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13:03
Grr
@MaryStar What is $\xi$ here?
@AlecTeal what are you growling about
Mary Star
@AlecTeal why do you growl at her?
She is helping me solve a hard problem
@AlecTeal you solve the problem?
@AlecTeal are you there?
Wait, it wasn't her asking a question?
no, it was i
@AlecTeal you type slow me thinks
13:07
The world doesn't revolve about you.
Now stop pinging me
@AlecTeal That's a rather strange response.
can u help me @alec?
he asked you to stop
sorry then
@Calculus I meant $z$. $f$ is the function you want to integrate.
and $L$ is the length of the circle, refers to the line length(circumference)?
13:11
Post a question if stuck
@AlecTeal 2 u or on the site?
again with the ping?
6 mins ago, by Alec Teal
Now stop pinging me
@skillpatrol but he was talking to me
@skillpatrol i think he was atleast
@skillpatrol i think he was annoyed b4
@skillpatrol but not now
13:15
are you trying to be a cyber bully?
@skillpatrol what do you mean?
that's the 5th time you have pinged me >8(
I am just trying to study really. But I am having trouble for some reason solving many important problems for my mid-semester examination.
this should be a baby problem but i cant get it. pls help!

show that $z^5 -6z^2 + 13$ has no zeroes inside the circle of radius $1$ around the origin
@Calculus It is the length, that means that if $\gamma : [a,b] \rightarrow \mathbb{C}$ differentiable, we have that $L(\gamma) = \int_a^b |\gamma'(t)| dt $
That's heck of pinging stuff lol@Calculus
13:20
@Rememberme haha oops!
@Rememberme i just send the messages and want the right person to know who its 4
@Calculus what's your question if it is related to topology I might come to some help
show that $z^5 -6z^2 + 13$ has no zeroes inside the circle of radius $1$ around the origin
@ShinKim I prefer the "non-standard notation". And why don't you stop complaining about stuff you can't change?
it should be really easy to solve that one right there but i cant get it
It's triangle inequality, @Calculus.
13:23
by looking at the magnitudes @Ted?
$0\lt |r^5-6r^2+13|\lt 1$?
Hey@TedShifrin
$|r^5-6r^2+13|\leq|r^5-6r^2|+|13|\lt 1$
Not like that, @Calculus. What is the largest $|z^5-6z^2|$ can be if $|z|\le 1$?
hi @Remember
@TedShifrin at z=1, so 5?
No. Remember that terms can be negative, too.
13:25
Strictly less than @Calculus
Just think $|a+b|\le |a|+|b|$ !
-7
ok one sec
7 at z=-1
Don't worry about whether it's the actual max. You just want to give a bound. $|z^5-6z^2|\le |z|^5 + 6|z|^2 \le 1 + 6 = 7$.
So can your total expression ever equal $0$?
Doesn't seem so...
Quiet, you, @Remember.
13:30
Fine....
i have a feeling, but i'll think about it more before being sure, that the +13 pushes us outside of the circle always
You should be able to give a one-sentence rigorous argument.
Ok ill write it down and get it
hope you don't mind me bothering you guys - I've got a probably really stupid misconception here - my binomial expansion for (1 + 2x)^-1 to three terms is different to the binomial expansion of (1 + 2x)^-2 with a coefficient of (1 + 2x) added after expansion (i.e. [1+2x]/(1+2x^2) with the denominator expanded) to three terms - why is this? :( I thought (1 + 2x)/(1 + 2x)^2 was equal to (1 + 2x)^-1
Sorry my i.e. should be "[1 + 2x]/(1 + 2x)^2"
So how are you doing $(1+2x)^{-1}$, @Ashley?
13:34
