Mathematics - 2015-10-22 (page 2 of 3)

Math Man

15:00

Either way it's funny :)

skull petrol

Yep

:D

evinda

Hi @robjohn !!!
Do you maybe have an idea if it is as I said above?

Daniel Fischer

@evinda Yes, it ought to be $\cos$, because $\sin' = \cos$.

robjohn

@evinda It should be $\cos(\xi)(y_2-y_1)$ where $\xi$ is between $y_2$ and $y_1$

evinda

Nice, thank you @DanielFischer @robjohn :)

robjohn

15:08

@evinda If you put parentheses around the argument to $\sin$ and $\cos$ then some confusion might be avoided.

Jessy Cat

15:25

Why the f&*^ won't people just give you a straight up answer after your comment back-and-forth has gotten longer than your original post?????

1

Q: How to prove complement of generalized Cantor set is dense in $[0,1]$

Related to a question I asked earlier.: Let $F$ be the subset of $[0,1]$ constructed in the same manner as the Cantor set except that each of the intervals removed at the $n$th iteration has length $\frac{\alpha}{3^{n}}$ with $0 < \alpha < 1$. I need to show that $F$'s complement, $[0,1]\backsl...

I hate people who say "it's easy to see that".

No. If it were easy to see it, I wouldn't be asking it, a-hole

Jessy Cat

16:00

@robjohn, can you PLEASE give me a more detailed answer than that????? I've been stressing for hours and hours and hours.

Don't get me wrong, I appreciate what you're doing; it just seems that the more hints I get to my question, the more questions I have, and it's extremely frustrrating.

Jessy Cat

16:19

Please somebody help me.

Agawa001

@MathMan i perceived an answer of you, quoting that from edison, and i corrected it with the very appropriate term "dispossession"

Balarka Sen

16:38

hi @TedShifrin

Ted Shifrin

Hi @Balarka

Balarka Sen

I sent you an e-mail.

And I think I also have the function you wanted.

Remember me

Hey@TedShifrin

Ted Shifrin

I haven't studied your email yet.

Hi @Remember

Balarka Sen

@TedShifrin $f(x, y) = |x||y|^5/(|x|^3+|y|^8)$, $f(\vec{0}) = 0$. This is unbounded around $0$ because if you approach by the curve $x = t^8, y = t^3$, then this looks like $1/2t$. $0$ directional derivative everywhere.

Plus, everywhere continuous away from $0$.

Ted Shifrin

16:42

Looks promising. I'll need to think.

Balarka Sen

Admittedly, though, I used algebraic manipulation to do this, not geometric thinking.

I want to get an example by fixing my $1/x$ construction, i.e., geometrically.

Ted Shifrin

Much as I love to be geometric, sometimes one needs to compute algebraically/analytically.

Balarka Sen

Fair enough.

Agawa001

who ever recently serial-upvoted me plz take back your upvotes im not a beggar

Balarka Sen

@MikeMiller Morning.

Remember me

16:48

Hello@MikeMiller

Ted Shifrin

Looks good. See if you can understand this genre of functions a bit better "geometrically." Start with the example in the book.

Balarka Sen

I have figured out the computation of $H^*(\Bbb{RP}^n)$ does, @MikeMiller. It's very geometric. One needs to interpret the boundary circle in the sketch Hatcher has as the $i$-th copy of $\Bbb{P}^{n-1}$, i.e., the one represented by the hyperplane $x_i = 0$.

Mike Miller

Ok.

Balarka Sen

(I have written up my understanding in a discussion with bananas here and here. But don't click if you don't want to).

@MikeMiller How are you?

@TedShifrin Hmm, ok.

Mike Miller

I'm fine.

