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00:06
@MaryStar If it were $w_{xt}$ rather than $w_{xt}^2$, but the exponent makes things difficult
So, it cannot be written in that form, right?? @robjohn
@MaryStar in what form?
$\int_0^1 \left (\frac{d}{dt} ... \right ) dx$ @robjohn
@MaryStar Not that I can see immediately. If that is all you have, at least.
Are you familiar with the energy method?? @robjohn
00:11
@MaryStar what are you calling the energy method?
@MaryStar Can you give an example of the energy method?
00:28
@robjohn @Committingtoachallenge

We use this method to show the uniqueness of the solution of a PDE.

For example, in my notes there is the following example:

$$u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \\ u(x, 0)=0 \\ u_t(x, 0)=0 \\ u_x(0, t)=0 \\ u_x(1, t)=0$$

$$\int_0^1(u_tu_{tt}-u_tu_{xxtt}-u_tu_{xx})dx=0 \tag 1$$

$$\int_0^1 u_tu_{tt}dx=\int_0^1\frac{1}{2}(u_t^2)_tdx=\frac{d}{dt}\int_0^1 \frac{1}{2}u_t^2dx$$

$$\int_0^1 u_t u_{xxtt}dx=-\int_0^1 u_{tx}u_{xtt}dx+[u_t u_{xtt}]_0^1=-\int_0^1\frac{1}{2}(u_{tx}^2)_tdx$$
@MaryStar You have extra information: $u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0$
@MaryStar I am off to the park with my dog.
@robjohn In the case that I am looking at it is as followed:

$$w_{tt}(x, t)-w_{xxt}(x, t)-w_{xx}(x, t)=0, 0<x<1, t>0 \\ w(x, 0)=0 \\ w_t(x, 0)=0 \\ w(0, t)=0 \\ w(1, t)=0$$

$$\int_0^1(w_tw_{tt}-w_tw_{xxt}-w_tw_{xx})dx=0 \tag 1$$

$$\int_0^1 w_tw_{tt}dx=\int_0^1 \frac{d}{dt}\left (\frac{1}{2}w_t^2\right )dx=\frac{d}{dt}\int_0^1 \frac{1}{2}w_t^2dx$$

$$\int_0^1 w_t w_{xxt}dx=-\int_0^1 w_{tx}\frac{d}{dx}[w_{xt}]dx=[w_t w_{xt}]_0^1-\int_0^1 w_{xt}^2dx=-\int_0^1 w_{xt}^2dx$$

$$\int_0^1 w_t w_{xx}dx=\int_0^1 w_t \frac[d}{dx}[w_x]dx=[w_t w_x]_0^1-\int_0^1 w_{tx}w_xdx=-\int_0^1 w_{tx} w_x dx=-\i
Ok... Maybe you have later a little time for that! :-) @robjohn
01:02
@KarimMansour Every time I hear that song, I think about drugs
lol why I like listening to this song while reading
it has a nice tone to it
I suppose I'm missing the point of this question: math.stackexchange.com/questions/1199089/… when students are taught about logic gates, are they not allowed to use XOR gates? (what I would have thought was the most basic of gates)
WTF am I saying
@JMoravitz it's not a "fundamental" gate if you will it can be composed from them though
@robjohn pictures of mars coming up (50mm lens)
Big red dot = mars, if you look close you can really see it.
Slightly longer exposure
@Committingtoachallenge still using it?
01:29
@AlecTeal Yes, nice and orange
@AlecTeal There is my picture of Mars
Wow nice
@ᴇʏᴇs I took that over my house, which I later found out is a great source of thermal turbulence. Thus, there was not much hope for the picture. This was before I knew about image stacking and before I had a decent camera. An 8" scope is capable of a much better image, but this was back when I was just starting in astrophotography.
01:49
Better than any pic I could have taken
Any close-up view of Mars is nice
02:31
@robjohn got any other pictures? Also it's amazing to think there actually are planets up there.
Like I know they are but seeing your picture makes it "real"
03:26
Hello everyone!
Hi pal :-)
@MikeMiller do you recall the name of that virtual DJ site?
03:44
12 hours ago, by robjohn
@AlecTeal I don't have any eclipse photos here, but I have some other astrophotos.
@infinitesimalsimplicio plug.dj
04:12
@infinitesimal how's life?
04:32
@infinitesimalsimplicio you changed your name?
Hey guys
Who can help with Gram-Schmidt orthogonal process?
0
Q: Finding the orthogonal projection $P_M(x,y)$ where $M=\{(x,y)|2x+3y=0\}$ using Gram-schmidt

MuskConsider the set $M=\{(x,y)|2x+3y=0\}$, in $(R^2,||.||_2)$. Given $(x,y) \in \mathbb{R}^2$, find $P_M(x,y)$. We've to use Gram-Schmidt orthogonal process. And how do I find the orthonormal basis e1 in M and subsequently $P_M(x,y)$. Any help will be appreciated!I'll be really grateful if anyone c...

