/sci/ - Science & Math

[Anonymous No.8299961]

Let's post (very) hard problems of our fields of study, but using images like this one that pretends it's actually easy.

Can /sci/ solve this one? It's from Hearthstone, if I remember correctly.

[Sage No.8299963]

-17

[Anonymous No.8300627]

>>8299961
We can probably show local uniqueness from the existence of a section and use properties of elliptic curves to extend that to global?
I don't know shit about elliptic curves senpai.

[Anonymous No.8300648]

>>8299961
ur family is trivial xd

[Anonymous No.8300662]

>>8299961
I stopped caring about math when I was introduced to the concept of elliptic curves. What a crock of shit. If your equation can only be solved by extending to a compactification that can't exist, like some kind of math deity , then you are fucking wrong and the math is flawed. Same for algebra solutions that basically say "the correct answer is whatever the correct answer is". Thats what the math said transcribed to words but god forbid if i wrote in down in english instead of the ancient math runes the professor word mark me wrong.

Math is logical and numbers never lie my ass. Math is just as flawed as any other human construct.

[Anonymous No.8300708]

[Anonymous No.8300723]

>>8300708
Nice try anon

[Anonymous No.8300757]

>>8299961
I want her to call me a filthy brainlet and then peg me in my boipussi while forcing me to admit my inability to solve problem after problem

[Anonymous No.8300826]

>>8300662
>I stopped caring about math when I was introduced to the concept of [simple mathematical concept]. What a crock of shit. If your equation can only be solved by [simple procedure] that can't exist, like some kind of math deity
So is this the newest ironic copypasta? I just want to make sure I have my memes straight.

[Anonymous No.8300854]

[Anonymous No.8301052]

This isn't really that hard, but for many /sci/entists this could be!

[Anonymous No.8301054]

Sorry guys, the first version wasn't hard but impossible to those who don't assume all topological spaces to be Hausdorff or atleast T1 by default.

[Anonymous No.8301078]

>>8301054
take the two points x, y
use the fact that there exists a continuous function f:X-> [0,1] with f(x)=0, f(y)=1
but if it's continuous then all values between 0 and 1 must appear for some argument
so X is uncountable

[Anonymous No.8301079]

This is the equation for the oscillation of a simple pendulum, you should be able to solve this.

[eqn]\frac{d^2}{dt^2} \phi + \frac{g}{L} \sin \phi = 0[/eqn]

[Anonymous No.8301085]

>>8299961

[Anonymous No.8301090]

>>8299963
It's - 12 dumbass

[Anonymous No.8301092]

And "efficient" means something better than the naive one that first comes into mind.

[Anonymous No.8301101]

>>8301092
CS fag

[Anonymous No.8301102]

>>8301054
Space is normal -> space is T1 -> all singletons are closed
Atleast two points -> two disjoint closed sets
Space is normal -> space is T4 -> Urysohn's lemma -> there exists a continuous map taking one singleton to 0 and the other to 1
Space connected -> the image is connected -> the image is the uncountable closed unit interval
The cardinality of the image is at most the cardinality of the domain -> X is uncountable

[Anonymous No.8301105]

>>8301054
countability is a retarded notion in constructive math

[Anonymous No.8301142]

>>8301105
calm down, wily burger
let the heathens have their degenerate fun

[Anonymous No.8301190]

>>8301105
That's just excusing your laziness using fancy words.

[Anonymous No.8301625]

[Anonymous No.8301653]

1/3

[Anonymous No.8301655]

2/3
not a problem per se, but with animu girl

[Anonymous No.8301659]

3/3 eh

[Anonymous No.8301666]

>>8301659
>God writes out the entire standard model Lagrangian just to create light
>doesn't know that the QED Lagrangian can be written with less than one line
>puts complicated looking formulas where they don't belong
Yeah it's made by a Christcuck alright.

[Anonymous No.8301672]

>>8301653
The inverse images are points [math]2\pi[/math] apart. Also the word "fibres" should be reserved for cases where [math]\phi^{-1}(m)[/math] is uncountable for each [math]m \in H[/math].

[Anonymous No.8301676]

>>8301672
>>8301666
>they think I have any idea about these solutions
>tfw I'm a chemfag actually
pic familiar

[Anonymous No.8301677]

>>8301666
>Doesn't understand the appeal of a single Theory of Everything
>666

[Anonymous No.8301681]

>>8301676
>solutions
Only one of the three you posted was even a problem. I know you're a chemfag, but you can't be this dense.

