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16:05
Nooooooooooooooooo, there is a flaw in my proof ....
I expect a counterexample or a proof. — user172903 6 mins ago
Miscommunication ahoy ^^
I get to change my username in a few hours. What should I change it to?
@JasperLoy
16:17
@WillHunting How about "Jasper Loy"?
@ILoveSGandNeverWantToLeave
@DanielFischer Cool. That is the name of the most awesome person on earth.
@nablablah OMG
I am very confused by the names of all the English dictionaries Oxford produces, but I think I have sorted it out now.
@WillHunting BigPanda
livinglanguage.com/languagelab/eslenglish/4361/… Can someone tell me if you can see the flash on this page, I can't.
Works fine for me in Firefox
16:22
@nablablah Do you see a video or an animation there?
No
Just audio
Hi @sarah, miss you, lol.
@nablablah So you do hear something?
Just a slideshow with audio
Yea, the man is reciting various vocabulary words
@nablablah OK, thanks. I don't see it.
@nablablah Oh I see it now, lol.
16:25
Hill Wunting sounds good.
@Sarah Oh I just sent you another email.
@WillHunting who is clarissa?
@Sarah OMG, you are so quick to notice. She is some girl I used to like, but I never had any girlfriend, lol.
Did she like math?
16:29
Or Still Hunting.
@Sarah Not really. Anyway she is history.
So she liked history
@nablablah LOL
@nablablah LOL
@Sawarnik LOL
@HillWunting That is pretty cool. I don't think I could write anything like that.
@Sarah I think I will change my username back to my name in a few hours when I can do so.
16:32
@JaperLoy ok
A question to upvote
0
Q: Prove that $\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$

Chris's sisWhat ways would you propose for getting the inequality below? $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$$ The left side may be written as $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}>\zeta(s)-1$$ but how do we prove then that $$\zeta(s)-1> \frac{...

@Chris'ssis Done!
If $x+y+z=1$, $x^2+y^2+z^2=2$ and $x^3+y^3+z^3=3$ what is $x^4+y^4+z^4$?
@WillHunting Thanks! I knew it was you! :-)
@Alizter Real.
16:34
5 upvotes within 2 minutes! :D
sigh
@Sarah How do you know, lol.
@WillHunting guessed :P
@Sarah I am pretty sure it is complex, because real numbers are also complex!
@WillHunting You and your imagination!
16:37
Yes, right now I am thinking of ... that is my secret ...
@Alizter Isn't it something called Newton formulas, not sure? :O
@JasperLoy tell me one of your secrets
@nablablah Let me think of which one first.
@Alizter I think 25/6
But thats just plugging in values and playing around with newtons formulae
ugly problem
Godddd, O got it!!!
16:40
thats not my problem
@Parth Hey!
literally I didnt make it
Hi @Saw @Parth
@Sawarnik Hello.
@ParthKohli Have you studied sequences
16:41
@Sawarnik Yes.
Who can follow my reasoning?
@Will I watched Good Will Hunting today.
Fred has 3 red balls and 5 blue balls in a bag. How many balls does Fred have? Answer: 10.
7
First, we know that $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}>\zeta(s)-1$$ that's clear.
@Sarah HOLY SHIT!
16:43
Then, we need to prove that $$\zeta(s)-1> \frac{3\sqrt{3}}{2}(\zeta(2s)-1)$$
right?
@Chris'ssis right
@ParthKohli it's base 8. I don't know what your dirty mind was thinking.
3
@Sarah That was amazing.
Agh..my connection :(
@Sawarnik What about sequences? I like sequences.
Sequences is love, sequences is life.
16:47
@ParthKohli Oh what .. are you sure?
@Sawarnik Definitely.
@DanielFischer can you follow my reasoning?
@ParthKohli I was studying them :) [was having some troubles with subsequence problems].
@Sawarnik Which kind?
It doesn't work ...
16:52
😿
17:02
@Chris'ssis let it rest for a while. Work on something else.
