Mathematics - 2017-03-11 (page 1 of 5)

Semiclassical

01:50

hi @ted

Vrouvrou

02:18

hello

Vrouvrou

02:46

someone have an idea about this:math.stackexchange.com/questions/2181343/…

Vrouvrou

03:09

Who knows Brezis-lieb lemma ?

Xam

03:43

Hello

Secret

I think I need a category theory of harmonic numbers (and other series defined number sequences) to better understand the symmetries within these numbers

Xam

@Secret hi

Secret

hi

skill patrol

03:58

hi

Zach Hauk

hi @skillpatrol

Artificial Intelligence Beach

►

skill patrol

Hi pal @ZachHauk

Nice vid.

Secret

04:14

If I understood correctly, I sometimes have this feeling when I do linear algebra and some abstract algebra investigations

But for me, at least for the moment being, it is to understand the patterns within mathematical objects and to use them

Abstract algebra is relatively easy on this regard because everything happens one step at a time.

Analysis, however have many things happening at the same time, thus hard to track

Integrals are somewhere in the middle. It is known that each integrand has some kind of symmetry that allows integrals to be evaluated without knowing their closed forms and integrals are linear functionals thus you can sort of treat it like a function with the integrand as its argument, which is another function and that can constrain the possible behaviour of the integral itself

If I understood correctly, algebraic geometry abstract this notions in the form of complex varieties and the Hodge conjecture then asked in some sense whether you can decompose an integrand

Infinite series are even harder, because unlike integrals, it has this extra complication of nonassociativity

(O and in case someone ask me why I don't ping, it is because I want to quote two messages at the same time)

If this

$$\sum_{k \in I}f(H_k)$$

forms nice algebraic structures like groups , rings etc. for some f, then the problem might be easier since e.g. groups encode symmetries of the underlying structure, and that is precisely what we need in order to to the pattern matching

Perhaps, an even more general question is whether we can find a $g$ such that:

$$g: \sum_{k \in I}f(H_k) \to \int_U f(H_k) d\mu$$

DHMO

@Secret hi

Secret

04:31

... I wonder if there is a topology where the harmonic series converges to a unique limit...?

...and whether it is always valid to start a problem in some topology, and then use a mophism to map its solution to another non homeomorphic topology in order to solve the original problem...?

DHMO

@Secret yes. the standard topology on the extended real line

if you want the topology to be purely on R, you can just set 0 to be infinity

you don't need 0 in this question anyway

Kasmir Khaan

hi DHMO

DHMO

@KasmirKhaan god dag

Kasmir Khaan

@DHMO good day !

You know about recursive relations?

DHMO

hmm... I saw your question above

Kasmir Khaan

04:40

:D

DHMO

let's use matrix

Kasmir Khaan

a_n = 12a_n-1 -27 a_n-2 +256 n for n>=2

a_0 = 38 , a_1 = -62

i found a solution for the case when i let n = 0

-88/3 9^n + 202 /3 * 3^n

DHMO

oh wow, I didn't see the n

Kasmir Khaan

that one agrees on a_0 and a_1

DHMO

I can't use matrix then

Kasmir Khaan

04:43

oh sory for typing this way , i cant do better

DHMO

varsagot

Kasmir Khaan

=p

DHMO

that's the wrong phrase isn't it

Kasmir Khaan

yes ><

DHMO

inga problem

Kasmir Khaan

04:44

now you got it :D

DHMO

tack

Kasmir Khaan

i calculated some value , a_2 = -1258

a_3 = -12654

DHMO

let's solve a simpler problem first: a_n = 2a_(n-1) + n, a_0 = 0

@KasmirKhaan you can't let n=0

Kasmir Khaan

it was given

where did you get that ?

a_n -12 a_(n-1) +27 a_ (n-2) = 0

that is when n = 0

but i found solution that agrees with the given values of n_0 and n_1

DHMO

when n=0, we have a_0 - 12a_-1 + 27a_-2 = 0

you can't just substitute one n without substituting other n

Kasmir Khaan

04:49

oh yes

-88/3 9^n + 202 /3 * 3^n

but that expression is correct

when n = 0 , a_n = 38

when n =1 , a_1 = -62

Harnoor Lal

Guys

please just put $$ around your expression to make it pretty and readable. $n_1$ instead of n_1

DHMO

@HarnoorLal we can read it without latex

Kasmir Khaan

sorry i did not learn yet how to use latex

I will focus on summer to learn it , now i have alot of courses

DHMO

@KasmirKhaan it would fail for n >= 2

Kasmir Khaan

yes i know but i got the homogenous solution from it

a_n = homogenous solution + particular solution

DHMO

04:52

I don't know this anyway

ursakta mig

Kasmir Khaan

hehe its okay

if you have other way its fine too

DHMO

talar du endast ingleska och svenska?

