Mathematics - 2014-02-19 (page 2 of 3)

anon

05:00

free abelian, yes

Pedro Tamaroff

Right, the point is to prove torsion free fin gen abelian implies free

anon

one can use induction on the number of generators needed

actually, hmm, let me review

gets out the green folder

Pedro Tamaroff

what colour are your folder(s)?

anon

all of them

Pedro Tamaroff

all green?

anon

05:02

no, all of the colors

Pedro Tamaroff

that's cool. so rainbow like?

anon

yeah

Pedro Tamaroff

mine are black and blue

sounds depressing!

Old empty bed...springs hard as lead
Feel like ol’ Ned...wished I was dead
What did I do...to be so black and blue

Even the mouse...ran from my house
They laugh at you...and scorn you too
What did I do...to be so black and blue

I'm white...inside...but, that don't help my case
’cause I...can't hide...what is in my face
(jazzman sounds)

How would it end...ain't got a friend
My only sin...is in my skin
What did I do...to be so black and blue

(instrumental break)

How would it end...ain't got a friend

anon

wonder if the plancheral measure on finite groups that aren't the symmetric groups have any interesting motivation or properties

Pedro Tamaroff

plancherel

lulz

anon

05:07

jeez

Pedro Tamaroff

hehehehe

anon

between our four tries, you had three

Pedro Tamaroff

you have to say it out loud like a frenchman

and only plansshhherrrel sounds right.

anon

if you're super drunk maybe

Pedro Tamaroff

sober works too

certified

anon

05:12

woah, more evidence of the duality between conj classes and irreps via random walks

head asplode

Pedro Tamaroff

@anon you readin paperz?

anon

browsin

Pedro Tamaroff

05:32

@anon is it generally true that for any sbgrp B of A tor(A/tor(B)) $\simeq$ tor A / tor B?

anon

yes: torsion elts of A/tor(B) are represented by torsion elts of A

(you could say it's an equality, really)

Pedro Tamaroff

Yes, true.

I was playing it safe. =P

@anon wait, a+tor B is torsion iff a has finite index in B.

not if a is torsion

DERP

@anon Sorry for the stoopid.

finite index in tor B is good enough =)

anon

05:48

(:

Pedro Tamaroff

@anon Wonder if your reaction is the same: who cares? =P

anon

actually my reaction was "trivial, next question" when I read the title on main

but indeed $1$ is an arbitrary argument and iterating trigonometric functions is rather unmotivated - the only reason these things crop up is because trig and functional composition are among the small set of tinker toys available to the littleuns (for example, one can "discover" them merely by playing on a scientific calculator)

Pedro Tamaroff

@anon true dat

I have a real problem now.

Not sillyness.

I hopw.

Lang proves that if one has an epi $f:A\to A'$ with $A'$ free (not $A$!) then $A=\ker f\oplus C$ for $C$ a subgroup of $A$, with $C\simeq A'$.

Now, suppose $A$ is a finitely generated abelian group.

Say, on $n$ elements.

Then we have an epi $f:F\to A$ where $F=\Bbb Z^n$ free on $n$ elements.

anon

haha, I just answered this like days ago

Pedro Tamaroff

OK, let me think for a second..

anon

05:58

see here

Pedro Tamaroff

I have an epi $A\to A/{\rm tor} A$

And $A/{\rm tor}A$ is free.

anon

I suppose I did it for R-modules and not abelian groups, but same diff

Pedro Tamaroff

Right?

So that's what he wants.

@anon Yeah, the proof is not crazy.

anon

tor(A) might be smaller than ker(f)

Pedro Tamaroff

But that was not my qustion.

anon

06:00

ah

Pedro Tamaroff

Yeah, no worries.

The point is I have to show there is a free subgroup of $B$ of $A$ such that $A=B\oplus{\rm tor}\; A$.