[Anonymous No.8301682]

>>8301677
>implying God would waste his time writing down all that shit just to say "let there be light"
God is a Pajeet who doesn't understand computational efficiency?

[Anonymous No.8301683]

>>8301681
>'twas merely an act

[Anonymous No.8301684]

>>8301666
Fuck off, Satan.

[Anonymous No.8301686]

>>8301655
Here's another.

[Anonymous No.8301689]

>>8301078
You're assuming path-connectedness, not connectedness.

[Anonymous No.8301862]

>>8301672
>>8301659 4/3
I managed to find some more in my fapfolders

[Anonymous No.8301866]

5/3
there's something very amusing about combination of animay girls and scientific mumbo-jumbo

[Anonymous No.8301867]

6/3
well, that's all I got for now which is anyhow relevant

[Anonymous No.8301875]

[Anonymous No.8301886]

>>8301655
That's a pullback!

[Anonymous No.8301893]

>>8301867

If we can have different kinds of infinity why can't we have numbers with infinitesimal components?

[Anonymous No.8301897]

>>8301893
You can but that's not [math]\mathbb{R}[/math]

[Anonymous No.8301904]

>>8300662

Math wankers have to atleast admit that they like to say simple shit with complex language

[Anonymous No.8301972]

Is it possible to use chords to divide a circle into equal area pieces with no two pieces congruent?

[Anonymous No.8302004]

>>8301625
This one wins

[Anonymous No.8302010]

[Anonymous No.8302098]

>>8301079
>muh non-small angle approximation
>muh elliptic integrals

[Anonymous No.8302330]

>>8300708
:^)

[Anonymous No.8302429]

[Anonymous No.8302462]

Not a day goes by where I don't wish I was a stronger and more confident person in life. Keep studying kids even on this fine Saturday night.

[Anonymous No.8302473]

>>8300662
Elliptic curves provide the basis for some real-world cryptography and algorithms for factoring integers.

[Anonymous No.8302537]

>>8302473
lol that's just silly. what would curves have anything to do with integers? XD

[Anonymous No.8302543]

>>8302473
lol why should i care about this mental masterbation with no practical utility

[Anonymous No.8302557]

>>8302543
>no practical utility

>cryptography

Did you even read the post, faggot? Do you legitimately not know what cryptography is?

[Anonymous No.8302570]

>>8302473

To make this conversation better, wordfilter each instance of "integer" to "nigger".

[Anonymous No.8302580]

>>8301625

I have a math degree and I legitimately cannot tell if this is gibberish or a real question

[Anonymous No.8302595]

>>8302580
>quantum field theory
Yep, it's gibberish.

[Anonymous No.8302600]

>>8302580

[Anonymous No.8302814]

>You should be able to solve this.

[Anonymous No.8302846]

>>8301866
I have a lot of these saved up.

[Anonymous No.8302852]

[Anonymous No.8302856]

[Anonymous No.8302861]

[Anonymous No.8302862]

[Anonymous No.8302869]

>>8300662
you were never introduced to the concept of elliptic curves

[Anonymous No.8303031]

>>8300662
>Math is the least flawed of any human construct.
FTFY

[Anonymous No.8303060]

>>8299961
For funzies

[Anonymous No.8303101]

>>8302846
(You)
I couldn't find much of these, so made one myself

[Anonymous No.8303122]

>tfw you will never have a qt anime grill read you a bedtime stem textbook
why haven't i an heroed yet mina?

[Anonymous No.8303388]

>>8301867 cont'd
7

[Anonymous No.8303389]

8

[Anonymous No.8303392]

9

[Anonymous No.8303395]

10
aaand once more I'm done for now

[Anonymous No.8303405]

>>8303388
Wait. If p is prime doesn't that imply that the square root of p is always irrational?

If so how are you going to work in the rational numbers?

[Anonymous No.8303409]

>>8303405
Okay, nevermind. You probably mean that the coefficients of the vectors can only be rationals.

