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11:00 AM
In addition, it is an integer sequence, as $G_0=2$, $G_1=4$, and we can proceed by induction.
In fact, we don't need that sequence stuff.
Let me start over
 
Why it's the same thing for $(3+\sqrt 5)$
 
@Dattier Does it work for $4+\sqrt5$?
 
no
try
 
Makes sense, since $|4-\sqrt5|$ isn't less than $1$
which is important for this to work
So here's a proof, for $2+\sqrt5$
 
Bravo
 
11:03 AM
$(2+\sqrt5)^n+(2-\sqrt5)^n$ is an integer, since it's the sum of something and its algebraic conjugate.
Now, $|2-\sqrt5|<1$ (it's like 0.2-ish)
so let's write $(2+\sqrt5)^n+(2-\sqrt5)^n=k$, and we know $k$ is an integer
Then $(2+\sqrt5)^n=k-(2-\sqrt5)^n$
 
I am trying to solve a four dimensional graph that has edges from A->B, A->C, B->D and C->D like Figure 8 on page 9 in this article: https://www.aaai.org/ojs/index.php/aimagazine/article/view/918/836

how do I compute to proof the distribution for this Bayesian graph to be valid? I have tried using the method below: P(A,B,C,D) = P(A) x P(B|A) x P(C|A) x P(D|B,C) but I don't seem to be getting the same distribution on both sides so does that mean that the distribution table depicting the graph is invalid?
 
And, for $n$ even and greater than $1$, we have that $(2-\sqrt5)^n$ is between $0$ and $1$
so $\{(2+\sqrt5)^n\}=1-(2-\sqrt5)^n$
and the RHS approaches $1$ exponentially
 
@AkivaWeinberger Bravo, bye
 
user84215
If you have no idea, I suggest event for Saturday, discussing the ways that new groups can be constructed from given groups. Good?
 
11:42 AM
ugh, working with arbitrary group morphisms is such a pain after only having bothered with automorphisms for ages >.<
 
Hmm, trying to look for a simple condition for finding if a point is in the limiting Menger sponge
The formal definition only gives an infinite chain of conditions
Although I guess I could try a few iterations and look at the remaining points' ternary expansions....
 
user84215
Because I have seen no disagreement, I will create the event: "The ways that new groups can be constructed from given groups".
 
user84215
It will start at ** 9 am GMT on Saturday, July 29, 2017** in Discussing Specific Topics and last 300 minutes.
 
user84215
11:57 AM
I don't know why ** ** does not work
 
There was a space beteeen the first set and the 9 @aminliverpool
 
user84215
Thanks
 
user84215
Because I have seen no disagreement, I will create the event: "The ways that new groups can be constructed from given groups". It will start at 9 am GMT on Saturday, July 29, 2017 in Discussing Specific Topics and last 300 minutes.
3
 
user84215
12:52 PM
You can register for the above event.
 
1:07 PM
Is there is a series test that says something to the effect that if $\sum n a_n$ converges, then $\sum a_n$? I believe there is, but I cannot remember the exact statement.
 
$na_n\ge a_n$ for all $n$
unless your series starts at 0, but a finite amount of terms won't change the behavious of the series
 
For positive $a_n$
 
There is
$a_n = o(na_n)$
 
@SteamyRoot Ah, right, I always think about series with positive terms
 
Wait that's not actually enough
 
1:12 PM
So, if $(a_n)$ is a positive sequence with $na_n \ge a_n$ for all $n$, then the convergence of $\sum n a_n$ implies the convergence of $\sum a_n$?
 
My guess is you have to quantify
 
@user193319 Comparison test
 
@SteamyRoot Oh. I see. That's how you prove it.
 
If the ratio test holds for $na_n$, then it also holds for $a_n$... But the ratio test can be inconclusive for $na_n$ even if it converges.
 
1:17 PM
@SteamyRoot I am somewhat confused by what you are saying. First you mention the comparison test, then the ratio test. Is the hypothesis $n a_n \ge a_n$ necessary?
 