Kasmir Khaan

some french and arabic too

how about you ?

DHMO

some french and spanish and limited german

Kasmir Khaan

nice :D

DHMO

04:55

I suppose german makes swedish easier, but I'm trying to learn it now

Kasmir Khaan

what is your native language?

DHMO

cantonese :p

Kasmir Khaan

nice :D

you got the hardest one so the rest would be easy

DHMO

lol

Kasmir Khaan

haha :D

DHMO

04:57

I kind of like how Swedish does not conjugate for persons

unlike French and Spanish and German

Kasmir Khaan

yes french is the hardest on that matter

DHMO

I agree

Kasmir Khaan

Can wolfram alpha solve such recursive eqations?

I really would like to know the answer

but grrrrrr so far i get many answers that none of them seem right

DHMO

yes

a(n) = 16 n + 14 3^n - 18 9^n + 42

Kasmir Khaan

hmm

where did the homogenious solution go ?

=p

or wait might be other way to solve that =p

@DHMO thanks for help again ! :D can you send me the window where you can plug expressions for recusive functions on wolfram alpha ? and intial conditions? :D

DHMO

05:06

I just typed "a_n = 12a_n-1 -27 a_n-2 +256 n, a_0=38, a_1=-62"

Adeek

hi

DHMO

@Adeek god dag

Kasmir Khaan

Oh okay , i really should learn how to use those things :D @DHMO

2017

@DHMO Hey, you there? I need some help with inclusion exclusion here! I was trying to prove the inclusion exclusion formula for when 6 different objects (say a,b,c,d,e,f) are to be distributed in 3 different boxes (say X,Y,Z). So to find the number of ways in which each box gets atleast 1 object I tried like: (3^6 (total number of ways) - 3 (they all go to one box) - 3C2 (2^6-1^6)) (they all go to exactly 2 out of 3 boxes). However, I seem to be getting a wrong answer. Can you spot the mistake?

DHMO

@2017 the complement of "each box gets at least 1 object" is not "they all go to one box"

2017

05:11

@DHMO I excluded the case when they go to exactly two boxes also

3C2 (2^6-1^6)

DHMO

oh ok

2017

Is that correct?

DHMO

it should be 3C2 (2^6 - (2C1)(1^6))

2017

@DHMO Oh oh right! I forgot that 2 :-P

Thank you :-)

DHMO

no problem

2017

05:13

How are you btw?

School going on?

DHMO

quite busy

2017

oh, exam season?

DHMO

ya

2017

mine too :-) All the best for your exams!

DHMO

thanks

2017

05:14

ttyl then :-)

Kasmir Khaan

@DHMO Good luck on your exams!

DHMO

tack sa mycket

Secret

05:41

[Chemistry] The DFT geometry optimisation is not converging. Need to try again using another integration grid

DHMO

05:57

@Secret did you see my answer to your topology question?

Secret

The extended real line one, or something earlier?

DHMO

the former

Secret

I am not sure how that will help. Say even if we had the harmonic series to converge to infinity in $\mathbb{R}^*$ which can be mapped to $0$ to stay in $\mathbb{R}$ that seemed to be still far from solving the original problem unless we can find some mophism that maps that 0 uniquely back to whatever the value of that series sum is

DHMO

oh I didn't answer your second question

I only answered your first question

Secret

right...

One of the interesting thing about the harmonic series is that according to my professors, it is kinda a "borderline" case where are series can go from convergent to divergent

DHMO

06:04

indeed

Secret

For example, one of my professors have shown that a variation of a p series, if p>1, no matter how small that is, it will converge. There's even one series where you only change one term in the harmonic series and it will converge

Adeek

@DHMO what is dog dag ?

god dag ?

DHMO

ignore it

Secret

it kinda reminds me of emergence that occured in statistical phenomenon, and also chaos theory. It seems the value of series is very sensitive to its components and their ordering

but I still don't quite understand how exactly that work, similar to integrals

I think it is safe to say that I don't really understand nonlocal maps

DHMO

06:30

@Secret but 1,1/2,1/4,1/8,... won't converge in my topology lol

Secret

Are the neightbourhoods of any point in the standard topology of the extened reals has the form of unions of open intervals?

