« first day (2317 days earlier)      last day (3000 days later) » 

09:00
Yeah, that's what we're trying to prove. What does this tell us about what the kernel of $\phi$ "should" be?
Sick, that random upvote gave me my linear algebra badge :D
kernel is $S \cap I$?
Indeed! Now justify to yourself / the grader why that's indeed the case (not too hard)
because the kernel is in
s and in I
Sort of. What's the zero element in $(S+I)/I$ ?
09:05
Elements in $(S+I)/I$ look like $x + I$ for $x \in (S+I)$. Think about how addition is defined here in the quotient ring
(a + I)(b + I) = ?
(ab +I)
Oh oops, I meant to write
(a+I) + (b + I) = ?
(a+b)+I
mhmm, so what is the additive identity?
For this quotient ring
I have an embarrassingly easy question. If $f:A\to B$ is a bijection and $g(f(x))=x$ then is $f(g(x))=x$?
09:08
additive identity?
yes
@sophie
Yeah, like if $(a + I) + (x + I) = a + I$, then what is $x$?
how do you prove it?
because if $g(f(x))=x$ then $f(g(f(x))) = f(x)$ and then you use surjectivity of $f$ o.o
09:09
and it shows that $f(g(y)) = y$ forall $y$
$(a + I) + (x + I) = a+x+ I$. How do we get $a+I$ out of this?
Yep, which means $0 + I$ is the additive identity. The "zero element" of $(S+I)/I$
of $(S+I)/I$
I don't think I like the notation $(S+I)/I$ very much
Neither do I @mercio
But it's what the problem statement used :/
09:12
I mean, it.. kinda works.. but...
What is wrong with that notation?
I did not write this
it's also $S/I$
Basically that
@mercio Only if $I\subseteq S$
09:13
@usukidoll, so what elements does $\phi$ map to $(0 + I)$?
Those elements are the kernel of phi
I kinda just assumed that $I$ was an ideal of $S$
@mercio Also, it is not just "notation". I mean, the symbols mean precisely the same as in most other contexts
$I$ is an ideal of a larger ring; $S$ is a subring
ah well, then it's fine
(and $S/I$ doesn't make sense)
Oh yeah, I guess that's right. I didn't think about it too hard.
09:25
do undergrads usually study a lot of number theory?
hi @KajHansen
I really don't understand here why is it 0 ?
and why we just compute dimension of $V^G$ ?
What is $\varphi$?
At any rate, I think this is a little bit beyond where I'm at unfortunately
@Sophie Depends a lot on the place and how much "a lot" is.
oh ok @KajHansen
@KajHansen $\varphi$ is just the average homomorphism.
Hi , can we without calculating a number in base 4 tell what it's first digit will be ? Like 17788 in base 4
09:38
@Adeek It is $0$ because an irreducible representation has zero-dimensional space of fixed points (unless it is trivial itself of course)
how to swap x and z
y
y=1/4a*(x^2)-a/2*ln(x/a)
?
@johnsmith Yes, just take the number mod 4
@johnsmith to write an integer in base n means to write is as a_0+a_1n+a_2n^2...
with all of 0<=a_i<n, so you just have to do repeated euclidean divisions
oh I see @TobiasKildetoft
09:57
Ford circles are pretty. They should be mentioned in any introduction to Farey sequences
"If $\{p_n\}$ is a sequence in a compact metric space X, then some subsequence of $\{p_n\}$ converges to a point of X." baby Rudin page 62. Does the theorem become false if you change compact to closed? I don't have any counterexamples
@TobiasKildetoft. 13 %4 =1 but 13 in base 4 is written as 31. And so the first digit is 3 not 1
@johnsmith Ahh, I would have called the one at the far right the first digit.
@johnsmith He meant the first from right to left
To get the one in the other end, first determine the number of digits using logs, then take the number minus its residue mod 4 to the power of one less than the number of digits
10:25
@TobiasKildetoft @KajHansen very good youtube.com/watch?v=FrNqSLPaZLc
@Adeek I am not convinced. He seems to be alternating between "simple terms" and "analogies". These are not the same and depending on the topic, either could be detrimental to understanding
Trying to teach something, in my experience, does a pretty good job at illuminating holes in understanding and also solidifying already-known concepts
That's part of why I really like this website
@TobiasKildetoft yeah those doesn't work sometimes in math it depends. But, I guess explaining the concept to yourself and how you understand it on a side note would help.
