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00:01
mmh, then i wonder why my prof made it as hard as possible to follow him :/
is the discussion of convergent sequence in $\Bbb N$ meaningful?
like, $((-1)^n+1)/2$ is convergent
it's 0 or 1, i dont see how this is convergent
(cesaro apart)
@Null what did your prof say that you find hard to follow?
@Null because we are considering $\Bbb Z$
@DHMO with the usual topology, a sequence in N is convergent if and only if it is eventually constant
00:03
apply the epsilon-delta definition
I'm talking about the topology in Z
what is $(-1)^{2n+1}$ in $\mathbb{N}$?
infinity, undefined, apples or oranges?
i meant Z instead of N
ah ok!
so in Z epsilon must at least be 1
so 0,1,0,1,... is convergent
to either 0 or 1
@arctictern i guess i have to see it myself. it was a proof about the minimal set of vectors that spans R^n, where n is in N.
00:07
right? @Null
@DHMO so, picking epsilon=1, there exists an L such that for all n>N we have |x_n-L|<1?
never mind i thought it was <=
bye
let $\varepsilon <0$
1
Q: linear programming for bus tickets

gilad sHi I work as a programmer at a bus company and I need to implement a ride initialization request. I think it might be a linear programming problem but I'm not sure and I ask for some help :) A passenger sends my server a request to initialize a bus ride. The request includes the different entit...

how are problems like this are called?
@arctictern don't understand your question :D
@Null Ew.
@Fargle maybe prizeminimalisation?
00:18
@Null Let 57 be an arbitrary prime.
hehe
welp, i finished taking my notes
thats my first time playing with categories :)
00:46
Hi all
hi @akiva
@AkivaWeinberger Howdy!
Hi DogAteMy
You know that old topology (but really algebra) problem about hanging a picture with three nails such that removing any one of the nails makes the picture fall?
00:49
Really? Well, you want to hang a picture with three nails such that removing any one of the nails makes the picture fall
I was gonna ask a related question
@AkivaWeinberger is it possible?
I don't see the first one yet.
@Null Yes @TedShifrin "First one"?
Original question ...
The question is essentially what shape does the rope take
00:51
@TedShifrin The picture is hanging by a string wound around the nails.
I didn't know what was meant myself. Google helped.
Ohh ...
Yeah, sorry, should have been more specific
So is this related to the Borromean rings?
Yes, I think
Yes
Heuristically it sounds like it is.
00:52
so i loop in some way my rope around 3 nails, such that removing 1 nail disolves the rope to the ground?
Yup, Borromean rings, then
Well, I think I see how that gets you two nails ^
is this possible for higher numbers of nails?
(for any number?)
00:55
mindboggling
Are we doing anything interesting?
yes
defining interesting
Borromean rings ... In disguise
@MikeMiller Have you heard of this one?
We're hanging a picture with a string wound around three nails
And we want the string to be in such a way that removing any one nail makes the picture fall.
Did anyone see The Man Who Knew Infinity ??
Oh, it's an old one, but I never thought about it any.
I was gonna ask a related question but it turns out no one else had heard of it
@AkivaWeinberger Unfortunately I don't know that I could actually construct the solution. I'm what you might call a not theorist.
jumps
smacks Fargle
00:57
LOL
@AkivaWeinberger Yep, seems like it
Dunno if it's more telling of you or my general jokester attitude that I knew a Ted-smack was coming.
This isn't really my game.
@Fargle I think we all knew it was coming
@AkivaWeinberger We'll call it both then.
00:59
Find x.
^ There it is
:D
lol
...What did the fish say when it swam into a wall
01:00
hi @Fargle
Also is there a sort of procedure to find bases of subgroups of free groups, knowing that they must be free?
Hello again @Adeek, @Semi
@AkivaWeinberger Dam
@ted I figured out how to use Macaulay2 to compute something. I'm reasonably happy about that.
01:01
What do you call a fish with no eyes?