DHMO

I believe so

except that [infty,infty] is also an open set @Secret

Secret

Then I don't see why that series cannot converge to 0 since at some point, the Nth term of the dynadic fraction sequence will be within any neighborhood of 0

DHMO

oh, we are referring to different topologies

DHMO

06:47

let's think about a topology on R where every monotone sequence converges to one limit

oh, easy enough, just the discrete topology with {0} removed, so every sequence converges to 0

let's add a restriction that there must be at least two sequences converging each to a different limit

DHMO

07:01

@AlessandroCodenotti buongiorno

do you have any idea?

Alessandro Codenotti

Hi

DHMO

can we make every sequence converge to one limit?

Alessandro Codenotti

Yes, in the trivial topology

DHMO

I mean, only one limit

I gave a topology above where every sequence converges to 0, but I added another restriction now

Alessandro Codenotti

What about constant sequences? I doubt it

DHMO

07:04

oh, that's a good point

so what should I do?

it's noteworthy that there are only as many sequences as the continuum

Alessandro Codenotti

That shouldn't be possible

DHMO

why not?

Alessandro Codenotti

If you want a topology in which every sequences has a unique limit just pick 2 different constant ones

DHMO

I don't understand

Alessandro Codenotti

A constant sequence always converges to its value (plus possibly other stuff too if the topology is non Hausdorff)

DHMO

07:14

I mean, what do you do after picking 2 different constant sequences?

They both converge to their own constant

so I am not seeing any contradiction

Alessandro Codenotti

Didn't you want all sequences converging to the same limit?

DHMO

oh, my question was misunderstood

I'm looking for a topology in which every sequence has exactly one limit

Alessandro Codenotti

Any Hausdorff one will do then

DHMO

@AlessandroCodenotti not all sequences have limit

0,1,0,1,0,1,0,1,...

Alessandro Codenotti

Every sequence alternating between $x$ and $y$ won't converge if they are separated by open nbhds

So you should fall into the trivial topology case again

DHMO

07:19

that doesn't satisfy "exactly" one limit

Alessandro Codenotti

I know, so it shouldn't be possible

DHMO

31 mins ago, by DHMO

oh, easy enough, just the discrete topology with {0} removed, so every sequence converges to 0

This works for all sequences except the eventually constant ones

so 0,1,0,1,... converges to 0 here

Alessandro Codenotti

Eventually constan sequences do converge to 0, but to another value as well

DHMO

I know

I just mean I gave an example that almost works

so how can you be so certain that it will never work

how many possible topologies are there?

I proved that it is between 2^R and 2^(2^R), but which one is it?

Balarka Sen

morning

DHMO

07:25

@BalarkaSen hi

do you have any idea?

9 mins ago, by DHMO

I'm looking for a topology in which every sequence has exactly one limit

Alessandro Codenotti

@DHMO the latter, but I can't prove it. The proof I've seen involves some ultrafilters stuff

Hi @Balarka

Balarka Sen

@DHMO As in, all the sequences converge to a single point?

DHMO

@BalarkaSen yes

the point can be different for different sequences

Balarka Sen

That seems contradictory to your "yes" :P

DHMO

misunderstandings

the English language is inadequate to convey mathematical ideas

let's all learn Lojban :p

Balarka Sen

07:32

No. If you want your question answered state it precisely. I don't care much about it otherwise.

Hi @Alessandro

DHMO

@BalarkaSen I am looking for a topology in which there is exactly one limit for every sequence

Secret

Is the notion of "minimal open set" make sense in topology. That is, a "minimal open set $S$ at a point $x$" will be defined as a open set containing $x$ such that every neighbourhood of $x$ must contain $S$?

DHMO

@Secret the intersection of all the neighbourhoods might as well be $x$ itself

Balarka Sen

@DHMO I am saying, does that mean "$\exists$ $p \in X$ $\forall$ sequences converge to $p$ in $X$", or what?

DHMO

@BalarkaSen that means, $\forall$ sequences $\exists$ p: the sequence converges to p

Balarka Sen

07:36

So just that every sequence converges?

