Mathematics - 2021-08-11 (page 1 of 2)

Ted Shifrin

00:00

Um ... open? So they agree on overlaps? Or the space is disconnected? Um ...

Thorgott

yeah, agreeing on overlaps

Ted Shifrin

So we have two formulas we have to verify agree on a large open set?

Thorgott

gluing on open sets behaves nicely, gluing along boundaries not as much

dc3rd

Even though I have come to grips with it, I guess it is just human behaviour to really try and look for the path of least effort.....in this case....grunt work will have to be done somewhere

Ted Shifrin

I reject this discussion.

You only glue abstractly in this situation, never concrete formulas.

At some point, @dc3rd, you want to think through inductive proofs that sums, products, quotients, compositions of smooth functions are smooth.

Thorgott

00:02

abstract, concrete, I don't see the difference

o.9

my man is colorblind

dc3rd

At a very high level I could imagine how the proofs would play out. They would be similar to composition of continuous functions that we've done, but just like you said at the inductive level.

Ted Shifrin

Where is induction coming in?

Thorgott

well, you can induct in the proof that composition of smooth functions is smooth

Ted Shifrin

Not you.

dc3rd

00:07

From you guys talking about gluing. I'm guessing that involves defining the function specifically at the trouble points?

well just briefly thinking about it as you ask...Im thinking the induction would be over the number of compositions

Ted Shifrin

No.

dc3rd

cue Price is RIght loser's theme....

Thorgott

yeah, that's something you can induct over too if you want to

Ted Shifrin

It could, but think through what smooth means.

No one is going to bother worrying about 4 compositions.

Thorgott

what I'm saying about gluing is just that if i have a smooth function $f$ on an open $U_1$ and a smooth function on an open $U_2$ such that they agree on $U_1\cap U_2$, they extend uniquely to a smooth function on $U_1\cup U_2$ that restricts to each of them on their respective domain

as a map between sets, this is obvious, and the smoothness works out since that's a local property, so may just be checked in either of the opens

dc3rd

00:10

Ah....induction would have to be on the partial derivatives.....since we are trying to show their existence

Ted Shifrin

I continue to say this is stupid except in the setting of sheaf theory. It never gets used in a concrete case.

Thor is being his stubborn, abstruse self.

o.9

You're too picky ted

dc3rd

From my limited knowledge at this level I would've thought along the lines of rstrictions of functions, but you're saying that's not the case Ted?

Ted Shifrin

No, I’m not. I spent 40+ years teaching and thinking about pedagogy.

Thorgott

I use this all the time, very concretely

Ted Shifrin

00:12

Bull.

Thorgott

granted, it might not be that relevant when you primarily care about rational functions on the plane and the like

Ted Shifrin

Smoothness is local. No more to say.

o.9

I don't like that argument

Ted Shifrin

@dc3rd Be explicit on the induction statement.

o.9

do I have to spend another 35 years to argue ?

Thorgott

00:14

gluing smooth functions on opens together is used everywhere in differential topology, Ted. and I know that you know this.

Ted Shifrin

If you’re going to call me picky, you’d better back it up with good reason.

o.9

I have to back it up?

How would I go around doing that?

Ted Shifrin

It really doesn’t, Thor, other than in cohomological issues. Saying smoothness is local, yes. Partitions of unity, of course.

You guys have totally interfered with the point I wanted dc3rd to understand. Congratulatioms.

See you another year,

o.9

@TedShifrin what does that mean?

Thorgott

saying smooth functions on opens glue together is pretty much equivalent to saying smoothness is local. I believe this is a good point to understand conceptually, even if one doesn't put it to immediate use

o.9

00:19

I was just joking around I had no good intentions for what it's worth

no bad intentions either tho

Thorgott

oftentimes, we glue manifolds together to obtain new manifolds. then, you can describe smooth functions on the glued result by giving functions on the ones we've glued agreeing on the overlap.

dc3rd

that induction statement isn't as clear cut as I first thought....hmmm

Thorgott

which statement exactly are you trying to prove rn?

dc3rd

induction over the partial derivatives to establish $C^{\infty}$

thinking about it right now out loud it means showing the existence of $\frac{\partial}{\partial x_{i_{k+1}}}\bigg(\frac{\partial^{k}f}{\partial x_{i_{k}}}\bigg)$

Thorgott

for which function under which hypothesis?

dc3rd

00:33

the $k+1$ partial derivative. Oh no specific function in particular. I was building off of what Ted was inferring to me a few mins ago.

