Mathematics - 2014-04-24 (page 2 of 3)

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9:08 AM

D: anyone?!

I found it!

@N3buchadnezzar In that number appears an interesting product you might like $$\prod_{n=0}^{\infty} \left(1+\frac{1}{2^{2^n}}\right)$$ (it's an elementary product)

@N3buchadnezzar Do you know where to find AMM papers older than 1973 ?

T_T

@r9m Jstor had the paper I was looking for..

9:17 AM

@N3buchadnezzar okay

@Chris'ssis Question

While doing repeated integration by parts the idea is to find a $u^k$, derivative such that u^k equals some constant.

Is it possible to find an expression where one integrates $v$, $k$ times such that $\int \cdots \int v$ is equal to some constant?

..

@N3buchadnezzar 1 over [0, 1].

@Chris'ssis Is that allowed?

@N3buchadnezzar Allowed to what? Maybe I don't get what you mean there.

9:25 AM

@Chris'ssis I can give an example, okay ?

@N3buchadnezzar by the way, don't miss that product above.

@Chris'ssis I wont ;)

Say $$ \int x^6 \cos x \mathrm{d}x$$

This is possible to integrate by parts (of course), since if we pick $u = x^6$, there exists a $k$'th derivative such that $\frac{\mathrm{d}^k}{\mathrm{d}u^k} u = 0$.

@Chris'ssis It is interesting that that product just came up in one of my answers yesterday.

@robjohn I just took it from American Mathematical Monthly, vol. 67 , 1960 that @N3buchadnezzar mentioned above. (page 693)

@robjohn Where is that answer?

@Chris'ssis actually, it is part of the answer that was redone, but the product is mentioned in this comment.

@Chris'ssis you can look at the earlier edit to see the product.

9:32 AM

@robjohn Thanks. That's the way.

@Chris'ssis of course, the product you give above is $2$.

@robjohn Indeed. It's one of my favourite products.:-)

@robjohn do you agree that it's elementary? I mean this could be easily given on high school tests.

@Chris'ssis what do you mean by elementary?

@robjohn say, that requires at most high school knowledge.

@Chris'ssis I don't know. Most students would look for a pattern after a few terms then try to prove the pattern. That might be a bit much, depending on the level of the course.

9:35 AM

@robjohn OK

@Chris'ssis It requires only high school knowledge, but that only means I might present it to a high school class, not that I would expect them to be able to come up with it on a test.

Hi all! Any help on approximating a quadratic form? Thanks a lot, and apologies for spamming...

@robjohn I remember I gave it to some students in the past and all were in trouble with it.

@robjohn In the same number of AMM it appears a cute integral $$\int_0^{\pi/6} \log^2(2 \sin(x)) \ dx=\frac{7\pi^3}{216}$$

@Chris'ssis I remember doing an integral similar to that in some answer...

@robjohn I rememeber that too. That's why I wanted to show it.

@robjohn At any rate, it's pretty easy. I think I'm going to attend it.

9:48 AM

@Chris'ssis the various parts of this answer should be enough to evaluate that integral

@Chris'ssis except there will have to be modifications for the limits.

@robjohn Right.

hi robjohn

@robjohn !!! hugs

ronjohn I was reading math.stackexchange.com/questions/746895/…

@usukidoll did I do something?

10:01 AM

robjohn I mean :)

ummm can you help me ^^ on this?
http://math.stackexchange.com/questions/767117/let-a-b-in-r-where-a-b-prove-that-there-exist-a-rational-number-c-and/767145?noredirect=1#comment1593788_767145

let me try again.... @robjohn

@user2179021 that is an answer to a question that was changed.

@robjohn The new version looks really hard!

@robjohn The answer on MO is all about log expansion it seems

D:

10:04 AM

@robjohn In any case.. I wanted to say I have really admired your answers on math.se. Are you a professional mathematician?

errrragh

@robjohn Do you have any idea about this question? math.stackexchange.com/questions/765803/…

I want to ditch decimal expansion... it's actually making my problem a lot worse

@user2179021 not any more. I was an assistant prof at UCLA from 1986-1988. Since then, I have just been an amateur mathematician.