Agawa001

16:56

@TedShifrin genre ?

the homologue of "genre" in french is "gender" in english or is it something in english i miss the meaning of

Balarka Sen

@MikeMiller Concise, as always. :)

Mary Star

17:14

Hello!! Could someone of you take a look at my question:

1

Q: Show that $\textbf{$\gamma$}$ lies on a sphere of radius $r$

Let $\textbf{$\gamma$ }(t)$ be a unit-speed curve with $\kappa (t) > 0$ and $\tau (t) \neq 0$ for all $t$. Show that, if $\textbf{$\gamma$}$ is spherical, i.e., if it lies on the surface of a sphere, then $$\frac{\tau }{\kappa }=\frac{d}{ds}\left (\frac{\dot \kappa}{\tau \kappa^2}\right ) \tag 1$...

differential-geometry curves spheres

?

Chris's sis the artist

@Agawa001 you're right, and I also starred your message.

Agawa001

@Chris'ssistheartist where? that vocabulary comment ?

@Chris'ssistheartist hey btw how are you

Chris's sis the artist

I also remember that I experienced that also here from people that never ever created 10 problems that can be published in a magazine or so, they were talking about my math.

My math? I talk about my math with anyone, human or god (see Namagiri) in the real life, it might be a very tough experience for any opponent (in case it's about having opponents - I hope it's never that).

Agawa001

@Chris'ssistheartist ah, i get it

Chris's sis the artist

@Agawa001 Better, thanks. You? :-)

Tien-Cheng Huang

17:19

Could someone explain the concept of manifolds to me?

Agawa001

@Chris'ssistheartist that bipolar disorder seems to have another iron grasp on me but im fine i ll get ove that like a boss as any time before

Chris's sis the artist

@Agawa001 Are you on treatment for that?

by the way, I was praising these days homeopathic remedies, but it seems things were not that great today, although I feel better than in other days.

Agawa001

@Chris'ssistheartist no i dont, and i wont

Chris's sis the artist

@Agawa001 Were you diagnosed with bipolar disorder?

Mary Star

Do you maybe have an idea about my question math.stackexchange.com/questions/1481897/… @robjohn ?

Agawa001

17:24

@Chris'ssistheartist yes, not by a doctor

Chris's sis the artist

@Agawa001 By yourself? :-)

Agawa001

dunoo sometimes i think people talk more than they breath, but sometimes i tend to think they are right

Chris's sis the artist

@Agawa001 That reminds me of an event I passed thorough some time ago when I hard time with sleeping after a long period of extreme overwork. Then I thought it might be a psychiatric disorders and then decided to go to a professor doctor.

Balarka Sen

@Tien-ChengHuang What, in particular, do you want to be explained?

Do you want to know the definition?

Chris's sis the artist

@Agawa001 I sat there for 2 hours or more or, and waited for a treatment, I was hoping to receive the pills and go home, take them and sleep well. Then he terribly disappointed me.

Agawa001

17:29

@Chris'ssistheartist bipolar persons dont sleep vis-avis how hard they work

@Chris'ssistheartist i wont take pills or any meds, they have side effects

unless it is serious enough

Chris's sis the artist

@Agawa001 The professor told me that I need no treatment, and said something like that "Take more rest and think less of things that trouble you".

Well, that was shocking, I paid a lot to go to him and I left with no treatment, no pill for sleeping. Then I bought some tea and slept well (after 2-3 weeks).

@Agawa001 The point is this: sometimes things are not that bad how we imagine them. It's better to visit a specialist and ask for an opinion than imagining you are confronting yourself with diseases you don't have.

Agawa001

hmm, i think so, lets not divulge more private informations about our selves in main. is there more nice integrals that may get me rid of this issue ?

Chris's sis the artist

@Agawa001 No worry, these days overwork often leads us to such trouble, and I often hear people having problems with the sleep from overwork. This is no private information, it's so common. ;)

Agawa001

not from your book, it ll be a lot over my head :D

Chris's sis the artist

@Agawa001 lol

Today I developed some solutions to some of my old problems, I mean cleverer solutions.