17:14
Question, I have the homogeneous linear second order differential equation as follows: $y''+8y'-9y=0$ With.. $y(1)=1,y'(1)=B$ I found the solution, but am having trouble with the following: 'Find B if y approaches 0 as t approaches infinity.' How would I go about solving this?
@Link The solution is a linear combination of an exploding exponential and a decaying exponential. Make $B$ so that only the decaying exponential is present.
@ccorn I think the decaying potential should be.. $Ce^{-8x}$ so I set this equal to B? Or..?
@Link Look at the exploding exponential. Figure out $B$ so that the initial conditions make the $C$ of the exploding exponential equal to zero.
Besides, the decaying one is $\exp(-9x)$
@ccorn I don't think it's exp(-9x) though
But I get the idea, thanks
17:44
No one answered my question here so far
5
Q: Prove that $\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$

Chris's sisWhat ways would you propose for getting the inequality below? $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$$ The left side may be written as $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}>\zeta(s)-1$$ but how do we prove then that $$\zeta(s)-1> \frac{...

Maybe it's hard?
@Chris'ssis Or not attractive. Inequalities probably score less than identities etc
@Chris'ssis Or it demands primarily opinion-based answers and thus risks being closed...
@ccorn In general inequalities are a pretty hard part of mathematics , one needs years of training for doing well.
@Chris'ssis undisputed
Omran Kouba seems a fan of them ... mayhe he is going to answer ...
Waiting for deoxygerbe to clarify his needs in this question
18:00
@Sarah I don't get why it is base 8?
@Alizter Yeah, maybe ...
@Alizter what?
@Sarah that problem that is pinned
@Alizter oh lol. dw
18:24
@ParthKohli Good for you.
I think maybe I should change my colour.
18:49
@robjohn I finally posted the previous inequality on main
6
Q: Prove that $\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$

Chris's sisWhat ways would you propose for getting the inequality below? $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$$ The left side may be written as $$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}>\zeta(s)-1$$ but how do we prove then that $$\zeta(s)-1> \frac{...

19:03
@Chris'ssis I posted a problem for you.
$$(1-t)^{1/2}\sum_{n\geqslant 1}(t^{n^2}-2t^{2n^2})$$ has a limit has $t\to 1$. What is this limit?
@PedroTamaroff I might have time to think of it if you prove the inequality I posted above. All I know about it is the fact it's elementary.
@PedroTamaroff IFF the limit exists, then the limit is $$\frac{\sqrt{\pi}}{2}(\sqrt{2}-1)$$
@Chris'ssis OK?
I gather that $$(1-t)^a \sum t^{n^a}\to \frac 1 a \Gamma(\frac 1a)$$
@PedroTamaroff Some of such limits are presented in Polya's book.
@Chris'ssis Yes, it is from P&S.
@PedroTamaroff Did you read it entirely?
19:16
@Chris'ssis I have both volumes, it takes some time....
But eventually, I guess I will.
@PedroTamaroff Yeah, it's a stuff that needs some time.
@DanielFischer
@DanielFischer I wonder if one can prove the following "elementarily."
Sure, @Pedro. Just ask Holmes. He'll tell you it's elementary.
19:23
Suppose we have a function $f:B(c,R)\to\Bbb C$ that can be expanded in a powerseries around every point of the ball, with a positive radius of convergence. Then the series around $c$ converges at every point of $B(c,R)$. This is a theorem of Hurwitz.
I haven't thought about it, it might not be too hard...?
This was mentioned in Remmert after we used the identity theorem to show that singularities exist in the boundary of convergence.
I downloaded Funktionentheorie
But I might not be able to read it...
Hehhehehe.
@PedroTamaroff $c\in \mathbb{C}$ and $B(c,R)$ is the open ball with centre $c$ and radius $R$?
And by "elementarily", you mean? No Integral formula and such?
@DanielFischer Right.
Apparently Hurwitz does this in his book with Courant.
I have a "copy".
But I cannot read German.
I haven't.
But I can.
19:26
gen.lib.rus.ec
Pirating books?
@DanielFischer Just some pages.