[Anonymous No.8303435]

>>8301867
See, but limx->(inf) of (1/10^(x)) is a limit. There's no point where (1/10^(x)) is actually zero. And 1-(1/10^(x)) is never equal to 1

[Anonymous No.8303440]

>>8303395
to be a pullback square, S must be a pullback and everything must commute
so S must be a subset of the product of X and 1 such that for some x in that product, phi(x) = true
(the product of X and 1 is essentially just X, which allows for this sort of abuse of notation)
assuming i is the inclusion map
phi.i:S->2 is always true, because the codomain of i is exactly S, and phi is true for elements of S
this makes S a pullback and makes the diagram commute

[Anonymous No.8303441]

>>8303435
and there's no point where 0.999... is anything but 1

[Anonymous No.8303449]

>>8303441
But there is a point where 1-(1/10^(x))=0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 and even more nines. To infinity at some point. But never really 1. So?

[Anonymous No.8303451]

>>8303435
fyi problem of 0.99.... =1? emerges only when flawed decimal notatoion is used, in base 12 there's no such problems, although there may be conundrum of similar kind dividing by, say, seven, which again in a different base doesn't yield recurring notations

[Anonymous No.8303454]

>>8303449
Yeah, and that point is 1.

[Anonymous No.8303457]

>>8303454
1=1-(1/10^(x))
(1/10^(x))=0
Solve for x

[Anonymous No.8303461]

>>8303457
>he didn't learn limit theory
kys

[Anonymous No.8303471]

>>8303435
>>8303449
>>8303457
I bet you are Zeno's """"paradox"""" defender too

[Anonymous No.8303473]

>>8303461
A limit is not necessarily the value of a function. It's what the function approaches. Try again

[Anonymous No.8303478]

>>8303471
The Achilles one? The theory might work but in reality that's not how shit works

[Anonymous No.8303482]

>>8303473
>value of function
ofc it's not, given a definition of a function
the number here discussed can be defined through previously mentioned function, where actual value is defined as a limit of that function

[Anonymous No.8303483]

>>8300662
>Math is logical and numbers never lie my ass. Math is just as flawed as any other human construct.
Pasta

[Anonymous No.8303584]

>>8303440
also acceptable
>characteristic function is subobject classifier in Set
>QED

[Anonymous No.8303597]

>>8303388
The Galois group of Q adjoin finitely many square roots of primes is isomorphic to (number of sq of primes) copies of Z/2Z.

[Anonymous No.8303611]

>>8302814
Use the primer number theorem

[Anonymous No.8303633]

>>8303597
It's not a proof if you use a result of the same order of difficulty without proof (although that will work). You might as well be saying "it's true, because it's true"

[Anonymous No.8303682]

>>8300662
Nice alteration of the pasta

[Anonymous No.8303701]

>>8301672
>Also the word "fibres" should be reserved for cases where ϕ−1(m) is uncountable
This is absolutely not the case.

[Anonymous No.8303717]

>>8303451
why don't we just use a base that is the product of all the basic numbers (1, 2, 3, 4, 5, 6, 7, 8, 9 and 10) then? we could divide everything without nasty results

[Anonymous No.8303736]

>>8303717
>he thinks product of 1 through 10 is sufficient
nah dog, multiple-digit primes exist and there's infinitely many of them
to be able to divide freely we'd need a base that is a product of all primes, which is improssible, OR we could use fractions for rational numbers that are found recurring in attempt to present them in decimal, that is with dot or comma, notation.

[Anonymous No.8304170]

>>8303633
Alright, the extensions over Q will be of degree at most 2^n as it is a composition of n extensions of degree 2.
Its Galois group contains automorphisms fixing all square roots of primes except for one that goes to minus itself. Hence the Galois group contains the group generated by all those autos which is isomorphic to n copies of Z/2Z.

But by degree considerations we found what the Galois group should be. Thus, a Q basis for the extension is given by the square root of product of primes, which gives Q-linear independence

[Anonymous No.8304215]

Newfag here and I wanna say you are hilarious, finding AT and AG problems on funpost make me laugh as fuck.

[Anonymous No.8304316]

>>8304170
You're still implicitly assuming that none of these extensions "collapse." For instance, it is not a priori obvious that [math]sqrt 3 \not \in \mathbb Q \left( \sqrt 2 \right)[/math]

[Anonymous No.8304353]

>>8302098
the fact that we still cannot analitically solve for the motion of something so simple as pendulum really makes you think.

on that note, the fact that you cannot solve for something as simple as the motion of a diatomic molecule without resorting to approximations also really make you think once you consider the real world (don't get me started on the universe) is made of extremely complex molecular and atomic structures.

physics is all about symmetries and approximations, and that's the best we can do.