And by the above, take the epi $A\to A/{\rm tor}\; A$ projection-

Then $\ker =\rm tor$.

anon

right

Pedro Tamaroff

And by the lemma yadda yadda

Mike

I always wonder when seeing questions like that why people don't ask about iterating cosines

why the functions dampen out, what the limiting value os

seems like a natural question, not which is greater

Pedro Tamaroff

@anon I don't know how far it spoils the answer, but I htink you could be a little more detailed.

anon

06:04

?

oh, you mean in the linked answer

Pedro Tamaroff

for example "Pick a basis $\{x_i'\}$ of $A'$, and pick $x_i$ such that $f(x_i)=x_i'$. Claim $\{x_i'\}$ is a basis for the submodule it generates.

anon

yes, I start out at a calculated vagueness and increase detail as I observe the OP's progress. if the OP doesn't interact or engage at all then it stops there.

Pedro Tamaroff

@anon ah, ok

anon

@Mike the haar measure is the unique invariant measure (with blah blah technicalities). is the plancheral measure unique in a similarly natural way? the definition dim^2/|G| seems so artificial.

you've been away for a bit @Mariano

Mike

the only representation theory I know is how to count irreps of $S_n$. I am the wrong person to ask

Pedro Tamaroff

06:09

@anon This is what I told you some days ago $$A\simeq {\rm tor}\;A\oplus\frac{A}{{\rm tor}\; A}$$

"Decompose $A$ into its torsion and nontorsion part."

But yeah nontorsion was a bad choice of words.

I guess.

@MarianoSuárez-Alvarez !!!

anon

I was saying there is generally more than one torsionfree subgroup B of A such that A=tor(A)+B

so one cannot speak of "the" nontorsion part ("part" implying "subgroup")

Pedro Tamaroff

Ah! OK.

Mariano Suárez-Alvarez

hello

anon

speaking of blue and black...

Pedro Tamaroff

@anon black and blue

Mariano Suárez-Alvarez

06:12

POLL: «there are more than one factor» or «there is more than one factor»?

Pedro Tamaroff

it has a nice ring

@MarianoSuárez-Alvarez IS

anon

there is

Pedro Tamaroff

Definitely.

Mike

is

Pedro Tamaroff

3-0

the crowd has spoken

Mariano Suárez-Alvarez

06:13

quite unconvincing :-)

Pedro Tamaroff

@MarianoSuárez-Alvarez Justification:

Mike

factor is singular, even though there's more
than one

Pedro Tamaroff

"Existe mas de un factor"

"there is" acts on factor, not on groups, so to speak

Mariano Suárez-Alvarez

ah english.stackexchange.com/questions/35389/…

Mike

you're acting on my nerves

Pedro Tamaroff

06:17

@MarianoSuárez-Alvarez Do you know what the Grothendieck group of a monoid is?

Commutative monoid.

anon

adjoin formal inverses

get a group

?????

profit

Mariano Suárez-Alvarez

@PedroTamaroff sure

Mike

not convinced about that last step

Pedro Tamaroff

@anon You take the monoid, make ${\bf Z}\langle M\rangle$, inkect $M$ into it, say $m\to \hat m$ and quotient by the group generated by $\widehat{x+y}-\hat x-\hat y$.

Mariano Suárez-Alvarez

that is a different way of constructing it

both give the same thing

Mike

06:19

@MarianoSuárez-Alvarez do you know why kids love cinnamon toast crunch

anon

you can also do the same thing to go from semirings to rings

and many other types of objects

it's functorial and is categorically free or universal or somesuch

Mariano Suárez-Alvarez

You can start from M, add a new set M' in bijection with M, and define on the free monoid generated by M\cup M' the least equivalence relation which maks the elements of M and of M' be inverse, and multiplication in M be that of M.

Pedro Tamaroff

@MarianoSuárez-Alvarez so it's that? "enlarge" $M$ to a group and preserve $M$s structure inside by taking such a quotient?