I suppose in this case it would be the hypothesis of compositions of smooth functions

Thorgott

I would suggest working with the total and not partial derivatives. The latter get a lot more clumsy notationally.

dc3rd

going to give it a gander and see what comes up

Thorgott

I also suggest doing the single-variable case as a special case first

that might make it structurally clearer how to set up this induction efficiently

dc3rd

00:49

that's a good suggestion...no need to jump straight to the complex

epsilon-emperor

02:58

hello, long time

Akiva Weinberger

03:32

I wonder how many people discovered as a kid that adding up odd numbers (1+3+5+...) gives you squares

It seems to be a very common first discovery

epsilon-emperor

i did!

leslie townes

i didn't

Russian Bot 2.0

@leslietownes indeed

i think i have got something

I don't even know who said this first lol

shintuku

03:50

probably franklin d roosevelt or yuri gagarin

Russian Bot 2.0

seems like every famous personality 'said' this

shintuku

it was definitely orson welles

leslie townes

thomas jefferson.

Koro

I want to know of one situation wherein $0^0=0$

leslie townes

i can't think of one. the binomial theorem is fairly persuasive on this.

Koro

03:57

then why is it considered "undefined"?

shintuku

probably will have to construct both the natural numbers and exponentiation to figure it out

leslie townes

you do have to choose which phenomena which you wish to preserve.

Koro

to that, I would say that I haven't encountered $0^0$ quite often as "defined or undefined" but as "indeterminate form", I have encountered it a lot.

Russian Bot 2.0

Zero to the power of zero

Zero to the power of zero, denoted by 00, is a mathematical expression with no agreed-upon value. The most common possibilities are 1 or leaving the expression undefined, with justifications existing for each, depending on context. In algebra and combinatorics, the generally agreed upon value is 00 = 1, whereas in mathematical analysis, the expression is sometimes left undefined. Computer programming languages and software also have differing ways of handling this expression. == Discrete exponents == Many widely used formulas involving natural-number exponents require 00 to be defined as 1. For...

leslie townes

unrelated but i'm at the point where i need to choose whether to shave or grow my beard out again. my wife is not a huge fan of the beard but most of my friends like it.

shintuku

04:02

for instance $0^0 = 1$ for cardinal numbers

leslie townes

i do believe that 1 is the most common value but not the only value you could think of. context is everything.

Russian Bot 2.0

yes

maa.org/book/export/html/….

Koro

@leslietownes right, I also think that context is the key here

:)

and that $0^0=1$ is not universally agreed upon

@shintuku number of functions from null set to null set is considered $1$?

I mean while defining a function $f:A\to B$, we do assume $A$ to be non-empty.

right?

in which case, $0^0=1$ for cardinals doesn't make sense, I think

Akiva Weinberger

04:23

There's a function from the empty set to every set

the empty function

This encapsulates the idea that $A^0=1$ (remember that $A^B$ counts functions from $B$ to $A$)

On the other hand, there's no function from a nonempty set to the empty set

(In category theory, the fact that there's a unique function from $\emptyset$ to any other set means that it's an "initial object")

(and the fact that there's a unique function from any set to $\{p\}$, a singleton, means that singletons are "terminal objects")

(in the category of sets)

Fun fact: in the category of vector spaces (where we only consider linear functions), the 0-dimensional space $\{0\}$ is both initial and terminal

Koro

04:36

how to define a function from nullset to say $\{1,2\}$ for example?

such a function won't have anything to map to {1,2}:(

($f:\emptyset\to \{1,2\}$) := (f(x)=1 $\forall x\in \emptyset$)

?

shintuku

it satisfies the definition of a function: for each x in the domain, there is a y in the codomain to which it maps

since there is no x, the statement is true

Koro

vacuously true?

shintuku

the function from nullset to $\{1,2\}$ is only the empty function

so $x^0 = 1$ for cardinals, including $0^0$

Koro

hmm ... I think it's well-defined (vacuously)

Let $x_1,x_2$ be two representations of $x\in \emptyset$, then $f(x_1)=f(x_2)$ is not false as $x_1,x_2 $ both don't belong to nullset hence vacuously $f$ is well-defined.

So $0^0=1=$ cardinality of set of functions from nullset to nullset

CowperKettle

05:19

robjohn

06:00

@Koro I define $0^0=1$, but note that $x^y$ is not continuous at $(0,0)$. Just because we define $0^0=1$, does not mean that $0^0$ is no longer an indeterminate form; that has to do with limits where $(x,y)\to(0,0)$, but $(x,y)\ne(0,0)$.

rain1

06:41

0

Q: How many flips to tell if a coin is biased?

