Mathematics - 2018-03-12 (page 1 of 4)

12:00 AM

Everyone seems to have different answers to this stuff so it's not that obvious

despite the tone of your response

Your question is so vague and undefined that the only possible answer is to be flippant; if you have an actual question, you should ask that question.

user539262

trying to understand the differences in terminology between things like sentence, statement, expression, atomic, predicate, proposition, formula, etc

"free variables", quantifiers and what they are actually modifying, what these modifications are called, etc

I want to get my terms straight so I can speak about these things more precisely

But online resources are woefully inconsistent

And even when I read something like Tao's Analysis he doesn't get into these details too much

Akiva Weinberger

1:46 AM

@user539262 Obviously an analysis text wouldn't go into this

You'd need a text on mathematical logic

or model theory or whatever

I don't think any logicians frequent the chat, but there are several logicians active on the main site (Mathematics)

0celo7

@AkivaWeinberger um, @AlessandroCodenotti ...

Akiva Weinberger

Is he?

0celo7

he does set theory

Akiva Weinberger

Yeah, but that's not the same thing

though there is a big overlap, I guess

0celo7

they're probably close enough for whatever that guy wants

@BalarkaSen you really like that one page from your dynamics book

Zee

2:44 AM

How do I show there is a canonical isomorphism between V tensor R1 to V

0celo7

@Zee what have you tried

Zee

Well a bunch of stuff , am suspicious the question is wrong

0celo7

it's not

Zee

Well one thing I was trying to do is show that V X R1 is isomorphic to the linear functionala from V to R1 but that can’t be right

0celo7

hint, a \otimes v "=" a.v, where . is scalar mult

or v \otimes a, dunno why you wrote it backwards

Zee

2:47 AM

Sorry , can you write this in English? I can’t compile latex

0celo7

You don't need to compile tex

Zee

Your not being helpful ...

0celo7

ok

the solution is there, so google what \otimes means

Zee

Otimes is the tensor product ...

0celo7

so what's the issue

Zee

2:56 AM

The correspondence you gave me is not a bijection

maybe my question wasn’t clear , let me restate it : V tensor R1 is ismomorphic to V , where V is a finite dimensional vector space and R1 is the real line

0celo7

I didn't give you the full solution

Shobhit

Hello. How can i construct a linear transformation which maps real axis to a circle? I think its a linear fractional transformation, but it was not mentioned explicitly in the text.

0celo7

@Shobhit this should be in any complex analysis book

Zee

The hint you have is useless , am not here to get spoon fed

0celo7

I'm useless :'(

Zee

3:05 AM

I didn’t say that , your hint is though

0celo7

why is it useless

Zee

It’s too trivial

0celo7

lol

this is a trivial problem

MatheinBoulomenos

If it's trivial for you, then what you're trying to prove should also be trivial to you

Zee

I don’t think you know the answer

Shobhit

3:06 AM

@0celo7 i am reading a complex analysis text, namely, complex analysis by lars ahlfors, but no hint is given as to how to achieve it. The author briefly mentioned cross-ratio to the reader, i think we are supposed to use that, but i dont see how.

0celo7

@Zee maybe, maybe not

but you're not being very nice

MatheinBoulomenos

The hint of @0celo7 is on point. He said that v tensor a "=" av, what if you take that "=" to try and define an isomorphism?

Zee

Am not being nice becouse I have 2 midterms in a week and three HWs to do and I don’t have time to listen to your half backed hints

0celo7

I thought you didn't want to be spoonfed

I can give you the solution if you need it

MatheinBoulomenos

@Zee you're the one who's trying to get help

Zee

3:08 AM

Yes and I regret it , wasted half an hour

0celo7

(btw google finds the full solution immediately)

Shobhit

@0celo7 help please

0celo7

@Shobhit Sorry not in a complex mood. Have you tried another book like Stein or Conway?

Shobhit

@0celo7 :( ok. I'll look into other books.

Zee

Can you post the link , I can’t find anything on google

0celo7

3:12 AM

Say the magic word

MatheinBoulomenos

@Shobhit consider the "Cayley map", $z \mapsto \displaystyle \frac{z-i}{z+i}$

Zee

please your highness

do you define the isomorphism by multiplying basis vectors by scalers ?

MatheinBoulomenos

@Zee yes! that's what I said by the way (and what @0celo7 was hinting at)

Shobhit

@MatheinBoulomenos ok, i'll google that and get back to you :)

0celo7

Take the map R x V --> V given by (a,v) --> av and lift to the tensor product via the universal property. This is an isomorphism.