04:47
@MikeMiller thanks pal :D
@JulianRachman still alive, thanks. How about you?
@StanShunpike yep, do you like it?
hey guys
I need some assistance from you guys regarding this one question?
Consider the set $M=\{(x,y)|2x+3y=0\}$, in $(R^2,||.||_2)$. Given $(x,y) \in \mathbb{R}^2$, find $P_M(x,y)$. We've to use Gram-Schmidt orthogonal process. And how do I find the orthonormal basis e1 in M and subsequently $P_M(x,y)$.
I got something but I'm not sure if it right or not?
I got $P_M(x,y)$ = $((2x+3y)/12, (2x+3y)/12)$?
05:06
@infinitesimal same. Still living. I finally got to the topology section of Simmons text
@infinitesimalsimplicio apparently there is a footballer named Fábio Henrique Simplício who earns €1.8 mil so you're in good company :D
I'll be really grateful if you could help me :')
Did you get the same answer?
05:56
Are these two expressions equal want to make sure my answer is infact correct to the answer given:
-18x +4x^2 +15x^3
and
4x^2 -18x + 15^3
mine is the first one, the answer is the second one
is there some sorta "best" practice about not putting negatives first in an expression ?
It is usually best to put powers of x in either increasing or decreasing order
what you say is "the answer" ends with $15^3$, but yours ends with $15x^3$
If you are asking US to check YOUR work, it will be ineffective if you type inaccurately.
what do you mean inaccurately
power of 3 didn't copy across so i had to use ^3 to represent it
oh wait
I mean if what you type is not EXACTLY what you have on paper, that mistake alone prevents us from rendering a correct judgment.
i typed the second expression wrong :P
-18x +4x^2 +15x^3
and
4x^2 -18x + 15x^3
i notice the answers never put negative numbers first in the expressions
don't know why
Yes, those are the same-has no on ever told you that a + b + c = b + a + c?
06:06
yeah but is there like a preference to the order in which people write it. like always better to put positive number first etc
There is no "one correct way" to write a sum, no
okay
thanks :)
at least that means i'm getting it correct then :D
Books often put negative terms "in the middle" so as to have to print one less character.
Why don't you factor out x?
well i gathered i couldn't get any simpler than what i currently got ?
06:09
ok
What does the question ask for?
to simplify the expression
Then you are correct :-)
what does factoring out x mean exactly?
ain't got that far in the learning module yet
ba + ca = (b+c)a
a has been "factored out"
aka the distributive property :-)
is that a form of simplification then ?
06:14
Your ans is simplest
its certainly simplified when reading it at least
so can mine also be factored out ?
06:57
hi guys
07:17
Hi pal :-)
@Dave yes, you can factor out the x; but if the question asks for the simplest form, then your answer is correct.
07:42
@Dave You can think of factoring expressions as like factoring integers into primes
Just as $2x + 4x^2 = 2x(1 + 2x)$, we have $6 + 36 = 42 = 6(1 + 6) = 6(7)$
We can factor $6$ further as $2(3)$. To see how these things are the same, suppose $x = 3$.
@DavidWheeler whats the question
@SayanChattopadhyay Dave is learning how to manipulate polynomial and multinomial expressions
08:34
@Dave: do you have ChatJax installed? I notice that you are not using latex, and also that David is using latex. Does what David writes appear rendered for you or as plain text?
Does $x$ have dollar signs around the x?
I see that I am late to the party :-)
-_- zzzz
I was just worried that Dave could not see the last bit that David wrote.
I find it interesting that he doesn't know what "factor out" means.
-_- zzz
hi @sayan (I'm back)
 