Mariano Suárez-Alvarez

You can't in general enlarge M to a group

for example, $\mathbb N_0\cup\{\infty\}$ with addition

Pedro Tamaroff

@MarianoSuárez-Alvarez Ah.

Mariano Suárez-Alvarez

06:22

any map from that to a group is constant

Pedro Tamaroff

Aha.

@MarianoSuárez-Alvarez I have a problem.

Proving $S_{n-1}$ doesn't embed in $A_{n}$.

Mariano Suárez-Alvarez

If n is odd that's easy

Pedro Tamaroff

yes

=)

so the problem is n=2m

anon

~~1, 2, 3~~ 4th person to note this :)

Pedro Tamaroff

the copy would have index m+1 in A_n

dunno if tht helps

@anon its a good observation!

Mike

06:27

I was first!

first in the history of mankind

Sush

I have a TeX file in notepad. How to convert it to pdf?

anon

apparently a book on quantum group theory has an index followed by a "$2$-index" :)

Pedro Tamaroff

@Sush TeXMaker?

@anon hmm, don¿t get it

anon

play on the idea of n-categories

so we have 1-things, 2-things, 3-things, etc.

Mariano Suárez-Alvarez

06:42

if S_{n-1} is contained in A_n, then inducing the sign representation from S_{n-1} to A_n gives a representation of dimension n/2, which is non trivial. Now for n>7, the smallest dimension of an nontrivial irrep of A_n is n-1, so we must have n<=7.

Pedro Tamaroff

@MarianoSuárez-Alvarez so one is left with A_2,A_4,A_6... and things are easy? =O I don't really know anything about representations though

Mariano Suárez-Alvarez

actually A_6's minimal irrep has dimension 5, so we have 6 too :-) (there are two in this case, but that doees not break anything)

A simple case: if $p=n-1$ is prime. For then $S_{n-1}$ has elements of order $p$ and $A_n$ doesn't

Sawarnik

@PedroTamaroff Hi

Pedro Tamaroff

@MarianoSuárez-Alvarez yes, I was told that by German, he was TA in analysis II

it's also a good observation

@Sawarnik Hello.

Sawarnik

@PedroTamaroff Which is greater, sin(sin(sin(1))) or cos(cos(cos(1)))? [in radians]

Pedro Tamaroff

06:57

cos cos cos

The sine goes to zero

Mariano Suárez-Alvarez

If $S_{n-1}$ is contained in $A_n$, then $A_n$ acts on the set $A_n/S_{n-1}$ which has $n/2$ elements, so we have a morphism $A_n\to S_{n/2}$

Pedro Tamaroff

has an attractive point

Mariano Suárez-Alvarez

This has to be injective as soon as $n\geq5$

Pedro Tamaroff

@Sawarnik sin sin sin is of order 10^{-6}, cos cos cos 1 is near 1!

Mariano Suárez-Alvarez

so $(n/2)!\geq n!/2$

Pedro Tamaroff

06:58

@MarianoSuárez-Alvarez Ah! That's very nice.

However $2m!<(2m)!$ for $m>2$

@MarianoSuárez-Alvarez I have to prove that, though.

I take it it involves that A_n is simple?

Mariano Suárez-Alvarez

sure

Pedro Tamaroff

For $n\geqslant 5?$

Mariano Suárez-Alvarez

you cannot do anything with A_n if you do not know that

Pedro Tamaroff

Hehehe.

@MarianoSuárez-Alvarez Something slightly tragicomic happened today.

My brother called me from Starbucks, woke me up. Said "You told me you wanted to buy coffee". I said "Yeah! Bring one." He brought me a cup of coffee. I mean coffee, brewed, from the store, not the grains. =P

Mariano Suárez-Alvarez

heh

Pedro Tamaroff

07:05

@MarianoSuárez-Alvarez How does one approach a problem like this? If $x,y$ are 3-cycles in $S_n$; $\langle x,y\rangle$ is iso to $Z_3,A_4,A_5$ or $Z_3\times Z_3$. I have some ideas. Say if they commute, it is iso to the last one, if they are the same 3 cycle, is to $Z_3$ so I can assume $x\neq y$ and $xy\neq yx$.

anon

"three cycles" $\ne$ "$3$-cycles"

Pedro Tamaroff

@anon I mean $3$-cycles.

