hiii

:l

how to use transfinite induction on $\alpha^{\beta} \leq \alpha^{\gamma}$? I know I have to do the zero case, successor case, and the limit ordinal case but who is the zero case here? I could try fix $\alpha = 0$ but then I will have $0 \leq 0$ :\ I know there is one property like $0^{\beta} = 0$ so I used that on $0^{\beta} \leq 0^{\gamma}$ becomes $0 \leq 0$ with that property .. weejfdkfsjdkfdljdslk

Ah, the "stop thinking about it for a while and try again later" technique of problem solving. Never fails. (Disclaimer: actually fails constantly)

yeah it does fail constantly degh

I kinda like moved on on another question... yaaaay >:\

I'm working on one I posted earlier. I think I have a solution, too.

It's like if I don't have an idea on how to attempt the proof I am screwed

I'm supposed to go by contraction about induction being valid on well-founded relations. well-founded...hmmm... there is a minimal element and irreflexive yaayyyy

It's not like $2 \uparrow 4 = s^

oooo noo

$2 \uparrow 4 = 2^{4}=16$

I just had to ignore the inequality proofs for ordinal exponentiation :\. Now it is a part 2 of some problem. :(

How do I go about expressing this sum? $\sum_{x \in 1,5,9, \ldots}(1-p)^{x-1} \cdot p$

Is there a way to use geometric series to express in a simpler form

@genescuba Yes, if the exponents are all of the form $4k+1$.

I see, so I can express this summation as $\sum_{i=0}^{\infty}(1-p)^{4i} \cdot p$

So I believe we get $\frac{p}{1-(1-p)^4}$

Ah, that pain when you think you've solved something, but the same mistake you made before shows up again.

I don't know why this one is giving me so much trouble. If I know that $\sum\limits_{i=1}^kn_ix_i=0$ and that $\text{gcd}(n_1,\ldots,n_k)=1$, then how do I remove an $x_i$?

I keep on trying and arriving at statements like "$g=g$"

@Rithaniel: Start with 2 and figure out how you use the Euclidean algorithm.

scp-wiki.net/scp-3380

Well sure, God in abrahamic religions do have something akin to the reflection principle, but to say that transfinite numbers are divine is a gross hyperbole

because infinity, as crazy its properties are, are not really godly powerful, for they themselves are limited in a host of ways by axioms

How to show that (3^20) +1 is divisible by 41

@mathsstudent is $3$ a quadratic residue mod $41$?

Answer this question and note that $20 = \frac{41 - 1}{2}$

@mathsstudent Hint: 3^4=81 is one less than a multiple of 41

heya, does anyone know how I can show that $f\colon\mathbb R\setminus\{-2,5\}\to\mathbb R\colon x\mapsto \frac{x-1}{(x-5)(x+2)}$ is surjective, without using any knowledge of analysis? (I'm teaching a course in "basic maths", and they haven't officially learned anything yet about continuity/derivaties)

nvm I found a way

well the discriminant of $(x-5)(x+2)-r(x-1) = x^2+(-3-r)x+(-10+r)$ is $(-3-r)^2-4(-10+r) = r^2+2r+49 = (r+1)^2+48 > 0$

@Rithaniel Sometimes you can sleep on a problem

There's one problem which you definitely cannot solve in your sleep, though:

"How do I know that I'm not sleeping right now"

Maybe that's why no one's ever solved it

> "How do I know that I'm not sleeping right now"

Actually, there are hints in some lucid dreams, such as the sensation of a soft mattress when touching some plants or furniture or other things in the dreamscape

There's also those common falling off building dreams where you felt you have landed on something bouncy very abruptly, and then the next moment, you knew you have been asleep

Trying to get my dreams to solve problems is very hard though because my dream induction works in an inverse fashion

Meaning, the more you consciously want something to appear, the less likely it will appear

Compounding that problem is this my dreams often use alien maths, which most of the time translating it to real life maths give gibberish statements

My dreams are better at coming up ideas instead of problem solving, I have only ever solved 3 problems in my 8 years worth of dreams

And...

I had a feeling that the alien maths may get even more alien now that I have been reading SCPs

Yo fellow math nerds

I need some help on 2A IV)

How is Ax a vector? Isnt A a matrix? So confused

A is a matrix and v is a vector, so Av is a vector (assuming that A has as many columns as v has rows)

matrix multiplication is a lovely thing

@Semiclassical How is Av a vector then?

if A is a matrix, the multiplacation of a matrix and vector will result in a matrix right?

A column vector is just a matrix with one column.

A is a 5-by-4 matrix. v is a column vector with 4 entries and therefore is a 4-by-1 matrix. ergo, Av is a 5-by-1 matrix i.e. a column vector

(another way to put it: the whole point of an m-by-n matrix is that it's a linear transformation that maps column n-vectors to column m-vectors.)

