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00:00
It’s wrong.
@TedShifrin. I appreciate feedback. Can you please hint where?
What $f$ do you have in mind?
It is confusing to read, but you just say that the final statement doesn’t hold. Where is your justification?
@TedShifrin. So, you are saying based on $f$ might and might not hold, so saying that $c$ might hold and might not. I got that. I should rather give an example.
What condition on $f$ is necessary for $f=O(f^2)$ to hold? Write actual mathematics.
@TedShifrin. Thanks. I will just give an counterexample. I gave very broad statement, which is wrong.
00:17
@TedShifrin. It depends on the function $f(n)$, so if take a counter example where $f(n)$ takes values in $ [0,1]$, then the statement is wrong. So it does not hold $\forall n\ge n_0$.
00:33
@Avra I commented, but my browser screwed up and I ended up editing your post rather than the comment. I fixed it all.
@Avra That is not all that is needed.
@robjohn. Thank you. That's fine. I am still young with posting answers, so I have long time to go to learn.
You need that $f(n)$ vanishes as $n\to\infty$.
$f(n)=1/x$ and $f(n)^2 = 1/x^2$ would work I guess
if we took the limit, we would have $\infty$
meaning that $f(n) $ is $\omega f(n)^2$?
$\lim_{n\to \infty} \frac{\frac{1}{x}}{\frac{1}{x^2}}$
that is the counterexample given in the other answer
Sorry!
 
1 hour later…
02:05
Imagine if english was phonetically consistent
02:20
angliche iz ghonetikallie konsistent
03:11
i love phonetically consistent langs
03:40
name the 3 users of the holy trinity of this room
i guess anybody can do that
 
2 hours later…
05:56
0
Q: Cardinality of the set $S=\{f:\mathbb R\to \mathbb R: f \text{ is no where continuous}\}$

KoroCardinality of the set $S=\{f:\mathbb R\to \mathbb R: f \text{ is no where continuous}\}$ Clearly $g:\mathbb R\to \mathbb R$ defined as $$ f(x)=\begin{cases}1; x\in \mathbb Q\\ 0; x\in\mathbb R\setminus\mathbb Q\end{cases}$$ belongs to $S$.Let $L=\{(rg):\mathbb R\to \mathbb R:r\in\mathbb R\setmin...

Apart from saying that card(S) is at least equal to continuum, what more can be said about cardinality of $S$?
Two special lectures will be open in the next semester. One is history of math which would be easy to study and but not much to gain. And the other one is topological methods in combinatorics which would be hard but sounds interesting. Which class would you take?
06:11
@Koro What’s the cardinality of all functions?
Hi @Ted :)
Did you mean "cardinality of all functions from $\mathbb R$ to $\mathbb R$"?
If yes then that would be $2^c (\gt c)$, where $c=$ the continuum
$\mathbb R^{\mathbb R}$ has cardinality $c^c$ which is equal to $2^c$ by Cardinal arithmetic.
06:46
So pretty unlikely your answer is right. This is actually a Baire category theorem application. Being continuous somewhere is a very “thin” condition.
my answer is incomplete. apparently more can be said about card (S)
card (S)= c as the answer to my post suggests, I am trying to fully understand the answer though
*card(S)=$2^c$
 
2 hours later…
08:50
@Koro Instead of $0$ for $x\in\mathbb{R}\setminus\mathbb{Q}$, map them to any point in $[-1,0]$, that gives $2^c$
yes!!
because set of functions from set of irrationals to set of irrationals has cardinality $2^c$?
09:05
@Koro actually, I was thinking the set of functions from $\mathbb{R}\setminus\mathbb{Q}$ to $[-1,0]$ is $2^c$, but either way
oh, that's what you said
I just misread
:)
 