in this context both solutions are using the formula 1 + nx + n(n-1)/2! x^2
While browsing through 1-answered questions, came across this question which had a wrong answer (got deleted when I pointed it out), and I tried answering.
Can someone please check the answer? I came upto convergence and pointed to Alpha. I feel it's incomplete. Can someone help?
That formula isn't right.
for three terms |x| < 1?
What is that the formula for?
binomial expansion of (1 + x)^n for |x| < 1 (it's an infinite series but I only need the x^0, x^1 and x^2 terms
13:36
@Soham tell me something ...
Rubik's cube has a connection with group theory right?
OK. So how are you applying that?
yes, @Remember
@Rememberme yes we have face rotations being elements of symmetric group
@Rememberme This article might have more detailed answer to that question!
@TedShifrin so did it give me an annulus of $7 \leq f \leq 20$?
no, @Calculus.
What's the smallest $|z^5-6z^2+13|$ can be?
13:40
Yes I was thinking of symmetric groups only ...
Can that provide a solution to Rubik's cube
What will be the generators and relations for that group.....
@TedShifrin 13?
No.
This doesn't sound like writing out a rigorous one-sentence argument.
thats because i am having trouble still
not sure why, i should have been able to do this last year probably
13:43
Probably.
i feel like i should be able to do minima but its been ages, take first derivative and look at 0
The most $z^5-6z^2$ could cancel off from the $13$, we agreed, was $7$ (because $|w|\le 7$ means you could have $w=-7$).
No, no, no.
Think of it like this @Calculus you will find the least value that will show that the unit circle will be engulfed in it (subset) and hence you will be able show... I think
You don't need the actual minimum. You just need a bound that shows the quantity cannot be $0$.
Sorry I didn't notice your message - I went from $ \frac{(1 + 2x)}{(1 + 2x)^2} $ to $ \frac{1}{1 + 2x} then found the three terms by 1 + (-1)(2x) + (-1)(-2)/2.1 * 4x^2 to get 1 - 2x + 4x^2
13:44
Ok so the minimum is 13-7=6, and the max is 20
so its an annulus of $6\leq f \leq 20$
that the zeroes lie in
OK, @Ashley. So now what is your issue?
First of all, @Calculus, you need modulus around $f$. You can't have inequalities with complex numbers. Second, your last line makes no sense.
Sorry, still getting used to LaTeX --- the desired answer was 1 + 6x + 20x^2
What desired answer, @Ashley?
They solved it via multiplying the expansion to (1 + 2x)^-2 (As opposed to (1 + 2x)^-1) and multiplying by (1 + 2x) after expanding rather than before
I'm doing an old paper for an exam I have and that's the mark scheme solution
okay so we have $6\leq |f| \leq 20$ in the unit disk, and hence we never attain the value of zero on the unit disk for $f$
13:47
That seems like a crazy way to do it. And there's apparently a mistake.
Correct, @Calculus.
Thank you Ted!!!
The solution is on page 17 of this document if it helps - mei.org.uk/files/papers/c410ja_4754.pdf - it's question 1 so at the very top of page 17
Oh god
I'm an idiot nvm they have 1-2x on the denominator
I'm so sorry for wasting your time :( I totally missed that
Right, I was about to point that out.
Now it makes sense, too :P
on the bright side at least my entire knowledge of exponents isn't invalid :D
sorry for that though!
No problem :P
13:54
Now another problem i couldnt get earlier today:

