's between Bohr's model and the modern quantum model.
22:06
VSEPR?
or the orbital model?
@ManishEarth In addition, @Chris'ssis is really a nice guy and you're wrong.
VSEPR is a model of space structure of atoms
Valence shell electron pair repulsion?
There is a chapter on the book about VSPR but it's not being taught.
@FreeMind I had a look at the history of both. I can see past flags/suspensions. He may be nice in general but he wasn't this time.
22:08
@UserX What are you exactly talking about?
38 mins ago, by Pedro Tamaroff
@FreeMind Do you know what a Fourier series is?
This ^ was rude
@ManishEarth When someone behaves in an inappropriate way, I think there is no reason to be nice.
@ManishEarth How is math chat nowadays vis-a-vis flags? I remember some days ago, it was a problem here.
@FreeMind an atomic model where the electron is of the form of an electron cloud, we have orbitals, spin, pauli's exclusion principle, hund's principle, least energy principle.
@JayeshBadwaik dunno, I popped in here after seeing some flags
22:09
What's the name of this atomic model?
@ManishEarth Ahh, I see.
@UserX Quantum model!?
@FreeMind When someone behaves in an inappropriate way, you don't retaliate, you back off and flag.
@FreeMind I'm in high school...
@skullpatrol that's ... not
22:10
:O
@UserX Dude, it's quantum model!
@FreeMind then greece has a really weird educational system...
@ManishEarth This sounds reasonable for you, but for someone who has been offended, the reasonable thing is to make the guy who doesn't watch his mouth understand what's going on.
We don't deal with schrodinger's equation and even if Heisenberg's principle is mentioned it's never used in exercises.
@FreeMind Doesn't give said person license to be rude on this site
22:12
@UserX It's taught everywhere, even in middle east, Quantum model should be taught in High school, US system is weird!
there are rules here
and rulers
who want to rule
@ManishEarth Rules are for evils in power to get the most out of innocents.
Okay, I'm out of here
@FreeMind oh that makes sense. But we're simplifying things I suppose. Where can I learn more? Also where can I learn more about iontic equilibrium(don't know if it's the correct word)
22:13
later pal
@UserX You can learn from books, what books? fundamental chemistry books? or university chemistry books, they've explained things from top to bottom.
@skullpatrol SE no fear, skull is here :)
@UserX have you tried wikipedia?
@FreeMind :D
@skullpatrol Wikipedia is not good for first time learning.
@FreeMind I just want to get a better understanding. Probably not university level as I won't understand them. @skullpatrol read the sentence before my last sentence.
@UserX Your problem is, you think whatever thought in the university cannot be understood, and you're wrong badly.
22:18
@FreeMind of course I don't think that. I've read some university level math(mostly calculus, analysis and multivar calc) and they can be understood because I can grasp mathematical contexts easily. Chemistry never came natural to me.
But on the last classes when we did Bronsted&Lowry acid/base model I started liking chemistry too
Wait till you see biochemistry :)
22:30
@UserX You know why the chemistry doesn't come natural to you? because the chemistry in basic levels is trying to explain advanced physics and advanced mathematics in plain words, in addition, for excuse, they try to prove things by only saying that "it has been proved by experience" , that's pretty dumb :)
@FreeMind I particularly hate that.
in Biology it gets worse...
The imitation machine is a great film.
@skullpatrol what I learnt till now(HS biology) is that biology is about definitions just to make sure biologists are talking about the same thing. Also some mechanisms that have huge leaps of faith.
...that's what I mean
"it has been proved by experience" gets replaced by "trust us we know"
22:44
None of the mechanism make sense according to the experiment unless you assume 90% of the mechanism
And I don't know if that's what biology is about
Biology is about training your memory.
@skullpatrol well physics does that too but in physics we usually avoid doing math we don't know yet. i.e for simple harmonic oscilation we didn't know calc yet and we were given $x=A\sin{\omega t}$ but then proved it myself. I can't do the same in biology or chemistry
Anyway, back to math. It's been two days and I still don't have an idea on how to find the general solution of $y'(x)+p(x)e^{r(x)y(x)}=q(x)$
That idea is definitely a brilliant substitution that has to do with logs but I can't figure it out
I'm just saying that the approach chemistry takes is a lot like the one biology uses :-)
@UserX @skullpatrol Science is so blanked up! vagueness and ambiguity are flowing in it.You just get used to it.
@FreeMind I disagree. Math is rigorous enough for me.
22:51
@UserX Believe me, the idea of infinity is one of the most mysterious things, it's not clear at all.
But, math is not an experimental science.
@FreeMind I disagree once more. Cantor cleared things up a lot
We just feel infinity!
No we don't. Depends on what context you see infinity.
@UserX That's a way to beat around the bush, we don't know it.
22:53
@FreeMind I don't understand what you're talking about.
@UserX I don't expect anyone to.
That's either a sign of a brilliant mind or a sign of shallow understanding.
Hi @AlexanderGruber
@AlexanderGruber have you read up on the fireworks we had in here :-)

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