Mathematics - 2021-07-22 (page 1 of 2)

copper.hat

12:04 AM

does fractional calculus have any current day application?

Ted Shifrin

Leslie will opt for fractious calculus.

copper.hat

fractious captious calculus maybe

Ted Shifrin

Captivating capriciousness

copper.hat

i am the very model of a modern major generaliser

Ted Shifrin

12:25 AM

Love G&S

copper.hat

many years ago my youngest brother was singing the pirates of penzance and i commented on how i was impressed by how erudite he was and he responded something along the lines of "oh yeah, i heard it on the simpsons". (to be fair, i think he was 13 at the time.)

Ted Shifrin

Typical. Just like so many people encountered classical music only on the Sunday cartoons.

leslie townes

12:51 AM

oh there's an enormous amount of G&S on the simpsons.

often in connection with sideshow bob.

copper.hat

Groening is a pretty talented fellow. Took me a while to appreciate.

Koro

@robjohn: for yesterday’s question on finding bijections, I thought of this way to skip rearrangement inequality:

leslie townes

mike judge is another one. king of the hill is full of wonderful stuff that hits in unexpected places.

groening gave a really nice blurb for my step sisters book. cool guy

when i was in grad school someone started a seminar called Many Cheerful Facts which always began with a taped recording of the line from G&S about that. i thought the tape was a little much but it was a nice seminar.

what's a tape, my daughter would ask.

Koro

Oh no, please ignore that needs more work. Argh

Amethyst Wizard

1:13 AM

1+1 = 1+1

copper.hat

king of the hill has its moments :-)

@AmethystWizard are you sure about that?

Amethyst Wizard

80% confidence

some people say 1+1 = 2

it has me shook

I could have sworn, 1+1 = 1+1

Ted Shifrin

1:32 AM

Not in today’s alternative US.

leslie townes

again.

Bob

2:01 AM

I have updated my post again. If anybody cares to look:

2

Q: Finding a best fit second order polynomial

Problem: Assume we have the following points: $(x_0,y_0), (x_1,y_1), (x_2,y_2), (x_3,y_3)$ where $x_0 = -3$, $x_1 = -2$, $x_2 = -1$ and $x_3 = 0$. Given the function $f(x) = Ax^2 + Bx + C$ find the constants $A$,$B$ and $C$ such that $f(0) = y_3$ and $$d = \sum_{i = 0}^{2} (f(x_i) - y_{i})^2$$ is...

not much chatting

good night

russian bot

6:06 AM

leslie townes

that is an image of the foot of richard astley.

copper.hat

not grandad?

leslie townes

grandad is depicted elsewhere.

copper.hat

6:21 AM

not part of the white shoe legal team

leslie townes

that's way more of a british thing than us. i mostly see brown and black shoes.

the white pants, well, we all wear those.

copper.hat

a few upscale ny legal teams such as wachtell et al where known as white shoe operations.

leslie townes

i have a good pair of shoes that i think the trade would refer to via the poetic name of oxblood shoes.

wachtell is something else.

copper.hat

i like nice (not necessarily expensive) shoes

allen edmonds, and the like.

leslie townes

i am exactly the same.

copper.hat

6:24 AM

unfortunately my cobbler (known locally as the shoe nazi) has retired

leslie townes

no ferragamos.

copper.hat

thats for those mafia folks

i like understated elegance

the jensen interceptor instead of the flambo lambo

leslie townes

i had a really great night with a lawyer from wachtell once, we had the same cultural touchstones and maybe 1.5x the advisable amount of alcohol.

they say they don't do cookie cutter stuff but a lot of their transactions are just, let's do this over again.

copper.hat

they certainly charge enough. a bit full of themselves

leslie townes

they have a phenomenal complex about how much they are worth.

shintuku

6:27 AM

i got $y = \pm \sqrt{e^{-x^2}}$, and I know it passes through $(0,1)$. How come I know the $\pm$ becomes strictly positive?