A fish which is not algebraically closed.
@Akiva Write down the corresponding covering space and pick a maximal tree inside it
would you guys like to to discuss the following question. Let H be a hilbert space , $\{x_k\}$ be orthagonal in H. Then $\Sigma x_k$ convergent iff $\Sigma ||x_k||^2$ is convergent.
Maybe a more algebraic solution? Maybe something similar to Gaussian elimination, but noncommutative
@Adeek I don't know much about Hilbert spaces, but maybe consider $\left\langle\sum x_k,\sum x_k\right\rangle$?
can you win at the Crapstables if the dices are biased? How biased they have to be?
also, martingal works. if you bet in base of the banks worth
(if you can bet, i might add lol)
yeah @AkivaWeinberger something along that lines I think.
01:06
I know neither of these games ^^
martingal is no game :D
its some fools strat, betting always double the amount as before
I guess one side is trivial
So we know that a sequence in a complete Banach space is convergent iff it converges absolutely.
Calculate f(blue).
@Adeek Yes--though a Banach space is already complete.
@Fargle ?
yeah right
01:11
It's interesting that the game "I win a dollar with probability $10^{20}/(10^{20}+1)$ and lose a hundred million million million dollars with probability $1/(10^{20}+1)$" is completely fair
and yet I feel like I have no reason not to play it as many times as I want
@AkivaWeinberger For a sufficient definition of "fair"...
@AkivaWeinberger not if money has deminishing value
@Null Doesn't it only affect the economy once I use the money? Until then, nothing changes, right?
I don't know economics
@AkivaWeinberger I think he means that, if you have already won 10,000 dollars, a dollar is not as valuable to you than if you've only won a hundred.
@AkivaWeinberger i mean. getting 1000 dollars, if you have 0, is far better then becoming 1000 dollars if you already have 10000.
01:13
That is to say, its marginal value diminishes.
@Fargle :/
@Null :)
also fail on me
"becoming a dollar"
that might be possible in some universe tho
@Fargle what do you think of the following reasoning. Suppose that $\Sigma x_k$ converges. Then since Hilbert spaces are banach spaces, so we must have $||\Sigma x_k||$ converges so this means that $||\Sigma x_k||^2$ converges but by pythagorean theorem $||\Sigma x_k||^2 = \Sigma ||x_k||^2$ which must converge as well.
I mean I will make it more precise by just considering partial sums but what do you think of it ?
01:15
@Adeek I think that line of reasoning holds.
Infinite dimensional Pythagoras? Wow
@AkivaWeinberger yeah it holds for hilbert spaces.
I couldn't remember for a bit the inclusion relation b/w Hilbert and Banach spaces, because I was being silly.
Interesting
Every inner product defines a norm, but not vice versa.
01:16
yeah
Correctamundo.
Something involving parallelograms?
Yeah
I could send you lecture notes @AkivaWeinberger
is laplace demon actually a true statement? IF you know the position and velocity of all particles in the universe, then you could calculate the future.
01:17
Sure why not my phone doesn't seem to be running out of memory anytime soon @Adeek
since the first is impossible, the statement itself is true?
For any $\langle\cdot, \cdot\rangle$ inner product, $\sqrt{\langle x,x \rangle}$ satisfies the norm properties.
Spoken too soon, probably, DogAteMy. ...
suppose now $\Sigma ||x_k||^2$ is convergent then we have $||\Sigma x_k||^2$ is convergent so we have $||\Sigma x_k||$ is convergent as H is a complete banach space $\Sigma x_k$ is convergent.
BTW, @Fargle, that's \langle ... \rangle.
01:18
what do you guys think ?
Gotcha, @Ted.
@AkivaWeinberger okay I will email it to you.
Karim: $\sum 1/k^2$ converges, so $\sum 1/k$ converges?
@TedShifrin My phone seems to have, like, 40GB of free space left
yeah that is a problem.
01:20
(Can't say the same for my computer...)
(...but then again my phone does not have Minecraft)
I think I have to construct a cauchy sequence somehow...
What are you trying to prove, Karim?
@AkivaWeinberger copied it. Will email it to you once I finish this question.
19 mins ago, by Adeek
would you guys like to to discuss the following question. Let H be a hilbert space , $\{x_k\}$ be orthagonal in H. Then $\Sigma x_k$ convergent iff $\Sigma ||x_k||^2$ is convergent.
Thanks!
01:22
I finished one side.
You'd better use orthogonality of the $x_k$, Karim.
That was in space Pythagoras
Are you sure you finished the one direction correctly?
*infinite-dimensional Pythagoras
01:23
@TedShifrin I think so atleast. I did the proof above.
OK, with partial sums.
Suppose that $\Sigma x_k$ converges. Then since we hilbert spaces are banach spaces we must have $||\Sigma x_k||$ converges so this means that $||x_k||^2$ converges but by pythagorean theorem $||\Sigma x_k||^2 = \Sigma ||x_k||^2$ which must converge as well.
yeah just have to change all of that to partial sum.
I would recommend using Cauchy, yes, for the other direction.
yeah. Hm I yeah I think it will get it.
Welp, I'm outta here. Later chat.
01:27
cya @Fargle
Bye @Fargle
@TedShifrin Why would you wanna finish one direction? Their albums are just too good
I'm outta here, too ...
cya ted
That one's over my head, @Krijn.
01:29
One Direction is a band
one direction lol
@TedShifrin Ohh, you lucky man
you guys should listen to good music like classical music
I do, Karim.
01:30
cool
Like Pachelbel
I don't really like pop music.
Who doesn't?
I really don't like people who dislike a whole genre
No, that's worded too strong
I really don't like it when people dislike a whole genre
Not so much Pachelbel ...
Well, you can not like me, @Krijn, cuz I really dislike rap music and some others.
@TedShifrin There's quite a lot of genius in rap nowadays, I would argue
01:32
Listen to Hamilton
Just now I listened vertigo ncs release,good it was
for good rap stuff
Anyhow, I was leaving. Have to cook dinner and go to bridge.
Careful I hear those things are flammable
LOL @TedShifrin
still here ?
I have a very simple proof.
@AkivaWeinberger what do you think of following.
Suppose $S_n = \Sigma_{n = 1}^{n = k} \ x_k$.
01:39
Sure
we are done.
Just use \$B\$. \$\Beta\$ doesn't exist.
if map(A,K) is the set of all functions between A and K. $r\in K$. What is $(r\cdot f)(a)$?
we consider cauchy sequence WLOG we can assume m > n. Then $||S_m - S_n||^2 = ||S_{n + 1} + ... S_{m}||^2 = ||S_{n+1}||^2 + ... + ||S_{m}||^2 = \beta_m - \beta_n$ Therefore $S_m$ is cauchy iff $\beta_m$ is cauchy.
i dont understand how functions can be scalars
is g(x)=5 such a scalarfunction?
01:44
Arright that makes sense ^^^
@Krijn I really don't people who don't like it when people dislike a whole genre :s (just kidding)
@Null What are all these things? $A$ and $K$ and $r$ and $a$?
A=nonempty set, K a field, r some element in K, a some element in A
i try to latex the excercise..
I would suggest that $(r \cdot f)$ is indeed defined pointwise, i.e. $(r\cdot f)(a) = r\cdot f(a)$
yes it is
01:49
So, what's your question then?
Let $K$ be a field and $A$ a non empty set.
Show that $V:=map(A,K)$ with the following binary operations is a $K$-vectorspace:

