Mathematics - 2021-06-27 (page 1 of 2)

ominous, esoteric, forbidding, useless ... anyone else for adjectives?

this integral's presumptuousness will amuse you.

it dribbles trippingly off the tongue?

there's a new yorker cartoon in this area. it did have something to do with wine.

my step sibling has published two cartoons in the new yorker. someday i'm gonna learn how to draw and get my own cartoon in there.

it'll involve integrals. i just need the right humorous angle.

1:03 AM

I think you should leave it to your daughter, Olivia, and ducks instead.

my daughter fell into that sludge.

I only see water, not sludge.

well, today my daughter became an honorary duck.

mallard mom didn't do a very good job of tending to my daughter when she was in the water, i must say.

She will need training.

there are bubbles in that water of uncertain provenance.

my daughter stopped crying after she dried out a bit and we found another mallard who had not five, but seven new chicks. we also watched a tern hunt in the water. really great dive bombs. my daughter is really good with bird ID these days.

one thing i miss about iowa city is watching bald eagles hunt the river in winter. there was a bend in the river near my house and they'd patrol it and then come down and grab fish out. i could watch that all day.

someone posted a solution to the ominous integral.

1:13 AM

Chris'ssis used to love all these things. I couldn't care less.

She has supposedly published a book, but I haven't tracked it down.

apparently it's from a book called (Almost) Impossible Integrals, Sums, and Series

whoever published such a book should probably be locked up

Oh, I wonder if that's her book. Is it within the past few years?

That sounds like it.

published 2019, looks like.

Yup. That's Chris'ssis, from Romania, but gender unknown.

Astounding that it got published with hardly any connections.

Google says he. Even though I am not interested in this stuff, I still respect the accomplishments.

there's always an audience for perversion.

1:17 AM

Career in business/accounting, but true love is mathematics.

published 2019?

i am sympathetic to that. this book is goofy.

I remember when she talked about it like 5 years ago

Yes, that's right. Not she, despite the moniker.

No question a talented soul. Just doesn't appeal to me.

the amazon page is interesting. comment in english, german, spanish, and italian. there is an international audience for this stuff.

1:20 AM

Well, think about all the stuff in Counterexamples in Analysis, Polya-Szegö, etc.

perverts, all of them

I'm sure this is a multiple duplicate, but I'm just too lazy to search.

So I just answered.

please read EOQS

Hell no.

you are supposed to look for a dupe for at least 72 hours

1:23 AM

oh, it definitely does feel like a dupe although my first search turned up nothing.

With these rules, I will quit entirely.

THe poster should look for dupes, not I. If I know I've answered something before, I find the dupe and so indicate. I'm not willing to do anything more.

i can find related questions but they have other stuff involved too. connecting the dots might be beyond the skills of the asker.

Certainly the families of functions I've listed have appeared dozens of times. Probably once or twice in my own answers.

there should almost be a tag for x^n sin (1/x). not seriously proposing this.

but it comes up enough.

You have to know to tag that. If you do, you're already done.

1:24 AM

does that include n=-1?

No.

positive integers only.

i've used that family in some of my answers. it's easier to use than a 'simpler' example that doesn't have a nice formula.

anakhro

2:02 AM

@leslietownes she's become amphibious?

yes.

anakhro

Congrats on the upgrade.

3:17 AM

LOL

5:14 AM

@robjohn I used that to conclude that the fraction expression involving N-1 is larger than that involving N but that is not enough yet because I want show to show that the difference is such that {$\sqrt {(N-1)^2+1}$} is in (a,b).

Am I going in the right direction?

if $\frac1{2\sqrt{n}}\lt b-a$, then $\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}\lt\frac1{2\sqrt{n}}\lt b-a$

the difference between any two adjacent square roots is smaller than $b-a$

I need to spend more time here

@robjohn I’m afraid but I don’t see how this will get me a {$\sqrt m$} in (a,b) as from this observation, I can use sequence of square numbers (that is numbers whose square root is integer) but number consecutive to a square number need not be a square number e.g 9 and 10

I've been spending a lot of time on Math Twitter, and the culture is just so much worse over there

By using square sequence, I can conclude that {$\sqrt {m^2+1}$} <b-a

5:28 AM

The question is showing fractional parts of square roots of integers is dense in [0,1)?

Yes @AkivaWeinberger

6 hours ago, by robjohn

@Koro consider that $\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}$ is a decreasing sequence.

