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8:01 AM
Don't take it so seriously @TheGreatDuck
 
but your name has the word "pissed off" nice
 
Think of it as "water off a duck's back" ;-)
 
~.^
 
@usukidoll that's why he reported me to the mods
 
haha
 
8:32 AM
Hello!
Can someone
help me with a question
I do not need the entire answer...just a minute solution
 
Mar 30 at 2:41, by Eric Stucky
"Just ask; don't ask to ask"
 
^
 
You have better chance that somebody will reply if you say at least approximately what your question is about. If it is related to a question already posted on the main site, do not forget to include the link.
 
I do not want the entire answer
THe first part
 
@robjohn Would it be reasonable to add "just ask; don't ask to ask" to Main Chatroom Etiquette Rules? (I am not sure whether anybody reads them, but since you maintain the guidelines, I guess suggestions like this are supposed to be addressed to you.)
 
8:38 AM
I know that you have to create a cumulative frequency table... I have made it
Now the median is thirteenth term
How to reach the thirteenth term
 
Median simply means that you find value which is in the middle of the list of values written in increasing order.
 
Could someone help me please
Yes but here
Its a different case
I have created the cumulative frequency table ...now what to do?
 
@Abcd You have 3-times 1, 6-times 2; which means that 9-th number is 2.
And 10-th, 11-th, 12-th, 13-th number is 3.
 
Its not the 9th number we need ... We need 13th number
Right?
That is the middle value
 
Basically 3+6+4=13 says that the block consisting of three's ends at position 13.
 
8:41 AM
Calculate cf
x f cf
0 0 0
1 3 3
2 6 9
3 4 13
4 7 20
5 5 25
 
cumulative distribution function?
 
f stands for frequency
 
oh nevermind
 
In other words this says that you have 1,1,1,2,2,2,2,2,2,3,3,3,3,....
 
cf stands for cumulative frequency
@MartinSleziak How to find from the table ???
WHich is the thirteenth term>
ACCORDING to the table>/
?*
 
8:43 AM
BTW you could also have a look at some similar questions from the past: google.com/…
@Abcd As I wrote above: You just notice that 3+6+4=13. So 13-th term is 3.
If fact you have already calculated that 13 in the last column.
 
hi guys
 
Ok, I will have to leave - I have some duties in the off-line world. Have a nice day!
 
hi dydxx
 
Okay
Thanks
hi @dydxx
 
do any of you guys have a strategy for finding 'nth' deriviatives?
 
8:45 AM
See if they oscillate
 
i always have trouble putting into a series form with factorials and stufff
 
successive derivatives of f(x) = sin(x) wrap back around after 4 times e.g.
 
for example y=x^2*ln(x+1)
 
Yeah, I'd probably do a few by hand
Hopefully a discernible pattern emerges, but it's not guaranteed.
 
if it's to compute a taylor series, I'd rather manipulate taylor series themselves
 
8:47 AM
its not to compute a taylor series , the question asks me to find the nth derivative of the function y=x^2ln(x+1)
 
ah then compute the first few derivatives until some order jumps at you
 
I found y'=2xln(x+1) x^2/(x+1) and y''=2ln(x+1)+4x/(x+1) - x^2/(x+1)^2
 
do a few more
you'll see it won't get much more complicated than $y''$
 
and y'''=6/(x+1) - 6x/(x+1)^2 + 2x^2/(x+1)^3
i struggly heavily finding patterns
 
and the next one ?
 
8:52 AM
(2(2x+3))/(x+1)^3 - (6(x^2+3x+3))/(x+1)^4
 
what
 
that was the fourth derivative
 
why can't you simplify it like the others ?
also at one point you might want to make a change of variables like u = x+1
 
u-1=x?
 
yes ?
 
9:16 AM
-im still extremely lost
 
can you show how you got such a garbled up expression for $y^{(4)}$
 
Greatest mathematician
Today is his birthday
 
: O
 
9:31 AM
In India today we have national mathematics day !
 
: O !!
 
9:48 AM
To find mean by Step Deviation method when data is in inclusive form do we need to convert it into exclusive form ?
 