copper.hat

the shoe nazi was near semifreddies off kensington circle

i liked him, he was yorkshire i think, no airs & graces

just straight up. that got to a lot of local folks who wanted to be appreciated

leslie townes

northerners generally get it

copper.hat

@shintuku there must be more context, otherwise it is impossible to know what you are trying to do

shintuku

I'm solving a differential equation, I start with $y\frac{dy}{dx} = -xe^{-x^2}$ with the knowledge that if $x=0$, $y=1$. I get to $\frac{y^2}{2}=\frac{1}{2}e^{-x^2}$

from which I get $y = \pm \sqrt{e^{-x^2}}$

apparently I should know that $y = \sqrt{e^{-x^2}}$, i.e. we can infer that we can do without the $\pm$

copper.hat

the solution is locally unique, and passes through $(0,1)$ so that sort of eliminates the other solution...

odes are full of technicalities.

shintuku

6:38 AM

hm, but why couldn't the function switch to $-\sqrt{etc}$ away from $(0,1)$?

copper.hat

the solution is continuous

Koro

you should also mention the region (like rectangular etc.) over which you are solving the ode

copper.hat

the solution is defined & continuous (differentiable even) for all $x \ge 0$, but you need to reason that.

it is not too hard to show that $y(x) \downarrow 0$.

shintuku

hm I haven't learned this, i'll have to do some reading

copper.hat

i don't know what your context is, so it is hard to help. are you familiar with the usual existence & uniqueness theorem for Lipschitz odes?

shintuku

6:43 AM

this was from a problem on khanacademy, and the video skipped over why this was possible heh

copper.hat

there are many varieties of the theorem.

i would guess khanacademy does not delve into the analysis side of things.

shintuku

i'm assuming you can find it in most intro books on odes? I'll go find that out

Koro

4

Q: Proving convergence of a sequence $\{x_{n}\}$ satisfying$|f( x_{n}) |=|f'( x_{n+1}) ||x_{n} -a|$,where $f$ is a function satisfying some conditions.

Theorem: Let $\displaystyle f:[ a,b]\rightarrow \mathbb{R}$ be a function differentiable on $\displaystyle [ a,b]$ such that $\displaystyle f( a) =0$ and that there exists $\displaystyle A\in \mathbb{R}$ such that $\displaystyle |f'( x) |\leq A|f( x) |$ for all $\displaystyle x\in [ a,b]$, then $...

real-analysis sequences-and-series proof-writing

copper.hat

almost certainly.

Koro

@shin, please refer the above

and then refer to exercise no. 27 in chapter 5 of Rudin's

shintuku

6:45 AM

wow thanks a lot for the references, I'll start on that right away

Koro

you'll see one of the sufficient conditions that guarantee uniqueness of ode.

copper.hat

hmm, the relationship to that question is not obvious to me

Koro

I must say @shin, in the link shared above, please refer the theorem which is being proved.

shintuku

@Koro noted, thank you

Koro

based on that theorem, you can get one sufficient condition for uniqueness of solution of $y'=\phi(x,y)$ under certain conditions.

"how?" is precisely why I suggested exercise no. 27 of chapter 5 of Rudin's PMA.

copper.hat

6:49 AM

actually, it is a little simpler here since we know $(y^2-e^{-x^2})' = 0$.

you need to be clear what it means for $x \mapsto y(x)$ to be a solution to the ode.

Koro

@copper, how is the lockdown situation there?

copper.hat

there is no real lockdown here.

there were some restrictions as in using masks, but they have been lifted

Koro

😮

copper.hat

travel to hawaii is a little more restrictive

Koro

i have one confusion: negation of "if p then (q and r)" is "p and not (q and r)" which is equivalent to "p and (not q or not r)" which should (intuitively) equivalent to "p and ((if q then not r) or (if r then not p))"

Am I right?

The confusion arose when I tried to negate the definition of linearly dependent set S in vector space V.

copper.hat

7:06 AM

i would say the negation is $p$ and ( not $q$ or not $r$).

Koro

The definition of linearly dependent (LD) set S in vector space V over field F: S is said to be LD if there exist distinct vectors $v_1,v_2,...,v_k\in S$ and scalers $x_1,x_2,...,x_k\in F$ (not all zero) such that $x_1v_1+x_2v_2+...+x_kv_k=0$

copper.hat

no. any set would be LD with that definition

$v_1 - v_1 = 0$.

i would define LD as not LI

Koro

In Hoffman and Kunze's for example, LD is defined first and then LI is defined as not LD

copper.hat

i missed the distinct

Koro

because I added that after seeing your comment :P

copper.hat

7:10 AM

i am off to sleep shortly, if you have a question...

even if you don't :-)

Koro

So I start negating the definition now: S is said to be LI if for all distinct vectors $v_i$'s in S

and for all scalers $x_i$'s in (all zero) ...

then I am kind of stuck here

copper.hat

why are you making the scalars all zero???