"Auf der Peripherie des Konvergenzkreises von $\mathfrak P(z/a)$ liegt immer mindestens ein singularer Punkt."
That I can guess.
@DanielFischer Let me see if I can sketch the proof.
It suffices we show that above, because if the radius of convergence was $\rho < R$, there would be a singular point in $B(c,R)$ which is absurd, since by hypothesis we can expand $f$ in powerseries around every point.
I mean, provide an elementary proof of this fact, which Remmert proves using the identity theorem (this I think is elementary, at least the id. principle for powerseries) and using Cauchy-Taylor.
What is Cauchy-Taylor?
That if we have $f\in\mathcal O(D)$, around each point we can expand it around a convergent powerseries with radius at least the distance from our point to $\partial D$.
In the case of a ball, we get the best fit, so the convergence is throughout the ball.
@PedroTamaroff Do Hurwitz and Courant use estimates on the coefficients for the Taylor series centered at other points?
19:34
@DanielFischer I think they do. They use the "rearrangement", to expand using binomials around other points.
@PedroTamaroff On page 50, near the bottom, "Folglich gilt $$\left\lvert\frac{1}{n!} \mathfrak{P}^{(n)}(a)\right\rvert \sigma^n \leq g",$$ How do they prove that? Is there something pertinent in §3?
I am feeling very restless now...
@JasperLoy you didn't change your username
@nablablah In a few hours, patient.
Sorry @JasperLoy
@JasperLoy Insomnia?
Hi @Saw
19:47
@nablablah Hi
@nablablah Not really. It is the restlessness I get from not having lived life at all. My real life begins when I get well...
@JasperLoy How do you get well
@nablablah By sorting out my thoughts. I intend to get well by the end of next year, as mentioned.
@DanielFischer I have no idea. =/
I cannot read German.
@JasperLoy Why does it take so long to sort out thoughts
19:49
I have to fly now.
@nablablah Because OCD is deep shit, bro.
@JasperLoy Is there a cure for OCD
@PedroTamaroff Bye. I'll think about proving that inequality without integration.
@nablablah It depends on how you define cure and how you define OCD. For me, there is a cure for OCD, and that cure is to sort out my thoughts on my own...
@JasperLoy Do you also count your steps to make sure they follow a pattern
19:51
@nablablah No.
I have to count by 2s in pairs
Otherwise I have to go back and start over if it doesn't match up
@nablablah Try to stop counting.
@sarah Are you still here?
what does the last part of: Let $E$ be nonempty subset of $\mathbb{R}$ and $f:E\to \mathbb{R}$, I don't know that notation. Does that mean the domain of $f$ goes to the reals?
@yiyi $f$ maps elements of $E$ to $\Bbb R$, yes
@yiyi You have never seen this notation before? Have you studied functions?
19:57
@WillHunting it is my first time studying analysis.
@yiyi OK, but you should have studied notation of functions before this, right?
@BalarkaSen just making sure I got the idea and terms correct so the
"Image" of $f$ is the reals.
OMG, what are they teaching kids in schools these days?
@WillHunting I took calculus, but never had this notation before.
@yiyi Yes.
19:58
@yiyi It depends on your definition of image.
@yiyi What kind of notation did you use for a function $f$ from $A$ to $B$?
@WillHunting Image is a fancy word for range.
@yiyi Some authors use domain and codomain. Range and image may mean something else you know.
OMG, this is terrible.
@WillHunting thought that domain was a set of the codomain.
@yiyi Huh?
20:01
@WillHunting that is why I am here asking questions to make sure I get the concepts correct.
Study basic set theory before jumping in and out of calculus, you know.
In mathematics, the codomain or target set of a function is the set Y into which all of the output of the function is constrained to fall. It is the set Y in the notation f: X → Y. The codomain is also sometimes referred to as the range but that term is ambiguous as it may also refer to the image. The codomain is part of a function f if it is defined as described in 1954 by Nicolas Bourbaki, namely a triple (X, Y, F), with F a functional subset of the Cartesian product X × Y and X is the set of first components of the pairs in F (the domain). The set F is called the graph of the function. The set...