I have a biased coin, how many flips do I need for you to be confident that it is biased? I think you could assume its fair and do a hypothesis?

i dont understand this closure at all

Koro

@robjohn I know that $0^0=1$ has nothing to do with $0^0$ being one of the indeterminate forms so the conclusion is that: it's not universally agreed upon that $0^0=1.$ :)

hi @copper!

copper.hat

Hi @Koro!

Koro

is it true that if $f'$ (derivative of $f$) exists on $[a,b]$ then it must attain its supremum/infimum on $[a,b]$?

intuition says yes as $f'$ follows ivt.

rain1

Any suggestions on how I might edit the question to get it re-opened?

Koro: Yes since it's a closed interval, extreme value theorem

Koro

06:57

but $f'$ may not be continuous so i think you can't apply extreme value theorem there

rain1

oh sorry i misunderstood the question

Koro

but i think that the conjecture is right. all we need is a proof now to prove that

copper.hat

@Koro I don't think so, I think the standard example of a discontinuous derivative (some $\sin $ thing) has an unbounded derivative on $[0,1]$.

Koro

the example that invloves $x^2 \sin \frac 1x$ ?

I think that derivative is bounded in a nbd of $0$ in this example

copper.hat

Try $x^2 \sin {1 \over x^2}$ then.

Koro

07:09

but you seem to be right as additional condition continuity may be required.

@copper.hat yes, this will be unbounded around $0$

thanks a lot :)

rain1

07:36

0

Q: How many flips to tell if a coin is biased?

Suppose that I have a coin which may be biased and may be fair. We can assume that each time it is flipped there is probability $p$ of it landing heads and $1-p$ of it landing tails. I do not know the probability $p$. I want to determine if it is biased or fair. How many flips do I need to perfor...

probability probability-theory hypothesis-testing

i have edited the question, can it be re-opened now?

Wolgwang

08:04

@rain1 Ask in CURED...

rain1

thanks

robjohn

@rain1 It still does not specify what "confident" means.

rain1

I edited it to say "to get a certain level of confidence"

robjohn

What does "a certain level of confidence" mean?

rain1

so that's just a parameter could be 80% could be 90%

it just means probability is greater than or equal to some parameter

does it need more editing to explain that in more detail?

added it

Do you think it can be reopened now?

Jasper

12:42

Is this an appropriate place to ask for advise? I've got two permutation groups $\Gamma, \Delta \leq \mathcal{S}_n$ that have disjoint supports (so the generating permutations of the two groups have no overlap; A point permuted by $\Gamma$ is a fixed point in $\Delta$). I want to specify the group generated by the union of the generators. Is this commonly called "group composition", or is this a direct product? Should I write $\Gamma \times \Delta$ or something like $\Gamma \circ \Delta$?

Oh there is no latex support here :(

Balarka Sen

This is a concrete realization of the direct product.

Jasper

How can I make clear that the direct product does not give a subgroup of dimension n^2 ?

I.e. distinguish between kronecker and direct products

Wait, I'm answering my own question. I'm using \otimes for Kronecker, so \times is not an overloaded operation on groups, so I can define it as such

Thanks!

Balarka Sen

Dimension of a subgroup is a nonsense terminology, but what you mean is that you're specifying a faithful permutation representation of $\Gamma \times \Delta$. I'd just say it verbally; the point is if the support of $\Gamma$ is $A \subset [n]$, $\Delta$ is $B \subset [n]$, $A \cap B = \emptyset$, then there is a natural embedding $S_A \times S_B \hookrightarrow S_n$ by composition.

The "permutation group $\Gamma$" is just the group $\Gamma$ equipped with a faithful representation $\Gamma \to S_A$, likewise for $\Delta$. So jointly $\Gamma \times \Delta$ has a representation in $S_A \times S_B$; you compose with the natural embedding above to get a representation in $S_n$ and not $S_{n^2}$.

Nobody uses "Kronecker product of groups". A mathematician on the street would be like wtf if you tell him that.

Jasper

I'm parsing your comment

Russian Bot 2.0

13:01

The more you know, the less you know

Jasper

My initial question is definitively answered; I can just use \times in the notation and it'll be all fine. I'm currently more reflecting on the remark that "dimension of a group is nonsense teminology". In my context I'm always working with groups generated by permutations acting on vectors of fixed size. Should I then always state "cardinality of the group support" rather than "dimension of the group", or is that equivalent if the representation is known?