3 hours later…
 
2 hours later…
13:23
Click bait I guess. The thumbnail is misleading.
>:(
13:47
lol
@CowperKettle YT really thinks I will like this video
but I refuse to click on it
how to derive $\frac{d}{db} ((\vec{a} \times \vec{b} ). (\vec{a} \times \vec{b} ) ) $ no rigor needed, its for a physics lecture (Cross product between a and b and dot product between () ()
14:02
I cannot parse that statement
by derive, do you mean differentiate
and what does d/db mean
what is b
14:24
"no rigor needed, its for a physics lecture" i think you're repeating yourself
(just kidding)
👍
Yea sorry i meant differentiate. its the partial differential to b, i thought that was clear?
Balarka sen, i agree, i will take all the trash talk that comes towards how physicists deal with mathematics, they deserve it, quite honestly its shameful.
@RyanUnger its a notation used, basically the gradient of that vector b
@MadSpaces See the Binet-Cauchy identity here.
Or here in more detail.
Hi @TeresaLisbon!
@BalarkaSen Hello!
14:34
Thanks ! i will check it out. By the way have you written a book? i think i read your name onc
Or did you appear in some math video on youtube
@MadSpaces No, not quite.
Ok i mixed you up then wtith another Lisbon.
Possibly!
15:32
people mix me up with other Towneses too.
15:49
does $$ \frac{r^n+1}{r+1}~ \forall~ r \in N, n \in 2m~ where~ m \in N$$ have a natural solution?
what do you mean by solution? is the question whether there is a natural number of that form?
what if the limit of ratio of two different sequence is 0 or negative or infinity in limit comparision test
I think it works of negative limit same as for original limit comparison test
there are a lot of solutions with r = 1 so i guess the interesting problem is if r is bigger than 1.
@leslietownes r is natural
sorry I see what you mean
yes I am interested in solutions for r larger than 1
16:01
Oh I forgot that terms of sums which are strictly positive doesn't result in negative limit lol
@leslietownes How many people are there in Townesia?
@EvilJohnRennie Except $1+2+3+4+\cdots=-\frac1{12}$ ;-)
i don't think there are solutions. is it possible that r^n + 1 = 2 mod r+1 whenever r > 1?
@robjohn wtf 0_0
The infinite series whose terms are the natural numbers 1 + 2 + 3 + 4 + ⋯ is a divergent series. The nth partial sum of the series is the triangular number ∑ k = 1 n k = n ( n + 1 ) 2 , {\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},} which increases without...
Is It related with rearrangement
16:05
It is a regularization of a divergent series. It is also the value of $\zeta(-1)$
@EvilJohnRennie No, not at all related to rearrangement since the terms are all positive
I keep forgetting the condition of theorems :-P
@robjohn I don't know anything about zeta stuff
consider that $\frac{r^n-1}{r+1}$ is always integral if $n$ is even
but n(n+1)/2 to infty as n infty
$\frac1{(1-x)^2}=1+2x+3x^2+4x^3+\cdots$ and letting $x\to-1$ gives $1-2+3-4+\cdots=\frac14$ (illegally)
The series 1-1+1-1+1-1+... has a name. Why?
16:13
then noting that $1+\color{#C00}{2}+3+\color{#C00}{4}+\cdots-\color{#C00}{(4+8+12+16+\cdots)}=1-2+3-4+\cdots=-3(1+2+3+4+\cdots)$
In mathematics, the infinite series 1 − 1 + 1 − 1 + ⋯, also written ∑ n = 0 ∞ ( − 1 ) n {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}} is sometimes called Grandi's series, after Italian mathematician, philosopher, and priest Guido Grandi, who gave a memorable treatment of the series in 1703. It is a divergent series, meaning that it lacks a sum in the usual sense...
then we get $1+2+3+4+\cdots=-\frac1{12}$ without zeta regularization
not legally of course
koro, i've never heard it called that. this might be an example of someone putting a very uncommon usage on wikipedia.
I never really understood what it's all about the divergent series 1+2+... being equal to -1/12
the sum is wrong as if it were true then nth term that is n would approach 0 as n approaches infty
@Thorgott hmm that is true. That is provable by Binomial Theorem
16:18
it's a direct exercise in factorization
hi leslie
hi thorgott
@leslietownes I've seen it before
the name, that is
@Koro hi
but how do I go about for the other situation?
hi robjohn
ok, so it's not entirely made up. some of the sources do refer to it that way.
16:20
yeah but why does this series have a special name? I mean it's divergent !
and its partial sums are bounded
what else
i'm with you. just because something can be associated with a person doesn't mean it needs to be named after that person.
@robjohn how do you get 4+8+12+16+...'s value?