Secret

@DHMO ah I see, and the problem is that a singleton is not necessary open in some given topology

DHMO

@BalarkaSen to exactly one limit

i.e. something like Hausdorf

Balarka Sen

Oh, so it can't converge to two things at once? Gotcha.

I was going to give the Sierpinsky two point space as example.

Hm.

DHMO

oh and ideally the topology is on R

but you can give me anything else

Balarka Sen

Suppose you take two points x, y from your space. {x, y} either inherits the discrete topology or the Sierpinsky topology, right? If the former x, y, x, y, x, y, converges to nothing so you're out. In the latter x, x, x, x, ... converge to x and y at once so you're out too. Doesn't that mean that can't happen?

DHMO

07:39

so we need to consider topologies on other sets :p

Secret

How can one have a neighborhood U of x that contains all points in a set, and not ending up having U being a neighbourhood of any other points that is not x?

DHMO

they can converge to values besides x and y for all intends and purposes @BalarkaSen

Secret

That is like as if $\in$ becomes a partial relation

DHMO

@Secret depends on definition of neighbourhood

some doesn't need neighbourhood to be open

Balarka Sen

@DHMO Sorry, what? I mean take a two point subspace of your space - you can either construct a sequence which converges to nothing or construct a sequence which converges to two limits. That contradicts "every sequence converge to a unique limit", right?

Secret

07:41

No, the issue is that the whole set is always a neighborhood of all its points

and that is enough to make constant sequences converge to something beyond the intended limit

DHMO

@BalarkaSen they can converge to other values

Balarka Sen

Ah, outside. Fair.

DHMO

@Secret [0,1] is a neighbourhood of 0.5 but not of 1

in some definitions

Balarka Sen

If x, y, x, y, x, y does converge to something (say z) wouldn't it also converge to x and y? That means any nbhd of z contains x and y both, I mean.

Secret

another question is that is it possible to have a topology where a constant sequence x,x,x,x,x,x, .... converge to anything but never x?

Balarka Sen

07:45

@Secret No, it always ever converges to x. It can converge to stuff aside from x (Sierpinsky two-space)

i.e., have multiple limits

DHMO

all neighbourhoods of z contains x and y. some neighbourhoods of x do not contain y. some neighbourhoods of y do not contain x. problem? @BalarkaSen

@Secret all neighbourhoods of x must contain all values in the sequence, so x must be a limit

Secret

ah ok

Balarka Sen

@DHMO Why're the next two sentences relevant? If my sequences converge to z they also converge to x and y, correct?

DHMO

@BalarkaSen i just explained why it is not correct

@BalarkaSen in the "discrete topology where {0} is not an open set", the sequence (1,2,1,2,...) converges to 0 but not 1 or 2

Balarka Sen

So {0, 1} and {0, 2} are open?

Just not {0}

DHMO

07:51

not really

Balarka Sen

Define your topology.

DHMO

wait

Secret

is the only open set containing {0} in that topology the whole set?

Balarka Sen

Then 1, 2, 1, 2, ... can never converge to 0.

Surely there's some other topology we are talking about.

DHMO

@BalarkaSen the basis is every singleton except {0}

@BalarkaSen what?

I'm starting to think that this is not a valid basis

Balarka Sen

07:52

@DHMO What's a neighborhood of 0?

Secret

ok so that means the smallest open set that contain each point x are the corresponding singletons {x}, but the smallest open set that contain {0} is the whole set

Balarka Sen

Yes, it is invalid as said.

DHMO

@BalarkaSen the whole set?

@BalarkaSen why not?

it's the excluded-point topology

Balarka Sen

But I guess you're right that my alternating sequence argument wasn't right. Every nbhd of x need not contain z (nor y), so the sequence won't converge to x actually.

@DHMO Ah, then that would indeed converge to 0.

Daminark

Hey guys

DHMO

07:56

@BalarkaSen what did you misunderstand?

Balarka Sen

I didn't understand the topology. I do now.

Secret

@DHMO actually wait a sec, I think that notion might be valid. This is because arbitrary intersections of open sets is not necessary open. Since the neighbourhood of x by definition must contain x, and if we use the open neighbourhood convention, then it should be possible to always find the smallest open set that contains x, such that all other neightbourhoods of x must contain it

These opens sets for each point, will then play the role of $\epsilon-$ open balls in any general topology

DHMO

@Secret what's the smallest open set containing 0?

Balarka Sen

@DHMO So the problem with this is constant sequences converge to two limits, just like Sierpinsky.