Agawa001

17:37

@Chris'ssistheartist like i have said before, bipolar persons dont sleep vis a vis the effort they spend

@Chris'ssistheartist good

Chris's sis the artist

@Agawa001 To have a disease is not something shameful or anything, just to know that.

Agawa001

@Chris'ssistheartist i just dont want to go further in details

Chris's sis the artist

@Agawa001 It's OK, no problem. Sure.

@Agawa001 Let me show you a nice elementary integral

$$\int_0^1 x^{n-1} \log(1-x) \ dx$$

Agawa001

i dont usually produce nice ideas when i am in "down" period but i ll see

Math Man

@Agawa001, Great, dispossession is the perfect term there. I like that! Great article!

Jessy Cat

17:44

For a long time they thought i had bipolar, but now theyre pretty sure it's adhd, which makes sooo much sense in retrospect.

Agawa001

@MathMan oh, i though you like edison :p

Jessy Cat

The depression I've had has been because it's so hard to fit into neurotypical society when you have adhd

you don't get tasks done in time, you do things to make people not like you, etc.

Math Man

@Agawa001 No I don't really like him. Why did you think so?

Agawa001

@MathMan it doesnt matter, he was someone who marked history though

Math Man

That's for sure!

Balarka Sen

17:46

Hello, @PedroTamaroff.

Agawa001

because of an answer i read from you

quoting from edison

about intelligence i think

Chris's sis the artist

@Agawa001 Don't think like " I have to calculate it", but rather "Let's play with it and have some fun". :D

Agawa001

@Chris'ssistheartist in progress ...

Pedro Tamaroff

@BalarkaSen Hey.

Tobias Kildetoft

@PedroTamaroff Hi. So, what are you up to currently?

Chris's sis the artist

17:48

@Agawa001 Mathematics is for fun! Turn the mathematics in a great time for you! ;)

Math Man

@Agawa001 Oh, I remember that one. I guess sometimes I quote people I don't totally admire!

Balarka Sen

@PedroTamaroff Started learning any algebraic topology? I mean, has the other half of your course started yet?

Pedro Tamaroff

@TobiasKildetoft Taking Topology, Algebra III and (Projective) Geometry.

Agawa001

@MathMan lol

Tobias Kildetoft

@PedroTamaroff what sort of algebra?

Pedro Tamaroff

17:49

Yes, @BalarkaSen, we started with the introduction to Algebraic Topology.

@TobiasKildetoft Algebra III is Galois/Field Theory.

Balarka Sen

Ah, nice. What have you learnt?

Tobias Kildetoft

@PedroTamaroff Neat

Pedro Tamaroff

First part is Galois Theory, second part is whatever the professor wants, I guess.

I am trying to learn some topics by myself, we're studying cyclotomic extensions in class.

@BalarkaSen I am reading Spanier's book, haven't gone to class really.

Math Man

@Agawa001 Some people write better quotes then anything else. "My pen is smarter than I am" -Einstein

for example

Balarka Sen

@PedroTamaroff okay. have you started on fundamental groups?

Math Man

17:52

although I don't think it applied to Einstein

Pedro Tamaroff

Yes, I started chapter two today.

Agawa001

@MathMan did he really say that ?

Balarka Sen

Fun (I haven't read Spanier, though).

Chris's sis the artist

'You have to go on and be crazy. Craziness is like heaven.' - Jimi Hendrix --- @Agawa001 I like this quote much.

Balarka Sen

What's on chapter 1?

Math Man

17:53

Who knows. Maybe just the internet said that but it attributes it to him. Good quote either way.

@Agawa001

Chris's sis the artist

I little dosage of craziness is always welcome! ;)

Pedro Tamaroff

(categories, functors) homotopy, retraction, deformation, H-spaces (H-groups, H-cogroups), suspension, fundamental grupoid, fundamental group, and showing $\pi_1(S^1)=\Bbb Z$.