Zee

3:14 AM

YA I thought about that

but I got stuck

0celo7

You just have to prove it is one. It suffices to find an inverse.

Zee

Couse I didn’t think of using basis vectors to show bijective

0celo7

You shouldn't have to use a basis

MatheinBoulomenos

you don't need to involve a basis at all

(sniped)

Zee

How else are you gonna show bijectivity ?

0celo7

3:16 AM

In this case you can write down the inverse map directly.

Zee

Ya , this seems too easy

I hate when this happens

MatheinBoulomenos

you can also show that it is injective and surjective

Zee

I can’t believe I waste a whole day on this , especially since I considered this hint at the beginning and convinced myself it can’t be true

Anyway thanks 0celo7 and mathein

Shobhit

@MatheinBoulomenos on the wiki page of cayley transform, its written " mobius transformation permute the generalised circle in the complex plane", what does it mean? And i get how they made the transformation using given points, but i dont see how it maps the real axis to a circle. Help please.

MatheinBoulomenos

generalized circles are circles and lines

Akiva Weinberger

3:26 AM

It means generalized circles map to generalized circles, and the same is true of the inverse map

MatheinBoulomenos

but you don't need to think about how Möbius tranforms act on generalized circles in general for this problem, this is more concrete. If $z$ is real, what is the absolute value of $\displaystyle \frac{z-i}{z+i}$?

Akiva Weinberger

$z$ gives the real line, $iz$ gives the imaginary axis, $iz+1$ gives the line of points with real part 1 (as $z$ varies through the real numbers)

Secret

I found it funny that every time someone in the math chat get converted from field A to field B, a corresponding starred message popped up

6 hours ago, by Alessandro Codenotti

Was Mathei transformed into an analyst? What kind of sorcery is this!

Shobhit

@MatheinBoulomenos i get 1. So since mobius transformation maps generalised circles to generalised circles, the real axis will be mapped to a circle or a line, but { 1,-i,i} do not lie on a line, so the real axis get mapped to a circle passing through these points. Is this correct? @AkivaWeinberger

Secret

Make me wonder, can we do a category theory of people swapping between maths fields

Akiva Weinberger

3:33 AM

If we invert that line (sending it through the mapping $z\mapsto1/\bar z$) we get $\frac1{1-zi}$, which would be a circle

The one centered on 1/2 with radius 1/2

MatheinBoulomenos

@Secret I'm not converted to an analyst. I just answered a question about analysis and Alessandro made a joke about it

Secret

Oooooooooo

Akiva Weinberger

Doubling it and subtracting $1$ gives us $\frac{1+zi}{1-zi}=\frac{z-i}{z+i}$, which gives the unit circle

Secret

Anyway, a category theory of people will probably be so complicated that nobody can verify the proofs

Akiva Weinberger

@Shobhit Yeah, that's the way Mathein was going for

(Those are the values at $z=\infty,-1,1$, I guess)

Shobhit

3:37 AM

@AkivaWeinberger got your analysis of the transformation, it was nice.

@AkivaWeinberger @MatheinBoulomenos thank you.

Yes the values are correct @AkivaWeinberger

Zee

@MatheinBoulomenos @0celo7 how can that be a bijection ? Take (1) (1 0) and (1/2) (2 0) these give you the same vector

MatheinBoulomenos

yes, but they're also equal in the tensor product

Zee

Couse the tensor product is modded by pulling out scalers ?

Akiva Weinberger

@Zee Remember that ax⊗y=x⊗ay

Silent

@orbit-stabilizer Thank you so much for this reply.

Akiva Weinberger

3:49 AM

=a(x⊗y)

Zee

I see

orbit-stabilizer

@Silent, no problem. Was this a question on your analysis homework?

Zee

I don’t like like the book Lee smooth manifolds , it’s terrible honostly

Akiva Weinberger

You're proving V⊗R=V for vector spaces V, right?

You should know that V⊗(R^n)=V^n as well.

Zee

Ya

Silent

3:51 AM

When I upvoted an answer in main site, a message in a blue strip was displayed, that went away so quickly, i could not read. some initial words were 'you haven't voted in long time ' or something. Any idea what does that say?

Akiva Weinberger

And (R^n)⊗(R^m) is isomorphic to R^(nm).