@BalarkaSen example $\sqrt x$ has a domain of $(0,\infinity)$ but the codomain is $\mathbb{R}$
Why does the chat have to make the Wiki thumbnail so huge
It covers like half the chat screen
20:02
@yiyi Oh gawd!
@nablablah It looks great though.
@WillHunting do you have another link, because wikipedia won't load in my country.
@yiyi May I know which country?
China maybe.
@WillHunting China.
20:03
@yiyi China bans Wikipedia???
Seriously^
@yiyi the domain and codomain are just any two sets in general; they needn't have any relationship or intersection with each other
OMG. This world is fucked up.
@WillHunting not exactly, but any https link has low frequency of working, and just in general wikipeida for where I am doesn't load often, but if I goto the fancy Hotel in town, it loads just fine.
You know, I am sick and tired of this world...
20:05
@WillHunting also, anytime something happens like in HK, the non-chinese web doesn't work so well.
Also, what exactly is $\delta$ in the $\delta$ $\epsilon$ proof. Is that $dy$ like in basic calculus.
Hmm. @yiyi Can't you be arrested for disclosing things like these then?
I mean the $\delta - \epsilon$ proof of a limit.
@yiyi I did not know China is as bad as the Muslim countries...
@Sawarnik I get a little leaway because I am not chinese, but if I say things that are not in line with the party on political issues, then yep. But saying the web sucks and general complaining they don't mind.
@yiyi I don't like my country either.
20:08
@WillHunting I only know China.
@yiyi I live in another country in Asia...
@WillHunting China can hide secrets that you can't imagine .. communist country, no democracy, state media ...
I am so sick and tired of this world...
Human beings can be so stupid and evil...
@yiyi Where are you studying real analysis from?
@WillHunting they can't be that bad, they do math.
20:10
@WillHunting Yup.
@yiyi Where are you studying analysis from?
@Sawarnik An introduction to analysis william r. wade.
@Sawarnik Only the Buddha can help me make sense of this world...
@yiyi It's a good book.
@WillHunting Amazon reviews don't tell so.
I like the book, just wish I got more feedback from the class, just turn in homework and get a grade. I download other books on analysis and they are seem different because they talk about metric space.
Many people don't even have a life. They just work from morning to night to make enough for a living. This is very sad.
20:19
I don't know, I just bought introduction to real analysis bartle and sherbert, and it seems good for me. I was grinding my way through it.
But any book, would tell you what delta or epsilon is? And functions?
@Sawarnik It's OK, but it contains very little material.
@WillHunting In comparison to?
@Sawarnik Well, it is expensive (I know you have the cheap version) and it does not treat multivariable case, lol.
@WillHunting Dammit, I am far from the multivariable case .. with my current pace it would take a year.
@Sawarnik It's OK. You are still young. You should enjoy yourself and get a girlfriend, lol.
20:23
Not really, still trying to figure out what delta is, as in what is the purpose.
@yiyi I am sure Wade treats epsilons and deltas...
@DanielFischer You are all I need, lol.
@WillHunting Girlfriends? I am tooo shy.
it says that $\delta$ is the tolerance allowed in the measurement $x$ of $a$ which will produce an approximation $f(x)$ which is acceptably close to the value $L$.
@Sawarnik You may study my 12 holy books to get a first class math education...
20:24
that is the only line i can find that explains what delta is.
@WillHunting No need, I can only study from what is reasonably priced.
@Sawarnik All my holy books have very very cheap editions...
what i understand is that $\delta$ is like $dx$ of $\frac{dy}{dx}$, the small change in $x$.
@WillHunting In my country?
@yiyi You should really read the book from page 1.
@Sawarnik abebooks.com has cheap versions at least.
@DanielFischer That's good news :-)
Thanks.
Please star this. abebooks.com has cheap international editions of most expensive math books. Otherwise get from amazon.com
@WillHunting I understand the previous two chapters, chapter one was basic stuff like inf and sup. Chapter 2 was series which was harder but not difficult, but chapter 3 the notation and the examples have been too simple, just need to check to make sure that I am correct.