Koro

Is there an example of a function $f$ (other than $f(x)=x$) continuous on $[0,1]$ and differentiable on open interval $(0,1)$ such that $f’(x)\le 1$ for all x in $(0,1)$ and that $f(0)=0=f(\frac 12)$?

Balarka Sen

@Jasper I would use "rank of the permutation representation is $n$", etc

This is borrowing terminology from linear representations of groups to permutation reps.

Jasper

I'll check that out, thanks!

@Koro

Koro

Jasper, I think you wanted to tag Mr. Balarka

Jasper

13:10

No I wanted to write an answer to you

Koro

Ahh okay. Please do give your suggestions. Thanks.

Jasper

I guess you're stating the question wrong, because "other than f(x) = x" does not have f(0) = 0 and f(1/2) = 0

Koro

It’s easily established with the given information that $f(x)\le x$ for all $x\in [0,1]$

By FTC.

@Jasper if no such f exists then I want to prove that function satisfying the conditions stated in the assertion is indeed $f(x)=x$

Jasper

But does f(x) = a sin(2 pi x), where a is sufficiently small?

Ah, it should be f(x) = x

but then it should say f(1/2) = 1/2, right?

Koro

@Jasper that won’t be acceptable as a is supposed to be fixed.

Jasper

13:14

You can pick a = 1/100 and the derivative will be <= 1; I just didn't think of how large a could be

Koro

All I could show was that $f=x$ on $[0,\frac 12]$ but for $x\gt \frac 12$ I had difficulty.

@Jasper yeah but then f(1/2)=1/2 won’t be satisfied. I tried sin/ cosine examples but alas they didn’t seem to work.

Jasper

why does "f(x) = sin(2 pi x) / 100" not work then?

Koro

Sorry, let me correct the question as $f(1/2)=1/2$ and f(0)=0

Jasper

Ah, then we're talking

On the interval (0, 1/2) it should be f(x) = x then;

You cannot cheat with piecewise linear functions? :P

Namely, if f(x) is continuous on (0, 1/2), f(0) = 0 and f(1/2) = 1/2, then the average slope must be 1, so if you're dropping below a slope of 1 at some time then you must compensate and have f(x) > 1 somewhere on (0, 1/2). There is a calculus theorem that states this, but I don't know the english name

Koro

If I define $g(x)=f(x)-x$ then my purpose will be to show g=0 on $[0,1]$. Noting that $g’(x)\le 0$ we conclude that g is monotonically decreasing whence for any t in (0,1/2) we must have g(t)=0 hence we have established that g=0 on [0,1/2]

Jasper

13:21

That is correct

Koro

But I guess nothing can be said about nature of f on (1/2,1] which means that we may be able to find a counterexample (that is a function which is not f=x)

Jasper

On (1/2, 1] all you need to show is that it can deviate from f(x) = x while maintaining f'(x) <= 1 right?

Koro

That’s why I was looking for such an example. I hope the background to the question is clear now :)

Jasper

A piecewise linear function would do this, but I think it's not the spirit of the exercise

Koro

@Jasper it won’t work!

Even a quadratic polynomial won’t do!

Jasper

13:23

Why won't it work? All you need is this:
f(x) = x for x in [0, 1/2)
and for x > 1/2 it should be a parabola that has slope 1 in the point (1/2, 1/2)

Koro

But that will not agree with f(1/2)=1/2 :’(

Jasper

One second

f(x) = x for x < 1/2
f(x) = (x^2 + 1)/4 for x >= 1/2

Oops no

Koro

Suppose we have $f(x)=ax^2+bx +c$ then we want f(0)=0 that is c=0 and f(1/2)=1/2 gives: $a/2+b=1$ and $2ax+b\leq 1$
These inequalities won’t be satisfied by any value of a and b for $x\in [0,1]$

Jasper

But you don't need f(0) = 0 if you make the function piecewise

But I'm not sure if that's the spirit of the exercise

Koro

@Jasper fair point. In that case, we can take our c to be non zero

Jasper

13:28

It should look a bit like a ski-slope in reverse; You start at (0, 0), move in a straight line to (1/2, 1/2), and then from (1/2, 1/2) you're entering the parabola-part