This suggests a generalization

Suppose we have any increasing sequence such that $a_{n+1}-a_n$ is decreasing

Hm, that's not enough

Hmm

I think we want $a_{n+1}-a_n$ to be decreasing and also approach $0$ in the limit

(without that last bit I don't think it's enough)

Mr. Akiva, let me share with you what idea (whether sufficient or not) I have in mind to prove the result

5:31 AM

Akiva, not Akita :)

It’s my iPhone autocorrection :’(

@Koro Consider all the $n$ between $\left\lceil\frac1{2(b-a)}\right\rceil^2$ and $\left\lceil\frac1{2(b-a)}+1\right\rceil^2$. Since $\sqrt{n+1}-\sqrt{n}\lt\frac1{2\sqrt{n}}\lt b-a$, one of the $\sqrt{n}$ must fall within $(a,b)$.

@robjohn Are $a$ and $b$ arbitrary?

Yes @Akiva

$(a,b)\subset [0,1]$

@AkivaWeinberger We are showing density of the fractional part of $\sqrt{n}$ in $[0,1]$, so $0\le a\lt b\le1$.

5:36 AM

6 mins ago, by Akiva Weinberger

I think we want $a_{n+1}-a_n$ to be decreasing and also approach $0$ in the limit

I don't think what I wrote there is enough either actually

'cause it allows for, say, $a_n=1-\frac1n$

Revision: $a_n$ is increasing without bound and $a_{n+1}-a_n$ is approaching zero

Conjecture: that's enough to guarantee $\{a_n\}$ is dense

That is correct and what I am using above

@robjohn FWIW, I'm not sure I understand your approach

So each step is less than the width of the target interval

but why does that guarantee one lands in the target interval?

@AkivaWeinberger and we span the entire interval $(0,1)$ with points closer together than $b-a$

Oh!

Oh OK then I see

Yeah so we're walking a unit distance, and the size of our footstep is smaller than the hole we're doomed to fall into

indeed

5:42 AM

and by footstep I mean "stride length" actually

That is how I would prove your conjecture

By the way

17 mins ago, by Akiva Weinberger

I've been spending a lot of time on Math Twitter, and the culture is just so much worse over there

No one on Twitter calls me Mister

so thanks for that :D

@AkivaWeinberger I saw that. I have never been on Twitter.

@AkivaWeinberger I think that is a cultural thing, so no one on Twitter is from the proper culture.

(@Koro Are you from India? I have a conjecture that people from India are more likely to call people Mister)

. o O ( I think that is correct, or at least in that area of the world )

5:45 AM

@AkivaWeinberger yes :-). I’m not sure about your conjecture though:-)

I have observed this phenomenon.

Some days ago, one of my office colleagues (very senior to me, age-wise and company experience wise) was asking me you should not call me sir, call me by my name :-)

In any case, I hope our discussion was helpful, Mr Koro :-)

But it’s just become a habit to call those elder than me sir :-)

@Koro are they senior in position?

5:46 AM

I think it's also common in the Southern US

Yes, they are

The Northeast US (where I am) is a lot less polite

@robjohn yes they are

@AkivaWeinberger really? I have not noticed that since the days of Mark Twain.

Maybe not

Is that where you are?

5:47 AM

@Koro then "sir" is not out of the question

@AkivaWeinberger no, I am in California (southern California, but that is not considered "The South")

@AkivaWeinberger I’m trying to understand the arguments yet. How fraction part of some n lies in target interval (a,b)

7 mins ago, by Akiva Weinberger

Yeah so we're walking a unit distance, and the size of our footstep is smaller than the hole we're doomed to fall into

The unit distance being between $\left\lceil\frac1{2(b-a)}\right\rceil$ and $\left\lceil\frac1{2(b-a)}+1\right\rceil$

17 mins ago, by robjohn

@Koro Consider all the $n$ between $\left\lceil\frac1{2(b-a)}\right\rceil^2$ and $\left\lceil\frac1{2(b-a)}+1\right\rceil^2$. Since $\sqrt{n+1}-\sqrt{n}\lt\frac1{2\sqrt{n}}\lt b-a$, one of the $\sqrt{n}$ must fall within $(a,b)$.