Find the maclaurn series for $\frac{1}{2+4z}$
and i think its:
$$\sum_{n=0}^\infty (4^n(-1)^n(2)^{-(n+1)}z^n)$$
with the radius of convergence about $z_0=0$
$|z|\lt \frac12$
$$\sum_{n=0}^\infty \left(\frac{(4z)^n(-1)^n}{2^{n+1}}\right)$$
@SohamChowdhury You are right. I cannot change it but why should i stop complaining? :D
Mew
Mew
hi
hi @Mew
Mew
Mew
can you help me with my question?
@TedS I am unsure how I can verify my answer, which will be scary on my midsem tomorrow
13:59
@Mew Depends. What is your question?
Mew
Mew
Is there a function that maps [0,1] to [0,1]
that looks like a sigmoid shape
yeah identity is one
but I want to penalize the low values and not amplify high values too much
that is why i want an S shape
sigmoid shape
Of course there exist such map.
Mew
Mew
are there some simple examples?
identity function
Mew
Mew
identity isn't sigmoid shape
14:01
so you are asking, is there a map from [0,1] to [0,1] and also sigmoid shaped?
Mew
Mew
yeah
of course there is one?
Hey@mew
Mew
Mew
doesn't have to be exactly "sigmoid" but just have that similar shape
of course there is one
but I wonder what is a simple one I could use
Hi rememberme
Seems like you doin some abstract stuff:p
Mew
Mew
14:02
this isn't abstract this is practical lol
you can also use sinonical fucntion
Mew
Mew
do you have a link shin?
Ahh I don't want that discussion now I am really tired
something like $f(x)=(\cos(\pi x)+1)/2$
Mew
Mew
hmm
I want the line to pass through (0,0) and (1,1)
14:07
ahh, it should have positive derivatives at all points but easy fix it anyways
you can just set the coefficient of cosine as -1
$f(x)=(-\cos(\pi x)+1)/2$
Mew
Mew
thanks it looks great
is there a way I can flatten it out more?
@Mew Ummm,, could you be more elaborate about 'flatten'?
Mew
Mew
as in, flatten the base and the top more, and make the middle section steeper
so that the low values are penalized even more
Ah, yeah sure. just power the cos term
Mew
Mew
ok thanks i'll have a go
thanks alot dude
14:17
wait
sorry, it's actually wrong if i power it
Mew
Mew
yeah power doesn't seem to work
i have a simpler example
Mew
Mew
hmm
lol
yeah that one didn't work
except for o = 1 when it was straight
anyone good at asymptotics able to help with math.stackexchange.com/questions/1321242/… ?
well, actually we can just stretch the ranges?
Mew
Mew
14:26
thanks alot for your input shin
I have also created a question on stack exchange here:
Yeah no problem
Mew
Mew
0
Q: What are some simple (or elegant) functions that satisfy these conditions?

MewI wish to have a function that maps [0,1] to [0,1]. I also require that f(0) = 0 and f(1) = 1. Also, I would like the function to be of sigmoidal shape, such as this: The above function is ok, but I would like the function to be flatter at the bottom and the top, and have a steeper ascent....

hope you will find a way to flatten that function
Mew
Mew
thanks bro
kinda looks like a bell curve
maybe I could shift the bell curve so that it passes through the piont (0,0)
then I can adjust flatness by adjusting the bell curve parameters
This is a bit contrived though so hoping for something more elegant lol
Hey @bolbteppa
14:29
Hey @BalarkaSen, what do you think of this Jordan-Holder (!) proof of the fundamental theorem of arithmetic! :D artofproblemsolving.com/wiki/index.php/…
I have seen it before, and I don't think it's of much use.
I mean, if that proof could be generalized to the fundamental theorem for general UFDs (integral closures of algebraic number fields, say), I'd have cared. No point in going high-brow about number theory over $\Bbb Z$ (which is easy, in general).
If you can prove the fundamental theorem of arithmetic over Z using Jordan-Holder for groups, I'm sure you can use Jordan-Holder for modules math.hawaii.edu/~lee/algebra/jordan.pdf on UFD' I'm sure?
Why can't it be generalized?
$R-\mathsf{Mod}$ is a much simpler category than $\mathsf{Ring}$. It's not apparent to me how you could apply things about modules to prove things about rings.
*sinusoidal fucntion
In any case, there are lots of non-UFDs out there, so I believe anything that's true in general for R-modules fails to say something interesting for rings. Of course, this is what I believe, and I'd be happy to know I am wrong.
(to emphasize what I'm saying : subobject structure of modules is much easier than subobject structure of rings. consider Noetherian modules and Noetherian rings, say)
OK, I gotta run.
14:40
@BalarkaSen Sure, I just mean the fact that it generalizes from groups to modules probably means the same idea can be used for UFD's, as that paper says
ok cool
 