10:00 AM
@MartinSleziak to what are you referring?
 
There is a cool exercice here : math.stackexchange.com/questions/2068339/… :)
 
I don't think that a compact cube can be covered by countably many translates of squares
 
@Ramanujan I understand the reference, but that does not explain why Martin said that to me.
 
11:17 AM
How can a dot in R^2 be uncountable?
correction: in R
 
I have no idea what that means
 
well, the basis of R\Q is a dot
 
$R^n$ is uncountable
 
The basis or R as a Q-vector space has uncountably many elements, I'm not sure what you mean with a dot
 
@AlessandroCodenotti the elements in the basis can be made 0 distance away from origin
 
11:26 AM
What does it mean to have 0 distance from the origin? How is that relevant
 
well because of dense?
any element in the basis can be made infinitely close to the origin
 
Define "distance from the origin", I don't understand what you're talking about, in my definition of distance there's a single point with distance 0 from 0
Ok, now that makes sense
I still don't know what a basis being a dot means
 
if you have a set in which every element is infinitely close to the origin
then isnt it a dot?
 
You can make them arbitrarily close, but if you fix a basis every element will have a positive distance from the origin
 
11:43 AM
Q is neither open nor closed?
 
I mean $(0,\epsilon)$ and $(\epsilon,0)$ is a basis of $\mathbb R^2$ as a vector space over $\mathbb R$ for every $\epsilon$, I have no idea what's this dot thing you're talking about
In $\mathbb R$ with the usual topology? Then it's neither
 
@AlessandroCodenotti but in the meantime i can fix the inf of the basis as 0 and shrink the sup indefinitely
 
See what I wrote above about $\Bbb R^2$
 
I can define a sequence of basis b_n such that lim(sup b_n - inf b_n) = 0
my mind is failing to picture how Q can be neither closed nor open @AlessandroCodenotti
 
@robjohn I am referring to Main Chatroom Etiquette Rules. And since this seems to be useful advice, I was wondering whether adding "just ask; don't ask" maxim to the guidelines would be helpful.
But on second thought, it is more of an advice/guideline than etiquette.
 
11:49 AM
Sure, you can do that even for $\Bbb R$ as a vector space over itself, I don't know what you're trying to do
 
I think he isn't talking about vector spaces at all ?
 
By "just ask; don't ask to ask" I mean that it is better to ask the question than asking: "Hey guys, may I ask a question here?"
But since it is mentioned in the room description and many people seem to ignore that, adding it there probably would not help at all. (Even if it were a suitable addition - it probably isn't.)
 
Does every poin in $\mathbb Q$ have an open ball around it entirely contained in $\mathbb Q$? No, so it's not open
Does $\mathbb Q$ contain all of its accumulation points? No, so it's not closed
 
I know, just that I cannot build a mental picture
 
@Ramanujan I didn't realize this was even a thing :o
 
12:09 PM
I know, I mean I cannot picture it mentally
 
It's a bit abstract. I don't really enjoy that stuff.
 
@AlessandroCodenotti In general a subset of a metric space, can be either open, closed, both open and closed, or neither open nor closed, correct?
 
but no non-trivial clopen sets in R
 
isn't it clopen closed and open ?
 
12:11 PM
That's because R is connected
 
real analysis...so harrrrrrrrrrrrrrd
 
Haven't you just said the same thing twice, @usukidoll?
 
it's after 2 in the morning
I'm dsalfjdslk32084092
 
@usukidoll, Everything is hard, until it's easy
 
:/
 
12:14 PM
@usukidoll, I'm just kidding, that some quote from some famous person. Once you get the hang of Real Analysis, you'll see that most of it (at the elementary level) is pretty easy
 
x_X what about senior undergraduate level real analysis
 
$\forall x \in \varnothing$, $p(x)$ is always true.
 
vacuous truth
 
@usukidoll, "Senior Undergraduate Level" - That means different things to students of different universities. Are you doing Measure Theory in your course?
 