@Adeek Sure, explaining something to others is a great way to learn, and trying to emulate that for yourself is fine
(I find the Feynman anecdote quite implausible btw)
Advanced math does not become easier by being broken into things understandable by someone who does not know the basics of the theory. It actually becomes much harder
yeah that is only for simple stuff like differential equations and linear algebra
10:34
idk how you can get away with replacing "hard' words with simpler ones
There are so many ideas stacked on ideas stacked on ideas it's pretty much essential
one thing that I do though is that write intuition for theorems rather than just
understand proofs and theorems etc I like to write my intuition as well
do you guys do that @KajHansen @TobiasKildetoft ?
@Adeek I hardly ever write notes at all when reading stuff
I see
Neither do I, but I do think it'd help for my memory
I usually am not writing because I'm laying on my back in bed and am lazy
I should probably sleep it is like 4 am here have to submit marking by 10 am today :S
10:41
lol
gl
I wanted to ask some thing though. This semester I am taking this class which because prof was bad I ended up teaching it to myself. But, I noticed sometimes I tell myself I have so many things to do and end up procastinating do you guys deal with such thing ?
I hate it when this happen. I should probably learn to deal with stress in a better way.
anyway gt
@Adeek, I took slightly subpar grades constantly
But generally was able to get it together in the final few hours
@Adeek what is your goal? To understand (and)or to pass examens?
If a function is linear and not injective,
is it uniquely determined by its actions on one argument?
my guess is yes
What do you mean by one argument?
one vector I plug in
i don't know how it's called in english
like, f is linear and not injective, therefore i only need one vector to know what it does. like f(1)=2
10:56
@Null No, why would that be enough?
I have some things about reduced homology that I don't yet understand. For example, let $f \colon X \to Y$ be a map. Is $H_n(X) \to H_n(Y)$ being an isomorphism for all $n$ equivalent to $\widetilde{H}_n(X) \to \widetilde{H}_n(Y)$ being an isomorphism for all $n$?
injective means, f(a)=f(b) is only the case if a=b. (a,b are vectors). I can't really imagine this in higher dimensions then $\mathbb{R}\to\mathbb{R}$
so not injective means: f(a)=f(b) for some $a\not=b$
and since it's linear, it means the function is constant?
that is my train of thought
@Null If $f:\mathbb{R}\to\mathbb{R}$ is linear and not injective then it is identically $0$
so this is a poor special case to consider
@TobiasKildetoft couldn't it be any constant, besides from 0?
no, those are not linear functions
11:01
@TobiasKildetoft i read, that in complex analysis you mean with linear function, one that goes through the origin, is this the case for algebra too?
@Null I mean linear as it is defined in linear algebra
@Null Btw, when dealing with linear maps between finite-dimensional vector spaces, this graphic here might help a lot once you understand it (which is basicly the rank nullity theorem) upload.wikimedia.org/wikipedia/commons/8/8f/Rank-nullity.svg
@TobiasKildetoft oh lol, that follows from f(a+b)=f(a)+f(b). Since then f(a-a)=f(a)-f(a)?
It visualizes $\dim(V) = \dim ker(f) + \dim im(f)$
@abenthy Yes, the functors are literally the same except at $H_0$ (and $H_{-1}$ for the empty space)
11:03
@Null Right
and in those cases it's all about path components and easy to see
thanks for clearing my misconceptions :)
11:15
@MikeMiller Ah, your remark about $H_{-1}$ is very helpful. So we have $$\widetilde{H}_{n}(\emptyset) = \begin{cases}0,&n \neq -1\\\mathbb{Z},& n=-1\end{cases}$$and for $X \neq \emptyset$ we have $$\widetilde{H}_{n}(X) = \begin{cases}H_n(X),&n \neq 0 \\ H_0(X) \oplus \mathbb{Z},&n=0\end{cases}$$
Wait, the last line is incorrect.
Can't edit it anymore, What I mean is $H_0(X) = \widetilde{H}_0(X) \oplus \mathbb{Z}$ for $X \neq \emptyset$.
11:39
does anyone know this continued fraction where you replace all the floors by round to the nearest integer? For example $\sqrt{61}=[5;\overline{-5,-4,3,3,-4,-5,16}]$
sorry $\sqrt{61}=[8;\overline{-5,-4,3,3,-4,-5,16}]$
 
1 hour later…
13:01
Is there a way to render mathjax here on a mobile?
I think im searching the wrong terms while trying to figure this out
0
Q: Bilinear proof of transformations

Aksel'sRoseI realize this question has been posted but it is on hold. It piqued my interest and now I cant figure it out so Im looking for some help. Question Let V and W be vector spaces over a field F and let $T \in Hom (V,W)$. For each $g \in Bil(W\times W)$, define $g(T): V \times V \to F$ by setting ...

any suggested keywords? Im having a hard time finding relevant info on google.