@robjohn Well we are all very grateful!!

3

10:07 AM

@usukidoll I think that is a good idea. I would not approach it that way.

good.

because I swear it could be simpler without it

and I've read some practice pdf examples that may sound similar to what I'm dealing with

hmm so how do I prove that c is rational and d is irrational?
like a proof by contradiction for d being irrational?
like if d is rational then $d + \sqrt(2)$ is irrational. Rational numbers are closed under subtraction, so $ d + \sqrt(2) -d=\sqrt(2)$, but we proved that d is irrational so

$a- \sqrt(2) <d<b-\sqrt(2)$ and then we have $a+\sqrt(2) <d<b+\sqrt(2)$ @robjohn

@robjohn I can answer your question re: the probability question

@usukidoll I would use that we know $\sqrt2$ is irrational and find rational $p$ and $q$ so that $p+q\sqrt2$ is in the given range.

@usukidoll precisely.

@robjohn Y is n dimensional. X has an infinite number of entries. So we can always do $\sum_{i=1}^n Y_i X_{i+k} for any k as X goes on forever. Does that make sense?

what about for c being rational ... hmmm...

there exists a rational c such that a<c<b xD

if we let $ a = \frac{d}{e}$ and $b =\frac{f}{g}$ for $e \neq 0 ; g \neq 0$ by the rational defintiion

for integers d e f g

10:18 AM

@usukidoll find an integer $n$ so that $(b-a)n\gt1$. Then, for some $k$, $k/n\ge b$.

errrrrrrr

@usukidoll find the smallest such $k$ and then show $(k-1)/n$ is between $a$ and $b$.

$ a < \frac{k-1}{n} <b$ ?!

@usukidoll yes

and then?!?! wow that's going to be hard a must be less than the middle less than b... eeekk

10:23 AM

@usukidoll no. it is pretty easy.

like 0.99999999999999<c<1

x_X'

how is this easy?

$(1-0.99999999999999)10^{15}=10\gt1$

o-o right

so do I use that as an example to err prove that c is rational?!

so find the smallest $k$ so that $k/10^{15}\ge1$

That would be $k=10^{15}$

k = 1?

oh ._______________.

10:28 AM

Then we have $0.99999999999999\lt\left(10^{15}-1\right)/10^{15}\lt1$

@usukidoll are we trying to prove the result with or without decimal representations ?

how did you get it so fast? let $c = k/10^15$? let k = the smallest integer which is $10^15$?!
So $0.99999999999999 <(10^15-1)/10^15 < 1$

@r9m without... if I do it with I will get letters and numbers everywhere ain't nobody got time fo dat

crud $10^{15}$

@usukidoll well if you want to do it with decimals .. one way is to chop off the tail of decimal representation after a suitable length (that gives you a rational) and add attach a tail of some decimal representation of say $\sqrt 2$ (that gives you an irrational number ) ..

that's what I've wrote when I replied to one person that firegarden guy

so the d part was easy.. I'm working on c

10:43 AM

@usukidoll I wrote an answer.

Karl Kronenfeld

@PedroTamaroff I noticed that, in fact, we have $$M/IM\otimes_{A/I}N/IN\cong M\otimes_AN/IN$$ which I discovered by symbol-pushing. Now I see that it follows from $$M/IM\otimes_{A/I}N/IN\cong M/IM\otimes_AN/IN\cong M\otimes_AA/I\otimes_AN\otimes_AA/I$$ and $A/I\otimes_AA/I\cong A/I$.

WHOA! how did my question got 5 upvotes XD I broke my record

@robjohn this one looks nice $$6\int_0^1 \frac{\arctan(x)}{x} \ dx-4\int_0^{1/2} \frac{\arctan(x)}{x} \ dx-2\int_0^{1/3} \frac{\arctan(x)}{x} \ dx-\int_0^{3/4} \frac{\arctan(x)}{x} \ dx=\pi \log(2)$$

11:16 AM

@robjohn The A_i can't be independent can they?