Koro

@copper.hat Good night copper. I'll think more the question meanwhile.

@copper.hat I am negating "not all zero" to all zero. Isn't that right?

copper.hat

it is for all distinct vectors and for all scalars (not all zero) then the sum is non zero.

Koro

ahh

"if p then q" is equivalent logically to "if not q then not p"

so if sum is zero then all scalers are zero

that answers my question

thank you @copper.hat

copper.hat

7:17 AM

roughly speaking the negation of "there exists foo such that bar(foo) is true" is "for all foo then bar(foo) is not true".

the quantifiers are crucial here, it is not a simple logical expression.

Koro

@copper.hat right

Meanwhile, I found this

5

Q: Logical formula of definition of linearly dependent

A subset $S$ of a vector space $V$ is said to be linearly dependent if there exist a finite number of distinct vectors $x_1, \ldots , x_n$ in $S$ and scalars $a_1 , \ldots ,a_n$ not all zero, such that $$ a_1 x_1 + \cdots + a_n x_n =0 $$ I want to translate this definition into l...

linear-algebra predicate-logic

I think that the answer is slightly misleading there. The linear dependence definition is not written correctly (in the notational form).

the definition of LD does not say that all $a_i$'s are non zero

copper.hat

why do they all have to be non zero?

not all zero is not the same as all non zero.

in any event, i need to go to sleep! good luck!

Koro

the definition only says "not all scalers $a_i$'s are zero"

and that's why the answer to this question, is misleading

6 mins ago, by Koro

5

Q: Logical formula of definition of linearly dependent

A subset $S$ of a vector space $V$ is said to be linearly dependent if there exist a finite number of distinct vectors $x_1, \ldots , x_n$ in $S$ and scalars $a_1 , \ldots ,a_n$ not all zero, such that $$ a_1 x_1 + \cdots + a_n x_n =0 $$ I want to translate this definition into l...

linear-algebra predicate-logic

@copper.hat right

@copper.hat good night copper.

robjohn

8:33 AM

@copper.hat that is a triple negative. That has to be against some rule somewhere.

russian bot

3

Adil Mohammed

9:59 AM

@robjohn

i cant find this "conic" tag?

robjohn

that would be because there is no "conic" tag

Does conic-sections not cover the question?

If you are looking for a synonym, look here

Adil Mohammed

10:42 AM

Yeah you are right

oh so they mean "conic section tag is also same as conic tag", my bad I thought it was a typo or some other issue

Koro

1:53 PM

If $|\frac{f^{(n)}(0)}n|\le K$ for all $n$, where $K\gt 0$ is a constant. What can we say about $\lim |\frac{f^{(n)}(0)}{n!}|^{1/n}$?

Argh... The correct hypothesis is $|f^{(n)}(0)|\le K$ for all $n\in \mathbb N$.

If $f(x)=e^x$ then the hypothesis is satisfied with K=2 for example and the limit under question is $0$

Infact, if $\inf_{n\in \mathbb N} |f^{(n)}(0)|\gt 0$ then also I have no problem because in that case I can do this: $|\frac{f^{(n+1)}(0)}{f^{(n)}(0) (n+1)}|\lt s\frac 1{n+1}$ whence ratio's limit is zero.

And then by use of Cauchy theorem, $n$th root also has limit $0$.

The problem arises when $\inf |f^{(n)}(0)|=0$

under this situation, can we say something about the limit in question?

AMDG

2:34 PM

I suppose things on Math SE should be kept more formal, yes, and using formal terminology is agreeable; is making something more succinct such as by using notations and replacing words such as "for all" with $\forall$, however, something that might be more frowned-upon given the Q&A style and for the sake of readability for general audiences who might not properly understand certain notations, even if a link explaining the notation is provided?