@robjohn I had noted that already but the way he has turned the question into a picture is truly amazing. I just need some time to process this 🌝

Let $A=\lceil1/(2(b-a))\rceil$. Then we're looking at all $n$ between $A^2$ and $(A+1)^2$, adjacent square numbers

so the fractional part of $\sqrt n$ "walks" the entirety of the unit interval from 0 to 1, in small steps

Therefore, there exists a smallest $n$ in that range such that $a\le\{\sqrt n\}$

Since it's the smallest one, $\{\sqrt{n-1}\}\le a$

Suppose by way of contracts took that $\{\sqrt n\}$ is not in $(a,b)$

Then $\{\sqrt n\}\ge b$

and so $\{\sqrt n-\sqrt{n-1}\}\ge b-a$

@AkivaWeinberger Shoulda kept this one as a strict inequality ($<$ not $\le$) but I can't edit it back

But $\sqrt{A+1}-\sqrt A<b-a$ by construction, and since it's decreasing, $\sqrt n-\sqrt{n-1}<b-a$ as well

Is it just me or is the window in which I can edit comments shorter than it used to be

I meant $\{\sqrt n\}-\{\sqrt{n-1}\}$, not $\{\sqrt n-\sqrt{n-1}\}$… though they should be equal anyway I think

@AkivaWeinberger I think $\le$ is just fine

I am working on “walks” the entirety argument. Rest all makes sense to me :-)

6:01 AM

$a\le\{\sqrt n\}$ and $\{\sqrt n\}\notin(a,b)$ isn't enough to force $\{\sqrt n\}\ge b$, because we could have $\{\sqrt n\}=a$, so I think I do need $<$

@Koro I'm thinking of the best way to write it

$n$ is between $A^2$ and $(A+1)^2$, so $\{\sqrt n\}=\sqrt n-A$ (unless $n=(A+1)^2$)

We want to show that there's an $n$ in that range with $a<\{\sqrt n\}$

so we want $a<\sqrt n-A$, or $A+a<\sqrt n$

or $(A+a)^2<n$

but $A^2<(A+a)^2<(A+1)^2$ so I think we can let $n=\lceil (A+a)^2\rceil$

Only problem is the "unless $n=(A+1)^2$" part

and also this argument feels kinda ugly to me

Mr Akiva, please save yourself the trouble, I’ll handle that part. It’s just $\frac1{2(b-a)}$ seemed extraordinary to me.

Mr. Rob comes up with such ideas often :-)

I think he got it by working backwards

$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}$ is a legitimately clever construction (is that the word? manipulation?)

and then $\sqrt{n+1}-\sqrt n<\frac1{2\sqrt n}$ from there is immediate

so if you want an $n$ such that $\sqrt{n+1}-\sqrt n<b-a$,

it's enough to find an $n$ with $\frac1{2\sqrt n}<b-a$

whence $2\sqrt n>\frac1{b-a}$ so $\sqrt n>\frac1{2(b-a)}$ and so we can take $\sqrt n=\lceil\frac1{2(b-a)}\rceil$

(remembering from the rest of the argument that we want the lowest $\sqrt n$ to be an integer)

6:19 AM

I have got it @AkivaWeinberger @robjohn

Yay ^_^

Thank you so much :-)

Entirely worth staying up from 1:30am to 2:20am for :)

:-) it’s 11:50 am here :-)

While discussing the question with Mr Rob yesterday (around 3 am) night, I fell asleep so saw his message in the morning

Time zones, man

But what can you do

6:22 AM

Yeah 😊

@Koro Fun fact:

Let $\phi=\frac{1+\sqrt 5}2$ be the golden ratio

Then $\{\phi\},\{\phi^2\},\{\phi^3\},\dots$ do not form a dense set

In fact, I want you to numerically (with a calculator) find out the behavior of that sequence

and prove a conjecture

How did you make that picture? @robjohn

with Mathematica

@AkivaWeinberger okay, will do that

Animal adoption has been allowed by a zoo here. Even tiger!

6:31 AM

@Koro: does that image make sense?

I thought it was {$\phi^i$}. I didn’t see it more carefully. @robjohn

I improved the image, if you click on it, it should be nicer

@AkivaWeinberger a lot of quadratic algebraic numbers do that

I got the image @Rob.

So $\sqrt {n+1}- \sqrt n \lt \frac 1{2\sqrt n}$

Very nice :-). I have used matlab at college a lot but never got to use mathematica

The distance between the points should have been $\frac1{2\cdot6}$. I fixed the image

@AkivaWeinberger: thanks!

6:48 AM

I put n=36 in the above identity and then the image is more clear.