[Anonymous No.8304399]

>>8304353
How does that make you think? The equation isn't linear lmao.

[Anonymous No.8304461]

>>8304353
Not really, when you think about how arbitrary the meaning of "solving analytically" is. Why does a Bessel function not count as an "analytic solution", when, say, an exponential function does? The fact that a certain differential equation can't be solved "analytically" isn't any deeper than the fact that a certain algebraic equation (like 3x=1) can't be solved over the integers.

[Anonymous No.8304578]

>>8300662
>>8302438
I stopped caring about memes once I was introduced to Harambe (pbuh). What a crock of shit. If you can save an ape by getting your dick out, like some kind of phallic deity, then you are fucking wrong and modern memery is flawed.

[Anonymous No.8304613]

>>8301085
fuck Haskell though

main = do
putStrLn "x1: "
x1 <- getLine
putStrLn "y1: "
y1 <- getLine
putStrLn "x2: "
x2 <- getLine
putStrLn "y2: "
y2 <- getLine
let slope = (read y2 - read y1) / (read x2 - read x1)
putStrLn $ "Slope: " ++ show(slope)

[Anonymous No.8304791]

>>8302462
Roid. Literally makes you stronger and more confident. Shorter temper though manageable.

[Anonymous No.8304808]

>>8301101
That's graph theory bud, i.e., math.

[Anonymous No.8304810]

>>8301625
how do I into hqft??

[Anonymous No.8304834]

>>8301653
But anon, this is just a standard exercise (see Dummit & Foote, 3.1.12)

[Anonymous No.8304842]

>>8304810
You don't lmao

[Anonymous No.8305122]

>>8304316
It is after I showed the degree of the extension is 2^n. The degree is multiplicative, so what you suggest is impossible: all the intermediate extensions must be of degree 2

[Anonymous No.8305131]

>>8305122
No. You haven't shown that these endomorphisms you've defined are actually automorphisms. Again, if the roots have some algebraic dependence then these will not in general define automorphisms, and you're just tacitly assuming that everything works out.

[Anonymous No.8305144]

>>8301625
Seems provable using some extension of the general Dold-Kan correspondence.

>>8302580
It's all real, check out HQFT on the nLab.

[God No.8305149]

>>8301666
This is literally why I had Michael throw your faggot ass from Heaven.

[Anonymous No.8305163]

>>8305131
I see! Thank you for pointing it out. Then you have to show it some other way

[Anonymous No.8305167]

[Anonymous No.8305200]

>>8305131
I agree with you that he hasn't solved the problem, but the issue is not that they might not be automorphisms; it's that they might not even be well-defined functions.

For example, suppose

[math]\sqrt{5}=2\sqrt{2}+8\sqrt{3} [\math]

Then evidently there can be no automorphism that takes [math] \sqrt{5}[\math] to [math] -/sqrt{5}[\math] but fixes [math] \sqrt{2}[\math] and [math] \sqrt{3}[\math].

Obviously the equation above is false, but the point is you have to rule out *all* possible equations like the one above, and asserting the existence of an automorphism like the one described in >>8304170
is just restating the problem. Because how do know that such a function is well-defined? At the very least, you'd have to rule out any non-trivial linear dependencies in the set of roots of primes, for the reason I explained above. But this is the original problem, so you've just gone around in a circle!

[Anonymous No.8305205]

>>8305131
I agree with you that he hasn't solved the problem, but the issue is not that they might not be automorphisms; it's that they might not even be well-defined functions.

For example, suppose

[math]\sqrt{5}=2\sqrt{2}+8\sqrt{3} [/math]

Then evidently there can be no automorphism that takes [math] \sqrt{5}[/math] to [math] -/sqrt{5}[/math] but fixes [math] \sqrt{2}[/math] and [math] \sqrt{3}[/math].

Obviously the equation above is false, but the point is you have to rule out *all* possible equations like the one above, and asserting the existence of an automorphism like the one described in >>8304170
is just restating the problem. Because how do know that such a function is well-defined? At the very least, you'd have to rule out any non-trivial linear dependencies in the set of roots of primes, for the reason I explained above. But this is the original problem, so you've just gone around in a circle!

[Anonymous No.8305242]

>>8301054
>>8305144
Actually, now that I think about it, all you need is the Cobordism hypothesis per Lurie. Then, your HQFT is determined by its value on the point, and I'm pretty sure the range Frobenius algebra is determined by the same data since Z is a dagger compact funtor, not merely monoidal.