I've got it; Use the standard representation
f(x) = ax^2 + bx + c
f'(x) = 2ax + b
f''(x) = 2a
At (1/2, 1/2) we want to have a slope of 1, and we want this slope to descend so that we have f'(x) < 1 for x > 1/2.
So the system of inequalities becomes
f(1/2) = 1/4 a + 1/4 b + c = 1/2
f'(1/2) = a + b = 1
f''(x) = 2a < 0
Choose a = -1, then b = 2 and c = 1/4. This does the job

f(x) = x; for x < 1/2
f(x) = -x^2 + 2x + 1/4; for x >= 1/2

I made a mistake somewhere. That parbola does not intersect (1/2, 1/2)

f(1/2) = 1/4 a + 1/2 b + c = 1/2
f'(1/2) = a + b = 1
f''(x) = 2a < 0
Choose a = -1; then b = 2 and c = -1/4

This does the job :)

Koro

14:02

@Jasper I was travelling while typing my earlier messages on my phone. In fact I was doing the same thing and about to hit send but due to transition during travel, could not do that. I was thinking of taking $b=1$ and giving $c$ also some value to get some appropriate $a$. Sorry for the delay in response. Now, $f(x)=-x^2+2x-\frac 14$ and I think that works !

Thanks a lot @Jasper :)

Koro

14:20

$f(x)=-2x^2+3x-\frac 12$ on $\frac 12 \le x \le 1$ also works :)

Red Banana

14:33

It is known that sounds like the bass in Benny Benassi's "Satisfaction" is produced with sawtooth waves (Hear [1] below). Yesterday, someone reminded me about Weierstrass function, I got curious: What sound does Weierstrass function makes? Now we have the answer! I made this: https://youtu.be/A4OTGF0QKJg where Rick Astley's "voice" is a sound produced with Weierstrass function.
The image in the video is an AI generated image made with Rick Astley's picture and a picture of Weierstrass function.

[1] : youtu.be/PyI_0mJrLlQ

leslie townes

15:24

i prefer the vocals of mr. astley, but interesting.

Russian Bot 2.0

hehe

maybe i dont

leslie townes

that's maybe the only meme anyone needs.

Russian Bot 2.0

yes

Kumar Shuvam

@RussianBot2.0 are you EUler2?

Russian Bot 2.0

yes

wait a minute

Kumar Shuvam

15:31

Ok! I'm waiting

troll bot

this is me

Kumar Shuvam

oh!

troll bot

yeah

Kumar Shuvam

long time no see ''intangible entity''?

troll bot

why is yt filled with these weird genshin impact ads

I don't play games

shintuku

15:37

yeah i dunno why but it does that with me too

Kumar Shuvam

Use ad-blocker

shintuku

i think the algorithm thinks i'm some sort of alt-right gun fanatic that hates feminists, i honestly never click on any of those videos not even by curiosity

troll bot

it is weird but I sometimes watch ads

shintuku

i keep getting ben shapiro and us military video suggestions

leslie townes

i think the algorithm just feeds people that at random because they know that people who bite on it watch a lot of them

i get a lot of video gaming ads because i think they have an estimate of my age and gender and are thinking, this'll work

the recommended videos also really seems to think i am a huge fan of the canadian band Rush

shintuku

15:43

the Algorithm works in mysterious ways

leslie townes

again probably based on some estimate of my demographic

Kumar Shuvam

If the derivative at a point is 0, what can be said about the polynomial function is it increasing or decreasing?

leslie townes

rush is OK but i don't need to watch live shows and 10 minute drum solos

kumar, consider x^3 and -x^3 at 0

Koro

@leslie, how did you know name is kumar and not shuvam?

:)

leslie townes

that may just be western ignorance, going with the thing that appears first.

Kumar Shuvam

15:45

haha! lol

don't remember me

leslie townes

even if the derivative is positive at a point, the function may not be strictly increasing on any interval around that point

Koro

one of my friends has similar name except that he has "bh" instead of v in Shuvam

leslie townes

french people list their last names first too. and write them in all capitals in professional contexts for some reason.

Kumar Shuvam

Actually my name is a mistake @Koro..lol, it should have been shubham..

Koro

his name is also written as Kumar Shubham and he also says that is by mistake

he says that it should have been Shubham kumar but is Kumar Shubham :)

leslie townes

15:48

my dad thought his middle name was francis until he joined the army and had to provide a birth certificate. it was frank. his parents thought it was francis too.

Kumar Shuvam

haha! yeah..can't do much about it, name doesn't matter anyways

leslie townes

when my daughter was born we didn't have a name. after about 12 hours they demanded that we name our daughter before we left the hospital.

i don't know what legal basis they had for doing that, but we came up with something.

i remember texting my dad a photo of her when she was born and my dad said 'what's her name?' 'i dunno, we'll tell you tomorrow'

we have a long standing tradition of doing everything at the last minute

in our family we tend to use middle names to pay tribute to other members of the family. with my daughter we just stole a first name from a grandparent.