$n$ goes from $6^2$ to $7^2$ the distance between points is between $\frac1{2\cdot6}$ and $\frac1{2\cdot7}$

We can increase $A=6$ to as big as we wish

Did you see what Akiva's sequence does? $\left\{\left\{\phi^k\right\}:k\in\mathbb{N}\right\}$

@robjohn A lot, but not all

sure

@robjohn not yet, I’ll do that soon

@AkivaWeinberger These behave this way because of Lucas Numbers

7:08 AM

@Astyx: welcome to the density party

Astyx

What's it about?

We've shown that the fractional parts of the square roots of the integers are dense in $[0,1]$. Now we're looking at the fractional parts of $\phi^n$

@robjohn @AkivaWeinberger: Here's the behavior that I found

hello

7:10 AM

@Koro You run into a precision error near the end

And by the plot, it looks dense to me

@Vrouvrou hi

Astyx

Oh, that's what those curly braces mean

but it's not dense as you say

@Koro dense? it is tending to $0$ and $1$

7:12 AM

i have a function $g: A\cup B \to \{0,1\}$ defined by f(x) for $x\in A$ and 0 for $x\in B $ is $g^{-1}(\{0,1\}=A\cup B$ or is equal B ?

Ahh, right. I got confused by the graph (somehow thought it was continuous due to the lines) @robjohn

@robjohn precision error?

@Koro it stops flipping between just above $0$ to just below $1$ because it doesn't see the number just below $1$ as smaller than $1$.

I think you mean excel is not precise enough

@Koro yes

right. mathematica will give more insights?

7:15 AM

@robjohn the idea is to prove that g is constant then f(x)=0

It's clear however that the set of fractional part of phi^n 's are not dense in [0,1$\rbrack$

well, you can set the precision in Mathematica. I don't know about Excel

@Koro yes. Now, we need to see why...

@robjohn okay. I have never used mathematica. Currently I have excel only so plotted it there. I'll get Mathematica too

@Vrouvrou are we talking about the same question?

@Koro you don't need to get it. You have access to Matlab, you say?

@robjohn yes, the exercise say if A and B are closed and $A\cup B$ and $A\cap B$ are connected then A and B are connected

7:19 AM

@robjohn okay. here's my intuition (based on the graph): there will be some N such that for all $n\ge N$, we'll have the fractional part much lower than 1 and so numbers close to 1 will not be limit points of the set.

There are only two limit points, but why?

@robjohn I had used it during my college. Right now, I don't have it in my system :'(

so I let a continuous function defined on A to {0,1}, as $A\cap B$ is connected then f on it is constant say f(x)=0, after that I defined $g:A\cap B \to \{0,1\}$ and i need to prove that g is continuous

@robjohn it will require rigorous proof on my part but I asserted my statement based on graph. :-)

@robjohn So please is $g{-1}(\{0,1\})=B $ or $A\cup B$ ?

7:25 AM

@Vrouvrou: in $\mathbb R$ or in some other metric space?

they are not in metric spaces necessarily

okay

@Vrouvrou what is the definition of connected that you are using

no not in metric-spaces

if any continuous $f: A\to \{0,1\}$ is constant then A is Connected

wher $\{0,1\}$ is withe the strong topology

Ah. Okay.

you mean the discrete topology? every set is open

7:28 AM

yes

so I want to prove that g is continuous that is the image inverse of each closed set is closed

@Vrouvrou: nothing more is given about A?

@Koro Yep :) That's the behavior I wanted you to find. It alternates between "very close to 0" and "very close to 1"

it is closed

The question is why :)

yeah. that will require a rigorous proof on my part :-)

i'll try and ask if i get stuck

7:33 AM

I have a hint prepared if you want one

not now. why because i already know the proof for proving {m+n$\pi:m,n \in \mathbb Z$} dense in $\mathbb R$ and it requires pigeon hole principal. I'll try using that first.

@AkivaWeinberger do you have a proof that doesn't use linear recurrence relations?

@robjohn Yeah

I mean I also have a "one line" proof but it's opaque

i like how every question on dense set is a completely new chapter.

(for me)

some one can help me ?

7:48 AM

@Vrouvrou so what you need to show is that any continuous function f on A is constant and that any continuous function g on B is constant. Right?

yes I start the proof for A I write it before

$f$ must be constant on $A\cap B$ and $g$ must be constant on $A\cap B$

g is not on B

no not like this

8:21

so I let a continuous function defined on A to {0,1}, as $A\cap B$ is connected then f on it is constant say f(x)=0, after that I defined $g:A\cap B \to \{0,1\}$ and i need to prove that g is continuous

Ok. I cannot see where your proof is going, so finish your proof and then I can look at it.