Also, if you have to change the branch of a curve as you vary the parameter in order to keep it defined, is that contour still called continuous?
Because it seems that's what happens if you want to use polar coordinates but the contour makes more than one rotation
@GFauxPas On Chrome on iPhone, yes. Otherwise, I don't know
Does "bilinear" mean "linear in all arguments"?
I'm on an android :(
Yeah: $f(a+b,c+d)=f(a,c)+f(a,d)+f(b,c)+f(b,d)$ @GFauxPas
13:08
The link on the side hangs before loading all the way
@GFauxPas android tends to use chrome so it should be possible on android as well
I'll try chrome
@Aksel'sRose Is there a reason we assume that the vector space $V$ has dimension $2$?
Note that a bilinear map from $U\times V$ to $W$ is not a linear map. Note that $(a,c)+(b,d)=(a+b,c+d)$ @GFauxPas
If a bilinear map were linear, we would have $f(a+b,c+d)=f(a,c)+f(b,d)$ for bilinear $f$, but we don't.
Instead, a bilinear map from $U\times V$ to $W$ corresponds to a linear map from $U\otimes V$ to $W$, where $\otimes$ means tensor product.
How do I open the bookmark without changing the tab in chrome?
Oh it didn't work :(
13:16
On iPhones, at least, there's a button on the top-right that looks like three vertical dots that opens a menu
and in that menu there's the bookmarks
Yeah same here
Make the second link a bookmark
To do that on iPhones, you star any page to make a bookmark, and then edit it to make it the right link
@GFauxPas Working?
$test$
No :(
could anyone help me with brackets please. In wikimedia.org/api/rest_v1/media/math/render/svg/… what is the sum over?
$e $
$e$
13:23
is it over everything inside the large square brackets?
To start ChatJax, then, be in this tab, open up bookmarks, and press the right bookmark
without leaving the tab
I give up, I'll try again later
Anyone want to answer my question about contour integrals? math.stackexchange.com/questions/2047334/…
13:28
@BoniLindsley not really
@Lembik which summation?
sorry guys, kiddo peed the crib so I was afk.
The first summation acts on what's in the parens
@GFauxPas How do we know this? I mean I can guess because the delta occurs in the last sum too. But what is the precedence rule?
Good ol' PEMDAS
Parens come first
13:34
@GFauxPas but there are no parentheses. It is sum (a_i) sum b_j basically
I think from you this is the same as sum ((a_i) sum b_j) right?
Are you talking about the one ranging over delta?
It bothers me that the second product doesn't have parens
Makes me question my interpretation
right, the sum over delta seems to include everything to its right
But why does it have parens around just the next expression? And why doesn't the author use parens after the second product symbol?
Seems sloppy on the part of the writer, I feel your pain
thanks :)
If I'm not sure if my writing is ambiguous I generally err on the side of extra parentheses
13:42
me too
What is A? Try to find another source
A matrix?
@TedShifrin I found a solution to the problem about a conditionally convergent series $\sum a_n$ such that $\sum a_n^3$ diverges. I don't think it's very elegant, though.
Let $b_n$ be a decreasing positive sequence such that $\sum b_{2n}^3$ diverges to infinity. ($b_n=n^{-1/3}$ works.) The alternating series $\sum(-1)^nb_n$ and $\sum(-1)^nb_n^3$ must converge by the alternating series criterion.

Define the series $a_n$ to be:
\begin{align}
(a_n)=(&b_0,-b_1,\\
&b_2,-b_3/3,-b_3/3,-b_3/3\\
&b_4,-b_5/5,-b_5/5,-b_5/5,-b_5/5,-b_5/5\\
&\dots)
\end{align}
That is, replace $b_{2n+1}$ by a group of $2n+1$ copies of $b_{2n+1}/(2n+1)$.

$\sum a_n=\sum(-1)^nb_n$ by grouping the terms, and so it converges. I will show that $\sum a_n^3$ must diverge, solving the problem.
14:07
CORRECTION: I meant, "that is, replace $b_{2n+1}$ by a group of $2n+1$ copies of $\mathbf{\color{Red}-}b_{2n+1}/(2n+1)$."
Also, $b_n=n^{-1/3}$ doesn't quite work since I had let $n=0$ (that is, I let $b_0$ be a thing), but that's an easy fix
14:20
hi chat
hi @Semi
Anything interesting this morning?
@abenthy Correct, and the functors $H_0$ and $\widetilde H_0 \oplus \Bbb Z$ are naturally equivalent.
@Semiclassical Something about infamy?
75 years to date, as it were. My grandmother also turns 80 today.
Hadn't thought about that, but you're right.