Yep, @Perturbative
I keep forgetting my ruler though
 
12:16 PM
not that I know of c.x
 
@Kari, I don't mean actual Measure Theory for the sake of Measure Theory, I mean't has he/she come across things like the Lebesgue Measure yet..
 
Oh, haven't we all at some point?
 
[0,1) is neither open nor closed?
 
Look at $\Bbb{B}(0, r)$ for being open (or not).
 
12:19 PM
@AlessandroCodenotti and it is also half-open
 
Consider the limit as $n \to \infty$ of $1-1/n$ for being closed (or not).
"In a topological space, a set is closed if and only if it coincides with its closure. Equivalently, a set is closed if and only if it contains all of its limit points." - @DHMO is a very useful condition
 
thanks
 
Inner Product $\subsetneq$ Normed $\subsetneq$ Metric $\subsetneq$ Hausdorff $\subsetneq$ Topological
That's a useful nesting of types of spaces
You'll study it a lot if you do more analysis
The more conditions you attach to a space, the further left your space goes in the nesting
 
It's important to notice that in general sequences are not enough to get all limit points though
 
@Kari I don't follow..
 
12:23 PM
Your ruler has a built-in Lebesgue measure
We've all measured intervals at some point :-b
 
Oh, lol
 
Oh, do you have a counter example, @Alessandro?
(I'm a bit rusty with this stuff)
 
$[0,$\omega_1]$ with the order topology is the standard example but it's not really intuitive if you're not familiar with ordinals
 
Yep, I have no idea
 
There are a few more examples here
 
12:31 PM
This is interesting stuff
 
@Perturbative so you are Indian?
 
12:44 PM
Sequences are enough in first countable spaces though @Kari (all metric spaces in particular)
 
First countable?
 
Every point has a countable local basis
 
Be surprised to learn that the definitions of convex (real) functions do not use limits/derivatives/calculus at all
 
Oh right, I didn't know that was a thing
Took it for granted with open balls and stuff in metric spaces
 
That is for every point $x$ there exist a set $\{B_n\}_{n\in\mathbb{N}$ of nbhs of $x$ such that for every nbhd $N$ of $x$ there is an $n$ with $B_n\subseteq N$
Yeah, in metric spaces you can just use the open balls with radius 1/n
 
12:48 PM
Like $\Bbb{B}(x, 1/n)$ in $\Bbb{R}^n$ or something
Yea, that's the one
 
There are first countable but not metrizable spaces though, like $\{a,b\}$ with the trivial topology I think
 
So the set of neighbourhoods is just $\{ a, b \}$?
(If I remember correctly that the trivial topology is just the empty set and whole space $X$)
 
can convex real functions be bounded?
 
Yes @kari
 
Bounded below probably, @DHMO
 
12:54 PM
@Kari what about above?
 
Restrict the domain then sure?
 
...
 
Something like $f(x) = x^2$ for $x \in (0,1)$ is certainly bounded
 
I'm talking about convex real functions whose domain is the whole real line
can they be bounded above
 
Oh ok
 
12:55 PM
do I really need to specify every single damn thing
 
Mathematicians wanna be free to do whatever they want. You gotta add restrictions
Do you consider $f(x) = 1$ to be convex, @DHMO?
 
cfp
Any ideas or comments on this question: math.stackexchange.com/questions/2066116/…
 
@Kari strictly convex
 
Oh, ok
In that case, I can't see it happening
 
[ramblings]
xe=e
ey=y
xy=a

y=ey=xey=xy=a
 
1:00 PM
@Secret ey, associativity assumed
 
 
@Secret a=e?
 
@Kari thanks
@Ramanujan I thought a=y
 
@DHMO and x=?
 
@Ramanujan x is unique
a,e,x,y are just elements of his new set
and the first three lines are his newly defined multiplication rules
 
1:05 PM
Is it from functions?
 
yeah, just randomly tossing stuff around. In order to do the analysis correctly I need to learn those theorems from universal algebra better
 
@Ramanujan no
 
but what's yx ?
 
well, given the rules here, it is not necessary defined
 
not defined
 
1:07 PM
(as mentioned, I am randomly tossing stuff around to chill off my brain from what is essentially a tedious combinitorics exercise)
 
1:30 PM
makes the following sense? for all elements that are not 0 the following holds: if a is a multiple of b. and b a multiple of c. then a is a multiple of c.
 