I start by the connectedness of A

7:51 AM

what is $g$?

Let $ f: A\to \{0,1\}$ be a continuous function

As $A\cap B$ is connected then f is constant on $A\cap B$, then I put f(x)=0 for all $x\in A\cap B$

let $g: A\cup B\to \{0,1\}$ defined by

f(x) if x is in A and 0 if x is in B

if I prove that g is continuous then as $A\cup B$ is connected then g is constant so f(x)=0 on A

you understand me @robjohn

how can you put $f(x)=0$ on $A\cap B$?

$f$ is any continuous function on $A$

because the restrictions of f on $A\cap B $ is continuous

and $A\cap B $ is connected

@Vrouvrou but why can you say that restriction is $0$?

so f is constant

8:00 AM

what if $f$ is $1$ on $A\cap B$?

f is constant on $A\cap B$ so it takes 0 or 1

it is not a problem if takes 1 I will just change the definition of g

yes. so the $f$ I am thinking of is $1$ on $A\cap B$

okay. Now you need to show that the $g$ you're defining is continuous

in this case I define g by f(x) on A and 1 on B

yes

@Vrouvrou show that is continuous

my problem is what is $g^{-1}(\{0,1\})$

Is it $A\cup B$ or B

8:07 AM

that is the whole question.

is $g$ constant

if it is continuous then it is constant

wait, $g^{-1}(\{0,1\})$ is $A\cup B$ since the range of $g$ is $\{0,1\}$

but that does not help you

what you want to see is what is $g^{-1}(\{1\})$

no it helps because $A\cup B$ is closed

$g^{-1}(\{1\})= B$ in this case and $g^{-1}(\{0\})=\emptyset $

why?

we have no 0 in the definition

of g

8:19 AM

no, you have that $g=f$ on $A$, why is $f$ not $0$ on $A$?

8:36 AM

It can't because we will have a contradiction

0=1

??

f is 1 on $A\cap B$, but why can't $f$ be $0$ somewhere on the other part of $A$?

the idea is to find that g is constant

yes, you have to prove that.

you haven't proven that yet

yes

but how to do ? to prove that g is continuous

without using inverse image

9:04 AM

we van say that $g^{-1}({1})=A\cap B$

$g^{-1}({0})= (A\cup B ) \setminus A\cap B$ ?

@Vrouvrou $A\cap B\subset g^{-1}(\{1\})$ because $B\subset g^{-1}(\{1\})$

9:26 AM

is there any solution ?

I am sure there is but there must be some property of continuous functions that you know that you are not applying. Using what you have given, I am not sure.

I use clopen sets in the definition of connected sets and using those, one can show what you are trying to show.

$A$ and $B$ being closed is used only to show that $A$ and $B$ are closed (and open) in the relative topology of $A\cup B$

9:41 AM

@Vrouvrou: You need that. For example, let $A=[0,1]\cup[2,3]$ and $B=(1,2)\cup(3,4)$. Both $A\cap B=\varnothing$ and $A\cup B=[0,4)$ are connected, but neither $A$ nor $B$ are connected.

@AkivaWeinberger: About as simple as I can get is showing $\phi^n+(1-\phi)^n=\phi^{n-1}+(1-\phi)^{n-1}+\phi^{n-2}+(1-\phi)^{n-2}$ using $\phi(1-\phi)=-1$. Then we can show the requisite inclusion by induction.

$\phi^n+\overline\phi^n$ is an integer simply because all the $\sqrt5$s cancel @robjohn

where $\overline\phi=\frac{1-\sqrt5}2$

If you want rigorous proof expand it with the binomial theorem

@robjohn

@AkivaWeinberger yes, but that gets more involved, I think.

or note that it's fixed under the field automorphism of $\Bbb Q[\sqrt5]$ that sends $\sqrt5$ to $-\sqrt5$

but that takes longer to explain

(Definitely the way I think about it though)

Now that is more advanced than either of the ways we've discussed.

we are talking about explaining this to a high school student.

I thought so, but maybe I am mixing people up.