@Semiclassical Oh wow i completely misunderstood matrix multiplcation my bad

The mantra is that matrix multiplication is done row by column.

One more question, when it says for any x in R4. Does it mean for any of the entries in all the 4 columns? just to make sure

(0,1,2,3) is an example of something in R^4.

(5,6,7,8) is another example.

Oh so when it says R(Something) its just a random vector of (Something) number of columns?

So for any x in R^4 means for any such example.

You just need to know what a typical example looks like.

it's a bit ambiguous at times whether RR^4 means a column vector or a row vector, alas

typically it'll mean a column vector if not stated otherwise tho

Now R^4 means 4-dimensional vector.

That means the first coordinate is real, the second is real, the third is real, the fourth is real.

Alright thanks

@usukidoll Hiii, I see you are still a doll. =)

One quick question if yall dont mind. This always keep trippin me up this chapter. What does it mean to be in the "span"

Suppose you've got a set of column vectors. You can take linear combinations of those vectors to get other vectors. The span is the entire set of such linear combinations

So, for instance, the vectors (1,0,0), (0,1,0) span the xy-plane in R^3

In other words, it is the set of vectors that can be generated by those vectors in the set. That is why you say they span the set.

Ahh

note, for instance, that the vector (0,0,1) can't be written as a linear combination of the vectors (1,0,0), (0,1,0)

so (0,0,1) is not in their span

One other subtlety: Suppose you've got the vectors (1,0,0), (0,1,0), (1,1,0)

the last vector can already be written as a linear combination of the first two, so you can't generate anything new with it

Ok makes sense. So in that explanation 2B should be false since the fourth vector doesnt span the first three vectors right? Cause its incosistent

so the span of those three vectors is the same of the span of the first two alone.

@Semiclassical Yeah i see why its not in the span!

Suppose that we have $nr$ balls where $n \geq 2, r \geq 2$. The balls are numbered $1,2, \ldots, nr$. The balls are placed into n cells with no restriction on the number of balls allowed per cell. Assume each ball is placed into a cell independently of other balls. Given that each of the $n$ cells has exactly $r$ balls. Find the conditional probability that the balls numbered $1,2, \ldots, n$ are in different cells.

Let $D$ denote the event that the balls numbered 1,2, ..., n are in different cells and let $E$ denote the event that each cell has exactly $r$ balls.

So we are looking for: $P(D | E) = \frac{P(D \cap E)}{P(E)}$

I have found that $P(E) = \frac{(nr)!}{(r!)^nn^{nr}}$

I am having trouble finding $P(D \cap E)$, does anybody have any ideas

@genescuba Well, your probability is discrete, so you can start with counting the number of ways to fill each bin, given $1$ to $n$ is already placed.

So you are saying assume that balls 1 through n are already placed such that they are not in the same bin, then count the number of ways to fill each bin?

Yeah

Im having trouble understanding why this works excatly

Oh wait basically I also have to account for the number of ways to place n balls which is like n! and then I count the number of ways to fill each bin

Yeah, I think you've got it.

Is it true that Dilworth's theorem gives number of swaps in problem given in stackoverflow.com/questions/57952706/… as it was said in artofproblemsolving.com/community/… ?

Someone know why a set of 6 vectors in R10 cannot span R10?

Doesnt set of vectors if they are not multiple of eachother span the whole Rx where x is a the # of col ofc.

How come in this question, from the equation below “Hence, $F$ is an n-multilinear form if...”, the indices aren’t given as $..., \textbf{u}_{i-1}, s\textbf{u}+t\textbf{v}, \textbf{u}_{i+1}, ...$ like they are in the preceding paragraph? math.stackexchange.com/questions/2144076/…

@amanuel2 yo, R10 has 10 dimensions so you need at least 10 vectors to span it

@rapasite Isnt R10 an infinite number of vectors with 10 columns?

So 6 vectors which span R10(Aka 6 vectors with 6 columns), should span the whole R10 if they are not multiples of eachother

right?

to understand this , one intuitive way is to see vectors as directions from the origin.Start to think about it with two vectors

take two randoms vector in the R2 for start easy

let say (1,2) and (1,1)

every points on the plane can be find with a combination of those two vectors

now you can subtract the first one from the second one so (1,1)-(1,2)=(0,-1)

(0,-1) is in the span so you can replace (1,2) with it , you are left with (0,-1) and (1,1)