2 hours later…
16:19
Hey guys
hey
anyone understand homology theory the basics?
Bib
Bib
17:17
is there any reference for the fact that all functors that commute with direct sums are additive functors?
MacLane proves it. The usual statement is "All functors between preadditive categories that preserve finite biproduct diagrams are additive".
Bib
Bib
in Homology? or Categories? Thanks by the way.
Categories
@Bib what does universal with respect to this property mean in this context: "In an additive category $\mathcal{A}$ a kernel of a morphism $f: B\to C$ is defined to be a map $i : A \to B$ such that $fi = 0$ and that is universal with respect to this property."
I mean @Mike
every other map $j$ with $fj = 0$ factors uniquely through $i$
17:28
thx!
@EnjoysMath You have proved that $Z_n(\bigoplus_i A_i) = \bigoplus_i Z_n(A_i)$ and $B_n(\bigoplus_i A_i) = \bigoplus_i B_n(A_i)$. $B_n$ is normal in $Z_n$, so $\bigoplus Z_n(A_i)/\bigoplus B_n(A_i) \cong \bigoplus (Z_n(A_i)/B_n(A_i)) = \bigoplus H_n(A_i)$.
lol
we need a better system than this latex
*mathjax
@BalarkaSen
how did you get the isomorphism?
it's a just an algebra fact.
i.e., if $M', N'$ are submodules of $M, N$, then $M/M' \oplus N/N' \cong (M \oplus N)/(M' \oplus N')$
(ps : ignore the normal thing I said above. I didn't note you were looking at $R-\mathsf{Mod}$-valued homology)
17:43
Are you a grad student?
me? no.
17:54
Night @BalarkaSen
I just started product topology in munkres
Its just the old gone Cartesian product..... Though there are few stuffs which I find pretty interesting
Like the Cartesian product of two bases is the base for the product topology on the two sets
product topology is certainly not just "old gone cartesion product". it's more subtle than you think it is.
@BalarkaSen his other book will make that distinction
Subtle @BalarkaSen... What do you mean by that?
Hi@AlecTeal
17:59
also, @Remember, I think you're going at a surprisingly high pace. have you done all the exercises before 2.19, and read all the theory?
i don't think you have done metric spaces in Simmons carefully.
@BalarkaSen that doesn't matter.
yes, it does.
skipping around the chapters would result in memorizing a bunch of phrase, with being able to do nothing with them.
It does matter just not to him
oh, I see.
Hey hey @BalarkaSen I am not doing anything fast...
18:00
@BalarkaSen I wrote software just so I could chapter skip and not forget what I missed.
Topology, Munkres, Me.
@Remember $U, V$ be open subsets of a top. space $X$. $f : U \cup V \to Y$ be a map between top. spaces. prove that if $f$ restricted to $U$ and $V$ are both continuous, then $f$ is continuous.
don't google.
I did all exercises in Simmons,till topolgical spaces.... Finished topological spaces in munkres ... Now doing product spaces which is just after topological spaces
@BalarkaSen in munkres
@BalarkaSen he's got two good book and no exams. It'll be boring as fuck later (which is why so much of Munkres is unread) but it wont harm
I wont @BalarkaSen
indeed :) it's called gluing lemma, @Remember. immensely powerful tool.
18:04
What do you mean when you say f is restricted @BalarkaSen
you should know what that means, if you have read the proof!
or did you just skim through the theorem and memorize it?
$f|_D:D\rightarrow Y$ for a function $f:X\rightarrow Y$ with $D\subseteq X$
$f|_D:x\mapsto f(x)$ BTW
This is in Simmons... If I am not wrong coz mu kres hasn't yet introduced continuity @BalarkaSen
what? no, gluing lemma is a specialty of Munkres. he certainly has introduced this before products.
Look at page 105
18:06
i have the book right with me, it's in chapter 2, $18
That's the hotspot on my reading
ok, I gotta go.
@BalarkaSen no I am doing product spaces it is after that I bet you
I think you have got something wrong homeomorphisms are introduced earlier in simmons@BalarkaSen
@AlecTeal tell me something pls check if I have got something wrong here ... Really I don't think he Introduces it before product topology @AlecTeal
@BalarkaSen currently I am on pg 86 chapter 2
Product topology on X \cross Y
@Rememberme just feel free to chapter hop. That is EXACTLY WHY I gave you access to my software
I chapter hop all the time. That's why I wrote it.
I didn't hop even one chapter and I don't like to do that@AlecTeal
18:15
Well you should
@Alec do you have the book
Can you please check that I am not missing something
No
Balarka is just trying to make sure you know the ins and outs, the product and box topologies are different in the infinite case. That's not important now. The gluing lemma lets you join two functions together given they agree where they overlap. It's not needed now either.
@BalarkaSen once again I am on pg86 chapter 2 and I am not skipping anything...
Skipping is GOOD!
Are you really saying there's a natural and good linear order to doing stuff in?
'cos there isn't. Jump to continuity! Make open sets matter. Understanding what you read is important
God knows @AlecTeal
18:20
There isn't @Rememberme - I wrote this software because I NEVER read books in order.
Chapter 1 is boring for example.
Anyways I have to wait to get this misunderstanding out of the window so that @BalarkaSen does not think o am skipping stuff
It doesn't matter if he does.
@AlecTeal In case I am not here will you please tell @BalarkaSen which page I am on....
I am on pg86
Chapter 2
Sorry for the trouble @AlecTeal
0
Q: Numerical integration with matrices