@BalarkaSen A rational point is a point within a triangle such that its distances to the vertices are rational. Prove or disprove that an equilateral triangle with side 1 has infinitely many rational points.

@user2179021 As I commented, I think the event that $A_i=0$ is independent. I am working on showing this.

@robjohn Thanks. I added a comment to show what is worrying me about that argument

@user2179021 I see... interesting.

11:55 AM

@robjohn I wouldn't like to give you an easy problem :)

@KarlKronenfeld What do you mean by symbol pushing?

@PedroTamaroff "using formulas whether understanding them or not" is what I would mean.

Karl Kronenfeld

@PedroTamaroff I used the following formula to switch the base ring from $A$ to $A/I$, $(M\otimes_A P)\otimes_B N\cong M\otimes_A(P\otimes_B N)$ and it did more than I expected.

12:14 PM

@KarlKronenfeld Ah.

I see.

12:32 PM

Look at the best answer here: answers.yahoo.com/question/index?qid=20080625054518AANd6kb !

Seems you kids learned how to behave. Good.

@PedroTamaroff Is it bad that I just had that line pop up on my sidebar and I knew who you were talking about?

@Sawarnik Makes one wonder what isosceles means.

@DanielFischer Yup.

@PedroTamaroff :D

@meer2kat May be.

Ridiculously gorgeous day today and my tennis student misses classed.

Classes*

12:46 PM

If $r(x)$ is decreasing function of $x$, then how to show that $r(x)\leq y\implies x\geq r^{-1}(y)$?

@Sush You asked this recently I think.

Yes, but didn't get proof, only example.@Sawarnik

Will you please prove it?

@PedroTamaroff I think it's more because I wasn't in this room earlier.

@Sawarnik Is that not a simple problem? AAA.

@ParthKohli No, not the problem, the best answer is classic!

@Sush Ask on the main.

30-60-90 is isosceles? Never knew that.

See, this is why I have never used Yahoo Answers ever.

12:55 PM

@ParthKohli I thought that one comment on the answer saying its wrong, given by "parth" was you.

@Sawarnik I'm not the only Parth.

@ParthKohli You don't even know such basic facts, oh no.

@Sawarnik Sorry, I'd keep the 30-60-90 fact in mind when doing other questions.

1:22 PM

in the following matrix $ X = \begin{matrix} x_{k} & x_{k-1} &... & x_1 \\ x_{k+1} & x_{k} &... & x_2 \\ ... & ... & ... \\ x_{m+k} & x_{m+k-1} &... & x_{m} \\ \end{matrix} $, which are the condition on the series $x_i$ for having X of rank k?

I mean if x$_i = \sum_{1}^p{\alpha_j x_{i-j}}$ would this affect the rank of X

Heya

How can I show that
$$ \sum_{n\geq 1} \left[ \log (n + 1) -\frac{1}{n+1} - \log n\right] = 1 - \gamma$$ ?
I tried a few tricks and mixing, but alas

$$ \sum_{n\geq 1} \left[ \log (1+ \frac{1}{n}) -\frac{1}{n+1}\right] = 1 - \gamma$$

But why

$$\gamma = \lim_{n \rightarrow \infty } \left(
\sum_{k=1}^n \frac{1}{k} - \log n \right)$$

This is the form I am used to seeing, it is quite close

1:37 PM

how is it called in english the $ for x->0, e^x = \sum_{n\geq 0} x^n/n!$

@n11 taylor?

yes also, was thinking to another term

@n11 power series ?

hmm not also, in French we call it "développement limité", but nvm, Taylor is perfect, maybe it could be used for your case

@N3buchadnezzar it can be done mentally.

@robjohn that log(sin(x)) integral is much harder than I initially thought.

1:44 PM

Mmm

@N3buchadnezzar Consider the partial sum, and write $\displaystyle \log(n+1)-\log(n) =\log\left(\frac{n+1}{n}\right)$. When you add up these logs, you get something nice ...