for $f,g\in V$ is defined by: $(f+g)(a)=f(a)+g(a)$ for all $a\in A$
for $r\in K$ and $f\in V$, $r\cdot f$ is defined by: $(r\cdot f)(a)=r\cdot f(a)$ for all $a\in A$
my question is, how do i show that scalar multiplication is closed. and what is a scalar in this context?
A scalar is an element of $K$, not the function itself
so if K=$\mathbb{R}$ any number would be a scalar?
So you need to show that $(r \cdot f)$ is still a map from $A$ to $K$
since K is a field, this is given or not?
01:53
Yeah, kinda
It's not hard
i mean, the result of function f will be multiplied
and K being a field guarantees closure
The other axioms probably require more thinking
i showed already that for addition everything applies
in another excercise
(ring axioms)
I'm sure you can do it, just check the axioms
I'm off to bed
G'bye any- and everyone
@Krijn good night
01:57
G'bilt
okay @AkivaWeinberger I will send you notes now
@AkivaWeinberger whats your opinion on using Aluffi's algebra for a second book in algebra?
I never read it
sent
02:05
Received
ty
@AkivaWeinberger sent it.
np
is this the right direction?
Distributive property

If $u,v$ are functions and $c,d$ are scalars, then the following has to hold for all $a\in A$:

$c(u+v)(a)=cu(a)+cv(a)$
or without (a)?
6
Q: Difference of fourth powers in three ways

Christian WollI'd like to find a number that's the difference of fourth powers in three ways or more. I.e.: $$k=a^4-b^4=c^4-d^4=e^4-f^4$$ Is this possible? There seem to be plenty of examples of differences of fourth powers in two ways. The smallest: $$310300575=134^4-59^4=158^4-133^4$$ I've checked numbers ...