T AbrahamI have a matrix integration problem. It is based on the first integral under the section, "energy transfer efficiency and transport time" in the article, environment-assisted transport. There is a function, $\rho(t)$ that is the most important in calculating the integral but is a time-dependent m...

My question wasn't getting any attention so could anybody here answer?
nobody here?
@Rememberme @AlecTeal what do u think?
18:45
@Rememberme which book?
Hello!!! Could you take a look at my question?
0
Q: Why does the inequality hold?

evindaLet $u(x,y), x^2+y^2 \leq 1$, a solution of $$u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0, x^2+y^2\leq 1$$ Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y) $. We suppose that $\min_{x^2+y^2 \leq 1} u(x,y) \neq \min_{x^2+y^2=1} u(x,y) $. At the solution it is said that since $\{ (x...

19:04
WAIT FOR IT
wait for what?
hi pal

 Algebraic Geometry

A room for anyone interested in algebraic geometry and nearby ...
19:21

 Algebraic Geometry

Discussion in informal spirit
19:34
Does anyone know a good source of recent statistical studies?
What do you mean statistical studies?
about what?
For example a study on effectiveness of a new medicine versus an old one, etc
Mainly I'm looking for data on two populations through which I can compare means
@ROODAY Just take pretty much any recent study
Do you know any good sites? Or what to google? Searching for "statistical studies" or "recent statistical studies" didn't yield anything very useful
19:42
@ROODAY Try pubmed
the problem is most of the things I find only show percentages whereas I need sample size, mean, sd
ok ill check that one out
You're describing almost literally every scientific study. Pick your favorite field, search "Journal of ____"
If you're on a college campus the articles should be available.
@ROODAY Note that these are not called statistical studies. They are studies in whatever discipline they are from, not statistics
I see. Are these journals publicly available, or when you said if I'm in college it should be available you meant as part of the college's archives?
Because I'm not in college
@ROODAY That is why I suggested pubmed
19:48
Alright I guess that's my best bet
thanks!
is this for a stats class?
@DanielFischer Hi, reading my exam on complexe analysis (what we talked about last time) it was to write the holomorphic function $f$ for the bounded domain $\Omega$ to $\Omega$ as $$f(z)=z+c(z-a)^k+(z-a)^{k+1}h(z).$$ I wrote that $h$ is holomorphic on a neighbourhood of $a$, but it was asked to prove that $h$ is, in fact, holomorphic on $\Omega$
yes, I need to find data to create a set of questions on
you need to narrow your question down from " study on effectiveness of a new medicine versus an old" to something more specific
@Gato With $f(a) = a$ and, as induction hypothesis $f^{(2)}(a) = \dotsc = f^{(k-1)}(a) = 0$, and $c = \frac{1}{k!}f^{(k)}(a)$?
19:52
Many online journals, or journal repositories, are accessible if you have institutional access. Being e.g. on college wifi gets you such access. That's all.
@DanielFischer Exactly.
Ah. Well I'm in highschool and currently at home, so can't use those particular journals
the public libraries have access