@Chris'ssis I thinkn I have an idea

@N3buchadnezzar This is the idea. Then, you're done.

@Chris'ssis If you do that with k parts you get $ \log (k + 2) $

for a partial sum $log(n+1)-log(n)$ is quite better

1:49 PM

@N3buchadnezzar $\log(k+1)$

$$ \sum_{1\leq k \leq n} \log(k+1) - \log k = \log (n + 1) $$

@n11 ah, yes!

@Chris'ssis Silly mistake

@N3buchadnezzar The left thing is to add and subtract $1$ to get $1-\gamma$.

@Chris'ssis This was the final step to evaluate $$ \int_1^\infty \frac{x - \lfloor x\rfloor}{x^2}\,\mathrm{d}x = 1- \gamma $$

1:53 PM

@N3buchadnezzar These integrals are pretty cute.

yeah, this is actually "homework"

@N3buchadnezzar did you see this one I proposed these days? $$\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \{ x_1 + x_2+ \cdots + x_{n} \} \ dx_1 \ dx_2 \cdots dx_n$$

@Chris'ssis See here.

@Chris'ssis What does $\{ x \}$ mean? $\{ x \} = x - [ x ]$ ?

@N3buchadnezzar It is the fractional part of $x$.

thats what I wrote =)

1:56 PM

@N3buchadnezzar btw, are you a math student now?

?

@N3buchadnezzar just asking ...

Well i do study mathematics and I will write my thesis in analysis. But no I am not a math student :p

applied math = physics

Teacher

1:58 PM

@N3buchadnezzar are you a teacher?

to be

OK. Good luck! :-)

@Chris'ssis Are you a teacher as well?

But yah, this year I am taking pure math subjects. Topology, CA and Analytic NT

@Sawarnik No, I'm just self-made with no background in mathematics.

2:00 PM

self-made = self-taught

Right.

what did the self-taught physicists say to the self-taught mathematician ?

we need to get out more often :D

@Chris'ssis Could you give me a hint for something related?

I do not remember the identities as good as you I think

@N3buchadnezzar For my question you mean?

$$ \lim_{s\to 1} \left[ \frac{\zeta'(s)}{\zeta(s)} + \frac{1}{s-1}\right] $$

2:08 PM

If $r(x)$ is decreasing function of $x$, then how to show that $r(x)\leq y\implies x\geq r^{-1}(y)$?@DanielFischer?

Contrapositive, @Sush. $x < r^{-1}(y) \implies r(x) > r(r^{-1}(y)) = y$, provided that $r$ is strictly decreasing. If it isn't, the conclusion need not hold.

@DanielFischer, thank you so much!

@N3buchadnezzar $\gamma$

@Chris'ssis yeah

@Chris'ssis $$ \frac{1}{1-s} = - \int_0^1 x^{-s}\,\mathrm{d}x = \int_0^1\frac{[x]-x}{x^{s+1}}\,\mathrm{d}x $$

2:27 PM

@N3buchadnezzar I wouldn't think of it but of the zeta function & derivative of the zeta function series around $1$.

$$\sum_{n\leq x} \frac{1}{n^s} = x^{1-s}/(1-s) + \zeta(s) + O(x^{-s})$$
Something like this?

I am very unfamiliar with the derivatives of zeta and even the logarithmic derivative

@N3buchadnezzar See $(47)$ here mathworld.wolfram.com/RiemannZetaFunction.html

@Chris'ssis well using that is clearly cheating

@Chris'ssis The problem asked us specifically to use the integral we derived

@N3buchadnezzar Well, where did you mention that?

@Chris'ssis I can believe that

2:36 PM

@Chris'ssis here

@Chris'ssis I think I got it

well no

fuuuuu, so close

@N3buchadnezzar I think that knowing $\zeta(s)=\frac1{s-1}+\gamma+O(s-1)$ should be enough...

22

@robjohn But how do you know that?

it's not 25. I found it again, guys. bit.ly/1lIFWjt

@N3buchadnezzar I have a post in ASCII about that... I don't think that I have moved that to MSE yet.