1 hour later…
09:58
Greetings
Salutations
@robjohn Have you seen the problem 1.80., (c) from Ovidiu's book?
$$\lim_{n\to\infty} n \prod_{k=2}^{n} (2-e^{1/k})$$
@Chris'ssis I don't have that book, so I don't know.
@Chris'ssis Ah, that looks familiar... I may have worked it somewhere...
@robjohn It's an open problem, but I'm not sure it has a closed form. I don't see yet why it would have a closed form.
@robjohn As far as I know, the author doesn't have a solution either.
10:34
Hi @infi.
This room seems more appropriate.
True @Lord_Farin
:-)
How are you doing?
Fine thanks, how are you?
I'm also fine, thanks.
10:38
@infi Any recent mathematical endeavours?
Not really, due to time constraints. How about you my friend? @Lord_Farin
@robjohn the major problem is that the $n$ in front doesn't have any impact on the first factors of the product, but far later on the remaining factors. Since we cannot control the first factors, we cannot find a closed form.
@Chris'ssis That is the problem with many products. The $n$ simply makes it converge.
I still haven't looked for that animated book on logic @Lord_Farin what was it called again?
@robjohn I don't know why the author considered it.
@robjohn Exactly.
10:44
@Chris'ssis considered what?
@robjohn Proposed that limit as an open problem.
@infi I've bought some books, they arrived yesterday. They're on set theory (forcing) and topology (homology).
I don't remember the book you're talking about :(.
@Chris'ssis because it is not easy, but simple enough to state to be interesting.
@robjohn I cannot work on such a thing.I would spend a lot of precious time dedicated to other part of my reserach.
@Chris'ssis Not all problems are for all people.
10:54
@robjohn The reason for that I won't work on it it's not because it is hard, but because the actual mathematics do not have solutions for the finite product I mentioned above. I'm not aware of such a product.
When $n$ is large enough we see the first factors have a word to say about the value of the limit and since they cannot be controlled, there is no sense to work on an illusion.
@robjohn I doubt any being on earth can finish that one.
The limit produces some illusions until one understands that one cannot do any progress on it without knowing the value of the product $$\prod_{k=2}^{n} (2-e^{1/k})$$
@robjohn I solved all the open problems from his book that were solvable (with one exception or so).
11:15
Hello, i have this set $A=\{(x,y)\in \mathbb{R}^2, x>0 xy=1\}$ you can prove that is closed using sequences ?
hello, off topic. if holomorphic functions have antiderivatives in open rectangles and discs, is there a nice way to see that they also have antiderivatives in convex open sets?
if i tke $(x_n,y_n)\rightarrow (x,y)$ i must show that $(x,y)\in A$ (x_n,y_n)\in A means that $x_n>0$ and $x_ny_n=1 $ but when i go to the limite i will obtain that $x\geq 0$ and $xy=1$ so (x,y) is not in A
@robjohn have you n idea please
11:32
@Chris'ssis If you've solved the open problems, then you should publish them.
@robjohn I sent one of them to AMM (some months ago).
@robjohn One of them I also showed you here on channel. The triple integral with fractional part. When I sent it to Ovidiu he told me no one sent him the solution before.
(referring to the fractional part integral)
There is an article under work with that one.
@robjohn in some cases the rigorosity level is not satisfactory enough to publish an article. I mean it's not really a good idea to send an article without clearly explaining all steps. They might not consider you a serious subscriber.
@robjohn some are intrigued by my attitude, sometimes, but there is a truth I wanna mention though, I worked incredibly crazy hard to get at this point, I cannot have small expectation from myself.
The fact that I solved the problems of Ovidiu it's only because I worked extremely hard and did a lot of research. It's not that I'm more gifted than X, Y, Z, no.
11:49
@Vrouvrou Try $x_ny_n-xy=(x_n-x)y_n+x(y_n-y)$
I mean I payed the price for that and did many sacrifices.
@Chris'ssis I'm pretty sure it's a combination of both.
But it's a truth that many who are gifted have issues with working "incredibly crazy hard".
@Chris'ssis Hard work for a given goal does not always produce the results one wants. If you've gotten what you want, then you are lucky.
@Lord_Farin Well, I don't consider myself gifted, but I only worked very much and did research and then I got the expected results.
@Chris'ssis It's a truth that many who are gifted have issues with acknowledging they are gifted :).
11:52
@Chris'ssis I found your paper
There are more papers.
@Alizter I'm not glad to be stalked. Congrats, btw.
@Chris'ssis It looks very interesting :)
@Alizter Thanks.
@Chris'ssis You know that series with the falling factorial?
@Alizter Yes, I sent it to the American Mathematical Monthly. The very sad part is that things there goes extremely slow.
I wonder if in this life they will manage to look over my work.
@Alizter I developed that one very much.
11:59
@Chris'ssis I see how you would evaluate it for small falling factorials but when you get to larger ones say $2011$, it becomes extremely difficult to evaluate. Is it then done indeirectly from there?