@user193319 The comparison test is only useful if $\sum na_n$ is absolutely convergent.
The ratio test doesn't require the absolute convergence, but may not work.
(of course, if the sequence $(a_n)$ is positive, convergence is equivalent to absolute convergence...)
 
@user193319 that's not an hypothesis, that's a consequence of the $a_n$ being positive
 
Actually, I think Abel's test may be conclusive here.
$\sum na_n$ is convergent, and the harmonic sequence is monotone and bounded.
Hence $\sum a_n$ is convergent
 
Oh. So the theorem should read "If $(a_n)$ is a positive sequence and $\sum n a_n$ converges, then $\sum a_n$ converges"?
 
No, being positive is not necessary.
But it makes the proof easier.
 
1:50 PM
[Random]
 
2:17 PM
hi guys
can anyone help me with my question?
 
hai
maybe? what is it?
 
hi guys, let $\omega_j$ a sequence such that $$
\sum_{j=0}^{+\infty} \omega_j \le \infty
$$ and for all $k$ we have $$
\omega_k \leq \sum_{j=k+1}^{+\infty} \omega_j
$$
and assume (but you can prove it is true) that for al $x$ in a certain interval $[-A,A]$ we have
$$
x = \sum_{j=0}^{+\infty} d_j \omega_j \;\;, d_j \in \left\{-1,1\right\}
$$
my question is it true that if I pick $x,y$ in such interval with $x < y$ then there's a $j_0$ such that $d_{x,j} = d_{y,j}$ for $j < j_0$ and $d_{x,j} < d_{y,j}$ for $j = j_0$
 
2:39 PM
@Secret More ordinals?
Is that $\omega^\omega$?
 
Not ordinals, but rather trying to construct some kind of topology by start with a converging sequence, such that zooming in any spacing, you get the same look, like a fractal
that is, imagine I have the sequence $10^{y},y\in \Bbb{Z}$, and then every two consecutive term, I put a copy of the orignal sequence in it
 
It looks like a thing with order type $\omega^\omega$@Secret
 
That I am not sure. However, the idea really came from previous night dream:
yesterday, by Secret
user image
such that pick any black space and zoom in, and you get the same grid pattern, no matter how deep you go. I wonder if I can wrote a metric for that...
 
@user8469759 If you add $\sum\omega_j$ to it and divide by two, it's the same as the question with $d_j\in\{0,1\}$ instead
 
therefore?
 
2:45 PM
I'm thinking it shouldn't work, if you have a term that satisfies $\omega_k<\sum_{j=k+1}^\infty\omega_j$
Note that binary expansion doesn't have this
In fact, double both sides of $\omega_k<\sum_{j=k+1}^\infty\omega_j$ and subtract $\sum_{j=0}^\infty\omega_j$ from both sides; that would give you your counterexample
 
hi chat
 
SBM
Hello
 
ohi
 
@Sem @SBM @Ste It's an S convention
 
2:55 PM
lol
 
Al-S-andro
 
Heh
If Astyx was around, you guys would have an A convention :P
 
Warring factions
 
He is watching
 
Oh no you guys have Secret you have us outnumbered
 
2:59 PM
more power to thes s
 
What if I change nickname and betray the A faction?
 
What nickname do you have? Ale?
 
They usually call me code
 
We'd need to persuade @Daminark to use his real name or something
(Don't actually do that Amin your username is cool)
 
Adminark
 
3:04 PM
No that's for after he gets promoted to admin
3
(can only imagine what Daminark would do with that much power)
 
This place would burn
 
1
Q: For given a subset of the set of all $m\times n$ real matrices can we find a Lipschitz function $s.t.$ a given condition hold?

Walter Tiban MalikLet $m,n \in \mathbb N$ and let $$\mathbb R^{m\times n}:=\left\{A : A \ is \ (m\times n) \ real \ \ matrix \right\}$$ Let $\ E \subseteq \mathbb R^{m\times n}$ be given and let $\Omega \subseteq \mathbb R^n $ be open. Does there exist a Lipschitz function $u: \Omega \to \mathbb R^m$ $s....