DHMO

I know

Secret

07:59

@DHMO That will be the whole set in your topology, since no union of any open set can contain {0} unless it contains {0}. but in that topology the only open set that contains {0} is the whole set

whereas for all other points, singletons contains itself

DHMO

i mean in the usual topology

@BalarkaSen the excluded point topology is just a proof of concept

Balarka Sen

@DHMO If x, y, x, y, x, y ... converges to z consider x, z, x, z, x, z, ... - that converges to both z and x in general, doesn't it?

Secret

@DHMO In the usual topology, it will be the set of open intervals (a,b) where a<0<b. The point is that we can set a and b arbitrarily small, and it will still be contained by all other neighbourhoods of 0 as long a,b is small enough

DHMO

@BalarkaSen not every nbhd of x must contain z

@Secret "arbitrarily small" is garbage

infinitesimal doesn't exist

every epsilon is a real number

@BalarkaSen consider the same example in the same topology

Balarka Sen

Yeah, I am not sure why I thought that. It's just the Sierpinsky example.

Secret

08:03

(Sometimes I really don't like that the rationals are not well ordered in the usual ordering, otherwise this would have been much easier)

DHMO

@Secret then consider the topology on omega_1

what is the smallest open set containing omega?

Balarka Sen

So it can only happen if no singleton in your space is open. If it is you can play this game to get a Sierpinsky space containing that as it's "1" and then come up with a bad sequence.

DHMO

i dont understand. what would happen if {0} is open?

Secret

Agh, yes, while $\omega_1$ is well ordered, $\omega$ is a limit ordinal thus it has no prececessors thus I cannot find a unique a such that $(a,\omega +1)$ contains $\omega$

DHMO

correct

so are we convinced that "minimal open set" is garbage?

Balarka Sen

08:07

@DHMO In your topology?

Secret

fine, in that case, I don't really know what I have been using in my reasoning when we do that left order topology with you a few weeks ago in figuring out convergence

DHMO

@BalarkaSen in any topology

you said no singleton can be open

@Secret is it few weeks ago?

"minimal open set" isn't garbage in some topologies

Balarka Sen

@DHMO If {0} is open pick 1. {0, 1} is either discrete or Sierpinsky. Cross the latter. If discrete, say 0, 1, 0, 1, ... converges to x \neq 0, 1. So any nbhd of x contains 0. {0, x} is then either discrete or Sierpinsky. But it can't be discrete because any nbhd of x contains 0 - so it's Sierpinsky. That means 0, 0, 0, .... converges to both 0 and x.

Daminark

@DHMO Well, you know the hyperreals... They're like...

Secret

@DHMO But it is for the standard topology, since the reals are not well ordered thus you cannot find (a,b) where a,b, arbitrarily close to c

DHMO

08:12

@Secret are we talking about the standard topology or the left-order topology?

Secret

which makes me wonder (barring that I miss out the oscillating sequences), why does this argument work

Mar 7 at 6:39, by Secret

@DHMO. I think there is no difference in the convergence behaviour of the sequence in the left order topology vs usual topology. This is because given any $s_N$, and for any given point $x$, in the usual topology $s_N \in (a,b)$ where $a,b$ is arbitrary close to x, and in the left order topology, $s_N \in (-\infty,c)$ where $c$ is arbitrarily close to $x$. Both type of neighborhoods are contained in any nonempty open sets of their respective topology,
therefore if a sequence converges to $x$ in the usual topology, it must also be in the left order topology

that's is the intuition I based this "minimal open set" idea from

(NB, the wrong conclusion above is because I miss out the oscillating sequences, which converges in the left order topology to many limits)

DHMO

@BalarkaSen nice.

Balarka Sen

So at least we know any potential counterexample can't have singleton open sets.

This is so darn confusing.

That's why I prefer spaces which are less painful to the eye.

DHMO

you mean to the brain?

Alessandro Codenotti

@BalarkaSen Like wedges of Hawaiian cones :P

Secret

08:16

DHMO: The idea is that if that concept works, then for any topological space, I can find the open set (or a "reasonable set of them") for each point x such that all neighbourhoods of x must contain them, thus the problem of whether some given sequence will converge to some point x will be completely determined by the size of these open sets for each point

Balarka Sen

Eye man. Topology is eye.

