Mathematics - 2018-03-25 (page 1 of 5)

Kasmir Khaan

12:00 AM

faust ! not there yet

Zee

Ya Faust knows algebra better than me , he can help you

Kasmir Khaan

doing some repetition of linear algebra

using Lang's Book

Zee

@JoeShmo you want a cookie bro ?

Faust

cause its saying that pick some $v_0$ it must be an eigenvector for the geneeralized eigenspace, but in this case you can show that the it is an eigenspace

Joe Shmo

@0celo7 its fine you can safely go back to being grumpy now. all is good with the world. until it isn't. then ill be back

Kasmir Khaan

12:01 AM

hmm

ill keep drilling on that :D

Joe Shmo

@Zee if you had a cookie, I'll take it

Kasmir Khaan

thnaks yall

0celo7

I'm busy

Faust

you need to show there is only one other eigenvalue other than 0

@KasmirKhaan

0celo7

writing up an awful proof

Zee

12:01 AM

@JoeShmo you gotta risk it to eat the biscuit

Joe Shmo

oh man

yuck. who has to read it?

Faust

@KasmirKhaan i am also slightly concerned you have the question missing/ copied down missing a piece of information one slight modification and the question is alot easier

Eric Silva

scrolls up and reads chat I've made a horrible mistake

Kasmir Khaan

@Faust the question is wrong faust, that is my opinion on that

or the author assumed other stuff he did not mention ><

@Faust do you have linear algebra book , author Lang ?

Faust

mm no but i can try googling it

Kasmir Khaan

12:08 AM

if so check page 58 question 16 +17

third edition

he is assuming stuff he is not mentioning

no its correct...

@Adeek Mabrouk

think of this way

@EricSilva thanks for the warning

Eric Silva

tfw youve met with a terrible fate

Kasmir Khaan

12:17 AM

@Faust it only makes sence if dim V = Dim kerF +1

Faust

say $ V=W \oplus U $ where W is everything sent to the kernel and U is everything that isnt in the kernel. Now assume that the dimension of U is 2 but we have that dim (R) is clearly 1 so U can't be sent to a 2 dimensional space it can only be sent to a 1 dimensional space.

That means that one dimension of U is sent to the kernel but thats a contradiction to our assumption that U has nothing sent to the kernel. if you assume the U is dimension 0 then $ V= W $ a contradiction so U must be dimension 1. you have some theoreom that says every 1 dimensional space is just an eigenspace.

@KasmirKhaan

so$ F(u) = \lambda u $ for all $u \in U $

or in your questions case $ c \in \Bbb R $ and $ u = v_0 $

spans U

Kasmir Khaan

hmm this makes more sense now

but hmm

why would have that every elemnt in V can be written as a sum of w+cv

ahaa

if we take an element in the kernel

then our c =0

Faust

yup

Kasmir Khaan

if not in the kernel then w=0

now it makes sense thanks faust :D

Faust

Precisely the hard part was showing theres only one eigenspace (sorry for going straight to generalized eigenvectors but its easier that way if you dont see the argument above)

wasnt your textbook jumping ahead that was my bad

Kasmir Khaan

12:23 AM

no that is fine

i just wanted a reasble argument

i was going crazy :D

the key part was the dimention here

that v_0 was generating V\W

Faust

theirs some formula that says $ \dim im T + \dim ker T = \dim W $ where $ T \in \mathcal L (V,W) $ thats what i used

once you know the $ \dim im F =1 $ then you can argue about the dimension of the subspace

and it leads to that contradiction argument i showed you

it looks like the argument just came from have a clever thought but the reality is its just from using a definition and following your nose.

<--- no cleverness to be had ^^

DarkVampiric AbstractArtist

@AkivaWeinberger Find and solve the appropriate congruences.
Anklebiters Childcare Centre bought t toys. The toys came in packs of 10. The next day, at the start
of the day there were n children at the centre. The children were evenly divided into groups of 4, and
each group was given 6 toys. There were 2 toys left unused. Two toys were destroyed by the children’s
brutal style of play and had to be thrown out. Later on, m more children were dropped at the centre.
The children were then divided into groups of 6, with 7 toys per group. There were 6 (intact) toys not

Akiva Weinberger

I can tell that whoever wrote this problem loves children

Faust

@DarkVampiricAbstractArtist do you know why n is at min 48 yet?