the wikipedia stuff on this topic seems to focus to an unusual degree on controversy among these people. some historians love that but it is useless for mathematics.
== Geometry and infinite zeros == === Grandi === Guido Grandi (1671–1742) reportedly provided a simplistic account of the series in 1703. He noticed that inserting parentheses into 1 − 1 + 1 − 1 + · · · produced varying results: either ( 1 − 1 ) + ( 1 − 1 ) + ⋯ = 0 {\displaystyle (1-1)+(1-1)+\cdots =0} or 1 + ( − 1 + 1 ) + ( − 1...
@napstablook what other situation?
16:22
the history of math does include people who were using different definitions, or no definitions, getting mad at each other. i just don't know why i ought to care about it. they used to burn people for being "witches." not a useful chapter in history.
i think that the series got its name before Riemann theorem on rearrangement
Guido Grandi is a cool name. i wish i had that name.
you can solve that using what I said
oh that seems to make sense thanks @Thorgott
16:25
so much is associated with history of Grandi series : en.wikipedia.org/wiki/Witch_of_Agnesi
@robjohn bruh but that is false :-P
May be I will meet the wierd paradox in future. For now I will not be obsessed with stuff that doesn't make sense to me lol.
16:46
A person is saying that evolution is a tool of white supremacists
Creationists are weird
if a_n approach 0 and b_n approach 0 as n approach infty does a_n/b_n approach 1?
o.9
o.9
no @EvilJohnRennie
o.9
o.9
let $a_n = \alpha b_n$
it will converge to $\alpha$
It needn't approach anything at all. Make up a bunch of examples for yourself.
o.9
o.9
16:51
sure that can also happen
or you can have $b_n$ be $0$ and then it's tragic
Hint: Take any bounded oscillating function $f(n)$ that does whatever it wants.
o.9
o.9
You can also first select $a_n /b_n$ first and make $b_n$ go to $0$ fast enough to that $a_n$ also does
oh I state it in wrong way again. If a_n approach l and b_n approach l as n approach infty then is a_n/b_n>0.
o.9
o.9
lol
I think you're following up on what I said, @o.9, but disguising it entirely.
o.9
o.9
16:53
sounds good @TedShifrin
sorry I am bit drunk today
@EvilJohnRennie Yes, you totally bungled it. Do you mean is $a_n/b_n>0$ for all large $n$?
o.9
o.9
Is that like ur homework or something
@TedShifrin yup
@o.9 nope
I just observe bunch of these stuff being as this
So what does $a_n/b_n$ approach as $n\to\infty$?
o.9
o.9
16:56
I'm confused
oh and a_n/b_n is real number
o.9
o.9
can't $l$ be $0$?
l is real number
I was thinking about l=0
o.9
o.9
sounds good
So @o.9 is jumping ahead. If $l\ne 0$, answer my question.
16:56
l=real number is obvious except for 0
o.9
o.9
Maybe I need to be drunk also
OK, then when $l=0$, we are back to what we were discussing before, exactly.
So go reread what we said.
Maybe wait 'til you're sober.
17:00
Write a paper when you are high
o.9
o.9
roll paper to get high
$a_n=b_n=0 $as $n\to infty$ imply $a_n/b_n>0$ for $n\in\Bbb{N}$ and $a_n/b_n\in\Bbb{R}$
There are many contradicting examples to this
This is hopelesssssssss ...
o.9
o.9
17:02
I've been getting trolled all day today :'(
Do you deserve a troll badge, @o.9?
I was this j^2/(j^3+1)/1/j goes whoos.. to 1 as j tries hugs infinity
o.9
o.9
the troll badge goes to the one that gets trolled?
The best way to counter a troll is 'no u'
You can decide what the badge means once you steal it.
17:04
and one (6+5j)/(3j^2+1)^2/1/j^3 goes to 5/9 too
also sin(1/j)/j---》1
@Evil You of course did not read what I told you earlier. Your examples are not the general case. Take $a_n = (-1)^n/n$ and $b_n = 1/n$. What is $a_n/b_n$? Or take $a_n=\sin(n)/n$.
@EvilJohnRennie Looks like garbage to me.
oops sorry
17:24
@leslietownes I have heard it called Grandi's series, but these being of limited use, it doesn't get as much press.
@EvilJohnRennie $\lim\limits_{j\to\infty}\frac{\sin(1/j)}j=0$. Did you mean $\lim\limits_{j\to\infty}j\sin(1/j)=1$
i'll start calling it Guido's series.
@TedShifrin I often feel that way.
Yes, you're usually more patient than I.
I answered a question about little-o a while ago that I wish I hadn't.
@TedShifrin You've had to put up with stuff while teaching for a lot longer. Other than online, I haven't really taught since 1998
Well, yeah, I taught 40 years in the classroom.
17:36
that will use up a lot of the patience store
Speaking of lack of patience, we wish @copper.hat bon voyage and safe travels.
Is he on his way to LOB now?
Land of Birth
oh, that was today? best of luck.
yeah, good luck
@robjohn oh yeah
good night
17:45
Yes, today was the duly appointed day.
@EvilJohnRennie later
 