@rapasite Yeah so (0,-1) is in the span of (1,2) and (1,1)

and (1,1)+(0,-1)=(1,0)

so you change your two vector

(1,2) and (1,1) became (0,-1) and (1,0)

so my point is you have news vectors with 0 in all positions except one who span the same space as before ==>R2

this way of using a combination of vectors to create the maximum number of zeros in every vectors who span the same space should if you play with it convince you that is not possible to span R10 with six vectors

if you take two random vectors in R3 let say a=(1,5,8) and b=(2,3,4)

you can do c=a-2b=(-3,-1,0)

d=b-2a=(0,-7,-12)

now with c and d you cannot continue caus when you try to get a new zero you will destroy the previous one

this is a very bad explanation from a math point of view

read en.wikipedia.org/wiki/Basis_(linear_algebra) this and click every link you find in the page

@rapasite Ok lol

Thanks tho i get it now

Read more about basis,vector space and you will get it, R10 is not only an infinity of vectors with 10 'real number' columns it is all of them

you cannot even create one with vectors from R6

you can find 6 "linearly independent" vector in R10 witch create a 6 dimension subspace of R10

@amanuel2 any more questions?

@rapasite Nah thanks a lot!

gregorybard.com/interacts/visualize_span.revised.html you can go there and try imagine the same stuff with more dimensions @amanuel2

@Secret Here's a question:

A function is monotone if it's either increasing or decreasing. Yeah?

Clearly, if a function is not monotone, then you can find $x_1<x_2<x_3$ such that $f(x_1)<f(x_2)>f(x_3)$ or $f(x_1)>f(x_2)<f(x_3)$.

How much do you have to remove from the definitions to make this false?

It's false for partially ordered sets I think

Is there anything else that can be said about this? (Maybe not)

Just random thoughts

(You can even find a poset with three elements that's a counterexample)

@AkivaWeinberger hi

(Looks like $\bigwedge$)

Hey

do you play chess?

I haven't in a really long time

do you want to play me?

Not at the moment, sorry

ok

I got addicted to it recently

@AkivaWeinberger How much do you have to remove from the definitions to make this false? what do you mean?

Basically, looking for near-counterexamples

There are no actual counterexamples because you can prove the theorem

but it's false if you have functions on a partially-ordered set instead of a completely ordered one

so they provide a near-counterexample to the original theorem

(This means, incidentally, that if you want to prove the theorem, you have to use the fact that it's completely ordered in some essential way)

A partially ordered set is one where you might have two things that are incomparable, i.e. neither $x\prec y$ or $y\succ x$

(A good example is sets ordered by inclusion)

you mean for exemple with the complex number < and > does not mean anything?

That also works

We can define $\prec$ to be whatever we want

yeah like the module of complex number

so we can also take $\{1,2,3\}$ with $1\prec2$ and $2\succ3$ and $1$ and $3$ incomparable

sure thing

Being a data structure, I imagine the concept has applications in computer science

that way you can discus every aspect of the definition even what its mean to be a function

I'm not sure though

@rapasite Yeah

In a sense, by the way, functions and posets ('p'artially 'o'rdered 'sets') are the same "type of thing"

They're both a relation

If you write $x\prec y$ as $(x,y)\in P$

and if you write $f(x)=y$ as $(x,y)\in F$

then you can represent each of these as just sets of ordered pairs

with some restrictions

(the "vertical line test" for functions for example)

(and transitivity for posets)

what if you allow your relation to have multiple answer for the same input?

You have that with posets

and completely ordered sets

(1,2), (1,3), and (1,4) are all in the usual order on the integers

i mean the same as your vertical line test

Yeah

but you can create whatever

I think the vertical line test is the only thing that defines a function

maybe the sum of the elements in the set is use for ordering

For a partially ordered set, you need transitivity (if (x,y) and (y,z) are in your thing then (x,z)), and also some other stuff

like you can't have (x,x)

or both (x,y) and (y,x) at the same time

Not sure what else

For a completely ordered set you need either (x,y) and (y,x) for every pair of distinct things x and y

(Every pair of things is "comparable")

what I mean is that you can chose how you will order your Xs and you can chose how f(Xs) will work too

Oh for the original question?

yeah if your Xs are couples and your f(Xs) are unique number and you define two ordering the result can invalidate your theorem

or can it?

lol

does completely ordered set accept equality for distinct elements?

howdy!

hows everyone?

Heya @JoeShmo. Long time no see.

yeah! how are you?

Doing fine, and you?

taking topology with Cappell this semester.

Oh, I thought you had already done that.

nope. i took undergrad topology in hebrew u in jerusalem for a semester

He just keeps on going. I mean, I retired a bit early, but some of these guys keep going into their 80s and ...

haha. its clear he is enjoying every moment

and i can't imagine how many times he gave these lectures because he has every word almost memorized

Oh, right, it was my former Berkeley undergrad friend here who took topology from him a few years ago.

its pure poetry

Well, say hi for me :)

will do! where do u know him from

Years running the UGA topology conference in the 80's. Maybe even before.

yeah so this semester he wants to jump into algebraic topology as quickly as he can

so he's skimming over the point set stuff pretty quickly. which is fine by me.