@Alizter For finding that one I used a new technique, say, I developed the last days. I mean I reached that point in a very curious way. You understand that when you see my proof.
@Chris'ssis When will I see your proof?
@Alizter I've been waiting for receiving an answer from AMM. I prepared an article for that.
@Chris'ssis Hmph journals can be slow :(
@robjohn Yeah, I might have had some luck.
@Lord_Farin And I missed one very important thing, the passion. :-)
12:05
I think gifted people are just people that are very good at looking like they do not work hard.
@Chris'ssis Passion is everything. Without passion, it would become a drain on one's energy, and one would quit eventually, be that sooner or later.
2
@Lord_Farin True!
@Chris'ssis I learned that the hard way. When pursuing PhD positions, I felt that drain, so I quit. That was just over a year ago. Only now, the passion has come back, and I have started maths again, buying books, contributing to MSE, loving to talk about maths.
@infi Ah, that book :). I remember now.
12:10
@Lord_Farin There are so many things that create the way to success, sometimes it takes us much time to discover them.
Besides that, I'm also very ambitious, I never give up, not matter how hard is the pressure on me. I'm there ready to die than to give up. Giving up is not an option.
In general, easy is not an option.
@Lord_Farin I followed some of my dreams in the moments when no one believed in me, I was the only one that was believing in myself 100%, being alone on the way. Then, nice things began to happen. I mean I payed the price for all I got so far, I was willing to pay it.
@Chris's Most commendable. I respect you for that.
@Lord_Farin I'd like my book also encourage people follow their dreams and never give up, no matter how hard the way might seem.
@Chris'ssis I see. It's good that you want to transfer your drive and determination to others. But many will not be strong enough.
@robjohn what i must see with this ?
@Lord_Farin As you mentioned, one also needs a burning passion.
@Lord_Farin I learned one important thing in the passed years: no matter what you do, try to see the beauty of the things you do. That might produce that flame of passion.
2
12:23
the thing is you can't teach anybody passion
it comes from within
@infinitesimalsimplicio I know, but trying to see the beauty in anything you do might help I think. I tell that from my experience.
@Chris'ssis A good advice, thanks. I will try and heed it.
I see a different picture in front of my eyes.
But it will be hard. Most consider me to have a negative mindset. Although I only try to keep in touch with reality.
@Lord_Farin No one is perfect, people might have different expectations from us. I'm sure there are also people that appreciate and accept us the way we are, those that share the same ideas with ours.
12:29
@Chris'ssis This is true. Although one can also appreciate and accept without sharing the ideas.
@Lord_Farin Sure, agree.
@Vrouvrou Use that with the fact that $(x_n,y_n)\to(x,y)$ to show that $xy=1$.
hi @Chris'ssis hows your progress on your book
@SayanChattopadhyay Very good, thanks. :-)
12:31
@Chris'ssis have you decided how many chapters you are going to have?
@infinitesimalsimplicio The book is almost done. Yes.
when is it releasing?
@SayanChattopadhyay It depends on the publisher. TedShifrin said this process might take even years. I hope it was more of a joke.
it is bad if it is years
12:35
Nah, not bad. It means @Chris'ssis can start on Book 2 :-)
but @Chris'ssis has get reviews for book 1
I'm out to buy some food for my pets. Then I have 1 tutoring hour. Back later.
$(x_n,y_n)\rightarrow (x,y) x_ny_n-xy=(x_n-x)y_n+x(y_n-y)\rightarrow (x-x)y+x(y-y)=0$ so $x_ny_n\rightarrow xy$ but when $x_n>0$ then $x\geq 0$ is $(x,y)\in A$ ?
@robjohn
Hello @DanielFischer!!! I am asked to write an algorithm and there is a hint that I have to use the 'allocation procedure'. What is the latter about?
12:54
@evinda I don't know. You should have something in your course that tells you what "allocation procedure" refers to. By the name, it has probably something to do with allocation, but whether it's an algorithm for allocation of memory as used in an operating system, or a given procedure that an application calls to have memory allocated, or maybe something else, I don't know.
$$\lim_{x\to\0} [\frac{1-{\sqrt{1-x^2}}}{x^2}=1/2$$
how to prove that this limit exists by epsilon delta method
13:16
I personally dislike the epsilon delta method, it seems to be more of a way to understand the concept of a limit than a method of evaluation...
i can see that this can be done without epsilon delta but how with epsilon delta @teadawg1337
@Sayan I don't know, I've never had a firm grasp on the epsilon delta method... Evaluating limits analytically is more intuitive IMO
@Sayan Sorry, I wish I could help :(
no prob @teadawg1337
Analytically, it's trivial @Sayan. Just multiplying by the conjugate of the numerator gives you $\frac12$, so I further question the effectiveness of the epsilon delta method...
13:40
Hello @robjohn!! Could you take a look at my question that I wrote in the edit??
0
Q: Energy Method to show uniqueness of solution of PDE