2:42 PM

@N3buchadnezzar these are well well-known series expansions ... - terrytao.wordpress.com/tag/riemann-zeta-function

@N3buchadnezzar let me look.

@Chris'ssis But you still need to take the logarithmic derivative of it

@N3buchadnezzar I just gave you a very useful link. Did you even open it? :-)

brb

@N3buchadnezzar If not, use this - math.stackexchange.com/questions/751948/…

@N3buchadnezzar I did move it to MSE. Look at this answer.

@robjohn A very good job! I agree it's enough to only know that $$\zeta(s)=\frac1{s-1}+\gamma+O(s-1)$$

2:55 PM

@Chris'ssis then $$\lim_{s\to1}\left[\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1}\right]=\gamma$$

@robjohn Yeah, BRILLIANT!

@Chris'ssis Ah, in this comment, I even show that :-)

@robjohn I was looking at that! :-)

@robjohn ;-)

goes off trying to prove this without that

@N3buchadnezzar so, there is the complete proof :-)

3:09 PM

..

@robjohn can one use this $$\lim_{x\to 1^+}\left[\zeta(s)-\frac{1}{s-1}\right]=\gamma$$ to prove that?

@N3buchadnezzar it certainly works the other way 'round... let me see.

@N3buchadnezzar yes, you can

just integrate

@robjohn I get $\lim_{x\to1^+} \bigl[ \log \zeta(s) - \log |1 - s| \bigr]$

@robjohn, please pardon me, but, does $G^{-1}(y)\leq z\implies y\leq G(z)$ if $G$ is increasing function (need not be strictly), $0<y,G(z)<1$?

3:24 PM

@FernandoMartin Hullo.

@Sush Which book are you studying from?

This was a recent question of mine:

$$\displaystyle \lim_{s\to 1} \, \left(\zeta (s)-\frac{\zeta '(s-1+\rho _n)}{\zeta \left(s-1+\rho _n\right)}\right)=\gamma -\frac{\zeta ''(\rho _n)}{2 \zeta '(\rho _n)}$$

@Sawarnik, Probability & Statistics, DeGroot, 2nd Ed.

@Sush Ah, both the things which I don't like.

Ok!

3:29 PM

@Sush Do you enjoy any else subject in maths? What?

@Sawarnik, yes.

Calculus.

@Sush if $G$ is increasing, it looks good.

@robjohn, do you mean that holds in general?

Please prove that!@robjohn, I am stuck at Probability integral transformation!

@Sush I guess if it is not strictly increasing you could have problems

@robjohn, Ok!

3:35 PM

@robjohn Can you recommend some online books/resources for studying very basic analysis.

sHALL i ELABORATE MY PROBLEM?@robjohn?

@Sawarnik Need be online, analysis or calculus?

Ok, Sorry.@Sawarnik

@Sawarnik I don't have any online references... I used Rudin, but I doubt that is online.

@robjohn Not a good book to learn from

I have it, and I love it. But it is much more usefull to look up theorems and proofs, than learning the material from scratch

3:37 PM

@MatsGranvik where $\rho_n$ is a root of $\zeta$?

I think this is good amazon.com/gp/product/0123877741/… found it elsewhere for 15 dollars

@N3buchadnezzar are you talking about baby Rudin?

@robjohn, here I am posting my problem as images:

user image

user image

@N3buchadnezzar you should get $\log(\zeta(s)(s-1))=\gamma(s-1)+o(s-1)$

@robjohn, here $G$ is not assumed to be strictly increasing, though $G^{-1}[F(x)]\leq z\implies F(X)\leq G(z)$ is assumed. (I think so.)

3:46 PM

@Sush an inverse has to be 1-1

@Sush by d.f., they must mean cumulative distribution function since they say it is between 0 and 1

@robjohn, so, because G is nondecreasing by being cumulative distribution function and also continuous as given, one to one guarantees that it is strictly increasing, too. Am I right?