Mariano Suárez-Alvarez

the 1st case also happens if xy=1

Pedro Tamaroff

@MarianoSuárez-Alvarez Right.

Ah. The previous exercises deal with 3-cycle with x\neq y^{-1}.

Mariano Suárez-Alvarez

an element in the subgroup generated by x and y will move, at most, the elements which are in the union of the supports of x and y

if that union has 6 elements or three, we know what happens

Pedro Tamaroff

07:08

right, so if they have something in common (don't commute) I am at most in S_5

anon

if x=(abc) and y=(def) consider whether |{a,b,c}\cap{d,e,f}| has 0, 1, 2, or 3 elements

Mariano Suárez-Alvarez

so we need only consider the cases where the union has 4 and 5 elements

Pedro Tamaroff

And the previous exercises has that. =P if they "are" in S_4, <x,y> is A_4

I have to show that

Mariano Suárez-Alvarez

it is enough the to describe the groups generated by (123) and (145), on one hand, and (123) and (124) on the other

Pedro Tamaroff

assuming x =/= y^{-1}

I did the above by counting, @Mariano.

The case things are in S_5 is a little more complicated.

But S_4 is easy

Sawarnik

07:13

@PedroTamaroff What do you mean by order of 10^-6

Pedro Tamaroff

@Sawarnik order of magnitude 10^{-6}

@Sawarnik This

Sawarnik

@PedroTamaroff So sin sin sin 1 is veery close to 0?

Pedro Tamaroff

At the physical level, I guess it is.

Sawarnik

@PedroTamaroff So close! 10 ^ - 6?

Mariano Suárez-Alvarez

For S_4 you can check that all 3-cyclees are in the group generated

so the group is contained in A_4

and sincee it is generated by even perms, it is A_4

Sawarnik

07:17

@PedroTamaroff I still did not get what you said completely

Pedro Tamaroff

@Sawarnik What part?

Mariano Suárez-Alvarez

Likewise, th one with support of order 5 is also contained in A_5 because it is generated by even permutations.

Sawarnik

@PedroTamaroff The 10^6 [my minus key along with many others is not working, I get frustated copying it from Sticky Notes, so I might omit that in obvious situations]

sin sin sin 1 is 0.67?

@PedroTamaroff Is this answer correct for math.stackexchange.com/questions/681324/tough-contest-problem :

We can easily compute the derivative of $f(x)=\cos(\cos(\cos(\cos(x))))$ by the chain rule, to get:

$$f'(x)= \sin(x)\cdot \sin(\cos(x))\cdot \sin(\cos(\cos(x)))\cdot \sin(\cos(\cos(\cos(x))))$$

And similarly for $g(x)= \sin(\sin(\sin(\sin(x))))$:

$$g'(x)= \cos(x)\cdot \cos(\sin(x))\cdot \cos(\sin(\sin(x)))\cdot \cos(\sin(\sin(\sin(x))))$$

If we can prove that the minimum value of $f$ is greater than the maximum value of $g$, then obviously there would obviously be no root of your equation. Since $f$ is continuous and has a period of $2\pi$, we can utilize the closed interval method on $…

Pedro Tamaroff

Oh, my bad.

I didn't do radians. =)

At any rate, I find the above really boring. =P

Sawarnik

Ok.

Pedro Tamaroff

07:25

But you can probably use a Taylor sieres if you want.

sin sin sin 1 is 0.67..... and cos cos cos 1 is .65

So if you verify that, you're done.

At any rate, any calculator does something similar to get an approximate value.

Sawarnik

@PedroTamaroff Should I post this as question on the main site?

Pedro Tamaroff

Do what you like. =)

I was going to tell you "I don't want to tell you what to do."