[Anonymous No.8305261]

>>8305205
That is very helpful, thanks. I now understand better what my Galois professor had to say about defining automorphisms correctly

[Anonymous No.8305264]

>>8305205
You're correct; I stated that completely wrong. Excellent explanation. This is one of the finer points of Galois theory that needs more time devoted to it.

[Anonymous No.8305281]

>>8302010
the sets contained as elements of
[math] \mathbb{X}[/math] are not well-defined,
therefore not sets

[Anonymous No.8305298]

>>8303471
>I bet
there is no wagering at 4chan, Grandpa

[Anonymous No.8305446]

>>8302010
is this the ubuntu font?

[Anonymous No.8305518]

>>8305298
ebin mene, gomrade

[Anonymous No.8305556]

[Anonymous No.8305643]

>>8305556
i-is that a manga guide to something?

[Anonymous No.8305648]

>>8305643
Yeah "Manga's Guide to Constructive Quantum Field Theory"

[Anonymous No.8305652]

>>8305648
sauce pls

[Anonymous No.8305666]

>>8305643
>>8305648
>>8305652
no, that's just yuri. This what the japanese actually cover in highschool.

[Anonymous No.8305737]

I bet you can't solve this basic calculus problem

[Anonymous No.8305755]

>>8305737
>0

[Anonymous No.8305768]

>>8305755
>>8305737
Sum should be a function of x you idiot

[Anonymous No.8305857]

>>8305737
such an ugly integral, pls recommend flashy way to do it that isn't elementary like mine

[Anonymous No.8305870]

>>8300708
where can I read about this? it's intriguing

[Anonymous No.8305877]

>>8305870
Collatz problem

https://en.wikipedia.org/wiki/Collatz_conjecture

[Anonymous No.8306333]

>>8305857
Lol, nice work but i think you have mistaken (2n)! with 2n!
That binomial coefficient is equivalent to 2n! hence your solution is wrong.
I'll try to post my solution

[Anonymous No.8306408]

>>8305737
>>8305857
[math]S(x)=\sum_0^\infty\frac{x^{2n}}{(2n)!}[/math]
[math]S'(x)=\sum_0^\infty\frac{x^{2n-1}}{(2n-1)!}[/math]
[math]S''(x)=\sum_1^\infty\frac{x^{2n-2}}{(2n-2)!}=\sum_0^\infty\frac{x^{2n}}{(2n)!}=S(x)[/math]
[math]S''(x)-S(x)=0[/math] This is second order differential equation. Solution is:
[math]S(x)=C_{1}e^x+C_{2}e^{-x}[/math]
From initial conditions we can find coefficients
[math]1=S(0)=C_{1}+C_{2}[/math]
[math]0=S'(0)=C_{1}-C_{2}[/math]
Finally [math]S(x)=\frac{1}{2}e^x+\frac{1}{2}e^{-x}=\cosh(x)[/math]

[Anonymous No.8306415]

>>8306408
fuck me hard ;*

[Anonymous No.8306424]

>>8301085
fite me irl cunt

[Anonymous No.8306584]

>>8305281
That's cheating, brainlet.

>>8305446
yes

[Anonymous No.8306762]

>>8305737
>I bet
>>8305298

[Anonymous No.8306769]

>>8302010
Because of the axioms of ZFC, you can only define sets based on a predicate if and only if you define a 'universe' (in the trivial) sense.

So you cannot say 'X is the set of all sets that don't contain themselves' because that is not a set but you could say
'X is the set of all sets that don't contain themselves and are elements of Y' where Y is another set and the universe we are working in.

To make this second statement (that does indeed define a set) equivalent to yours we would simply have say that Y is the set of all sets but this leads to a contradiction.

From the axioms of set theory it can be proven that no set has itself as an element and therefore this Y does not exist.

Therefore you have a set of the form 'X is the set of all sets that don't contain themselves inside this set that does not exist' and as is the case with the usual contradictory by definition sets, X would be the empty set.

Therefore X does not has itself as an element simply because X = {}

wew.

[Anonymous No.8306798]

>>8306769
>restricted comprehension
Lmao

[Anonymous No.8306807]

>>8306798
I mean, we are working in ZFC right? Anything not ZFC is not worth my time anyways so I only care to answer this in the context of ZFC.