02:21
Distributive property

If $u,v$ are functions and $c,d$ are scalars, then the following has to hold for all $a\in A$:

$c\underbrace{(u+v)(a)}_{\text{some element in $K$}}=cu(a)+cv(a)$
Since $K$ is a field, this has to be true.
right?
@AkivaWeinberger here?
if $C$ is a category with object $A$, are the morphisms in category $C_A$ simply morphisms between morphisms between objects of $C$ and $A$?
think about it @meow-mix
that is, are the morphisms in $\mathrm{Hom}_{C_A}(X,Y)$ defined as morphisms from the morphisms in $\mathrm{Hom}_C(X,A)$ to morphisms in $\mathrm{Hom}_C(Y,A)$?
what are the objects of $C_A$?
02:27
any one know how to get wolfram alpha to calculate the solutions of sin^1(o/h)? don't know how to tell it to calculate and return the two solutions
morphisms from objects of $C$ to $A$
yes
@WDUK solutions? thats not an equation, it doesnt have solutions
so you need to connect those $C's$ in a natural way.
maybe im using the wrong wording
02:29
This natural way satisfies the hypothesis of category as given in the book.
i am trying to find two solutions between 0 and 360 degrees of an equation in the form of sin theta = (a/b)
or least how to tell wolfram to find the two solutions
what, so you have the a and b?
yeah
and you want to find theta? your solutions are $\sin^{-1}(\frac{a}{b})$
or, if youre not using mathjax, sin^{-1}(a/b)
so just plug that into a calculator :]
i got the answers as 135 degrees and 45 degrees of sin ^1(19/27) just trying to get wolfram to also give me the same solutions in degrees
those are rounded
02:32
what do you mean "in degrees"? theyre already in degrees
@Adeek so umm morphisms in $C_A$ are the actual diagramS?
this is what its giving me
and i dont want that i want a simplified degrees answer
add "in degrees" to the end
@meow-mix specail kind of diagrams
02:34
@Adeek how are diagrams considered rigorous mathematical objects?
no i want 135.275... degrees and 44.724... degrees not that overly complex solution its providing
seems it wont give it any simpler than that though
lol those arent complicated
what was your search query
paste it here, the exact thing
degrees sin theta = 19/27
They satisfy certain conditions of the category given above.
@Adeek are diagrams categories or something?
or are they something else
02:36
well, if you want to think of that way here is how I think of it. So we need morphisms connecting $C_1$ and $C_2$ which satisfy certain property.
we have objects of the category are just morphisms given from $C_i \rightarrow A$ where $C_i$ are objects inside of $C$.
yes that i understand
@meow-mix the diagram $A\xrightarrow{f}B$ in a category $\cal C$ is formally defined as a functor $J\to\cal C$ where $J$ is the category with two objects $X,Y$ and a single nonidentity morphism $X\to Y$. the functor sends $X$ to $A$, $Y$ to $B$, and $X\to Y$ to $A\xrightarrow{f}B$
are the morphisms of $C_A$ commutative diagrams?
yeah @meow-mix
@arctictern they don't introduce functors yet.
yeah, functors are in chapter 4 i believe
02:38
yeah
you asked if diagrams have a formal definition.
you shouldn't be unnerved by not having a formal definition
i was just wondering, lol
tons of students work with real numbers without being exposed to cauchy sequences or dedekind cuts
because real numbers are intuitive
02:39
ok ok i get it
@WDUK stop being afraid of radians
better yet, stop being
afraidians
i understand radians fine i had to get the result in degrees
beyond my control
saying that i dont know what the "n" is in the solutions
it feels unclean
$n$ is just an integer
the whole $2\pi n$ thing is because angles are equivalent every $2\pi$
or at least, apply the same rotation
so is it common to have the inverse function as the answer for a solution
02:43
haha @arctictern
rather than the actual output of the inverse function
imagine people working with real numbers as dedekind cuts
well yes