Kumar Shuvam

So Leslie you mean to say we can't say about monotonicity at a point for which derivative is zero

leslie townes

yeah. monotonicity is a property of a function on an interval, and a derivative at a single point isn't enough to tell you what happens on an interval.

Kumar Shuvam

I see.

leslie townes

15:57

if you imagine two parabolas, one upward facing and one downward facing, that meet at a single point. you can scribble any curve you want between them and it will have whatever derivative the parabolas do at the common point of intersection. but you can make it oscillate back and forth as much as you want around there and change the increasing/decreasing behavior at will.

robjohn

@KumarShuvam consider $f(x)=x^2\sin(1/x)[x\ne0]$. It has $f'(0)=0$, but the derivative oscillates wildly between $-1$ and $1$ as $x\to0$.

leslie townes

the classic example of the phenomenon i mentioned. note it fits the above schema with x^2 and -x^2 being the two parabolas.

that function is a good example of so many things.

Koro

this example works as counter-example in many cases

professor Rob, I think you should say for non zero x and for $x=0, f(x)=0$

because $f(x)$ as stated is not even continuous (nor discontinuous) at $x=0$

robjohn

@Koro I guess that is necessary; however, since $|\sin(x)|\le1$, it would be hard to assign any other value to $x^2\sin(1/x)$ than $0$.

$0\cdot[-1,1]=\{0\}$ :-)

Koro

But given only $f(x)=x^2\sin (\frac 1 x)$, we can't comment on its continuity at $x=0$ unless we define it first at x=0 . Right, professor Rob?

robjohn

16:06

all we have to say is that it is continuous,then we can comment on it.

if it is continuous, $f(0)=0$

Koro

ahh, right. reverse-engineering :)

leslie townes

in a math classroom i think it would go without saying. in a physics classroom sometimes you have people say weird stuff like "and it's infinity at 0" and it's like, what? why? what is happening?

robjohn

@leslietownes that goes for the delta function, of course

leslie townes

yeah, of course.

Kumar Shuvam

It makes sense now

Thanks a lot, all the professors :)

leslie townes

16:09

i don't mean to insult physicists. when they do weird math stuff it's because it works, and some of them even have instincts for how it ought to work. it's not lazy or being wrong.

it's just sometimes informal.

Koro

i really like physics' rough and ready proofs related to divergent theorem etc.

Stokes' law

shintuku

Given $\{x, y\} \in A$, we infer $x \in \{x, y\} \in A$. what's the axiom that allows me to infer the second statement?

is it comprehension?

leslie townes

i had a physicist on my thesis committee. normally the 'outside member' (= non mathematician) on the thesis committee is something of an afterthought who is just there to provide a signature. he asked me deeper questions about some of my results than any of the mathematicians did.

Koro

axiom of choice @shin ?

leslie townes

frankly it scared the hell out of me.

robjohn

16:11

you have to watch out, because a small tweak to a correct theorem that involves divergent series or integrals can go badly wrong.

shintuku

i was thinking axiom of choice might have something to do with it

Nicholas Roberts

i have a basic question, can someone give me a hint or a step in the right direction? Suppose I have a strictly increasing function with $\lim_{x \to -\infty}f(x) = 0$. Can we deduce that $f(x) > 0$? Seems like it should be true

leslie townes

i wanted to say 'you're the outside member, you know you don't have to do this?' he found several weak points in my dissertation after reviewing it for five minutes.

shintuku

i'll check it out thanks Koro

leslie townes

then he complained about the government for 20 minutes. he was an interesting guy.

Kumar Shuvam

16:12

@robjohn @leslietownes Can you recommend me a book to just read for gaining deep insights on Caclulus just for fun as a high school student?

Koro

@robjohn I haven't read those in detail yet. Professor Ted suggested me a book once but I have not completed it yet. I saw "rough and ready" proofs for curl divergence related stuff :)

leslie townes

kumar sylvanus thompson's 'calculus made easy' is full of insights although it is very informal and somewhat antiquated. thomas koerner's 'calculus for the ambitious' or 'companion to analysis' are a little technically deeper.

shintuku

@KumarShuvam spivak

if you already have done high school calculus

Thorgott

@shintuku whatever your definition of $\{x,y\}$ is, this is by definition

Kumar Shuvam

Alright :)

Koro

16:27

@NicholasRoberts suppose that $f(y)\lt 0$ for some $y$ then there exists $r\lt 0$ such that for all $x\lt r$ we have $f(y)\lt f(x)\le -f(y)$ and hence choosing $x$ smaller than $y$ we'll have $f(x)\gt f(y)$ which is a contradiction. So $f$ can't be negative. Can you take it from here?