@Sush this doesn't make sense. there must be more context that is not stated here

@Sush it's inverse is strictly increasing.

@robjohn, so doesn't that imply that G is also strictly increasing?

@Sush no, consider a uniform distribution on $\left[0,\frac13\right]\cup\left[\frac23,1\right]$

the cdf is constant on $\left[\frac13,\frac23\right]$

Ok,, but if G has its domain 0<z<1, then?@robjohn?

3:57 PM

@Sush that cdf does have domain $[0,1]$

I think cdf has domain entire real line and range is $[0,1]$, @robjohn, sorry!

@Sush their cdf possibly. My cdf was a distribution on $[0,1]$

Ok! still confused! How does that inequality, hold?

☻

@Sush which inequality?

4:17 PM

@N3buchadnezzar this one $$\int_0^1\frac{\sin(\pi s)}{\zeta(s)} \cdot\zeta(1-s) \Gamma(1-s) \ ds$$

4:30 PM

@Chris'ssis Mathematica says $$2\pi\frac{1+4\log(2\pi)}{\pi^2+4\log(2\pi)^2}$$

@robjohn Yeah. :-) (I just created it)

hi, the following set R+ = {x$\in$ R | x > 0} , g:R+ $\rightarrow$ R+. g(x)=(x+1)^2-1, is g one to one functions and onto? by my calcaulations I can get any number in R+, and exactly once so it is, but what is the way that I can be sure?

$\ g(x) = (x+1)^2-1$

R is real numbers set

@SethKeno $g(0)=0$ and $g'(x)=2x+2\gt2$

g(0) is not possible since R+ is a real number > 0

@SethKeno then take the limit.

@SethKeno or consider the function $f=g$ on $R+$ and $f(0)=0$

4:37 PM

@SethKeno If you see it graphically, both the roots are non positive, so yes.

Algebraically, $(x+1)^2-1=(y+1)^2-1$, gives $x=y$. So it is one to one, if I understand your question correctly.

ok thanks I don't understand it myself yet, but I'll go study about functions limits I think the answer lies there...

Since $f$ is continuous and $f(0)=0$, and limit as $f$ goes to infinity is infinty, $f$ crosses every positive no, by IVT. @robjohn Am I doing everything correctly?

4:54 PM

..

@robjohn hi!

@Sawarnik that statement is not true without qualification (e.g. $y=-x-2$)

@robjohn Why not? We know x and y are positive.

@Sawarnik that is why I said without qualification.

Ok :)

@meer2kat have you seen kung fu panda?

5:01 PM

@Sawarnik as they said $f(0)$ makes no sense.

@Sawarnik It depends. Is it going to lead to some ridiculous argument if I say yes? :P

@meer2kat What kind of ridiculity you are expecting? :P

@robjohn Ok then we have to resort to limits, right?

@Sawarnik Yes I've seen Kung Fu Panda

@Sawarnik or something. I gave two options above.

@Sawarnik KFP = Kentucky Fried Poultry?

@Sawarnik any

5:06 PM

:D

ack... more (removed) comments

2

@robjohn i'm tempted to remove my hello to you since you didn't answer :(

@meer2kat you can't, now :P

@Sawarnik i could do it again though

i'm just sad i didn't receive a hello back from him

@meer2kat He only replys to sensible people :P

5:09 PM

@Sawarnik Hmmm he'll be awfully silent then with present company.

@robjohn I can't get a hello? :(

@meer2kat Nopes, don't you understand? :P

@meer2kat Haha, its true. And he is. :D

Well that's quite rude

@Sawarnik watch your tongue kid

@meer2kat Ok, madame.

@robjohn To reply to your message a while ago when I was away from my computer. Yes $\rho_n$ is a zero of zeta. Later Raymond Manzoni gave a very nice answer to a related question.

5:39 PM

@meer2kat hello... don't be so touchy. Things come up all the time and I don't get to reply right away. I didn't see your "hi" because I was elsewhere and Sawarnik's ping was after yours. I didn't see yours until reading my inbox.