Can u please give some hints for this
please please
 
excuse me? you only asked 5 minutes ago
give it some time please, btw I find your avatar to be in bad taste :(
 
3:19 PM
One of the very useful things to remember about this site: You don't have to answer a question if you're not interested in it. And if you don't want to listen to a user, there's always the option to hide their posts.
Hmm. Today's XKCD makes me wonder:
 
We have so many user(insert number) around recently, it is a bit difficult to tell who is who (especially at least 1 of them is an alt of skullpatrol)
 
@Semiclassical thnx for the advice
 
How 'sharp' is the the boundary of the path of the eclipse? The last image makes it out to be pretty binary, but I don't know if that's correct.
(Though maybe this is more an astronomy/physics type question)
 
I imagine it's pretty sharp, given how spectacular and sudden the "diamond ring" effect tends to be. Though not perfectly sharp, of course.
 
Right.
 
3:28 PM
Hello guys
 
I should look to see if there's some pictures out there.
 
And, damnit, now I'm thinking of physics yo mama jokes again >.<
 
lool
...huh. i just noticed that "lool" looks like a pair of glasses.
 
Does the leading coefficients have to be 1, in a matrix that is in Row Echelon Form?
 
Specifically, the "yo mama so fat I can see behind her" one :P
 
3:30 PM
black hole mamas probably fatter. you can definitely see her back
<s>Help me construct a metaphor for the event horizon in this context such that it does not sound sexist... before the edit timer is up</s>
 
Wut
 
Guys
 
Any object that's heavy enough will bend light rays near it, you don't need something the scale of a black hole for that
 
@user685252 Yeah of course take time
 
@Semiclassical Thanks for sharing
 
3:47 PM
The exterior derivative on bundle-valued forms depends on the choice of connection, right?
Yes, it does.
 
That feels like something which should have a physics implication/connection.
Something something QCD.
 
4:24 PM
Hey guys
Hey guys
@Akiva @Astyx mwahahahaha hahahah AHAHAHAHAHA
 
Yoyo @Daminark
 
How goes it?
 
Pretty well actually, my flu's getting better too
How's things going with you?
 
4:51 PM
Things are going well, it's been rather busy
 
5:12 PM
Problem: In a homogenous system of 5 linear equations in 7 unknowns, the rank of the coefficient matrix is 4. The maximum number of independent solution vectors is...
By independent solution vectors, does the author mean linearly independent?
Is the question just asking for the dimension of the null space?
 
@AkivaWeinberger Ordinals?
 
@user193319 sounds like it to me
 
@user193319 Yes
 
Thank you both!
 
Speaking of which, if you want, you can try this function out. In terms of the fast growing hierarchy, it grows at about $f_{\omega^{\omega^2}+\omega^\omega}(n)$.
 
5:15 PM
How's it going @TobiasKildetoft and @Simply?
 
@Daminark Fine
 
Idk @Daminark
@AkivaWeinberger In other words, I've surpassed the goodstein sequence's length when starting at $2^{20}\approx10^{6}$.
 
Isn't the Goodstein function $\approx f_{\epsilon_0}$?
 
Yeah
But $\epsilon_0$ reduces down to $\omega^{\omega^{\dots}}$
And the truth is $G(2\uparrow^2n)\approx f_{\epsilon_0}(n)$
So it grows a bit slower than you might first imagine
Lol, when your arrays really get nested: [3, [2, [[-1, [[[[[3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [3, [4]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
 
Where have I gone wrong in my solution?
I have to find the value of cos (22.5)degrees
Identity used by me: $cos(2A)= 2cos^2A -1$
 
5:26 PM
@Abcd That was a poor idea
 
@TobiasKildetoft Why?
 
Just draw the triangle corresponding to that angle
 
@TobiasKildetoft I would only have a pen, pencil and eraser during the exam (It's an OMR based exam where calculators aren't allowed)
 
you get a right triangle were the hypothenuse has length $1$ and the two other sides are equal
Exactly, no calculator needed
 
sketch it
 
5:28 PM
@TobiasKildetoft I wouldn't have protractor
 
What's wrong with your solution anyways
 
@TobiasKildetoft He's finding cos(22.5), not cos(45), no?
 