I would have assumed that the grad course there assumed students already knew point-set.

At UGA we took the point-set out of the graduate topology 40 years ago. :)

apparently last year he didn't get into algebraic topology until the second semester. whereas this year he is planning on getting into the good stuff by mid october latest

That's crazy.

we even took point-set out of undergrad lol

oof. well arent you fancy :)

we just do 2-3 weeks of point set in undergrad alg top

Hey everyone!

Well, that's nuts, @Mathein. It's actually needed for a lot of mathematics, and it certainly teaches certain skills better than a lot of other mathematics courses.

hey everyone

but then when do u do point set top?

They don't.

you do a bit in analysis, a bit in alg top, a bit in diff geo when you need it

interesting

A lot of mathematicians think it's a waste of time. I think a solid semester of it is good for undergrads who're going on in mathematics.

You can't do it in graduate manifolds courses. There's way too much to do in there.

I mean, you use it to prove stuff.

yeah. i thought munkres was a great course fwiw

I agree it's important to do know a bit of point set, but the idea here is just that you can pick up what you need along the way

some profs do it in analysis 3

or analysis 2

Well, why not pick up basic group theory and ring theory, too, then?

haha

I don't want to prove the Tube Lemma every time I use it in differential topology or algebraic topology or other more advanced courses.

Anyhow ... it shows a certain — shall we say — prejudice.

we did the tube lemma as an exercise on the second sheet in alg top I think

It's used in point set to prove product of compact spaces is compact, but it shows up everywhere ... complex variables, diff top, alg top, ....

my personal experience has been is that if we you know some basic notions like connectedness, compatcness and separation axioms, but you don't need a whole course for that

So let's say you have to give a math talk, but you know the audience doesn't know much about your field, but you want to get across that the content is quite difficult (because they're going to grade you on it), do you try to lean towards watering the talk down to the point everyone knows what's going on (but then it seems like you've done something easy) or do you try to show them how difficult it is and risk losing connection with your audience

ted whats the counting trick for finite topologies up to homeomorphism? something with posets and trees

we just have one topology prof/lecturer he usually teaches alg top or diff top. If he did point-set that would mean less alg top or diff top courses offered

I don't ever think about questions like that, @JoeShmo.

As I said, @Mathein, there's a prejudice ...

Other people could certainly teach it.

@Perturb: How long is the talk? The rule of thumb is to make the first 40 minutes accessible and get technical for the last 10-15 minutes (assuming an hour talk, of course).

And ALWAYS remember to do lots of examples.

heya @Jasper

@TedShifrin Hello! I saw a ping when I was scrolling upwards. I only realised it is a new message after a long time, LOL.

Saw a ping or heard a ping? How did you see it if you were scrolling back? :D

@Ted I have exactly 13 minutes to explain spectral sequences to an audience that most likely hasn't encountered any algebraic topology before (I'm not complaining though, I'm very grateful for the opportunity to be able to do so)

@TedShifrin Oh, I saw a 1 with a circle around it at the bottom. So one can actually see it. =)

Oh geez. 13 minutes is impossible. Maybe just spend the whole time on an example. Say with deRham cohomology (most people know about closed and exact ... sorta).

Oh, at the bottom. That's sneaky, @Jasper :P

I hated giving 10 and 20 minute talks at research conferences, @Perturb. You really have to prepare everything ahead on slides. There's no time for writing or thinking.

10 is absurd. 20 you can breathe a little.

@TedShifrin Thanks for the advice, I'm in the process of completing the slides

Practice a few times, @Perturb. Usually things go way slower than you possibly imagine.

I could think about what I'd do in an hour lecture, but I have no good ideas for 13 minutes, other than what I just mumbled.

Yep definitely, I'll make sure to go over everything thoroughly so that I can push out most of what I plan to in the 13 minutes.

Don't forget examples. The biggest mistake newbies make is not doing examples.

I think for something like this having the whole talk be an explicit example might be good.

But I'm repeating myself.

I won't be able to go over just one example in the talk though, the talk is supposed to be a presentation on our Bachelor's thesis, and it's supposed to sort of encapsulate what we did in our Bachelor's thesis, so we have to give an exposition of what the central focus of our thesis was (in my case it was the spectral sequence of a fibration) as well as the extent of the work we did, which makes it a bit harder.

Most of them won't even know what a fibration is, let alone the topology. Stick with a fiber bundle.

we have 90 minutes to present our bachelor's thesis. Sounds a lot easier than 13

Yup. 13 is laughable.

Even worse because undergraduates have little experience giving talks on papers or teaching.

definitely in a critical mood today

I'm trying to understand how this answer was gotten but can't seem to comprehend it