Zee

This one I knew but the one before it seems interesting

Akiva Weinberger

Take n=2

Silent

@orbit-stabilizer No, no. Rudin's problem 4.8 says that 'f uniformly continuous function on real bounded set, then image is bounded.'

So, i had that question.

Akiva Weinberger

(v,w), an element of V^2, can be mapped to v⊗(1,0) + w⊗(0,1)

where (1,0) and (0,1) are elements of R^2

(Note that it's not a "pure tensor")

orbit-stabilizer

3:54 AM

If f is a continuous function defined on a compact set, is the image bounded?

I'd like to say yes

We have uniform continuity due to the compactness

Zee

If F is real valued then yes

Continoues function takes compact to compact and by hein Borel compact is bounded

Akiva Weinberger

(In the space V⊗W, things of the form v⊗w with v∈V and w∈W are called pure tensors. Things of the form v1⊗w1+v2⊗w2 are not necessarily equal to pure tensors, but they're still in V⊗W.)

Zee

Thanks akiva , tensors are new territory

Akiva Weinberger

(Those are meant to be subscripts.)

Zee

Are we talking about tensors as an algebraic structure or as something more specific ?

Akiva Weinberger

3:57 AM

@orbit-stabilizer Yes, it's part of the reason compact sets are "almost finite" (not a rigorous term)

@Zee Tensor products of vector spaces

orbit-stabilizer

Okay. But, if we know that for f, the preimage of every open set is open. So find some open cover of Im(f), look at preimage of that, that gives us an open cover for the domain of f. By its compactness, we have a finite subcover. Now map that back, and we get a finite subcover of Im(f)

Does this work?

Akiva Weinberger

Yeah

orbit-stabilizer

Okay. So that gives us compactness. And apparently in some topologies, compactness -> boundedness?

I don't know much topology

Can we even talk about boundedness without a metric?

Akiva Weinberger

In Euclidean space $\Bbb R^n$, something is compact iff it is closed and bounded.

orbit-stabilizer

Yup, by hiene-borel

Akiva Weinberger

4:00 AM

In a general metric space, I think something is compact iff it is complete, closed, and totally bounded?

orbit-stabilizer

But, I'm interested in maps of the form f:X->Y where X and Y are topological spaces

Zee

There is a Version for metric spaces but it has weaker conclusion

Yes akiva

Akiva Weinberger

"Boundedness" on its own is not a topological property

so you can't define it on just topological spaces without extra structure

orbit-stabilizer

Gotcha. It's interesting that compactness -> completeness

Akiva Weinberger

For example, (0,1) is bounded and R is not, but they're homeomorphic

orbit-stabilizer

4:01 AM

Right

Zee

Compact is complete only in metric spaces

Akiva Weinberger

Well, "complete" is only defined on metric spaces

orbit-stabilizer

How can you talk about convergence outside of metric spaces?

Akiva Weinberger

and ordered spaces as well actually

Dedekind completeness, the least upper bound thing

orbit-stabilizer

fair

Silent

4:03 AM

@orbit-stabilizer I think you can talk about convergence in any topological space, Munkres does the same.

Akiva Weinberger

@orbit-stabilizer A sequence $(x_n)$ converges to $x$ if every open set containing $x$ contains infinitely many elements of that sequence

orbit-stabilizer

Okay that works. But you can't have cauchy sequences though, right

Akiva Weinberger

"Completeness" isn't a topological property (take (0,1) and R again)

orbit-stabilizer

Or @Akiva, every open set of $x$ contains an element of that sequence. Same thing

Akiva Weinberger

Oh, actually I think we're both wrong

orbit-stabilizer

4:04 AM

Really?

Akiva Weinberger

It needs to contain all but finitely many elements of the sequence

orbit-stabilizer

Yeah, I think mine covers that

Akiva Weinberger

This way, (0,1,0,1,0,1,…) doesn't converge to anything, like it's supposed to

orbit-stabilizer

Oh

I see

You're right.

Akiva Weinberger

The other definitions would have it converge to both 0 and 1

orbit-stabilizer

4:05 AM

Man, the subtleties

Akiva Weinberger

But yeah, as long as you give the compact set a metric, it will be complete and totally bounded

Silent

Why we can't define cauchy sequence in arbitrary topological space @AkivaWeinberger

?