@robjohn lol i don't really care :P

@meer2kat oh, so now you don't care :-p

@robjohn ;) :P

@robjohn I was talking about RCA, but for Calculu I think ruding might still be too tough

@N3buchadnezzar oh, just calculus, yeah. I was thinking of analysis.

5:45 PM

I really like the style of Rudin though. It is very compact and straight forward

5:58 PM

Hi @me

"added examples and changed false information about Bambi" lol

Can someone give me a little help with measure?

user image

I'm not quite sure how to parse this...

6:26 PM

To what extent do you apply the path of least resistance when doing mathematics, and interacting on the forum?

Judging on the non-existent replies to this question, you do apply the path of least resistance.

Which is a good thing.

6:44 PM

@N3buchadnezzar I know that RCA is often too slick and this sometimes obscures the idea behind the proof, but PMA is a bit better.

@robjohn Indeed, like I said I love it as a reference. But would not want to learn the material from it

Can you explain something to me ?
\begin{align} \log(\zeta(s)) &=\log\left(\frac1{s-1}+\gamma+O(s-1)\right)\\ &=-\log(s-1)+\log\left(1+\gamma(s-1)+O(s-1)^2\right)\\ &=-\log(s-1)+\gamma(s-1)+O(s-1)^2 \end{align}

@robjohn How does the third line follow from the second ?

@N3buchadnezzar $\log(1+x)=x+O(x^2)$

That makes sense, thanks

@robjohn Hmm, should it not be $O(s-1)^4$ then, since you square the error ?

I'm going to propose this for a contest (newly created) $$\int_0^1 \int_0^1 \left \{\frac{e^{\large x^{\Large\cos(x)}} - e^{\large y^{\Large \cos(y)}} }{e^{\large x^{\Large\cos(x)}} + e^{\large y^{\Large \cos(y)}}} \right \} \ dx \ dy$$

where $\{x\}$ - is the fractional part of $x$

@Chris'ssis 1!

6:59 PM

The previous one was prettier.

@Studentmath What about challenging beauty?

That one always tends to annoy me, until (and if) I get it right at least.. then it's prettier, yes.

@Studentmath I challenge the problem to a beauty contest. I think I'm going to lose though ;)

Just since the judge's have no taste, @meer2kat

@Studentmath i see

7:08 PM

@Chris'ssis My estimate is that it is less than 1

@robjohn Could you help me?

@robjohn True.

@Chris'ssis actually, I'm going to venture that it is $\frac12$

@robjohn Very well said!

I now understand that the for something to be measure zero it needs to have a countable cover- so how can I show that any finite cover, of say [0,1], has length at least one?

7:09 PM

@Chris'ssis no paper needed...

@robjohn Yes! No need for paper! :-)

@Anthony what is the set you are looking at?

@robjohn [0,1]! I know it's covering compact, so every cover has to have a finite subcover.

@Anthony list the set of intervals, and show that their length must add up to at least 1

What do you mean the set of intervals?

7:16 PM

@Anthony an open subset of $\mathbb{R}$ is a union of open intervals.

so an open cover of a compact set is a finite union of open intervals

But so all I know is that given any cover, I can construct a finite subcover. I don't know what that subcover is, right? I mean, is it just very straightforward?

It's also strange how when you have a countably infinite number suddenly the length doesn't have to stay the same...

@Anthony the definition of compact specifies open cover

@Anthony even a countably infinite open cover of $[0,1]$ has to have length totaling at least 1

Thanks for the help.

7:52 PM

intervals? I must be in the right room

8:18 PM

$$\lim_{n\to\infty} \left(\int_0^n \arctan\left(1+\frac{2}{x(x+1)}\right) \ dx - \sum_{k=1}^n \arctan\left(1+\frac{2}{k(k+1)}\right)\right)=\frac{\log(2)}{2}$$

8:58 PM

I don't think I'm going to make it to class

@user127001 which class?

I hate commuting 4 hours every day. I always miss my stop because I'm reading my math books on the bus and it takes even longer

@robjohn modern algebra

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