shintuku

@Thorgott seems like it. if for any property there is a set whose elements have this property, $\{x,y\}$ is the set for which an element is either $x$ or $y$

$\forall \gamma (\gamma \in \{x,y\} \iff \gamma = x \lor \gamma = y)$

this is either the axiom of comprehension or a definition

shintuku

16:54

any way to search only the titles on mathse?

oh there's a search faq

you can do it writing title:(query)

leslie townes

i exclusively use google to search mathse.

it's cool that you can do that, however.

shintuku

argh you need the axiom of restricted comprehension actually

leslie townes

do we need it? i could do without it. let's dump it.

Nicholas Roberts

@Koro

Do you mean to fill in the missing details? Because that seems like it proves the statemewnt

statement*

shintuku

but then $x \in x \iff x \notin x$ :'(

robjohn

17:14

@shintuku this has nothing to do with it.

shintuku

hm

the untrained eye might potentially mistake that axe of choice for the axiom of choice

robjohn

Oh, yeah, it caught me.

a pick-axe might be more appropriate ;-)

Koro

@NicholasRoberts yeah and you may consider one case separately when $f(y)=0$ for some $y$ and note that you again get a contradiction. This will complete the proof. :)

Nicholas Roberts

@ko

@Koro thanks, i filled in the details of what you wrote. will work on $f(y)=0$ now

just to make sure, you took epsilon to be $-f(y)$ in the assumption of the limit being 0 at negative infinity, right?

Koro

glad that helped :)

@NicholasRoberts yes because i assumed f(y)<0; and then I used definition of limit at -infty

Nicholas Roberts

17:27

yup, just making sure

@Koro Suppose $f(y) = 0$ and take any $x < y$. Then $0 < \frac{f(x)-f(y)}{x-y} = \frac{f(x)}{x-y}$. Since the denominator is negative, and the quotient is positive, the numerator must also be negative, that is, $f(x) < 0$ which is a contradiction to the first part of the proof. Does this work?

KeithMadison

I have a (I think relatively simple) statistics problem-- I have an instrument which records the current time every time it observes the "tick" sent by another instrument. I want to look at the percentage of ticks measured out of the total number of ticks, as a function of time. So, I bin the data in 10 second bins. I compute this percentage for each 10-second interval, and plot these percentages as a function of time. How, then, do I compute the statistical error for each point?

schn

@robjohn , just out of curiosity, would you say the following notation is sensible (equation 1 in the question): $$\mathbb{E}[f_n(t)]=\int_{\mathbb{R}}k(u)f(t-hu)\mathrm{d}u = \lim_{c\to\infty}\int_{-c}^{c} s(u;t,n) \mathrm{d}u , $$ where $s(u;t,n)=k(u)f(t-hu)$. Possibly even $\alpha(t,n)=-c$ as well as $\beta(t,n)=c$..

KeithMadison

I recall it being something like sqrt(D - N)/N (where D is the "denominator", and N the "numerator"), however I'm not sure and don't know what to search for to find an answer.

Koro

@NicholasRoberts It looks fine. What I had in mind was: if $f(y)=0$ then for any $x\lt y$, we must have $f(x)<0$, which is a contradiction as we showed in the first part :)

Nicholas Roberts

oh yeah, much easier lol

pretty much the same as what i wrote

Koro

17:42

yeah :)

robjohn

@schn the stuff with $\alpha$ and $\beta$ looks a bit odd, but I think otherwise it looks reasonable.

schn

Yeah, I agree.

user10930711

Hi is anyone here good at stochastics?

geocalc33

18:02

0

Q: Prove this lattice a discrete subgroup of $\Bbb C$ and has an Eisenstein series

Let $(e^x,e^y)$ be a lattice of points in $\Bbb R^2_+$ for $x,y \in \Bbb Z/\{0\}. $ Consider a transformation with real parameter $a,$ acting on the lattice: $(e^x,e^y)\mapsto (e^{ax},e^{y/a})$ Is there an Eisenstein series for the lattice? Is this a discrete subgroup of $\Bbb C?$ I know that...

complex-analysis group-theory integer-lattices

robjohn

18:24

Is what a discrete subgroup of $\mathbb{C}$?