Yep, transitivity is a fancy word for what you described
 
mmh, but transitivity for what?
 
The relation $a \sim b$ defined by $a$ being a multiple of $b$ i.e. there's some $k$ for which $a = kb$, @Null.
In this case, $a \sim b$ and $b \sim c \implies a \sim c$.
 
@Null I don't think the restriction regarding 0 is necessary
 
it is, at least for b
was lazy :d
 
1:32 PM
All users, in the last post of Terence Tao, What's new, he tell us that the AMS has a repository of open-access mathematics lecture notes. Good afternoon.
 
a is a multiple of b <=> a = kb
b is a multiple of c <=> b = mc
therefore a = kmc <=> a is a multiple of c
 
$ a = k_1 b = k_1 (k_2 c) = (k_1 k_2) c $ i.e $ a \sim c$ with $k = k_1 k_2$, @Null.
 
@Null for example?
 
(or exclude 0 as a factor)
 
Do you have a link, @user243301?
 
1:33 PM
@DHMO ah right. it is not neccessary
altho, without 0, it is a slightly stronger statement
 
but the proof is the same
 
yep
 
How on earth should I prove that $\displaystyle\frac{1}{\ln x}+\log _x\left(\ln x\right)-1 \le 0$?
 
ok, wikipedia's article on variety (algebra) and universal algebra are not useful. Other than Brihckoff theorem and the use of signatures to characterise an algebra structure, I don't see anything else that can be used,

Perhaps that means I am still not at the level to understand fully the stuff, and/or that I need to find an actual text on universal algebra to cover all the details
Wikipedia is good for introducing the topic in question, but to go deeper often books are needed.
Which makes me wonder why for group theory, wikipedia's seemed to be quite self contained
(probably because it has such a long history and such a popular branch of abstract algebra)
 
1:50 PM
@Kari the link is in the last post of Terence Tao Blog What's New, seems that is a project that started, and to be continue in the next future.
 
I ended up finding it pretty quickly!
Yea, it seems to be an ongoing collation of lecture notes which is nice
 
2:26 PM
Hello!! I am looking at the following:

Let E/F an extension, S = {α1, . . . , αn} ⊆ E algebraically independent over F
and T ⊇ S a subset of E, that spans E algebraically over F.
Show that there exists a set B between S and T, that is a trancendental basis of E/F, as follows:
Let T\S = {β1, . . . , βm}.
If T = ∅, then B = S is the trancendental basis.
Otherwise, we define S_0 = S and for i = 1, . . . , m,
Show that B = S_m is a trancendental basis of E/F.
I don't understand the case T = ∅. We have that S is a subset of T. When T is the empty set, how can S be non-empty?
 
Where's the stipulation that $S \neq \varnothing$, @MaryStar?
Doesn't $T = \varnothing \implies S = \varnothing = B$ already deal with the case where $S$ is empty?
 
What would C/~ where a~b <=> a^4=b^4 look like?
 
is $C$ the set of complex numbers ?
 
yes
 
then it looks just like $C$
 
2:35 PM
I thought it would look like the first quadrant of $\Bbb C$
 
The "two edges" in the quadrant piece glue togather though
 
@Kari Ahh... Does the empty set span E algebraically over F ?
 
@BalarkaSen that is right
so C/~ where a~b <=> a^n=b^n where n is natural number basically divides C into sectors
now what the hell would it look like if n is not an integer?
 
the fundamental domain is a sector, sure
 
please don't talk about undefined things
 
2:37 PM
@mercio ??? C is closed under power
 
can someone please verify if I'm wrong here? math.stackexchange.com/questions/2067233/…
 
I'm not sure, @MaryStar
 
no
 
z^n doesn't make sense if n is not integer in general
 
see Mark Hurd's comment
 
2:38 PM
only integer exponents
 
@Kari We have that T ⊇ S a subset of E, that spans E algebraically over F, this holds also when T is the empty set?
 