Zee

Look into Nets and filters

Akiva Weinberger

In R, every Cauchy sequence converges. In (0,1), not every Cauchy sequence converges. But they're homeomorphic

In addition, "uniform continuity" doesn't make sense on topological spaces either

orbit-stabilizer

That's not really getting at the heart of the question though

Secret

4:09 AM

1,1/2,1/4,1/8,1/16,1/32,... converge to 0, but 0 does not exist in (0,1)

orbit-stabilizer

Yes, I think we understand that

It's more: let's try to define Cauchy sequences without a topology, what doesn't work

without a metric**

Secret

Without a metric, how can we have a notion of consecutive difference, which is a key property of Cauchy sequences?

Akiva Weinberger

Same thing that goes wrong when you try to define uniform continuity, another thing that can't be defined on topological spaces

Essentially, the problem is the "uniformly" bit

orbit-stabilizer

We can still have a sequence of points, can we not? @Secret

Secret

but a Cauchy sequence is one where the difference between terms will vanish as the term number gets large?

So it is stronger than the notion of a converging sequence?

orbit-stabilizer

4:13 AM

Yeah

No

Every convergent sequence is cauchy, but the converse is not always true

Akiva Weinberger

@Secret In (0,1), the sequence you gave (2^(-n)) is Cauchy even though it doesn't converge to anything in the set

orbit-stabilizer

I think. I could ble wrong

WDUK

how do you solve for the second value of t for 0 = t(25-4.9t) . i know t = 0 but the other value am unsure

Akiva Weinberger

To take the nonstandard analysis perspective: we can ask whether something in $X^*$ is infinitely close to something in $X$, but we can't always ask whether something on $X^*$ is infinitely close to $X^*$

orbit-stabilizer

if $a\cdot b = 0$, then $a=0$ or $b=0$. This is not the exclusive or. @WDUK

Akiva Weinberger

4:15 AM

The topology has no idea that $N\not\approx N+1$ but that $N\approx N+\frac1N$

WDUK

so -5.1 i get roughly

@WDUK, seems right

since 25 = -4.9t

yup

See:

In the mathematical field of topology, a uniform space is a set with a uniform structure. Uniform spaces are topological spaces with additional structure that is used to define uniform properties such as completeness, uniform continuity and uniform convergence. The conceptual difference between uniform and topological structures is that, in a uniform space, one can formalize certain notions of relative closeness and closeness of points. In other words, ideas like "x is closer to a than y is to b" make sense in uniform spaces. By comparison, in a general topological space, given sets A,B it ...

I once looked at these but then forgot most of it

orbit-stabilizer

4:17 AM

@AkivaWeinberger, hmmmmm I see what you're saying. Here, X* and X are disjoint?

Akiva Weinberger

but I think the idea is that these have an intermediate amount of structure between topological spaces and metric spaces

and they let you define uniform continuity and completeness

@orbit-stabilizer No, they're nonstandard analysis thingies, it would take me too long to explain them

I guess I shouldn't have mentioned them

WDUK

@orbit-stabilizer thing is in my book regarding a falling object it seems to not calculate it that way... see image i.imgur.com/rdQX3Cq.png why do they switch it to a positive surely since its -5.1 its not a real solution and should be disregarded?

Akiva Weinberger

Essentially, $X^*$ is $X$ "with infinitesimals added"

orbit-stabilizer

Haha yeah, I don't know non-standard analysis. Uniform spaces look interesting though.

Akiva Weinberger

and $N\in\Bbb R^*$ was taken to be an infinitely large element of $\Bbb R^*$

orbit-stabilizer

4:19 AM

@WDUK, my bad. You calculated it wrong.

Akiva Weinberger

(Nonstandard analysis is unrelated to the uniform spaces thing)

Silent

I found this article relevant to our discussion.

WDUK

oh ?

orbit-stabilizer

$25-4.9t = 0 \iff 25 = 4.9t \iff t = 25/4.9$ which implies that $t > 0$.

Akiva Weinberger

Is a "linear space" an old-timey name for a vector space? @Silent

WDUK

4:20 AM

ah okay

i see it now

thank you

0celo7

@AkivaWeinberger Yes

orbit-stabilizer

@AkivaWeinberger, I think so. I saw it in my calculus of variations book as well

0celo7

It's a better name than vector space

Akiva Weinberger

It does make sense, doesn't it

0celo7

it's a Bourbaki word, hence a good word

Akiva Weinberger

4:21 AM

Hah

orbit-stabilizer

"A vector is an element of a linear space".

Doesn't have that same ring to it though.