For example, I would like to edit this answer to make it a bit clearer and easier to understand, and not everyone who visits SE (myself included) has, y'know, a doctorate in mathematics to be able to understand all terminology and notation. I believe it would be better to instead prefer jargon over notation, and hyperlink certain jargon that might not be so clear to the average reader. Would the regulars in here agree, or what do you suggest?

Also, how would one restate "you can write the number in base $b$ as" more formally? I know what it is in my mind, but I don't know how it would be termed: "there exists a value $x$ that agrees with base $b$:" Something like this, but agrees obviously isn't a very good choice of wording here.

leslie townes

that answer looks fine to me, although i do have a doctorate in mathematics. :) i think that $\forall$ and demon sister $\exists$ should be deprecated outside of purely symbolic mathematics. they don't help. although i wouldn't edit a question to remove them.

AMDG

Hm, ok, so I see we share the same thoughts about $\forall$ and $\exists$.

leslie townes

basically all abbreviations, symbolic and otherwise (e.g. "s.t." and "wlog") are bad to me. like nails on a chalkboard bad. but some people like them.

AMDG

2:49 PM

I mean I believe the whole purpose is to compress a statement due to the frequency of these terms or phrases.

leslie townes

my advisor had a thing against people who used too many greek letters. we're all a little bit crazy.

if that were all they were used for i might be OK with it. i think a lot of people lean on them as a kind of substitute for understanding what they are talking about. if it looks like alien language it must be math even if it isn't organized well.

or they're writing to appease some external authority instead of to convey information.

AMDG

Well my motivation and thoughts on improving this answer are somewhat small little things like changing $\{ 0,1,\dots,b-1\}$ to a closed interval and things like that which is a bit easier to read and requires less thought.

robjohn

@leslietownes $\theta\alpha\tau\varsigma\ \sigma\iota\lambda\lambda\upsilon$

AMDG

$\{0,1,\dots,b-1\}$ takes me a hot second to read. $[0, b)$ I can understand in an instant.

leslie townes

with a closed interval you might have some people wondering if you can use non integer inputs.

robjohn

2:52 PM

@AMDG the latter looks like a range of reals, however, to me.

AMDG

Well then you can just state that the interval is in the naturals, right?

leslie townes

that's more confusing to me than just (implicitly) listing them.

robjohn

I agree

AMDG

$[0,b) \text{ is a half-open interval over the naturals.}$

Interesting... I must think rather differently, then.

leslie townes

nothing wrong with thinking differently, but it does suggest this is a style thing. outside of typos and formatting errors, if it's just style, i leave it.

robjohn

2:55 PM

Yes, I don't mess with people's style even when editing their $\LaTeX$.

leslie townes

are there any bots that go around any SE making routine formatting fixes? i imagine it will happen eventually.

robjohn

I doubt there will be. That would bump posts unnecessarily

AMDG

Well if we're talking about being more concise, perhaps I'm just not used to it, but I find $\{0,1,\dots,b-1\}$ to be ambiguous. There's no guarantee about what the dots mean because it's defining a set, so you have to assume rather than infer that this is defining an interval over the naturals. For all we know, the sequence is 0, 1, 2, 4, 8 and so on.

leslie townes

oh, interesting.

i guess you couldn't have a third party or user do it. if something were implemented internally maybe that could be turned off. unless it's impossible to turn off.

and before anyone asks, yes this is about langle and rangle.

AMDG

Vectors are based

Koro

3:15 PM

Any suggestions here?

1 hour ago, by Koro

If $|\frac{f^{(n)}(0)}n|\le K$ for all $n$, where $K\gt 0$ is a constant. What can we say about $\lim |\frac{f^{(n)}(0)}{n!}|^{1/n}$?

Now, mod diamond looks silver color

leslie townes

should be at most 1 although this is more of a vibe than a solution.

some of my vibes are negative vibes

Koro

hmm… what if $\inf |f^{(n)}(0)|$=0$?

And not all derivatives are zero at 0

Then I am not able to comment on the limit in question

leslie townes

i think it's generally clear that the limit might not exist.

i could be wrong about this.

i would defer to anyone who has been in a math class in the last 10 years before deferring to me.

Koro

I can currently think of only $f(x)=e^x$ fitting the description of the question…

leslie townes

oh is it n in the denominator in the first part and n! in the second?

i missed this aspect of the problem

Koro

3:23 PM

Leslie, I’m sorry for causing confusion but there is no n in denominator in hypothesis.