@Gato Well, $$h(z) = \frac{f(z) - z - c(z-a)^k}{(z-a)^{k+1}}$$ is evidently holomorphic on $\Omega\setminus \{a\}$. The previously mentioned things show it is holomorphic at $a$.
Morning @MikeMiller.
Morning.
20:01
@DanielFischer Ok, I did this but It was asked in another question, so perhaps I missed something else. Another question : to compute the maximum of $\sin(z)/z$ on the unit circle, I used fist the maximum principle to say that it's at $\vert z \vert =1$ but how can I continue ? I now that $\sin(z)=z-z^3/3+\cdots+ o(z^{2n+3})$
@Gato The triangle inequality gives you $$\lvert \sin z\rvert = \left\lvert\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+1}\right\rvert \leqslant \sum_{n=0}^\infty \frac{\lvert z\rvert^{2n+1}}{(2n+1)!} = \sinh \lvert z\rvert.$$
@DanielFischer So I have $\max_{|z|=1}|\sin z|\le \sinh(1)$,not sure how this helps..
@DanielFischer Ok I need to remember that $\sin(z)=\sinh(iz)$
@Gato Right. In the inequality above you have equality for purely imaginary $z$.
@Gato Well, $i\cdot \sin z = \sinh (iz)$.
20:16
@DanielFischer So the maximum is at $z=i$
@Gato We need an absolute value somewhere, for a complex-valued function, it doesn't make sense to talk of a maximum, so $\max \left\lvert \frac{\sin z}{z}\right\rvert$ is attained at $i$ and $-i$.
@DanielFischer Right, badly expressed. So to write correctly a solution is : By the maximum principle $$\max_{z\in D}|f(z)|=\max_{|z|=1}\frac{|\sin z|}{|z|}=\max_{|z|=1}|\sin z|.$$
Furthermore $\lvert \sin z\rvert = \left\lvert\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+1}\right\rvert \leqslant \sum_{n=0}^\infty \frac{\lvert z\rvert^{2n+1}}{(2n+1)!} = \sinh \lvert z\rvert$ by the triangle inequality, then I have equality for purely imaginary z, so the maximum of $\vert\frac{\sin(z)}{z}\vert$ is attained at $i$ and $−i$.
Is it sufficient to get all the points in an exam ?
21:11
can someone tell me how did they get $||x_n - y_n|| = \frac{4}{3}$ in the following proof
example 3.1
@Gato Depends. If the grader is strict, you need to say a little about why you have equality for purely imaginary $z$.
What do you lot think of the diagram at the top of maths.kisogo.com/index.php?title=Types_of_set_algebras - I've made it so the arrow directions represent the same relationship, does that come across?
Part of me thinks "Dynkin system" should be vertically above sigma-algebra
21:32
@DanielFischer Ok, thanks. I am working with the following exercise : Let $f$ be an entire function, such that $\vert Im(f(z))\vert -\vert Re(f(z))\vert <0$. I need to prove that $f$ is constant on $\Bbb{C}$, pretty sure I need to use Liouville's theorem. I plot the graph of the inequality but not sure how I can use the hypothesis.
@Gato You need to transform it a bit before applying Liouville's theorem. If $f$ is entire and there is a $w\in \mathbb{C}$ and an $\varepsilon > 0$ with $\lvert f(z) - w\rvert \geqslant \varepsilon$ for all $z$, what can you do to make Liouville's theorem applicable?
@DanielFischer Just by considering $g(z)=\frac{1}{\vert f(z)-w \vert}$
Right. Does $\lvert \operatorname{Im} f(z)\rvert < \lvert \operatorname{Re} f(z)\rvert$ give you $w$ and $\varepsilon$?
@DanielFischer Ok I got it : $w=i$ and $\varepsilon=\sqrt{2}/2$
For example.
21:45
which one would you choose?

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