Akiva Weinberger

I don't know what makes something a "space", though. What class could have both vector spaces and topological spaces as subclasses?

orbit-stabilizer

There doesn't need to be a parent class

Akiva Weinberger

I guess they're both generalizations of $\Bbb R^n$

0celo7

@AkivaWeinberger the category $\mathsf{Space}$

Akiva Weinberger

4:24 AM

Thanks.

0celo7

You are welcome

orbit-stabilizer

I don't think there are any conditions on something being a space. I think it's pretty arbitrary.

The dual space to the dual space is called the duel space where all the yugioh battles occur.

Akiva Weinberger

Not to be confused with Atari's Space Duel

0celo7

that's the double dual

Corellian

@TedShifrin I was thinking if the straw has width, i.e. its walls enclose an interior.

0celo7

4:27 AM

(dual space)' = space duel

Akiva Weinberger

Double Duel is not a bad name for a video game

Thing is I think you'd need four players

0celo7

it's like a double date but for losers

Akiva Weinberger

There's something called the dual numbers, called that 'cause… I dunno

It's the ring of things of the form $\{a+b\epsilon:a,b\in\Bbb R\}$, where $\epsilon^2=0$

The weird thing is that, if $f$ is a polynomial, then $f(a+\epsilon)=f(a)+f'(a)\epsilon$

Same with rational functions actually, so long as they're defined at the input.

orbit-stabilizer

$\epsilon \not \in \mathbb{R}$?

Akiva Weinberger

Yeah it's just an extra thing

orbit-stabilizer

4:31 AM

Is this from nonstandard?

Akiva Weinberger

Nah

It's just a symbol, where we're giving it this rule of squaring to zero

Shobhit

How can i find a linear fractional transformation which maps 0,i,-i to 1,-1,0

Akiva Weinberger

If you know about quotienting rings, this is $\Bbb R[x]/\langle x^2\rangle$

or $\Bbb R[\epsilon]/\langle\epsilon^2\rangle$ I guess

orbit-stabilizer

I know about quotienting groups :P

Secret

Affine space is very different from vector space as far I recall

orbit-stabilizer

4:34 AM

yeah

Akiva Weinberger

It's essentially a hyperplane in a vector space of one dimension higher, that doesn't go through the origin

Shobhit

I know that for a linear transformation (Tz1,Tz2,Tz3,Tz4) = (z1,z2,z3,z4) where the R.H.S is the cross ratio.

But i cant find such T

Secret

Yeah, lacking the zero vector actually change some of its properties

orbit-stabilizer

@Secret, hows your phd going

Akiva Weinberger

@Shobhit Honestly I would do this by trial and error.

$iz$ maps 0,i,-i into 0,-1,1

Secret

4:36 AM

I am currently analysing my second batch of data, there is a trend slightly different but nothing interesting yet. I am trying to get to the 3rd set of data which I knew something interesting happened there besides a trend

orbit-stabilizer

What sort of data is it?

Akiva Weinberger

$2iz+1$ maps them into $1,-1,3$

$-2iz+1$ maps them into $-1,1,-3$

$\frac1{-2iz+1}$ maps them into $-1,1,-\frac13$

Secret

One thing that is clear, though, one of the ligands seemed to control the whole set of trend while the other two just modulates it. The data are molecule geometries and energy. Since my molecule has 3 ligands to vary, and there are 3 important geometry patameters each, it is at least 6 dimensional data

I am planning to get the trends analyse it before finding some hints on the explanation of the trend by analysing their electron density

orbit-stabilizer

Interesting. How are you going about analyzing it? If you're talking about a trend, is there a time component?

Akiva Weinberger

If $a:=-2iz+1$ and $b:=\frac1{-2iz+1}$ then $a-9b$ maps them into

$8,-8,0$

Secret

4:40 AM

Currently I make scatter plots with colour gradient which give me how two of the parameters are correlated (because the literature said they are correlated). This allow me to extract the ordering of those effect

Akiva Weinberger

So then an eighth of that

Secret

Two different plots with the colour gradient in two directions will allow me to visualise 4 dimensions at a time

Akiva Weinberger

Probably not the best method, I'll admit

Secret

and no they are all ground states molecules, thus no time component stuff yet

orbit-stabilizer

It's interesting how covariances are everywhere. Surprising how many things are linearly correlated. Or can be modeled as such.

What area of chemistry is this?

Akiva Weinberger

4:42 AM

Oh, wait, are Möbius transformations closed under addition?