Mary StarIn my notes there is the following example about the energy methhod. $$u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \\ u(x, 0)=0 \\ u_t(x, 0)=0 \\ u_x(0, t)=0 \\ u_x(1, t)=0$$ $$\int_0^1(u_tu_{tt}-u_tu_{xxtt}-u_tu_{xx})dx=0 \tag 1$$ $$\int_0^1 u_tu_{tt}dx=\int_0^1\frac{1}{2}(u_t^2...

Do you have an idea??
13:50
@SayanChattopadhyay Do you really NEED an epsilon-delta proof, or can you use the rules of limits?
see @DavidWheeler the question says prove the limit exists....i did it by taking the conjugate but i felt like using the epsilon delta proof but then i got stuck at a point
If you already proved it analytically, there's no need to prove it again @Sayan
You see, then, that $\lim_{x \to 0} \dfrac{1-\sqrt{1-x^2}}{x^2} = \lim_{x \to 0} \dfrac{1}{1 + \sqrt{1 +x^2}}$
@David \to :P
13:55
and that equals: $\dfrac{\lim_{x\to 0} 1}{\lim_{x\to 0} (1 + \sqrt{1+x^2})}$
$=\dfrac{1}{\lim_{x\to 0} (1 + \sqrt{1+x^2})}$
$=\dfrac{1}{\lim_{x\to 0}(1) + \lim_{x\to 0}\sqrt{1+x^2}}$
$=\dfrac{1}{1 + \sqrt{\lim_{x\to 0}(1+x^2)}}$
David i wanted epsilon delta method i can do this normally
Well, you can do the epsilon-delta thing by working backwards.
And changing the form of the lmit algebraically, doesn't change the $\delta$ required.
14:12
Hi @DavidWheeler
from here we can work it out
@DavidWheeler hello
Hi @Vrouvrou
Zou Bisou Bisou
hi @ᴇʏᴇs
Ok, here is my thought (I hope it works). Let $\epsilon > 0$ be given. Choose $\delta = \min(1,\sqrt{\epsilon})$
14:23
morning, chat
We begin by converting our limit to $\dfrac{1}{1 + \sqrt{1-x^2}}$, since that's just multiplying by 1.
We suppose that $|x| < \delta$
Morning @Semiclassical
Note that since $|x| < 1$ we have $\sqrt{1 - x^2} < 1$
Note also that: $0 < \dfrac{1}{1+\sqrt{1-x^2}} < 1$
14:29
@Vrouvrou $A$ is where $xy=1$ right? so use the equation I gave to show that $(x_n,y_n)\to(x,y)$ and $x_ny_n=1$ means that $xy=1$.
Now $\left|\dfrac{1}{1 + \sqrt{1-x^2}} - \dfrac{1}{2}\right| = \left|\dfrac{1}{1 + \sqrt{1-x^2}}\right|\cdot\left|\dfrac{2 - 1 - \sqrt{1-x^2}}{2}\right|$
Good job @DavidWheeler
$< \left|\dfrac{2 - 1 - \sqrt{1-x^2}}{2}\right| = \left|\dfrac{1 - \sqrt{1-x^2}}{2}\right|$
$=\dfrac{1}{2}(1 - \sqrt{1 - x^2})$
Since $0 < |x| < 1$, we know $0 < 1 - x^2 < 1$. so $(1-x^2)^2 < 1 - x^2$
Ah, @Alizter is not here.
@Chris'ssis Was that monstrosity an answer to one of the questions going into your book? :P
14:40
@robjohn ok i understand that x_ny_n\rightarrow xy and so xy=1 but x_n>0 means that $x\geq 0$ you see my problem
@teadawg1337 Yes. :-)
And thus $1 - x^2 = |1 - x^2| = \sqrt{(1-x^2)^2} < \sqrt{1 - x^2}$
anyone know some good references on painleve transcendents? i'm trying to find a precise statement of a claim made in passing in a physics article
@Chris'ssis Are you organizing the problems in order of increasing difficulty?
@teadawg1337 It will be hard to do that, I lost the sense of the difficulty. But, of course, it's good to take into account such an order. I try to do that.
14:44
This, in turn means that: $\dfrac{1}{2}(1 - \sqrt{1-x^2}) < \dfrac{1}{2}(1 - (1 -x^2)) = \dfrac{x^2}{2} < (\sqrt{\epsilon})^2/2 = \epsilon/2 < \epsilon$
Of course, difficulty is relative @Chris'ssis
And....he left.
But I was still interested in the solution @DavidWheeler so your work wasn't for nothing
@teadawg1337 There is such a plesure discovering new things, I live intensely these events. It's like a drug, you want more and more.
@DavidWheeler hello
14:47
Hi @Vrouvrou
@Vrouvrou Hello