Balarka Sen

Seems way too much categorical. But each to his own taste.

Agawa001

@MathMan it applies to this era as: "my keyboard is smarter than me"

@Chris'ssistheartist yes, limited craziness

not to go all wild i mean

Balarka Sen

@PedroTamaroff If $X$ is an $H$-space, then $\pi_1(X)$ is abelian. Have you, by any chance, seen that result?

Pedro Tamaroff

17:56

@BalarkaSen Yes, Spanier proves that.

Math Man

@Agawa001 Nailed it!

Balarka Sen

Cool. Using this, and some result you will know later on, you can show - say - that wedge of two circles, $S^1 \vee S^1$, is not an $H$-space (thus not a topological group)

Agawa001

@MathMan sometimes you code things that are indeed smater

kasparov had been alwayes beaten up by his own chess-program

Pedro Tamaroff

@BalarkaSen Well, yes. That has nonabelian fundamental group.

Balarka Sen

Well, proving that requires some work. You can directly do van Kampen, or you can construct a cover such that two loops $f * g$ and $g * f$ lift to paths with different endpoints.

I presume you have done one of the two?

Pedro Tamaroff

17:59

Everything requires some work.

You can use van Kampen's theorem to show the fundamental group is $\Bbb Z\ast \Bbb Z$.

Balarka Sen

Fair enough, I thought you weren't familiar with that as it's not listed in among the things you wrote.

Jack

Can somebody tell answer this? math.stackexchange.com/questions/1491746/… why is r treated differently across different sequences??????

I mean series *

Pedro Tamaroff

@Jack $(-1)^n (x+6)^n = (-(x+6))^n$, so your $r=-(x+6)$.

Balarka Sen

You can alternatively construct the cover $E \to S^1 \vee S^1$ where $E$ is the $x$ and $y$-axis in $\Bbb R^2$ with a circle attached at each integer point. The covering map just wraps the $x$ axis inside into the second circle and $y$-axis into the first circle.

Then the loops $f * g$ and $g * f$ lift to paths with different endpts, where $f$ is the look which winds the first circle and $g$ is the one which winds the second circle.

Thus, as a corollary of homotopy lifting, $f* g$ and $g* f$ are not homotopic.

This is easier than using SvKT.

Pedro Tamaroff

Why do you think it is easier?

Balarka Sen

18:04

Proving SvKT is a lot of work.

Agawa001

@Chris'ssistheartist $x^{n+1}log(1-x)=\int (n+1)x^nlog(1-x)-\int \frac{x^{n+1}}{1-x}$

$\int \frac{x^{n+1}}{1-x}=\int \frac{x^{n+1}-1}{1-x}+\frac{1}{1-x}$

Chris's sis the artist

@Agawa001 you have a definite integral from $x=0$ and $x=1$.

By the way, in my book I'm going to present 3 or 4 solutions to this integral, and although elementary it's important as a tool for calculating some other series.

Here is a more advanced one

$$\int_0^1 x^{n-1} \log^2(1-x) \ dx$$

or even $$\int_0^1 x^{n-1} \log^3(1-x) \ dx$$

Agawa001

@Chris'ssistheartist evaluating is simple

Pedro Tamaroff

@BalarkaSen What have you been studying lately?

Agawa001

@Chris'ssistheartist eww thats too much

Chris's sis the artist

18:10

@Agawa001 :D The squared version is cuuuutttttttteeeeeeeeee! :-) I also present it in my book with 3 solutions at least.

Balarka Sen

@PedroTamaroff I worked hard for a week to understand the computation of the graded cohomology ring of $\Bbb{RP}^n$ with $\Bbb Z/2$-coefficients, and I have been successful. I have put Hatcher in my shelf and studying calculus for now.

Pedro Tamaroff

@BalarkaSen Calculus? As in...?

Balarka Sen

Multivariable calculus.

Pedro Tamaroff

Ah.