1 hour ago, by Koro

Argh... The correct hypothesis is $|f^{(n)}(0)|\le K$ for all $n\in \mathbb N$.

This is the correct hypothesis.

leslie townes

feels like it ought to be zero then. maybe by stirling's formula, or some weak version of stirling's formula.

you never need full stirling.

i used to have a document with three or four approximate versions of stirling's formula. i should dig that up.

Koro

Leslie, I tried to make use of this inequality also: $\liminf \frac{a_{n+1}}{a_n}\le \liminf a_n^{\frac 1n}\le \limsup a_n^{\frac 1n}\le \limsup \frac {a_{n+1}}{a_n}$ , where $a_i\gt 0$ for all $i\in \mathbb N$

leslie townes

the ratio and root stuff. well covered in rudin. i think it might need a little more than that but i could be wrong.

Koro

Leslie, this inequality follows from one theorem in Rudin’s PMA book.

leslie townes

you are clearly a well educated person. :)

Koro

3:30 PM

Following the inequality, existence of limit of ratio implies existence of limit of nth root.

leslie townes

i don't see how to control the ratios from the given hypothesis but i agree that the inequality does imply that.

Koro

Leslie, if $\inf |f^{(0)}|=r\gt 0$ ( this infimum exists because bounded below by 0), then we use ratio test on $a_n=|\frac{f^{(n)}(0)}{n!}|$.

leslie townes

inf f^n? OK. you can control the ratios if those values do not get too small.

Koro

And conclude that $a_n^{\frac 1n}=0$

leslie townes

i do think the answer is zero. i note that the smarter members of this chat have not chimed in to tell me that we're wrong.

Koro

3:34 PM

But $\inf |f^{(n)}(0)|=0$ is the problem :’(

leslie townes

there i think it's a distraction to look at ratios.

although i'm still thinking about it

Koro

Leslie, two options are there for this limit: 1) the limit tends to infinity, which has already been dismissed by example of e^x ,2) limit tends to zero (this seems correct but doubtful still)

SAJW

is circle:ellipse the same as square:rectangle?

leslie townes

roughly, at the level of that analogy, yes.

copper.hat

is that part of an iq test?

the word analogy

leslie townes

3:52 PM

i failed my iq test.

Koro

How to take an iq test?

SAJW

no, was just curious if a true ellipse is never a circle or can be a circle.

leslie townes

good question. i think some of the standard ones are copyrighted and not easily distributed online.

Koro

How do they say that x’s iq is so and so etc.?

copper.hat

long time ago they were part of an interview for large orgs

Koro

3:53 PM

How do they measure it?

leslie townes

i took one in the third grade. i remember it mostly involved moving around blocks to make patterns. dumb test.

circles are definitely ellipses.

copper.hat

like anything, there are some tests on which they base their pronouncement

SAJW

the average IQ of any population is by definition 100

copper.hat

no, its much lower than that

SAJW

nope

copper.hat

3:54 PM

yes, definitely

leslie townes

i made a butterfly out of some colored blocks. what a great test.

Koro

Is single digit iq possible?

leslie townes

it's not by definition but by design that is the intent.

i don't even know why i took that test or what it was supposed to do for me.

SAJW

then iq is just a meaningless measure

leslie townes

the school district's psychologist was a nice guy.

copper.hat

3:56 PM

we had to take them in high school

leslie townes

SAJW, yep.

koro, yes although i think in practice you would probably not administer the test to someone who lacked the capacity to score higher.

copper.hat

my dad used to like puzzles, so he had some (penguin i think) iq test books at home.

if you know some of the tricks your iq magically improves

like any evaluation

leslie townes

there's a section of the LSAT that reminded me of an iq test. lots of puzzles.

copper.hat

people put too much faith in a single metric

SAJW

I should have answered in those "what is the next number" the correct answer: the first number, since repitition is the easiest pattern

copper.hat

3:57 PM

next number in sequence: 1,2,3,4,?

i put 0

occam be damned

SAJW

i like that

copper.hat

i'm in a bird mood this morning

leslie townes

that's really weird, i have almost no memory of elementary school but i do remember the iq test.