@Sawarnik, that is what I was thinking, would have been great to include that in the book.
@yiyi Oh, this diagram's not even in the book?
@Sawarnik there are no diagrams really, just a few of some of some graphs. The one for the sandwitch theorem looks like a child drew it.
20:32
Under the Einstein notation convention, why isn't the quantity $\partial_{[a, v_b]} = \frac{1}{2}(\partial_a v_b - \partial_b v_a)$ always zero?
im going back to study.
@yiyi Bye :)
@Sawarnik Rudin's books have no pics, lol.
I should go to.
Bye.
@WillHunting eyebrows
@DanielFischer What English-German dictionary do you recommend?
@yiyi You misspelled sandwich.
20:39
@WillHunting Google Translate
@WillHunting Oh yeah!
Witch.
@nablablah Was my previous profile pic better or this one?
@Saw I think your handsome face is much better
@nablablah Oow, how do you know I am handsome?
@sarah Are you still here?
@Sawarnik Because he saw your pic?
@WillHunting It was too small to judge whether I am handsome or not though.
20:48
(-:
@iceboy likes to repeat after me, lol.
.
..
...
..
.
@anon is very quiet.
I am thinking of changing my pic, what should I use?
It is Hari Raya Haji here today.
@WillHunting What?
@Sawarnik A public holiday.
21:00
@WillHunting In Singapore? Is it something religious?
@Sawarnik It's for the Muslims.
You mean something like Bakri-eid, which I think is to be on Monday?
@WillHunting Strange name though.
21:17
On the last line, where is $a$ gone for $\frac{1}{a-1}$?
looks like they forgot it
But the limit is $\frac{1}{a-1}$
Wolfram confirms it
Any idea ?
well, the difference between a/(a-1) and 1/(a-1) is 1, so presumably there's a missing 1 somewhere
I don't see where it is though
But wolfram does say that the limit is $\frac{1}{a-1}$, and @Chris'ssis confirmed it. The proof is by @BalarkaSen from a while ago.
But he's not here
@TheGame What did I confirm? I didn't read that. Now I'm working on something else.
21:31
@Chris'ssis The limit
1/2013
For$\sum_{k=0}^\infty\dfrac{1}{\binom{2014+k}{2014}}$
@Chris'ssis Could you please read the pic above ? It should be fast
I don't see where the $a$ is gone in the last line
@TheGame Something looks weird there.
What is it ?
@TheGame What you pointed out above.
@Chris'ssis I mean, what exactly is wrong in the proof ?
I've been searching for a while but I can't find the bug
The bug? Is this software? LOL
21:41
@WillHunting Stop laughing and help us >:c
@TheGame Sorry, but I work on something else now. I don't have time for searching the mistakes in Balarka's proof.
@Chris'ssis Ok :)
I'll be off then
@BalarkaSen Ping ping ! Look at the discussion above, where is the trick ?
Have a good day/night/whatever if you live on Mars like me :D
@robjohn you need to see something interesting though (as regards that question)
@TheGame Good night!
@robjohn $$\frac{1}{x(1-x^2)}\ge \frac{3\sqrt{3}}{2}, \space x \in (0,1) $$
@TheGame See you in your dreams!
@robjohn Anyway, I'll think of it tomorrow ...
This inequality must be the key ... (the author also uses it for other inequalities, pretty frequently)
21:53
I am going to bed.
22:05
Hello Professor @TedShifrin
"Calculate $1.05^x$ by hand" Geesh, highschool students sure get fun problems these days
*x should be 45 silly me
22:24
without a calculator?
some teachers are just old-fashion, I guess
$$(1+x)^a \approx \exp\left( \sum_{k=1}^N \frac{(-1)^{k+1}}{k} x^k \right) \ , \quad N \geq 1, \ |x|<1$$
I guess
23:16
@N3buchadnezzar your rhs is missing the $a$ in front of the sum
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