As far as I can tell, its correct.
 
@Abcd You would not need one. I meant draw it on your head using that you actually know what a 45 degree angle looks like
 
@TobiasKildetoft I am finding cos 22.5 not cos 45
 
5:30 PM
@Abcd That was not what you wrote at the top of that page
 
@SimplyBeautifulArt Answers don't match. The answer given is: $0.5(\sqrt{2+\sqrt2})$
 
@Abcd Try rationalizing the denominator
 
@Abcd Then you should reduce your answer to that one.
 
@SimplyBeautifulArt How?
 
You have $\sqrt2$ in a denominator
 
5:31 PM
@Abcd Ahh, I see now what you meant
 
@TobiasKildetoft Look at my message above
5 mins ago, by Abcd
I have to find the value of cos (22.5)degrees
 
so multiply by $\sqrt2$ in the numerator and denominator
 
@Abcd Hint: $$\frac{\sqrt{2+\sqrt2}}2=\sqrt{\frac{2+\sqrt2}4}=\dots$$
 
Oh. It was simple :(
Thanks!
 
np
$\eta\rho$
:D
 
5:35 PM
$\not p$
 
p
:-)
 
Bob
5:51 PM
If a student is taking a math course and he is assigned homework problems and before he turns them in to the professor he posts the questions and his answers to stack exchange for somebody to check them, is there any reason why somebody on Math Stack Exchange should have a problem with that?
 
@Bob Unless the professor is fine with this, it might well be considered cheating
 
Bob
Assume the professor is fine with it
 
Then I think the attempted solution should not be posted as an answer since the person will not be certain that it is a correct answer.
 
Bob
@TobiasKildetoft thanks and have a nice day
 
@AkivaWeinberger I think I can easily exceed the Goodstein sequence starting at about a googol in base 2 lol
 
6:06 PM
"Assume the professor is fine with it"?? Having been a professor for about 40 years, I would not be fine with it and would consider it cheating. @Bob
 
@TedShifrin really? i ask probably one question from every assignment that i can't figure out on stack exchange to get a hint unless its like a take home test or something.
hows it different from getting hint form a prof or anthor student?
Someone just posting the answer to my question and not explaining how to do it properly is just going to make me do alot worse when test times comes >.>
 
6:22 PM
@Faust7: If you tell your professor you're doing this (or put on your paper that you got significant help from ...), then it's one thing. Copying from another student is no better. When I gave students help in office hours, then I knew what I was telling them and I didn't just do the problem for them. (I also counted homework as a very high component of the grade — often 35% — and put a detailed discussion of acceptable behavior in my syllabus.)
 
Hmm i guess i should be more careful and ask my specific prof what they feel is ok to ask here. I miss the old days when you used to be able to just post this is a homework problem can i just have a hint on here. nowadays that doesn't seem to stop people from just giving the whole answer.
But they even took away the tag for that i think :(
 
@Bob it is unethical.
 
@Faust7 well, I'm pretty sure that people will respect it if the individual asking the question states "Please give me a hint and don't just answer"
 
@Daminark you'd be surprised
 
hi chat
 
6:36 PM
Plus, the incident above isn't one of asking for help when one doesn't know the answer, this is literally proof verification. So it's also possible to just say "Did I make any illegal jumps? If you intend to spill out the answer, please just don't say anything"
 
i could understand proof verification when u first start that business its a difficult egg to crack. but im not sure any prof would be fine with posting your entire hw assignment on the internet to have it answered for you
if they were fine with it, it may be time for them to consider a new career imo
 
Why should one student have that advantage?
 
is every finite subset of the reals bounded?
 
sure. it's finite, so it can be ordered. hence there's a max and min. (there's probably a nicer way to say that but w/e)
 
@Daminark Not true! I've gotten in fights over this.
BTW, I'm not capitulating on your Demonark name. Just the link :P
Howdy, @Semiclassic.
 