But I just did.

Sawarnik

@PedroTamaroff Since you find this boring, may I ask you what you like from what you might get from me?

Pedro Tamaroff

@Sawarnik Come again?

Sawarnik

@PedroTamaroff What?

Pedro Tamaroff

07:29

I don't understand what you said

Anyhow... I should sleep now.

Sawarnik

@PedroTamaroff Oh, I meant that you found this thing boring. So what you like, within my level?

Pedro Tamaroff

@MarianoSuárez-Alvarez Byes! The other day I was meaning to ask you something the other day but I could not -- it's nothing serious, I'm just looking for advice. Would you mind me dropping by some day this week or the other?

@Sawarnik I like Analysis.

But that's quite a vague statement.

Sawarnik

@PedroTamaroff Within MY level?

Pedro Tamaroff

I don't know what your level is. Calculus is a nice subject, specially if one studies it carefully. There are some nice theorems in there. For example, Darboux's theorem on derivatives (that they satisfy the intermediate value property) is not that widely known.

Sawarnik

@PedroTamaroff My level, is non rigourous understanding upto integrals. Seems that I have to study a lot to interest you.

Pedro Tamaroff

07:36

@Sawarnik There is a nice book called "Understanding Analysis" by Stephen Abott.

If you search carefully you can even find a .pdf neat copy.

Sawarnik

You can sleep, I wont disturb you now.

Pedro Tamaroff

That's an awesome book.

Also Spivak.

Spivak is good.

Good is an understatement actually.

Fantastic. And with a british accent.

Sawarnik

I will search thanks :)

Theta30

@Sawarnik without checking all sounds a valid proof

and the graph from original question sustains your assertion

the graph of Sammy Black

Mike

@Pedro Weren't you going to sleep?

Sawarnik

08:01

@Theta30 Just the last bit, sin sin sin 1 is greater or cos cos cos 1 , gonna ask on the main site.

Mike

i guess he was

Sawarnik

@Theta30 math.stackexchange.com/questions/681850/…

Steven Stadnicki

08:18

@Sawarnik Without trying to put words too much into Pedro's mouth, I think the reason that this feels 'boring' in that at first glance it doesn't teach anything interesting or new. Perhaps if some magical method for comparing iterated sines and iterated cosines appears then there might be something interesting there, but it seems unlikely that you'll add new methods to your toolbox with a problem like this.

I freely confess to being old and grumpy, though: I generally have the same feeling about most of the 'contest style' three-way inequalities that seem to pop up so frequently.

Mike

I think inequalities are boring

But, I am a dirty analysis nonliker

Sawarnik

@Theta30 I have a feeling my proof is wrong.

Steven Stadnicki

@Sawarnik (Also, FWIW: I can see the chat but many people who are on the main site may not have quite-so-easy access, and/or may not be able to dig back through the chat logs readily. I would encourage making the question self-contained in terms of motivation; you don't by any means have to, but it might bump interest in the question.)

As far as the original question goes, it would take a lot more poking, but I think the easier approach might be to just go through the doubly-iterated functions; with $f(x)=\sin(\sin(x))$ and $g(x) = \cos(\cos(x))$ then it certainly appears that $g(x)\gt f(x)$ everywhere, and there might be enough there to show from some combination of that and bounding ranges that $g(g(x))\gt f(f(x))$ everywhere.

Chris's sis

08:47

Greetings

@N3buchadnezzar I've just created a marvellous double integral :-)

N3buchadnezzar

=)

Chris's sis

@N3buchadnezzar $$\int_0^1 \int_0^1 \frac{y}{x} \left \lfloor \frac{x}{y} \right \rfloor \ dx \ dy$$

@N3buchadnezzar then you may try this version $$\int_0^1 \int_0^1 \frac{y}{x} \left \lfloor \frac{x}{y} \right \rfloor \left\{\frac{x}{y}\right \}\ dx \ dy$$

nullgeppetto

09:08

Hi all! Let $A$ be an $n \times n$ matrix. What could be a reasonable choice for measuring the size of the matrix? What I am looking for is a mapping $\mu: \mathbb{R}^{n \times n} to \mathbb{R}$ that could "measure" the size of the matrix. Something like a norm for vectors. Could determinant be such a measure? What else? To be more specific, having two multivariate Gaussian distributions (two covariance matrices), I would like to be able to talk about "small" and "big" covariance matrix.