DarkVampiric AbstractArtist

Lol. My NT lecturer gives unusual questions...

Akiva Weinberger

12:33 AM

@Faust You mean 6(n/4), no?

'Cause that's the quantity that's t-2

DarkVampiric AbstractArtist

I think 6*4

Faust

24x -10t -2 =0

DarkVampiric AbstractArtist

Because 4 groups of children were given 6 toys each.

Faust

tells you n must be at least 48

DarkVampiric AbstractArtist

24n=2 mod 10 (as two were unused)

Akiva Weinberger

12:35 AM

@Faust Why 24?

Groups of 4, 6 per group, means 6/4=1.5 per child on average

Faust

n=6x we know that 6x * 4 mod 10 = 2

Akiva Weinberger

No... What's your x?

DarkVampiric AbstractArtist

I think Akiva is right, because the question states that the children we're evenly divided into four groups.

Faust

why am i so bad at explaining things

DarkVampiric AbstractArtist

Usually in number theory we work with divisors not multiples. Yeah, I also suck at explaining, I'm dyslexic.

Faust

12:37 AM

you have 6 toys

and 4 groups

Akiva Weinberger

n/4*6 mod 10 = 2

@Faust Not 4 groups

Groups of 4

Faust

the number is divible by 4

n= 4x

DarkVampiric AbstractArtist

so 6n/4= 2 mod 10?

Akiva Weinberger

We don't know how many groups

Faust

so you have 4x *6 -10t -2 =0

Akiva Weinberger

12:38 AM

@DarkVampiricAbstractArtist That's what I'm thinking, yeah

@Faust It's not four groups

Faust

i dont mean theres 4 groups

DarkVampiric AbstractArtist

We'll make 6n=8 mod 40 in that case, Akiva.

since 1.5 is not an integer.

0celo7

Appendix E of the thesis is done.

Only one proof left.

Faust

is the answer 252 incorrect?

Akiva Weinberger

@Faust If x is the number of groups, then n=4x, yes, but then 6x-10t-2=0

not 6*4x

DarkVampiric AbstractArtist

12:40 AM

You could be right, Faust. But I have to work with congruences, not the ones in linear forms.

Akiva Weinberger

32 makes sense for the first half @DarkVampiricAbstractArtist

Faust

the onyl work u need to do on n is to find the minium number that makes sense

Akiva Weinberger

becauae then it's 32/4=8 groups and 6*8=48 toys

Faust

otherwise its completly unneeded

who cares about the number of groups!

Akiva Weinberger

but it could also be 12 @DarkVampiricAbstractArtist

Faust

12:41 AM

well at least u got the right anser

the minium number of toys used is 48

Akiva Weinberger

because then it's 12/4=3 groups and 3*6=18 toys

Faust

in the first session

Akiva Weinberger

so we need the second half of the problem to decide between 12, 32, 52, etc

Faust

6*7y = 42y -10t =8 is the other relation

Akiva Weinberger

72 maybe? Lemme try that

First half: 72/4=18, 18*6=108, works

Second half: 72/6=12, 12*7=84

…not quite

Faust

12:46 AM

its 252...

DarkVampiric AbstractArtist

let m=6x, 6x-10t=8?

Akiva Weinberger

Oh I forgot about the extra kids!

Faust

not m=6x

m+n = 6x

you start with n kids and you add in m more kids

so your bucket has n+m kids in it

Akiva Weinberger

It's 32 at the start, then 4 are dropped off so it ends with 36.

Check it; it works. 50 toys at the start, two are destroyed, 48 toys at the end.

@DarkVampiricAbstractArtist

Anklebiters Childcare Centre bought 50 toys. The toys came in packs of 10. The next day, at the start of the day there were 32 children at the centre. The children were evenly divided into 8 groups of 4, and each of the 8 groups was given 6 toys, for a total of 48. There were 2 toys left unused. Two toys were destroyed by the children’s brutal style of play and had to be thrown out, leaving the Centre with 48 toys. Later on, 4 more children were dropped at the centre, with 36 total. The children were then divided into 6 groups of 6, with 7 toys per group, for a total of 42. There were 6 (in…

^The problem with real numbers substituted in

DarkVampiric AbstractArtist

I backtracked from 46 because the second time six toys weren't used.

i mean 42-6=36.