[Anonymous No.8306811]

>>8306807
>Anything not ZFC is not worth my time
>can't prove continuum hypothesis
>can't prove Riemann hypothesis
>implying it's worth anyone's time
Rofl

[Anonymous No.8306816]

>>8306811
>>can't prove continuum hypothesis

Literally no one cares. If it was important enough then we would already have the ZFC-CH axioms of set theory and even though some people tried adding an axiom that implied it, no one cared enough to listen to those faggots.

>>can't prove Riemann hypothesis

Who says we can't? We definitely can. ZFC is practically perfect.

[Anonymous No.8306820]

>>8306807
>Anything not ZFC is not worth my time anyways
hahaha oh wow

[Anonymous No.8306834]

>>8306820
The axioms of ZFC are like a basketball team.

Imagine you have a team with Lebron James and other professional players that are also pretty famous for being literally too fucking good. Lets say you have 4 of them.

That is a pretty good time, right? Now imagine you add 2 fat 10 year olds kids to your team. You still have your 4 master players but now by adding that little extra you made your team shitter.

Also notice that if someone asked you to take out one of your original 4 players then that would also make your perfect basketball team worse.

That is ZFC. The axioms of ZFC are the Lebron James of mathematics.

You take some of them out (like morons who don't want to believe choice) and it becomes shit.

You add some little extra (like the fags who want to add axioms that imply the continuum hypothesis) and it becomes shit.

Just take what you have and exploit them for a lot of money and theorems like the owners of basketball teams do.

[Anonymous No.8306901]

>>8300627

>formalizing a rigor-demanding mathematical question using disgusting physics notation
>>>/out/

[Anonymous No.8306914]

>>8306834
{{{{{}, {}}, {}, {{}}}, {}, {{}}},{{{}, {}}, {}, {{}}},{{{{}}}},{},{{},{{}}},{{{}, {}},{{{{{}, {}}, {}, {{}}}, {}, {{}}}}}

[Anonymous No.8306918]

>>8306834
Even fucking ETCS is better than ZF.

[Anonymous No.8306921]

>>8306834
No, you don't get it. Powerset, pairing, union, infinity - these are the Lebron James of set theory. Extensionality and restricted comprehension are those guys that never score but they're pretty solid for passing to the Lebron Jameses.
Replacement and regularity are the fat 10-year-old kids that don't do shit.
Choice is like that guy who is pretty mediocre overall, but makes clutch shots every once in a while.
Large cardinal axioms are like some Mexicans that only play soccer and don't speak English but could be good at basketball if they manage to learn the rules.

[Anonymous No.8306925]

>>8304353

>define a function as the solution to the equation
>it is now analytically solved

wow amazing! see >>8304461

also friendly reminder that our models are just approximations to reality in the first place. anyone who doesn't live by this mindset sounds like an undergrad baby desu

>For such a model there is no need to ask the question "Is the model true?". If "truth" is to be the "whole truth" the answer must be "No". The only question of interest is "Is the model illuminating and useful?"

[Anonymous No.8306936]

>>8301893

This is called nonstandard analysis. See https://en.wikipedia.org/wiki/Hyperreal_number .

If by "numbers with infinitesimal components" you mean "real numbers", then no, the we've defined the reals precludes such notions.

[Anonymous No.8306987]

>>8306834
What are the bad things that happen if you add the continuum hypothesis?

>>8306921
?
You can't prove shit without extensionality. Hardly any of the other axioms even uses the symbol "=".
Powerset is your creepy uncle that you rather not invite, but sure, he's part of the family.

[Anonymous No.8307015]

>>8306936

wouldn't it make more sense that .9 repeating is that function in the proof AT infinity instead of the limit as it approaches infinity

[Anonymous No.8307022]

>>8306987
>Powerset is your creepy uncle that you rather not invite, but sure, he's part of the family.

>this is what finitists actually believe

[Anonymous No.8307042]

doubt anyone will get it without googling]

kind of mathematics related

[Anonymous No.8307258]

>>8302010
If you use any logic on it and a consistent definition of sets, then there's no answer because you'll get a paradox

Russell's paradox

[Anonymous No.8308154]

>>8306584
what you call "cheating" is what
others call "analysis", anthropoid

[Anonymous No.8308182]

>>8308154
>the sets contained as elements of
>X are not well-defined,
>therefore not sets

This is factually untrue. In ZFC you can give many examples of sets that do not contain themselves: {}, {{}}, {{}, {{}}}, etc. These sets are perfectly well-defined.