copper.hat

sorry, fowl

leslie townes

that's pretty bad. even for you

copper.hat

3:59 PM

it really bugs me when neighbours blow the horn instead of getting out of the damn car

leslie townes

oh i do hate that.

copper.hat

splitting headache and i didn't even drink

Koro

@copper.hat haha, Copper another one: simplify (x-a)(x-b)…(x-z)

0

because (x-x) is also a factor

copper.hat

to simplify i would take $a=b=\cdots=z =x$.

leslie townes

at one apartment i lived at, there was an apartment across the way, and a guy who visited would announce his arrival by whistling. one day i lost it and told him just knock on the f'in door i don't want to hear this s blank blank blank.

copper.hat

4:00 PM

jk

Koro

I saw that somewhere :-)

leslie townes

honking is worse than that.

copper.hat

@Koro i like that sort of puzzle

we have a neighbour who clearly does not believe that the remote locking works, so she has to do it a few times.

i think i am just peculiarly sensitive to noise

leslie townes

i have a real problem with it. at the old house the neighbors were having a loud party and i went to their power box and shut everything off. i would have fought anyone who had a problem with it. i was fit to be tied.

some people don't really think of noise as capable of giving offense, but to me it's no different than if you walk into my house and take a crap on the floor.

copper.hat

then there's the leaf blowers...

leslie townes

4:05 PM

i don't mind those so much although they are annoying. it fades into the background. we live in a flight path so there is a lot of droning noise anyway.

i used to live in a neighborhood where a motorcycle gang would go by with all of the noise that came with that. that was annoying.

copper.hat

right now there is a cable/at&t? truck right outside working on the telegraph lines, shouting like they are in the middle of an f*ing dockyard. blocking the street. when my daughter was young they would come by in the middle of the night.

albany seems to have some eternal construction going on.

leslie townes

in situations like that i think, me not flipping out right now is my gift to the universe.

you're welcome, everybody.

copper.hat

yes, definite good mood today

leslie townes

we had some people repairing wood on the house the other day and they were swearing constantly in spanish. it was not annoying but funny, literally every noun had a spanish f-word in it. i was glad my daughter wasn't home because she understands spanish and has to speak it at school.

i don't even know how they understood each other, every noun was "that f---ing thing."

Koro

Leslie, the limit is indeed $0$.

copper.hat

4:12 PM

the collection of wires on the poles is a total eyesore.

yes, i spent a lot of time in construction and with said workers.

leslie townes

our neighborhood put a lot of that underground, which i understand is the standard in europe.

copper.hat

i don't understand why people don't care about the sheer ugliness.

i realise there are real problems as well, but some care for eyesores seems reasonable too

leslie townes

i think stuff is easier to maintain if it's underground. you don't have 50 million pieces of independently potentially rotting wood holding your infrastructure up.

but what do i know

Koro

Here is how: $|f^{(n)}(0)|\le K$ so dividing both sides by $n!$ gives $\frac K{n!}$ on RHS of inequality, and then taking $n$th root on both sides gives $|\frac K{n!}|^{\frac 1n}$, which tends to $0$ because limit of ratio $\frac{1}{n+1}$ (K gets cancelled here) tends to $0$ and there the limit in question tends to $0$.

@Leslie

copper.hat

i think folks in the us are used to shorter timescales than in europe.

Koro

4:17 PM

Here I have also used inequality I stated above:

1 min ago, by Koro

Here is how: $|f^{(n)}(0)|\le K$ so dividing both sides by $n!$ gives $\frac K{n!}$ on RHS of inequality, and then taking $n$th root on both sides gives $|\frac K{n!}|^{\frac 1n}$, which tends to $0$ because limit of ratio $\frac{1}{n+1}$ (K gets cancelled here) tends to $0$ and there the limit in question tends to $0$.

49 mins ago, by Koro

Leslie, I tried to make use of this inequality also: $\liminf \frac{a_{n+1}}{a_n}\le \liminf a_n^{\frac 1n}\le \limsup a_n^{\frac 1n}\le \limsup \frac {a_{n+1}}{a_n}$ , where $a_i\gt 0$ for all $i\in \mathbb N$

I used this inequality

leslie townes

that feels right. even very basic versions of this stuff i would classify as approximations to stirling's formula.