N3buchadnezzar

09:40

@Chris'ssis Here?

N3buchadnezzar

10:05

I`ve been looking at this question math.stackexchange.com/questions/430276/… for a few hours now..

$$
\int^{ na }_{ma} \cfrac{ \log(x-a) }{ \, x^2+a^2 \, } \,\mathrm{d}x
=
\cfrac{\log 2a^2}{2a^2} \bigl( \arctan n - \arctan m \bigr)
$$

I know this result to be true, I have checed it numerically for a few hundred values and they all seem to hold.

Now, the integral can possibly be show by using the clever substitution

$$
x \mapsto \frac{at + a^2}{t - a}
$$

By using this and simplifying one finally gets that
$$
I_1 = \int_{ma}^{na} \frac{\ln(2a^2)}{(t^2+a^2)}\mathrm{d}t - I_1
$$

nullgeppetto

@DanielFischer, goodmorning! Could you consider for a bit my question above? About the measure of a matrix. Thanks a lot!

Mike

what properties do you want this to have

N3buchadnezzar

This is a simple equation that can be solved for $I_i$, but doing so leads to something strange
$$
I_1 = \frac{\log(2a^2)}{2} \int_{ma}^{na}\frac{\mathrm{d}t}{(t^2+a^2)}
= \cfrac{\log 2a^2}{2a} \bigl( \arctan n - \arctan m \bigr)
$$
Which is not what one wanted to show! There is an $a$ factor missing, and I can not for the life of me figure out where this error snuck in.

Daniel Fischer

@nullgeppetto There are various matrix norms. Probably an operator norm would be more useful to you than the Frobenius norm.

nullgeppetto

10:24

@DanielFischer, thanks a lot! Have a nice day!

Daniel Fischer

Thanks, I'm trying to have one. Have one yourself too.

nullgeppetto

@DanielFischer, :-)

Chris's sis

11:44

@robjohn I think I've found a way to finish that.

robjohn

@Chris'ssis Good... I have not had time to look into it. Sorry.

Chris's sis

@robjohn I was so depressed last evening I felt I'm completely lost. I couldn't bear that situation. I simply saw no way.

@robjohn I still have to work on it. The way is now clear to me.

robjohn

@Chris'ssis There are times I've just had to step back from a problem for a while. I just keep banging my head on the same walls. Take a break and get a fresh start later.

2

Chris's sis

@robjohn That's a great piece of advice.

Chris's sis

12:08

@robjohn I think this is the most beautiful question I attended this year, but it's so evil.

techno

12:19

hi

can Dijkstra;s SP algorithm be applied for

non directed graphs

doe the graph needs to be directed?

Huy

13:07

Let $A, B$ be two Hermitian matrices and $B$ positive semi-definite. I want to find the eigenvalues of $A-tB$ for $t \geq 0$. As those are the solutions of $$\det (\lambda - (A-tB)) = 0$$ for $\lambda$, I thought I will have $n$ polynomials as a solution. However, in a paper I am currently reading, it is stated "The $n$ functions $\lambda_1, \dots, \lambda_n$ are restrictions to $\mathbb{R}_+$ of branches of the solution $\lambda$ of the polynomial equation $\det(\dots) = 0$.

Is that statement due to different branches of taking the complex root? How would I find "the" solution $\lambda$ of the polynomial equation?