@arctictern can you help me with the following question.
@Adeek sorry for being stupid lol
i think theres a missing $ /s
02:48
don't put \ in front of $ unless you want to make a dollar symbol
Let H be a hilbert space, $\{e_k\}$ be orthagonal sequence in H such that $e_k \neq 0$ for all k and that $\{a_k\}$ sequence in $\mathbb{K}$. Then there is $x \in H$ such that for every k one has $a_k = (x,e_k) / ||e_k||^2$ iff $\Sigma |a_k|^2 ||e_k||^2$.
yeah @arctictern.
@meow-mix don't worry about it we are all here to learn :)
> orthagonal
read this in brooklyn accent
lol
I proved before that for orthagonal sequence we have $\Sigma ||x_k||^2 < \infty$ iff $\Sigma x_k$ is convergent.
iff what exactly?
ah, you mean $\sum |a_k|\|e_k\|^2<\infty$
yeah
02:51
double bar signs are \| btw
oh
${\huge \checkmark}$
the conditions on $a_k$ force $x=\sum a_k e_k$
well, I guess that plus things orthogonal to all of the e_k's
hm
like, if A is the closure of the span of the e_k's, and B is the orthogonal complement, then we can decompose x=a+b, in which case the e_k's are a basis for A so we may write a=c_1e_1+... and then solve getting c_i=a_i. then |x|^2<infinity gives (sum |a_k|^2 |e_k|^2) + |b|^2 < infinity, so just delete the |b|^2
02:55
I think << is more obvious. If $\Sigma |a_k|^2 \|e_k\|^2 < \infty$. Can I use the result before somewhere?
If the sum is finite then just set $x=\sum a_k e_k$
oh I think I have an idea. We have $\Sigma |a_k|^2 \|e_k\|^2 = \|a_ke_k\|^2$ so by the result that I proved before we have $\Sigma a_k e_k < \infty$.
As you mentioned we just set $x = \Sigma a_ke_k$ and that will get it I think.
one, I already did the whole problem. two, $\sum |a_k|^2 \|e_k\|^2=\|a_ke_k\|^2$ makes no sense. unless you mean $\|\sum a_k e_k\|^2$
@arctictern but it is Hilbert space so Pythagorean theorem holds.
three, $\sum a_ke_k<\infty$ makes no sense since the sum is an element of $H$
@Adeek what is your point?
03:01
@arctictern sorry I was confused. I agree with you.
I guess we will just have to deal with partial sum and put $x = \Sigma a_ke_k$ as you mentioned.
can anyone explain me this?
when i watch pdf's of my profs, some signs get "too small"
such as "=" gets to "-"
but when i make my own pdf all is fine
@arctictern I don't understand why do we need the sum to be less than infinity ? I mean we didn't use it anywhere in our proof ?
21 mins ago, by Adeek
Let H be a hilbert space, $\{e_k\}$ be orthagonal sequence in H such that $e_k \neq 0$ for all k and that $\{a_k\}$ sequence in $\mathbb{K}$. Then there is $x \in H$ such that for every k one has $a_k = (x,e_k) / ||e_k||^2$ iff $\Sigma |a_k|^2 ||e_k||^2$.
yes
you meant to have $<\infty$ at the end there right?
03:09
yeah
so, first, you have to show that if there is such an x, then that sum is < infinity. and second, you have to show that if the sum is < infinity, then there is such an x.
I have told you how to do both directions.
I don't understand so for the direction < we construct the $x = \Sigma_{i}^{k} a_i e_k$ and using orthagonility we prove that $a_k$ is given by the formula above right ?
03:26
@arctictern ?
I already did the whole problem for you.
also, $\sum_i^k a_i e_k$ is nonsense
I don't understand the solution.
so why $x = \Sigma a_k e_k$ makes sense ?
If $\sum |a_k|^2\|e_k\|^2<\infty$ then we let $x=\sum a_k e_k$, and it satisfies $a_k=(x,e_k)/\|e_k\|^2$ for all $k$. Conversely, if there is an $x$ such that $a_k=(x,e_k)/\|e_k\|^2$ for all $k$, then we may write $x=a+b$ where $a$ is $x$'s projection onto $A$, the closure of the span of the $e_i$s, and $b$ is $x$'s projection onto the orthogonal complement of $A$. Then $\|x\|^2=\|a\|^2+\|b\|^2$ and $\|a\|^2=\sum |a_k|^2\|e_k\|^2$ must exist.
@Adeek it's a hilbert space, so you can add those things up
but why $\Sigma a_ke_k$ converges ?
because it's a cauchy sequence
03:31
oh I see.
I see thanks a lot i understand it now.

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