@BalarkaSen C is closed under power!
 
so what?
 
no
 
so a^n is defined even when n is not an integer
 
2:39 PM
sqrt(z) is not a well-defined thing. you got to choose a "branch"
 
no it's not
 
Hi @Balarka
 
@BalarkaSen doesn't make it "undefined"
 
please provide me with a definition
 
well it is an equivalence class
 
2:39 PM
whom are you quoting? :P
 
@BalarkaSen not you
 
you just said that C was closed under power
 
Hi @Alessandro
 
yes it is
 
and elements of C are not equivalent classes
 
2:41 PM
@DHMO Either you choose a branch for z^(1/n) or that doesn't make sense.
 
@BalarkaSen sure
 
what's $\sqrt{1+i}$ ?
 
@mercio $\sqrt{1+i} = \sqrt{\sqrt2\exp(i\pi/4)} = 2^{1/4}\exp(i\pi/8)$
 
whats' $\sqrt{-1+i/100}$ ?
 
come on
 
2:44 PM
wait no
I have a better question
 
@mercio i
 
Depends what you mean with square root.
 
@Ramanujan no
 
what's $\sqrt{\sqrt 2 \exp(9i\pi/4)}$ ?
 
@mercio yes this is a better question
that demonstrates the need for branch
 
2:45 PM
no that demonstrates the need for not writing it EVER
 
First one also has multiple roots....
 
:34286276 we're asking about sqrt(1+i) not (1+i)^2
 
I'm asking you to provide a definition of $\sqrt z$
 
There isn't any proper definition.
 
then don't talk about it
 
2:47 PM
$\sqrt{z} = \{x\in\mathbb{C}|x^2=z\}$
 
On the reals, we agree that $\sqrt{x}$ means the "positive" root and $-\sqrt{x}$ the negative one.
But on the complex numbers, when you have two roots, you can't name one positive and negative.
 
@mercio Proofwiki provides a branch
 
you can define it using $\exp(\frac{1}{2}\ln(z))$ where you take the principal branch of the complex logarithm. Does that not work?
 
yes that works
 
then $\sqrt a = \sqrt b$ is equivalent to $a=b$ so your equivalence relaion is... not very useful
(and all of those are still ugly)
 
2:49 PM
Sure, but then you have to agree what the principal branch is first
 
I don't agree about any branch
 
Then you can't do any complex analysis or any sort of mathematics on Riemann surfaces.
 
I can
 
This became a rather tedious discussion
3
 
Yup.
 
2:50 PM
sorry
 
Good morning chat.
 
morning chat
 
Good morning cat.
 
good morning
 
@Fargle @DHMO @BalarkaSen hi
 
2:51 PM
hai
 
Hi @Adeek, @Fargle
 
How goes it everyone?
 
@Fargle I am solving allufi
 
Watching anime, @Fargle. You working?
 
@Kari Not at the moment.
And that sounds fun, @Adeek.
 
2:55 PM
yeah it is quite fun @Fargle that book is awesome no wonder people gave it 5 stars
it is much better than Dummit and foote.
 
Aluffi's algebra?
 
I guess if you have the right muturity
@Kari yeah
also the algebra flavoured with category theory can be abstract not so much though if your used to things.
 
Algebra: Just add water.
 
3:09 PM
Hey! Landed
 
You've landed, @Akiva?
Also, is it you who Ted keeps calling DogAteMy?
 
I'm in Israel now!
Yes
 
Oho, for a Christmas break?
 
oy vey oy gevalt
 
Yeah! (Well, Chanukkah, but same thing)
 
3:11 PM
Chrismukkah, @Akiva :-)
 
Wi-Fi connection is spotty, by the way
 
@AkivaWeinberger how is the weather in israel
such amazing performance
 
@robjohn Only now I noticed that you asked why I addressed that message to you. Well, the post on rules of this chat room has your name under it and you are one of the room owners. So I suppose you maintain that post.
 
3:44 PM
honestly the guidelines could use a rehaul
I floated the idea like a year ago but then never got around to it
 
hi frands
 
@GFauxPas thanks for the edit lol
 

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