Balarka Sen

13:24

@PedroTamaroff I think it's : Find a non-zero $\mathcal{C^{\large 1}}$ function $f(x)$ over $\Bbb R$ with non-zero derivative such that $$f(f(x)) = 1$$

Dunno, though, have to check later whether this is the only condition imposed on it.

(PS : My believe is that there are none)

@N3buchadnezzar Limit superior?

What's on RH, @Mats?

Mats Granvik

@BalarkaSen Nothing, as impossible to me as always.

N3buchadnezzar

@BalarkaSen ?

Balarka Sen

@N3buchadnezzar limsup, or more compactly, "sup".

Toxicated American English.

@MatsGranvik Are you familiar with Conrey's method of pushin Hardy-Littlwood's proof of "a positive proportion of zeta lies on critical line" to "the positive proportion is 1/3"?

I am looking for it a while.

N3buchadnezzar

@BalarkaSen Check out an hour ago or so in chat, simple problem. Nobody bothered to answer :p

Balarka Sen

Banana comes to the party.

Mats Granvik

13:35

@BalarkaSen No I am not familiar with it. I have heard about the result, but that is about all I know.

Jasper Loy

@BalarkaSen A.

Balarka Sen

@MatsGranvik Are you familiar with Hardy-Littlewood's result?

Mats Granvik

@BalarkaSen No, not really familiar with Hardy Littlewood. My approach to Riemann hypothesis has not ever been to be able to prove that zeros are where they should be. Instead I set my self the goal of finding a formula for the first zeta zeros a few years ago. But at the time I said that, I knew that it is impossible. It is not that we don't know what the zeta zeros are, they are the zeros of the Riemann zeta function. But I hope there would be a simple formula for the zeros, as I said.

Balarka Sen

@MatsGranvik I believe there isn't a simple formula for zeta zeros. In fact, it is a mere consequence of that of the primes that zeta zeros are deterministically random.

@JasperLoy A

Sawarnik

14:12

@BalarkaSen You there?

user52932

16:13

Hey guys I was wondering if anyone knows an answer to this

2

Q: Function integrable over the interior but not over the set

Let $S$ be a bounded set in $\mathbb{R}^n$; let $f:S\rightarrow\mathbb{R}$ be a bounded continuous function; let $A$ be the interior of $S$. Give an example where $\int_Af$ exists and $\int_Sf$ does not. There's a theorem that says if $\int_Sf$ does not exist, then for a set $E$ of measure n...

THe answer posted is a fat cantor set thing

but I was wondering if a simpler example like $\mathbb{Q} \cap \mathbb{R}$ might work just as well

The measure on boundary would be non-zero

d.i...

Hello....Can anyone help me express the statement " Any mammal that has long ears has at least one of its predators with yellow eyes having all of its cubs that cannot fly" in the logical mathematical way.

user52932

and we can let our function be lets say 1

Kareem Mesbah

16:38

Hello, guy! Can you please help me figure out what's objectionable with this proof?

it's a proof that the arithmetic mean is at least as large as the geometric mean

for all real nonnegative integers a and b

(a+b)/2 >= sqrt(ab)

a+b >= 2 * sqrt(ab)

a^2+2ab+b^2 >= 4ab

a^2-2ab+b^2 >= 0

(a-b)^2 >= 0

which is true since squares of real numbers are nonnegative

what could be objectionable here?

Alexander Gruber

16:56

@KareemMesbah well you didn't prove it

Kareem Mesbah

@AlexanderGruber actually, this is not my proof

this is a bogus proof

Alexander Gruber

the proof assumes it is true and then shows it leads to something else that is true

but that doesn't imply that the original assumption was true

for example i could assume that 1=2, multiply both sides by 0, and conclude that 0=0, which is true. but that doesn't mean 1 really equals 2.

Kareem Mesbah

@AlexanderGruber, well, could you please help me fix that?

just gimme a hint not the answer

Alexander Gruber

@KareemMesbah you could try reversing the steps: you want to finish the proof with $(a+b)/2\geq \sqrt{ab}$

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$Mathematics$ Mathematics