for my taste, the straightest solution to that limit is to make an $x=\sin(2t)$ substitution (which is nicely behaved in the neighborhood of $x=0$) and observe that trig identities simplify the function to $(2\cos(t))^{-2}$.
I got my first gold badge today
Just out of curiosity, is it incorrect to simplify $\lim_{n\to\infty}(2y^{n+2}-y^{n+1})$ to $\lim_{n\to\infty}(2y^n-y^n)$?
@DavidWheeler Me too (recently)
14:49
@Semiclassical Finding the limit wasn't the problem-he wanted an epsilon-delta
Hi @JulianRachman
From rationalizing the denominator, you can just "plug in 0" (it's a continuous function)
right, right
which is part of the point of studying continuous functions, they have limits
14:51
Guys, is my above simplification incorrect?
@DavidWheeler i have these two sets $\{(x,y)\in \mathbb{R}^2, x>0, xy=1\}, B=\{(x,y)\in \mathbb{R}^2, y=0\}$ i know that these two sets are connected, closed and disjoint , and $A\cup B$ is closed because the finite union of closed sets is closed but what about the connectedness of $A\cup B$ ?
What do you think?
How come they are connected but disjoint
each ONE is connected, no one said they were connected together.
Oh I thought the claim was both were disjoint sets and connected so I was confused
14:54
if A \cap B \neq \emptyset then A\cup B is connected but A and B are disjoint
so i would say that A\cup B is not connected
@teadawg1337 I don't think it is, although the two limits are the "same".
I recently learned a fact that would say these two sets are connected but I could be wrong
but how to prove this @DavidWheeler
if $xy=1$ and $x>0$, then $y>0$ and so $y\neq 0$. so i don't see how there could be any intersection between the two sets
Assuming that $y$ is a function that permits such an operation, such as a convergent recursive sequence or something of the like @DavidWheeler
14:56
@Chris'ssis I am now
You just said $A \cup B$ could be partitioned into two disjoint closed sets.
$$\sum_{i=1}^{\infty} \frac{(1)_{39}+(2)_{39}+\cdots+ (i)_{39}}{ i^{60}}=$$
$$\frac{3231335901102978103703268646579456842357417355434581295104 \pi ^{58}}{2483218575550195467684461870283566015625}$$
$$+\frac{3256897906883967917451234841843013734577488558070941789093746719064064 \pi ^{56}}{110267221955231080977646736839709173245458642578125}$$
$$+\frac{57944093844318409598299938458048299637915996738124310777940217823232 \pi ^{54}}{391489065602002037573079023882089434822802734375}$$
$$+\frac{6179471257497264986709337653078562206021337790396989997148464340992 \pi ^{52}}{229120676527652738947035
omg lol
@_@ sorry
Your first limit is $y^{n+1}(2y -1)$, the second is $y^n(2 - 1)$
Those don't look the same to me....
14:59
@Chris'ssis Heheeh. Did you generalise this as a function $f(x, y)$? Then you could talk about it easier. Ask AMM how long they are going to take and they reply $f(39,60)$.
If you mean a sequence, you should write $y_n$.
@DavidWheeler so A\cup B are not connected, what about d(A,B) please
@Alizter I generalized it in a specifical way, but I cannot tell you the details now. I doubt they can do it (in a short period of time).
What's your definition of $d(A,B)$ @Vrouvrou
@DavidWheeler Never mind, I see my mistake now
Sometimes I ask questions before thinking :S
15:01
d the distance on \mathbb{R}^2 and $d(A,B)=\inf_{a\in A,b\in B} d(a,b)$
Oh, so the infimum of the distance between elements of the sets
I don't know how to prove it, but intuitively it seems the infimum will be $0$