0celo7

12:53 AM

@BalarkaSen If I want to be a geometry apologist, the thesis is done

Akiva Weinberger

48-6=42 @DarkVampiricAbstractArtist

0celo7

What's left is some technical elliptic regularity and then that's it

Faust

wierd so 42 is a solution

why can n<48

Akiva Weinberger

32, you mean? @Faust

n=32, m=4, t=50

DarkVampiric AbstractArtist

6 of those 48 toys were unused, Faust.

Akiva Weinberger

12:55 AM

The point is that t is 2 more than a multiple of 6, and that t-2 is 6 more than a multiple of 7.

t=50 is the smallest number that satisfies those.

Once we solve for n and m and see that they are positive integers (as they should be), we're done.

Faust

we have 6 *4 = 24 toys oh

Akiva Weinberger

???

Faust

i dont get it

if n can be less than 48 then 42 is a soltuion

but i dont understand how n can be less than 48

Akiva Weinberger

There are 32 kids. They are divided into groups of 4. Each group is given 6 toys. How many toys were given? @Faust

Answer: (32/4)*6=48

At no point do you multiply 4 by 6

Faust

4x *6 = 24x with 2 left over from to so x=2

ah

n=4x

n=8?

Akiva Weinberger

12:59 AM

WHY ARE YOU MULTIPLYING 6 BY 4 I DON'T UNDERSTAND

Or n by 6

n is the number of kids. n*6 would only make sense if you gave each kid 6 toys.

Faust

you have to have at least 4 groups?

Akiva Weinberger

No one said that

There are "groups of 4", that means each group has 4 kids in it

There might be 2 groups, there might be 100 groups, we don't know, all we know is that each group has 4 kids

There are n/4 groups

Faust

thats the same thing as im saying x is the number of groups

n=4x

eveyr group gets 6 toys

Akiva Weinberger

So then you look at 6x

6x+2=t

Faust

oh because not every child gets only one toy

they fat asses and get more than 1 i get it

Akiva Weinberger

1:02 AM

Each child, on average, gets 6/4=1.5 toys

Faust

i assumed each child got 1 toy

Akiva Weinberger

In any case, solution: n=32, m=4, t=50

Leaky Nun

why am I seeing numbers everywhere

am I in the wrong room

Faust

i had the coltions down to 42z -10t - 2 =0 but ruled out z=1 as a solution

cause i thought eveyr chidl had to get at least 1 toy :(

DarkVampiric AbstractArtist

Number Theory @LeakyNun

Faust

1:04 AM

anyway thanks

Leaky Nun

@DarkVampiricAbstractArtist put an "elementary" in front of it. don't insult proper number theory.

DarkVampiric AbstractArtist

Sorry lol.

Leaky Nun

lol

Faust

number theory is appropriate elementary number theory is really hard

DarkVampiric AbstractArtist

How do I work n=32, m=4, and t=50 into a linear congruence equation? 6n=4 mod 50?

Oh wait, you've already solved it for n and m, I think.

anakhronizein

1:11 AM

@LeakyNun I take it you have never looked at springer.com/gp/book/9783540586555

Leaky Nun

@anakhronizein what do you mean?

anakhronizein

You can put whatever you want in front of it, doesn't make it any less trivial. ;)

Leaky Nun

lol

but when you say "elementary" we know you're talking about olympiad stuff

anakhronizein

Seriously, take a gander at that book whenever you can. Great book apparently, but I'll never know!

Joe Shmo

hey for the following row reduced matrix -- wolframalpha.com/input/…;*x_3,+x_1*x_3,x_1*x_2,+1%7D%7D

the claim is that it's rank is 2 for all x != (t^-2,t^-2, t, t^-3)

why?

how did they figure out x != (t^-2,t^-2, t, t^-3)?

DarkVampiric AbstractArtist

1:22 AM

@JoeShmo it depends on the number of pivots. For example, if the matrix has (1,0,0), (0,1,0), and (0,0,1) then the rank is 3.

Joe Shmo

yuh,

the question is where does x != (t^-2,t^-2, t, t^-3) come from?

actually, i think i know

if x == (t^-2,t^-2, t, t^-3), then i think the rank is 1

since if the matrix is as above , then A(t^-2,t^-2, t, t^-3) will give you the same entry on the top and bottom

DarkVampiric AbstractArtist

Yes, I think the rank is 1 because one of the entries on second row and third column is a nonzero.