Dammit, I think they aren't

Secret

Organometallic and computational chemistry

Shobhit

I do like what you did, unfortunately i wont be able to do it, nor might the professor will award me marks for that. Any algebraic method? @AkivaWeinberger

Secret

Catalysis sector

orbit-stabilizer

wow, sounds cool

I cannot do chemistry for the life of me. I hate it with a passion.

Akiva Weinberger

@Shobhit I think the $a-9b$ step doesn't actually leave you with a linear fractional transformation anyway, sorry

orbit-stabilizer

4:43 AM

I think I got scarred by the first year chem labs.

APerson

what is the notation for when $\mathcal A$ is a subcategory of $\mathcal B$?

Shobhit

Oh ok

Akiva Weinberger

@APerson I think just $\subset$?

orbit-stabilizer

im so tired of school

can't wait for summer

APerson

@AkivaWeinberger hm, ty - I think many people just write it out too, but I thought there must be some alternative

The Testosterone Fanatic

4:46 AM

@orbit-stabilizer I hate the summer

I LOVE the snow

orbit-stabilizer

$\mathcal A <_{C} \mathcal B$

The Testosterone Fanatic

I'll prolly settle somewhere in Siberia if given the chance

orbit-stabilizer

driving in the snow is not fun

The Testosterone Fanatic

not if you get the relevant tires

orbit-stabilizer

Still. Traffic just runs slower.

Akiva Weinberger

4:48 AM

@Shobhit New try: We have 0,i,-i. We want 1,-1,0

$z+i$ gives $i,2i,0$. $\frac{z+i}i$ gives $1,2,0$.

APerson

also, a coreflection arrow is a homomorphism of the supercategory right?

Akiva Weinberger

$\frac i{z+i}$ gives $1,\frac12,\infty$.

$\frac{2i}{z+i}$ gives $2,1,\infty$.

$\frac{4i}{z+i}$ gives $4,2,\infty$.

The Testosterone Fanatic

@orbit-stabilizer en.wikipedia.org/wiki/Snow_chains

Akiva Weinberger

$\frac{4i}{z+i}-3$ gives $1,-1,\infty$.

orbit-stabilizer

Yeah, I understand. I live in Canada :P

Akiva Weinberger

4:50 AM

Simplifying, $\frac{-3z+i}{z+i}$ gives $1,-1,\infty$.

Inverting, $\frac{z+i}{-3z+i}$ gives $1,-1,0$.

The Testosterone Fanatic

@orbit-stabilizer I see. I'd love going there some time. But I'd not want to live there if given the chance

Shobhit

How in the world did you come up with that? @AkivaWeinberger

orbit-stabilizer

Haha, why? I think it's amazing.

Akiva Weinberger

Also equals $\dfrac{1-iz}{1+3iz}$

The Testosterone Fanatic

it's a beautiful country no doubt, but it has something err... wrong with it. I prefer not to mention those things here at the risk of getting banned from here

Shobhit

4:53 AM

I can read what you did, but i wont be able to come up with something like that. Its very cool. But i have to do this in my exam, how can i find the transformation algebraically

@AkivaWeinberger

orbit-stabilizer

Compared to the states, I like our healthcare and education :P

The Testosterone Fanatic

I suppose I'm just not liberal enough for living in Canada, in a nutshell

Akiva Weinberger

I guess I was helped a lot by the fact that one of the targets was 0

orbit-stabilizer

Oh. The political correctness stuff? It gets overblown. A lot of people here aren't like that.

Akiva Weinberger

I don't know if this sort of thing would work in all cases

The Testosterone Fanatic

4:55 AM

@orbit-stabilizer What the general population thinks doesn't matter much, only what the law-makers do, who only listen to those whose voices are the loudest, who in turn aren't very moderate folks

Shobhit

Yes, there may be a case where such a transformation may not exist, and we can be stuck on this method for 5-10 minutes manipulating ratios XD @AkivaWeinberger

Secret

Strange phenomenon no 1: it seems that everytime I recently acquired a new interest on something to explore its knowledge, for the coming 2 months the frequency of seeing discussions involving that thing increases above average levels

orbit-stabilizer

I guess. I'm just speaking personally though - I haven't felt the effects of these things

@Secret there's a term for that

Baader-Meinhof phenomenon

0celo7

aka glitch in the matrix

agent smith

The Testosterone Fanatic

@orbit-stabilizer You haven't experienced its effects because there's only a probability, which is certainly a lot less than 1, but you becoming the subject of the abuse of perverted laws does carry a probability much higher than what it should be

Transcript for

$Mathematics$ Mathematics