@Alessandro Beautiful space

Secret

but since the concept does now work, I am going to figure out how to deal with them

Balarka Sen

It can be seen, unlike some non-T0 garbage

DHMO

@BalarkaSen there are 2^2^2^Z possible topologies on R, yet we can only define Z topologies. There are many many many topologies that one can never define

Balarka Sen

True. Lots of restrictions given by the axioms.

Secret

08:19

Take the sierpinski space as an example

Balarka Sen

I think of Sierpinsky as [0, 1) mod (0, 1), which is a little easier to visualize

DHMO

@BalarkaSen that is nice

but did you mean (0,1] mod (0,1)?

Balarka Sen

Yeah, thanks.

Those are homeomorphic anyway.

DHMO

but we prefer to map 1 to 1

Balarka Sen

Mhm.

Secret

08:22

This set has only two points, thus my idea works. the smallest open set containing 0 is {0} but the smallest open set containing 1 is {0,1}. Therefore, a sequence can only converge to 0 if it is eventually 0, but a sequence can always converge to 1 because 0,1 \in {0,1}

The issue, as pointed out by DHMO, is that this idea don't work for arbitrarily topological space

Balarka Sen

@DHMO So, now, smallest open sets you can have are doubletons. None of them can be Sierpinsky too. So let's say choose {0, 1} - we know 0, 1, 0, 1, ... converges to x. Consider {0, 1, x}. Any nbhd of x contains 0 and 1, so an nbhd of x looks like the whole space here - excluded point topology on the discrete set with 3 points. Hmm.

DHMO

@BalarkaSen if (x,y,...) converges to x, then (y,y,...) converges to both x and y. If it converges to z instead, then (x,x,...) converges to both x and z. Therefore, such a space is impossible.

I'm not sure why you never made that step

Balarka Sen

@DHMO Is the first sequence x, y, x, y, x, y, ... ?

DHMO

yes

Balarka Sen

Oh, you're just claiming it's not sierpinsky in the first two sentences. OK

@DHMO Oh, duh, yeah. {x} is open in {x, z} so it is Sierpinsky.

DHMO

08:29

I'm just doing it case by case

Balarka Sen

Yeah, it was just the same argument. I assumed more than what is needed.

Good call.

You don't need that {x} is open in the whole space, just that it's open in that two-point space which is, like, definition of subspace top.

So we're happy.

DHMO

yes

thanks for providing the main argument

Secret

Actually, while I knew the definition of convergence, how can I check ALL the neighbourhoods of a given point x to ensure that some given sequence S converge to x given there is no minimal open set

Balarka Sen

No problem. The thanks goes mutual.

Secret

ok

DHMO

08:33

@Secret use the basis

Secret

11

Q: does every topology have a basis?

This might be a silly question, but i was wondering, is there any topology that cannot be generated by a basis? if not, given a topology, is there a reliable way of figuring out a basis for it? it probably matters if the set $X$ the topology is on is countable or not, right? Would it matter if th...

general-topology

Ouch

There are topologies where the only basis is itself

Balarka Sen

@Alessandro So, what's new?

DHMO

@Secret you can't; that's why i deleted it

Secret

Agh..., I will worry about this convergence problem later, better get back to see how my DFT calculation goes

2017

@DHMO Is there any online site where we can plot parametric equations like $x=A\sin(wt)$ and $y=A\sin(wt+\phi)$ which varies with time. I'm trying to visualize Lissajou's figures. Can desmos do it somehow ? Or anything else?

DHMO

08:42

yes

2017

how?

desmos?

DHMO

yes desmos

2017

okay lemme check

how should I type the equations?

like x=Asin... and y=Asin...

?

DHMO

yes

2017

It is showing the two graphs separately...

Not the superposition

:/

DHMO

08:45

(sin t, cos t)

Secret

[List of maths that I am aware I suck in]
1. Topology
2. Infinite series
3. Special functions
4. Number theory
5. Analysis
6. Combinitorics
7. Statistics and probability theory

Anything else you can assume I am not aware I suck in, but it might be true that I suck in it

and... to make matter worse:

It seems my memory on most physical chemistry stuff have decayed a lot

2017

@DHMO Ah, it is working now

Thanks a lot :-)

DHMO

@Secret list the areas of math which you are aware that you do not suck at

Secret

Based on user response, probably linear algebra

(however I still get the advanced stuff wrong sometimes)

DHMO

can you prove that "for a 3x3 matrix where each row, column, and diagonal sum to N, its determinant is divisible by 3N"?

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$Mathematics$ Mathematics