So the rank can't be 2, but it's 1.

Joe Shmo

something doesn't add up. the row reduction is not defined at x1 = x2

any point that satisfies that. so why go as far as x != (t^-2,t^-2, t, t^-3)?

Akiva Weinberger

1:47 AM

So I just realized:

One of the principles of flag design is that a flag should still be immediately recognizable when really small or far away

which means that most of our user icons are good flags according to that

(Well, one of the other principles is "it should be simple enough that a child could draw it from memory" but whatever)

DarkVampiric AbstractArtist

Are you talking about the "flags" that are used to report "spam"?

spam messages*

Leaky Nun

2:07 AM

@BalarkaSen ok so I watched the recording of the last lecture of the course

"last" as in "there's none after it"

and it ends on the equivalence between simplicial and singular homologies

so, ugh, not even degree

0celo7

hey @EricSilva

Akiva Weinberger

2:26 AM

Why is 3B1B consistently high art

Secret

[Philosophy] I need to reread this again before proceeding:

Ultrafinitism

In the philosophy of mathematics, ultrafinitism, also known as ultraintuitionism, strict-finitism, actualism, and strong-finitism, is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers. == Main ideas == Like other strict finitists, ultrafinitists deny the existence of the infinite set N of natural numbers, on the grounds that it can never be completed. In addition, some ultrafinitists...

Eric Silva

sup

Secret

Skewes's number

In number theory, Skewes's number is any of several extremely large numbers used by the South African mathematician Stanley Skewes as upper bounds for the smallest natural number x for which π ( x ) > li ⁡ ( x ) , {\displaystyle \pi (x)>\operatorname {li} (x),} where π is the prime-counting function and li is the logarithmic integral function. These bounds have since been improved by others: there is a crossing near e ...

But basically, in order to get a better idea to speculate what kind of future technology that the concept of both potential and actual infinity can provide us, we need to first ignore their existences and see how things fall apart (if any)

Very recently in the logic room we have deduced even potential infinity that is the (constantly growing) collection of all natural numbers may not be realisable in a physical universe because a computer will ran out of memory

This speculation, while may not lead to fruitful results because it is so hard, is inspired by how something as abstract as logic lead to one of the most important transformation of the modern society: The computer

Since infinity should be a bit less abstract than logic (as you can easily define it using logic), we suspect it should be easier to extrapolate the technology it promised to give us in the future

orbit-stabilizer

Wait....you're saying you deduced in the logic room that countable infinities aren't realizable because computers will run out of memory? Isn't that obvious?

Secret

no, it's more like this:

orbit-stabilizer

2:37 AM

Is analysis more 'applicable' than algebra?

Secret

in Logic, Mar 16 at 3:37, by user21820

@Secret What does it mean by "they cannot prove"? The point is that any practical formal system you design will essentially be equivalent to a proof verifier program that must be run on an ideal computer (one with unbounded memory and time and so on).

in Logic, Mar 16 at 3:55, by Secret

I am trying to understand whether actual infinity is needed to talk about the existence of a model of PA, in order to understand why it is circular, but it seems I have been misunderstood about potential infinity the whole time, thus I think I need to revise everything and reformulate the question later.

From what we have discussed and I have comprehended so far, it seems potential infinity corresponds to an ideal computer thus even that is circular, let alone actual infinity

orbit-stabilizer

I know that's a very vague question

Secret

in Logic, Mar 16 at 6:55, by user21820

@Secret Sorry I was away. Yes, potential infinity corresponds to an ideal computer (of Turing's sort, namely not that we have an infinite tape but we have an indefinitely extensible tape).

orbit-stabilizer

Oh gotcha.

Secret

@orbit-stabilizer Well not necessary, for example, finite fields give us cryptography which runs most of the security of our current systems (and also future quantum proof encryptions), and semigroups are often involved in finite state machines and robotic programming

also don't forget about groups: They are literally everywhere from maths to other disciplines since they govern most symmetry in systems

orbit-stabilizer

2:40 AM

It seems like potential infinity is the common infinity 'model' used in basic analysis. Things growing without bound.

Secret

Indeed it is, this is why it is technically speaking not a number, but a process

orbit-stabilizer

Interesting. And how do we use actual infinity in math?

The extended reals? 1 point compactification?