[Anonymous No.8308196]

>>8304808
That has nothing to do with graph theory.
Who let /g/ in?

[Anonymous No.8308276]

>>8307042
I can solve it!!!!!
thread don't die pls. just give me 8 hours to get back from work, and I will post the solution

[chemfag from this thread No.8308280]

>>8308276 cont'd
inb4 "solve"
it's not so hard t/bh f/am, but I'll draw it anyway.

till later

[Anonymous No.8308595]

>>8308280
actually now that i look at it again the assembly seems more obvious than at first

[chemfag from this thread No.8308683]

>>8308595
now you try and solve this
>>8301676
it's not really hard 2bh too

[Anonymous No.8308819]

>>8308683
Cu(I) explains the tetrahedral copper corners

so basically like a cylinder

[Anonymous No.8308829]

>>8308683
it was me who asked the question btw, want the follow up question? i was checking through the previous years exam of my upcoming courses

[chemfag from this thread No.8308970]

>>8308829
I'm thinking to change "core" to pentasubstitued cyclopentadiene type to make metallocene-like links, while the bottom one terminates with halogen
>pic familia
>>8308819
fuck, you beat me at that
I wanted to try and build in in acd/chemsketch, but the stupid program fucks up the geometry

[chemfag from this thread No.8308984]

>>8308970 cont'd
not necessarily halogen, it could be NHC or any other donor, really

[Anonymous No.8309034]

>>8307042
Hey that forms a MOF doesn't it? There is software that can solve these problem. The automated assembly of secondary building units might be able to solve that. Wait are those phenyl groups?

[Anonymous No.8309040]

>>8308970
theres a metallocene button in chemdraw by the way

but I think the idea you have might not work depending on what core ligand you have in mind, if the nitrogen donors are close to the cyclopentadiene core their donating capacity will be impaired by the core becoming more electron poor upon metallation

[chemfag from this thread No.8309058]

>>8309040
I'm thinking about bulky pentaaryl Cp's

[Anonymous No.8309080]

>>8309058
oh interesting never knew those existed

what about the orientation of the terminal acetylacetonate? if it isn't flat then your geometry gets fucked up

[chemfag from this thread No.8309093]

>>8309080
I just came up with it myself
considering there's five of them, one can't really predict what would happen.
so can't chemsketch

[chemfag from this thread No.8309105]

>>8309093 cont'd
perhaps, if we used a benzene analogue of this compound the orientations of acac's would be alternating, and no metallocene-alike links would be necessary

provided such compound exists, of course

[Anonymous No.8309126]

>>8309105
yeah but what if you would want an assembly of cylinders

i would chuck in an extra ring in the linker and modify that with something

[chemfag from this thread No.8309147]

>>8309126
like this?

[Anonymous No.8309150]

>>8309126
Why do you want an assembly of cylinders?

[Anonymous No.8309153]

>>8309150
the question states to use the self organized assembly as a building unit

[Anonymous No.8309161]

>>8309147
Well if you would want to keep the copper corners intact and prevent interference, you ideally would want a ligand in which the angle forces something that isnt a tetrahedral geometry

so

no idea

[Anonymous No.8309171]

>>8309126
Like a periodic structure with cylinders? How about pic related?

[chemfag from this thread No.8309174]

>>8309161
oh right, I messed up, there must have been an extra link

still though, I haven't gotten any better idea than metallocenes

[chemfag from this thread No.8309686]

another idea.
modify the original core by doubling it and connecting the doubles by (CH3)3 at places originally occupied by Ph's

[chemfag from this thread No.8309693]

>>8309686 cont'd
as you can see, the difference in internuclear distance it rather drastic, so the only way of connection of these, given Cu+ cation would be the formation of polymer chain

[Anonymous No.8309704]

Leave it to a retarded chemfag to derail the thread.

[Anonymous No.8309722]

>>8309686
pretty lit strategy actually

[Anonymous No.8309736]

>>8309704
gonna derail with more chemistry with this anon

[Anonymous No.8309744]

>>8309686
reminds me of this molecular cage for purifying fullerenes:
http://www.nature.com/ncomms/2014/141126/ncomms6557/abs/ncomms6557.html

[Anonymous No.8309975]

>>8309744
>a buckyball net

[Anonymous No.8310057]

>>8309975
poetry

[Anonymous No.8310113]

>>8303122
*minna

[Anonymous No.8310117]

>math major
>having fun working these tough math problems
>thread invaded by chemistry wizards

post more math problems i feel left out

[Anonymous No.8310126]

>>8310117
Then you suck at math cause chemistry = applied physics = super applied math.