2 hours later…
19:54
Compartmental models are a very general modelling technique. They are often applied to the mathematical modelling of infectious diseases. The population is assigned to compartments with labels – for example, S, I, or R, (Susceptible, Infectious, or Recovered). People may progress between compartments. The order of the labels usually shows the flow patterns between the compartments; for example SEIS means susceptible, exposed, infectious, then susceptible again. The origin of such models is the early 20th century, with important works being that of Ross in 1916, Ross and Hudson in 1917, Kermack...
In 2014, Harko and coauthors derived an exact so-called analytical solution (involving an integral that can only be calculated numerically) to the SIR model
What is the meaning of 'can only be calculated numerically'? Solvable only via computer?
20:20
Have you heard, for example, of Simpson’s rule?
I've heard of it in calculus class.
20:42
That's a numerical means of evaluating integrals for which there is no antiderivative in elementary terms.
It was done long before computers.
hoo boy.
Is that your way of critiquing my behavior? :D
If someone doesn't agree with what you say, say it again louder and louder ...
That means you're right.
no critique from me.
just the usual internet goofiness.
I knew you would not want to agree with me, as a matter of principle. :D
 
3 hours later…
23:27
@TedShifrin WELL, OF COURSE IT DOES!
23:39
Isn't it redundant in the definition to state that it must be a bijection since for the inverse to exist it must be?
definition of homeomorphism that is
depends on how you set it up. for some, injections have inverses too.
I like to say bijection with both directions being continuous.
Or smooth for diffeo.
continuous linear bijections have bounded inverses if the space is nice enough, but not always.
just throwing that out there.
ah yes, the forgetful functor from the category of Banach spaces with bounded linear maps to the category of sets reflects isomorphisms
yeah, you need something like completeness.
i had a really cool result once and then it turned out that the space i was using wasn't complete, so all the cool theorems went away. that was a bummer.
23:57
Greetings all. I need some assistance on (ii). In order to show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic. I attempted to factorise the cubic. If $x=1$ is a solution, then the cubic may be expressed as follows with no remainder. $x^3+2bx^2-a^2x-b^2 \equiv (x-1)(Ax^2+Bx+C)$. Comparing coefficients, $A = 1, C = b^2, B = a^2+b^2$ $\therefore (x-1)(x^2+(a^2+b^2)x+b^2)$. Finally, if we sub $x=1$ into the resulting quadratic, we get $1+a^2+b^2+b^2 > 0$
Why is my solution incorrect?

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