14:26
It'd be weird if you had remembered my grandmother's birthday, @Semi. ;)
pah, you know what i meant
Yeah, I know — 99 years and 1 day since Finland's independence from Russia!
Where's Ted with his thwaps when you need him?
I dunno; I need him here to verify my (long) answer above
To find a convergent series that diverges when every term is cubed
or to prove it impossible
A simple counterexample would be nice :)
14:34
Well, I found a complicated counterexample
I figured.
I'm not awake yet, so I'm defaulting to snark over actual intelligence
2
15:05
Could someone of you take a look at the induction that I have written in math.stackexchange.com/questions/2047712/… ?
I'm more than a bit annoyed by the response I got to a comment of mine
(Paraphrased) Answer: "It's this result because of theorem X. " Me: "You haven't explained what theorem X means, so this isn't a helpful answer." Response: "If they're interested in the subject then they can google it."
Is a contour integral well defined if the contour wraps around the origin completely but the integrand requires a branch cut to be single valued?
Hi there
I have a silly question:
What is the remainder of the following division: 1/2?
Is it 0 or 1?
1 = 0 * 2 +1
1 = 0.5*2 + 0
When doing division with integers, one typically requires that the quotient and remainder both be integers
so 1=0*2+1 works (quotient 0, remainder 1) but not 1=0.5*2+0 (quotient 1/2, remainder 0)
15:20
Otherwise the division algorithm would last forever
Thanks, so the question must emphasize at that point?
Could
Or always end in zero
i mean, if you did the second version, you could write pretty much any combination
e.g. 10 = 5*2+0 = 2.5*2+5 = 3.3*2 +3.4...
you only get a unique answer if you require that the quotient be an integer and be as big as possible.
@Semiclassical I see. So, the "norm" is to consider integer quotient.
(otherwise you could say 10=4*2+2)
15:22
@GFauxPas do you have an example in mind ?
@Semiclassical I see.
Yeah. Otherwise division wouldn't be a well-defined operation.
usually you can make a change of variable ($x = \exp(t)$) to "uncurl" the singularity
@Semiclassical Was it a comment or an answer?
An answer.
15:23
but then a closed contour gets ripped open
If it was a comment I'd be fine with it.
you know, I'm heavily engaged with mathematics in Engineering, I forgot the basics. I know that may sound silly, but that is the status quo, unfortunately.
Thanks.
For example, I consider 1/2= 0.5 almost immediately.
Well, keep in mind
15:24
well that depends on the context
There's division as just p/q, and in that case I'd say 1/2=0.5 is appropriate.
@Mercio, $t^{-.5 + 3i}\exp (-t - 2it) $
engineers usually don't work much on number theory
But there's also the remainder of p when divided by q, and that's an integer
As is the quotient of p when divided by q.
@Semiclassical nods
15:25
I'm sorry but you just wrote $t^{-0.5+3i}$
so if I was a dictator I would put you in prison
I'd say that p/q corresponds to a rational number rather than an integer. to indicate the quotient, I've seen p % q.
and for the remainder, people typically would say p mod q.
anyway you want to do $t = \exp(u)$
Better now?
because $a^b$ is ONLY defined for $b$ natural integer or $a$ positive real
and if you do a contour integration you want $t$ complex
For example, when it comes to answering petty questions like GRE quantitative questions, I'm jumping to the conclusion so as to save time, but the true story is that I have made very blatant mistake in conforming to the basics.
@Semiclassical I see.
15:29
so just replace $t$ with $\exp(u)$ and you will get $\exp((-0.5+3i)u) \exp(-t-2it)dt$
I'm defining $t^{-0.5 + 3i} = \exp ((-0.5+3i)\log t) $
yes but how are you defining $\log(t)$
That's the problem, if the contour wraps all the way around the origin, there's no one branch cut that works for the whole contour
@Semiclassical Ah, OK; I misunderstood who said what. I added my own comment.
I saw, yeah. Thx.
You can see why I was a bit annoyed.