6:42 PM
heya
 
Funnily, about 3 years ago, I found some of my own differential geometry students posting some of my (harder) homework problems here. You would have chuckled at what I said in class the next day @Faust.
 
whuh-oh.
 
lol i never bothered posting any dif-geo questions on here there too hard no one wnats to answer them
 
Speaking of hints, I'm rather pleased that no one has yet given this one away completely to the OP.
Faust, I've answered several hundred of them.
Answered to various degrees of detail ...
 
Why? Like I could see it if the question just asks, but if the OP were to literally say "If you intend to spoil the whole answer, please do not answer this question", that people aren't gonna be jerks about it
 
6:45 PM
Wrong. People have been jerks.
More times than I can remember I was in the process of discussing with the OP in comments, and the OP was figuring it out, and then someone wanting to show off just posted a complete solution.
One removed it, as I recall, when I flamed him about it. But plenty didn't.
 
Trust me i have put this a hw question hints only please and had just an answer no explanation posted on several occasions...
 
I know I've helped you several times in here, Faust. I don't recall if I answered any of your posts.
 
i find i dont learn anything if someone just gives me the answer ^^
 
After years of experience, I'm pretty good at asking leading questions or giving slight hints, but it takes experience to learn to do that.
What's best is for you to post something you've tried, and then usually there can be a direct suggestion about what was wrong or what you were not seeing.
Anyhow, what're you working on now, Faust?
 
some basic real analysis theres one question i can't seem to remember how to do.
Prove by mathematical induction that every non-empty finite subset $ S \subset \mathbb {R}$ of real numbers has a maximum element.
actually im confused why it says to use induction
 
6:49 PM
How are you approaching it?
 
posts the proof
Lol jk
 
Well, officially you only know how to define $\max(a,b)$.
That's why they say to use induction.
 
well i started with every finite subset of the real being bounded
 
yeah, uh, don't trust what I say as a rigorous argument :P
 
then i said that any bounded subset has a lub in R
then i wanted to show that lub was the max of S but it didnt work out
 
6:51 PM
It may be right, but anything out of my mouth should be taken with a grain of salt as far as rigor goes
 
Well, it will work, but how to justify boundedness? It sort of assumes what we're trying to prove.
 
yeah
 
Let's go with their induction suggestion.
 
thats when i realized i had a problem before i started
 
@Semiclassical SemiHint?
 
6:52 PM
lol
 
What should we do induction on, Faust?
 
when you say define max (a,b) do you meant everything in that interval? how do you know you can use the welll ordering principle to geta max and a min?
 
No, no, I mean the maximum of two numbers.
 
ok
 
Check to see that your book defined that?
 
6:53 PM
it does slightly diffrent notation
 
There's actually a formula for it. Or you can use a definition by cases.
 
(Side note @Ted: they have the details up for the RTG finally, and it looks pretty fun )
 
You're going too fast, Semiclassic.
 
one last question can you use the well ordering principle on a finite subset of the reals ( i thought no)
 
Don't be so fancy, Faust.
The principle you were taught is just for positive integers.
 
6:55 PM
thought so
 
Sounds like you still have plenty to learn, Demonark. :)
So did they define the max function for two real numbers? What did they do?
 
lemme find my textbook it appears my couch has eaten it since i went to sleep
 
LOL
That's expensive couch food.
 
Unless it was a Dover book
:P
 
6:56 PM
and I just noticed that "lol" looks like a person throwing up their hands.
 
Not when I type it, Semiclassic.
 
\('-')/
 
I have trouble reading most of these emojis.
 
I've never gotten into Eastern-style emojis
 
Oh that's an emoji that I've never seen used elsewhere
 
6:57 PM
besides possibly >_<
 
@TedShifrin They are emoticons. Emojis are when the emoticon is replaced by an actual image (yes, this is vital information)
 
I just kinda thought it looked funny once I saw it and started using it. Also this: ( ._.)
 
See ... I truly flunk internet 101.
 
or maybe they are just emotes actually
 
(>_>)
 
6:58 PM
¯\_(ツ)_/¯
 
WTF is that, Demonark?
 
That's someone shrugging
 
shrugs
 
/me hardly ever use anything but :)
 
I'm someone who played a text rpg for a long time, so doing stuff like shrugs is much more familiar to me
 

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