If you are asking about the lattice, since $(1,1)$ is not in the lattice, I don't see how it can be a subgroup, but maybe I don't understand.

geocalc33

18:41

yes, is the lattice a discrete subgroup of $\Bbb C$

@robjohn

robjohn

18:59

Both $(e,e)$ and $(1/e,1/e)$ are in the lattice, but their "sum" $(1,1)$ is not (I assume the addition of this group is componentwise multiplication, if not, you need to specify what the group operation is).

leslie townes

i don't think lattice is the right word for this. image of a lattice under something, maybe.

robjohn

@leslietownes looking at it on log-log paper it might be a lattice minus a couple of lines

Avra

Hello. Given 3 sets definitions as follows:

A, the keys to the left of the search path;

B, the keys on the
search path; and

C, the keys to the right of the search path.

Question: why set $A$ is empty please?

The item we are looking to search for is 3.

robjohn

because there is nothing to the left?

Avra

@robjohn. Yes. Thanks. So def of set $B$ is what exactly?

All items to the key we are looking for should be the search path?

robjohn

19:06

the nodes on the search path

Avra

Can you please define search path in your own words?

robjohn

I am assuming that whoever set this problem defined search path and the nodes that lead to the item found are on the search path. Look to the place where this problem was given.

Avra

@robjohn. This is the question. Professor Bunyan thinks he has discovered a remarkable property of binary search
trees. Suppose that the search for key k in a binary search tree ends up in a leaf.
Consider three sets: A, the keys to the left of the search path; B, the keys on the
search path; and C, the keys to the right of the search path. Professor Bunyan
claims that any three keys a ∈ A, b ∈ B, and c ∈ C must satisfy a ≤ b ≤ c. Give
a smallest possible counterexample to the professor’s claim.

I am not sure if there is a definition of the search path here? This is why I got confused.

robjohn

how are the keys numbered? Are we only considering this tree? there is not enough information as far as I can see.

Avra

I thin I got it now.

@robjohn. I think the search path is the set of all nodes along which we traversed to get to the key. In case we are looking for 3. 3<8 -> 3<4 -> search path to the key {8,4,3}

robjohn

19:13

@Avra well, yes, that seems to be what the diagram is saying

@Avra there seems to be no logic to the numbering of the nodes other than they go up to the right and down to the left. Then how was the search path chosen?

Avra

In the above example, if we are looking for 6, then search path: 6>2 -> 6<8 ->6>5 : search path is{ 2,8,5,6}

@robjohn. I think it's based on comparing nodes. We add all nodes included in our comparisons until we get to the key we are looking for

geocalc33

19:39

@robjohn okay i have now specified that in the post thx

hopefully it gets reopened

robjohn

20:06

@geocalc33 Still, how can it be a group when it doesn't have an identity element?

That would be $\left(e^0,e^0\right)$, which is explicitly not in the "lattice"

I am not sure what an Eisenstein series is, but the group part seems questionable.

Joey

20:24

Say you ask me to exhibit a family of functions $(e_k)_{k \in \mathbb{Z}}$ that is linearly independent in $L^2[0,1]$ but not orthonormal in $L^2[0,1]$. So I give you a family of such functions $(e_k)_{k \in \mathbb{Z}_{\ge 0}}$ i.e. indexed over $\mathbb{N}$. Why is this not enough?

geocalc33

@robjohn yeah i realized that (1,1) needs to be in the lattice so i added that part back into the post

thanks

0

Q: Prove this lattice is a discrete subgroup of $\Bbb C$ and has an Eisenstein series

Let $(e^x,e^y)$ be a lattice of points in $\Bbb R^2_+$ for $x,y \in \Bbb Z $ with group operation as componentwise multiplication. Consider a transformation with real parameter $a,$ acting on the lattice: $(e^x,e^y)\mapsto (e^{ax},e^{y/a}).$ Is there an Eisenstein series for the lattice? Is th...

group-theory integer-lattices

robjohn

@Joey Did I ask? Did I say this was not enough? Perhaps this is in reference to some answer somewhere, because I think the answer to my questions is "no".

It would help to get the full context, rather than passing judgement without all the facts.

Joey

Haha fair enough.

ibb.co/0YS3zCS

ibb.co/sRcChvQ

This the context. The question they are answering is in the second link (sorry it doesn't let me upload images). They didn't exactly say it wasn't enough but then why did they need to construct the bijection at the end? @robjohn

shintuku

20:44

this set theory book is so cool. the chapters i've read are just a list of definitions and theorems with proofs

makes it easy to find what i want

robjohn

@Joey because they want it indexed over $\mathbb{Z}$. That was their requirement, don't know why. It doesn't seem important.

Joey

@robjohn Thank you.

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