15:30
Thank you guys. That's a great room. good for you.
yes that's why you do a change of variable that expresses everything in terms of something that would act like $\log(t)$
if it existed
That assumes the integral is well defined in the first place
Which is my question
yes but it isn't because $t^{...}$ isn't well defined
but since you can define $\log$ locally
you can do it if you bother to give a starting value for $\log(t)$
It is if you integrate different segments of the contour individually and change the branch cut as you go along
yeah but if you do the change of variable I'm telling you to do, you transform something ugly that is not well-defined into something beautiful that makes sense and has no problem at all
15:33
I want to know if I'm allowed to do that, because if that's true I can homotopically move the contour into a simply connected domsin
If you have a different $\log$ for each segment you can do the change of variable separately on each part
and hopefully the values of $\log(t)$ will connect to each other at the extremities
:o
This integral is related to the gamma function btw
but if your contour is a circle around $0$ then it won't connect and instead the contour will be, after the change of variable, a straight vertical line of height $2i\pi$
Which has $z^{a-1}$ in the integrand
which tells you that indeed there is a potential problem
15:36
So I'm trying to understand the gamma function integral
ah visual studio installation finally failed after 1 hour : D
Grats
It says there is not enough space on disk even though he said it only required 8 gb
and I told him to put it on the drive that had 30 gb free
so I'm very happy
Oh I've had that error before I forgot how to fix it
because apparently it has to download and install a BAZILLION thingies
15:38
So @Mercio I'd very much like the gamma function to make sense
and it puts them on C: drive without telling me
the gamma function is $\Gamma(s) = \int_{\Bbb R^+} t^s e^{-t} \frac {dt}t$ ?
after a change of variable $t = \exp(u)$
you get $\Gamma(s) = \int_\Bbb R \exp(su) \exp(-\exp(u)) du$
umm
that can't be right can it ?
ooh i forgot $e^{-t}$
And it's shifted so the pole is at the origin but that's not important now
and after the change of variable you have no problem whatsoever
That definition is valid when $s$ has positive real part.
hmm really ?
But in any event you have $x^s$
But you've got $\Gamma(z)=\Gamma(z+1)/z$, so that makes it easy to analytically continue.
no, I have $\exp(su)$
see ?
all better
$\exp$ is your friend
it is well-defined everywhere
well, $e^{-e^u}$ is a bit less nice
agreed
but still well-defined
no logs, no cheating, no branch cuts
15:44
tbh i'd probably stick with the first definition, just with the understanding that $t^s=e^{s\log t}$
For one because doing the saddle point approximation is easy in the first presentation
well if you don't pick $t$ in the complex numbers it's fine
so now my C drive is full of ghost useless stuff
It's just that when I graphed Re vs Im of $t^a e^{-st} $ for certain combinations of $s,t$, I got a contour that wrapped around a few times before settling down
I don't even know where it put the things in there
Sounds like you need to hire an exorcist
I'm tempted to melt the thing
15:47
Trial by fire.
out of frustration
Or, better yet, trial by water. If it sinks, it's gone; if it floats, it's a witch drive and should be burnt.
I bet if I free 3 gbs of space and try again it'll fill it up with the same thing and fail again
So however I choose to define $\log$, the contour will need to have a different branch cut at some point
(I don't actually advise either of those disposal methods)
15:48
well in the definition of $\Gamma$ you are given the contour and it's just $\Bbb R_{>0}$
Eh.
That's not the contour you do if you want to analytically continue it.
you don't really need a branch cut unless you plan on pulling the line around
I was graphing the contour you get naively if you parameterize using $\theta (t) = t $
For analytic continuation, you'd want a contour that runs back and forth along the positive real axis, but with positive/negative imaginary parts.
(I would do the analytic continuation with the functional equation)
15:50
True.
@mercio I'm looking for intuition of the integral thi
I'm not quite sure what you are trying to graph
honestly
Amusingly, $1/\Gamma(z)$ also has a contour integral representation: en.wikipedia.org/wiki/…
I introduced a Parameterization for $z^t e^{-t} $
okay, I can do this problem if my new question is right. Can someone check out my added question at the bottom of this, please:math.stackexchange.com/questions/2047210/…
15:52
If you replace $t = \theta (t), \theta'(t) = 1$
It becomes a contour integral
I feel like an idiot every time I post about this....hopefully this is the last time you all will see this question!
hi @Semiclassical, do you know Rudin's introduction to analysis book?
You don't need to feel bad about asking questions
I don't understand $t = \theta(t)$
I did a substitution
15:54
$t = \theta(u)$ ?
To make it look like the definition of a contour integral
Nope. @Sophie
wait.. what's a contour integral for you ?
@Semiclassical thanks anyway. I'm trying to understand the motivation to all this topology
$\int_a^b f (\theta (t)) \theta'(t) \, \mathrm dt $ If the contour is smooth
$\theta (t) $ is a parameterizaton of the contour
15:57
but the contour in the definition is $\Bbb R_{>0}$
so what parametrisation do you take ?
and what bounds $a$ and $b$ ?
But doesn't the contour depend on $z $?
If you want to do the Gamma function as a contour integral, you'll need to use a different definition than the $\Bbb R_{>0}$ one. The usual modification is a Hankel contour.
Oh...

« first day (2317 days earlier)      last day (3000 days later) »