[Anonymous No.8310135]

>>8309704
Atiyah-Singer index theorem

[Anonymous No.8310140]

>>8307042
>>8308819
>>8308970
>>8309058
>>8309040
>>8309034
>>8308984
>>8309093
>>8309080
>>8309105
>>8309126
>>8309147
>>8309150
>>8309171
>>8309174
>>8309686
>>8309693
>>8309744
>>>/x/

[Anonymous No.8310141]

>>8299961
Explain as best as you can, the four peaks in this H-NMR. (Stampname is filename)

[Anonymous No.8310182]

>>8310135
Prove it.

[Anonymous No.8310189]

>>8310126
i actually agree with you

[Anonymous No.8310394]

>>8302557
it can't help me pass integral calc

[Anonymous No.8310516]

>>8310141
>caffeine
Integral at ~ 4.7 is at least 5-6 times larger than the others, caffeine lacks that amount of H symmetry/coupling.

[chemfag from this thread No.8310523]

>>8310141
what do you mean by "explain"?
everything looks fine to me except that large peak at 4.8, and with no data on intensity and multiplicity I can't tell shit

[Anonymous No.8310635]

>>8310117
Why not both?

Prove or disprove the applicability of the molecular conjecture to d-dimensional periodic body hinge frameworks

Here is the proof of the molecular conjecture for non-periodic d-dimensional systems:
http://arxiv.org/abs/0902.0236

[chemfag from this thread No.8310842]

>>8310516
could it be residual solvent peak?
I see no other explanation here

[Anonymous No.8310852]

>>8310842
I think ~7.7 is the solvent because:
1) solvent peak is usually very small
2) solvent peak is generally nowhere near the other H environments

[chemfag from this thread No.8310934]

>>8310852
>solvent peak is usually very small
depends on how concentrated a specimen is
~7.7 must be a lone aromatic proton that is present in caffeine

[Anonymous No.8310965]

>>8305877
thanks m8

[Anonymous No.8310970]

>>8310934
Ran a chemical shift prediction, you're correct about the aromatic proton

[chemfag from this thread No.8310995]

>>8310970
name of the program please?

[Anonymous No.8311002]

>>8310995
IQmol
http://iqmol.org/
It's a visualization package / GUI for Qchem. Wish I knew about it in my earlier undergrad years.

[Anonymous No.8311019]

>>8310934
man i remember we got a few spectra and a chemical formula and had to work out a structure in my first NMR class

it had these peaks below 1 and if you don't know that is from a cyclopropyl fragment the spectra become real fucking puzzling

[Anonymous No.8311036]

>>8311019
How much reference material did you get?
We always had nasty complicated molecules, but we had pages and pages NMR shift data on hand in exams (like pic attached).

Also, doing a double bond equivalency calculation before you start is always a good to get an idea of the number of double bonds/rings

[Anonymous No.8311040]

>>8311036
usually it was H nmr, C nmr and MS spectra

yeah and we had reference tables but no cyclopropyl fragment

the prof MEMED us

[Anonymous No.8311049]

>>8311040
Yeah every so often you'll get memed pretty hard by a question with an unexpected answer.

[Anonymous No.8311055]

>>8311049
I remember a lot of my classmates actually got memed by having to explain this graph. The filled and empty circles denote 1000 and 2000 rpm stirring, respectively. What can you learn from this?

[Anonymous No.8311061]

>>8311049
>>8311055
oh the reaction is catalyzed by Ru/TiO2 catalyst by the way

[Anonymous No.8311074]

>>8311061
I have no idea what to make of that graph. A quick google of ethyl pyruvate to ethyl lactate gave me this though: http://pubs.rsc.org/en/content/articlehtml/2014/gc/c4gc00191e

[Anonymous No.8311118]

>>8311074
yeah thats where i took it from

hint it tells you something about where the selectivity is determined in the catalytic cycle

[chemfag from this thread No.8311332]

>>8310141
>>8310516
>>8310852
~4.8 is heavy water (deuterium oxide at table)
>http://www2.chem.umd.edu/nmr/reference/isotope_solvent.pdf
other peaks are at the right places

[chemfag from this thread No.8311976]

bumpin