ok @DavidWheeler have you an idea please
@EYES hello
I don't see any ones from MSE @skullpatrol
15:28
hi @DanielFischer
15:39
@DavidWheeler is not here ?
@Alizter the idea is to put all in a very nice form, not in a form with very ugly numbers. This is possible, my generalization is based upon that.
So, to finish my work for any of the cases I need no computational systems to compute crazy large numbers.
It's enough to only use some pen and paper.
15:59
There is no infimum
I mean it's 0
The distance never gets there, though
you can't actually go all the way out to infinity.
16:18
@Vrouvrou
@DavidWheeler If the distance is defined as the infimum, how can the distance not be 0 if the infimum is 0?
@ᴇʏᴇs Um, the distance is 0, but the sets do not touch.
@DavidWheeler Because they are separated right
I think the exercise is trying to show you that connected $\neq$ $d(A,B) = 0$
Ohhh
I had that misconception earlier
Now I know
connected $\Rightarrow\mathrm{d}(A,B)=0$ but $\mathrm{d}(A,B)=0\not\Rightarrow$ connected
boy, those characters don't mix well
16:23
Also $d(A,B) > 0 \implies$ separated but separated $\not\implies d(A,B) > 0$ right
$\mathbb{R}/\{0\}$ is probably the simplest example.
Hi @Semiclassical
Hi @Semiclassical
Hi @SayanChattopadhyay Did you get @DavidWheeler's $\epsilon - \delta$ proof
@Chris'ssis Such simplicity -_-
@robjohn :-)
@Vrouvrou are you worried that $x=0$? (it would be nice if you said this so I don't have to guess). Since the $y_n\to y$, you know that the $y_n$ are bounded, say $y_n\le M$, and so the $x_n$ are bounded away from $0$, say $x_n\ge 1/M$.
@DavidWheeler but how to prove that d(A,B)=0
@robjohn if x=0 then (x,y) is not in the set
@robjohn The amazing part is that one needs to compute no ugly large numbers as written there.
The closed form can be expressed in a very nice way.
Consider the points $(n,0)$ and $(n,\frac{1}{n})$
you are with me @DavidWheeler ?
16:43
@Chris'ssis You mean that enormous sum is not necessary?
@robjohn Exactly.
Clearly, $d(A,B) \leq \inf\{\frac{1}{n}: n \in \Bbb Z^+\}$
i don't know
@Chris'ssis Then why did you post that horrendous thing?
@robjohn Well, to emphasize the difference between such an answer and a brilliant answer. That thing, as I said, can be nicely done with no need of such computations.
@robjohn If you like that much I can place the result in db. :-)
16:51
@Chris'ssis I'm always interested to see these things. Feel free :-)
@TimDavids Hi.
@robjohn OK. Besides that, I trust you. ;)
can someone explaine me d(A,B)=0
@DavidWheeler I didn't understand..if we have the points $(0, n)$ and $(n, \frac{1}{n})$, then we have to prove that $\inf{d((0, n), (n, \frac{1}{n}))} = 0$
@robjohn check that.
16:53
@Vrouvrou It means that no matter how small $\epsilon\gt0$ is, you can find an $a\in A$ and a $b\in B$ so that $\mathrm{d}(a,b)\le\epsilon$
i know but how to prove this
Pick $n > 1/\epsilon$, then the two points I gave will work.
@Chris'ssis Okay, so that is just the statement. It is much simpler
@robjohn Yeap. Well, I used an interesting approaching way. You'll see when I release my book.
$d(a,b) = \sqrt{(x_1 - y_1)^2 + (x_2- y_2)^2}$ right
16:58
$d((x_1,y_1),(x_2,y_2)) = \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2}$
@robjohn Keep in mind this: APPLY S TO I family TO GET I family. That will explain the solution. Just remind me that when I release the book. I'll show you all the details about this way.
Oh yea

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