Right, groups are everywhere. But it seems like rings are mainly used for number-theoretic purposes.

Secret

Actual infinity is the postulate of the existence of something larger than all finite numbers. In set theory for example, $\omega$ which is the set of all natural numbers in its usual well ordering, is an actual infinity since $n < \omega $ for all $n$ and any increasing function $f$ will not allow us to reach $\omega$ e.g. $f(n) < \omega$, unless it already includes $\omega$ in part of its specification

orbit-stabilizer

Wait a second. We have the axiom of infinity tho in ZFC, don't we?

Why do we need that if we don't use actual infinity in analysis?

You mean $\aleph_0$?

Secret

We normally don't see that distinction of these infinities in set theory (ZFC and ZF) because of the Axiom of Infinity, which basically postulate the existence of an infinite set hence $\omega$. Thus combining with the argument of ideal computers, basically as far we know we cannot have actual infinity in a system except adding it in from axioms.

$\aleph_0$ is the cardinality of $\omega$, but yeah it is also an actual infinity

orbit-stabilizer

2:46 AM

Ohh. I see...

What happens if we get rid of the Axiom of Infinity?

Akiva Weinberger

ZFC+~AoI is equivalent to PA

ZFC plus the negation of the Axiom of Infinity

Secret

@orbit-stabilizer This is a question heavily debated in the logic room. While from studying the various MSE and MOs as well user21820, Leaky's etc. insights, we seemed to be doing mostly fine by getting rid of uncountable well orderings and even lesbegue integration, I think the question on whether we can do analysis without potential infinity is still an open question

Akiva Weinberger

I think the right word is "bi-interpretable"?

orbit-stabilizer

You need choice for PA?

So Lesbegue integration goes away, but Riemann doesn't?

Akiva Weinberger

No, but ZFC+(neg AoI) is equivalent to ZF+(neg AoI)

With no infinite sets you don't need choice

orbit-stabilizer

2:49 AM

Ah, right.

Akiva Weinberger

You essentially map hereditarily finite sets to numbers, to show that it's bi-interpretable with PA

Secret

Sorry I typed slow, I will catch up with you guys soon enough

orbit-stabilizer

So, we could do analysis without AoI, but we're unsure if we can do analysis with potential infinity

Akiva Weinberger

If I recall correctly, {} is 0, and $\{x_1,\dots,x_n\}=2^{x_1}+\dotsb+2^{x_n}$

is how the mapping works

using $x_i$ to mean both the set and the number

This works if there are no infinite sets

0celo7

I am doing the final proof of the thesis

much excite

orbit-stabilizer

2:53 AM

I did my first thesis when I was 9. Good times.

Daminark

:thinking:

0celo7

@Daminark tell me, wise one who just took FA, is the set of bounded invertiple operators $X\to Y$ open in operator norm

I am reasonably sure it is

Akiva Weinberger

Puzzle: Assume there is no continuous map from an $n$-ball to its boundary, fixing the boundary. Show that any polynomial in the quaternions with one term of highest degree has a root

(Meaning $ixi+x+j$ doesn't count)

(but $x^2+ixi+x+j$ does)

This is essentially FTA but weird

0celo7

It is true if $X=Y$ using the Neumann series. I don't know if that trick works here.

Secret

@orbit-stabilizer whenever we are adjoining some algebraic structure with some infinite element (often defined to be larger than any element in the structure) that infinite element is an actual infinity. Thus yes, one point compactification and extended reals use actual infinity

Leaky Nun

2:58 AM

@AkivaWeinberger where is induction?

Akiva Weinberger

I dunno exactly but it's somewhere

Some combination of axioms

Daminark

@0celo7 I remember that this is true

Akiva Weinberger

Prolly Foundation in the end

Thing what prevents x={x}

0celo7

@Daminark I can't figure out how to prove it for $X\ne Y$. Quite frustrating.

Leaky Nun

@AkivaWeinberger the proof of induction inside ZFC uses axiom of infinity

at least the proof that I know

DarkRunner

2:59 AM

Just wanted to clarify the following question:
Let P (n) be the product of all the non-zero digits in a number n. For example P (22) = 4 and P(207) = 14. Let S be the set of all positive integers with at most 2018 digits. Find the sum of the values of P(x) where x ranges over the elements of S.

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$Mathematics$ Mathematics