Koro

i didn't explore right hand side of inequality at first, as this example: $\frac 1n\to 0$ but $(\frac 1n )^{\frac 1n}\to 1\ne 0$ was in my head..

:-) anyways it's all sorted now.

copper.hat

@Koro the limit is zero because $\sqrt[n]{K \over n!} = e^{{1 \over n} \log{K \over n!}}$ and ${1 \over n} \log n! \to \infty$.

leslie townes

windows 10 is now reporting the air quality index on my taskbar. this is not something i asked for.

Koro

yeah in my win 10 also

how to get rid of that?

copper.hat

4:23 PM

for a short while (nadella) my views on ms softened a little, but as of late my hatred is returned

leslie townes

i really, really, really don't like turning an OS into a service.

copper.hat

not a fan of effective monopolies

AMDG

I don't see any air quality stuff on my taskbar...

copper.hat

pay more get less. as for customer service. muahahahahh

i need a break :-)

Koro

@copper.hat that is one nice way. What I did was use this nice way: Let $a_n=\frac K{n!}$, then $\frac {a_{n+1}}{a_n}\to 0$ and so it follows by the inequality I linked above that $a_n^{\frac 1n}\to 0$

:-)

leslie townes

4:26 PM

apparently it's part of "news and interests," if you right click you can disable it.

Koro

^<^

@leslietownes thank you! I turned it off.

leslie townes

you and me both.

i live near a port, i know the air quality sucks. i don't need updates.

Koro

:)

leslie townes

in the peak of the (last) coronavirus epidemic it was actually fairly funny. the sea outside the port was just littered with cargo ships honking at each other.

i don't fully know why they couldn't deliver. maybe there weren't enough dockworkers or regulations around how many dockworkers there could be.

four in the morning, BWAAAAAAAMP. thanks, cargo ship.

AMDG

@leslietownes I dare someone take this out of context. This statement would make it seem that the coronavirus is a regular event.

"Better prepare for the next annual Coronavirus Epidemic:tm:"

leslie townes

4:34 PM

i just don't know if there's going to be another one. i'm fully vaccinated but for some people they still get sick, and there are new variants, i just don't know.

Koro

i have only taken 1 dose

AMDG

I don't know what they're expecting. The thing will just mutate... much like every microorganism. Everything microscopic seems to mutate regularly.

Koro

it's fun to type like leslie

in lowercase letters

:)

AMDG

So I question what they think rushing a vaccine out into the public will do. "Oh no, controversial topic."

Ted Shifrin

@Koro Does that Mean that You will be Passive-Aggressive like Leslie?

leslie townes

4:37 PM

it's all i was taught. you should meet my father.

copper.hat

i think covid will become a seasonal event like the flu.

sometimes i get tired of fighting and passive aggressive is the way out

leslie townes

the first time my dad met my wife he spent approximately 90 agonizing minutes interrogating her about her life while making fun of her at the same time.

that's what i was raised in.

that was 20 years ago and we're one happy family, but, i had weird role models.

Koro

i don't know about that professor @Ted. :)

@copper, did you see this

14 mins ago, by Koro

@copper.hat that is one nice way. What I did was use this nice way: Let $a_n=\frac K{n!}$, then $\frac {a_{n+1}}{a_n}\to 0$ and so it follows by the inequality I linked above that $a_n^{\frac 1n}\to 0$

copper.hat

@Koro is there a question there?

Koro

no, I used this way to calculate the limit instead of taking log.

AMDG

4:41 PM

@copper.hat "Aight, guys, we developed some $\gamma$-covid vaccines. Oh, sorry, there's now a $\delta$-variant. You should probably take another vaccine for that one too. We've also speculatively begun production of vaccines in anticipation of $\epsilon$-covid which might or might not work. Better take it just to be safe than sorry."

SAJW

doublepoint=colon (":")

copper.hat

i think they should call it the $\Delta_n$ variant, just to set perspectives

leslie townes

great point. it's mostly a question of framing

AMDG

@copper.hat Nah, the best one is obviously $\kappa\omega\upsilon\upsilon\iota\delta$-variant.

Oh, nice, my edit was approved.

copper.hat

maybe $\kappa o \beta\iota\delta$ :-)

AMDG