9:56 AM

The one line proof I was thinking of was: "$(\frac{1+\sqrt5}2)^n+(\frac{1-\sqrt5}2)^n$ is an integer because the $\sqrt5$s cancel, and the second term is $(-0.618)^n$ which goes to $0$"

OK, maybe not actually one line

and there are details to fill

I was just saying that $\phi^n+(1-\phi)^n$ is an integer by induction.

and of course $|1-\phi|\lt1$

Once you get the idea to consider $(\frac{1+\sqrt5}2)^n+(\frac{1-\sqrt5}2)^n$ in the first place you're pretty much in the home stretch

The hard part is coming up with that step

yes, but I don't think that would occur to some

Right, that's why it's a challenge

Using Excel, it won't happen ;-)

9:59 AM

It might

We'll see

He might think of considering the sequence of distances to the nearest integer

and might think of looking at the successive ratios of that sequence

and realize that it's a geometric series

These are computational steps, not logical ones

so Excel could help

Well, he did get exposed to the idea that $\left\lfloor\sqrt{n^2+1}\right\rfloor=n$ today

I'm not grading him on this

I don't care how long it takes

I'm patient. I've been on this site for about 10 years and I've seen some pretty slow progress. That they keep with it is the good part.

eventually, things click into place, as long as they do the thinking and don't have others do the thinking for them.

10:04 AM

Do you know the infected squares puzzle?

not by that name, if I do

20bits.com/article/the-infection-puzzle

@robjohn Thoughts?

(This one took me a while…)

Hmm, a simpler rule than Life. So far, I haven't see anything less than $n$ on an $n\times n$ board.

I think I can see a way to show that it must be $n$, but not sure yet

Much simpler: there is no death (black cells never turn white)

the applet does not run on my computer

10:17 AM

Shame

The code page is not valid either.

you can't go past the furthest in a given direction than the starting infection

left or right

up or down

lyme

is there a way to bump my question?

Martin Sleziak

10:43 AM

@lyme Do you mean something like: Legitimate methods for attracting attention to an old unanswered question and How to grab users' attention on an old question?

You can find there a lot of links to other related discussions, one of the links posted there is this posts on Meta Stack Exchange marked faq: How do I get attention for one of my own questions without a good answer?

wklm

11:06 AM

Hi folks! I'm trying to calculate the characteristic function of the binomial distribution
https://imgur.com/a/NeKlm21

I was trying to plug it onto the taylor expansion, but it doesn't seem to work. Could you please provide me with some hints?

epsilon-emperor

imgur.com/a/niSs1fG

So, in (9), why is $f\in A$?

To get (8) they applied Riesz Representation

shintuku

does anyone have a tip on showing that, if a sequence has an upper bound $B$ but $M$ is not an upper bound, there is an item of the sequence greater than $M$ but smaller than $B$?

does this follow from $!(s \leq M) \rightarrow (s > M)$?

12:11 PM

@AkivaWeinberger I am familiar with cube roots of unity and I think that one who’s familiar with cube roots of unity and binomial theorem, will naturally think of this (somehow cancelling the sqrt)

And I have observed that all powers (positive integral) of $\phi$ can be written as $m+n \phi$, where m,n are integers (non negative)

And this is because, $\phi^2= 1+\frac {1+\sqrt 5}2$.

shintuku

@shintuku solved

So we keep on multiplying both sides by phi and can verify my observation is true. How this will (or will not) lead to a final answer, I don’t know (yet)

xan

12:51 PM

Hi all. Is anybody here able to help me with this one? math.stackexchange.com/q/4183477/169914

2:17 PM

not i

schn

2:52 PM

@robjohn In a conversation we had a couple of days ago, you wrote "...characteristic function of $[0,1]$...". What is the characteristic function of an interval? Is this simply a function that returns 1 if a variable is in the interval and 0 if it isn't?

schn, yes

schn

cool

thanks

anakhro

3:06 PM

Are we beating people up yet

... i hadn't gotten that memo, so no

schn

3:19 PM

@robjohn, you wrote "$k_\alpha$ is just a family of functions gotten by squeezing their domain and scaling to keep their $L^1$ norm constant. The characteristic function of $[0,1]$ scaled to maintain it $L^1$ norm.” Could you elaborate how the $L^1$ norm of $k_{\alpha}(x)=\frac{1\{x\in [0,1]\}}{\alpha}$ is maintained? Also, what do you mean by squeezing the domain?

If $-1\leq x \leq 1$, then $\int_{-1}^1 |k_\alpha (x)| \mathrm{d}x$ takes on different values for different $\alpha$.

shintuku

3:58 PM

anyone has a favorite exposition of the proof of multivariable chain rule?