Agawa001

@MathMan i despise idea-stealers but i cohabit with them, so i understand you quoting from someone you dislike

Chris's sis the artist

18:21

@BalarkaSen Since you mentioned multivariable calculus ... I think I just created something new in 2 variables ... $$\int _0^1\int _0^1\frac{\log ^2(1-x y)}{x y (1-x y)} \ dx \ dy$$

It looks so brilliant!

Agawa001

mind-blown! seeking mercy in @MathMan's question

Chris's sis the artist

@Agawa001 What about a selfie with that double integral? :-)))))

Agawa001

xD

@Chris'ssistheartist i really wished to reach that level in maths

Chris's sis the artist

@Agawa001 Change your way of thinking. If you wish that, just do it! It's that simple! :-)

Agawa001

@Chris'ssistheartist not much time left :(

i wish too that i can live twice

Chris's sis the artist

18:26

@Agawa001 Think that I'm self-educated, I'm not coming from a system where I was taught by great professors, absolutely no, and I had time for that. Do it every day, and then it happens.

Trust yourself, do it every day, make progress, never get discouraged, never give up, crawl yourself if needed, with passion. If the whole world says "You fail!", then you are even more motivated to excel.

Balarka Sen

@Chris'ssistheartist Multivariable calculus is hardly just about computing complicated multivariable integrals, though.

But thanks.

I don't have a clue how to compute it.

Chris's sis the artist

@Agawa001 There is nothing that can stop you from success, but you. It's not just nice words, it's the truth.

If you wanna succeed, you can do it.

@BalarkaSen I was just saying that it crossed my mind that integral to create when you wrote 'multivariable calculus'.

Balarka Sen

Ah, I see.

Agawa001

@Chris'ssistheartist i wasnt talking about Encouragement speech i was talking about being old enough to stop learning and begin practising

Chris's sis the artist

@Agawa001 There no such a thing. Learning is a process that never stops, it lasts till the last day, just to accept that and happily do the work to do.

@Agawa001 It's a great blessing we can learn the whole life, it makes life really beautiful, you enjoy the miracles of the life depths every day.

sajjad

18:35

@Agawa001 we assume that it is periodic function and using the same method but with separate solutions for even and odd state

Math Man

@Agawa001 Yep, I wrote a rock paper scissors game that always beat me in the long run.

Agawa001

@Chris'ssistheartist yes, life is a bless and we must take advantage of it to learn and turn off those lame comments we hear from real people more than duplicate accounts.

Math Man

@Agawa001 @Agawa001 Which one?

Chris's sis the artist

@Agawa001 well-said!

Agawa001

@MathMan hmm ?

Math Man

18:45

You said "seeking mercy in mathman's question." curious which question?

Agawa001

2

Q: Are there Latin squares with no repeats on the diagonal not of the form 2y+x+1(mod n)?

I am looking for a certain kind of latin square (nxn). Rules: No repeats in any column or row (Definition of Latin Square) No repeats in any diagonal including others than the main diagonal. So $\begin{matrix} 1&2\\2&1 \end{matrix}$ would not qualify nor would $\begin{matrix} 1&2&(3)&4\\4&(3)&2&1\\2

matrices modular-arithmetic latin-square group-homomorphism

Math Man

@Agawa001 Ah yes. It's a fun one. I am not getting any answers or even comments on it though. Is it too hard for people??? ;) or does it need a bounty?

Agawa001

@MathMan it would be kind action

Math Man

Or is it just too time consuming for anyone to feel like trying?

Agawa001

@MathMan generating a general formula is indeed time taking

Math Man

18:51

@Agawa001 True.

Chris's sis the artist

$$\int _0^1\int _0^1\frac{\log ^2(1-x y)}{x y (1-x y)} \ dx \ dy=\frac{5}{2}\zeta(4)$$

However this is not the type of integral I study these days (it's not an advanced integral), but it's pretty good for some fun.

Also it's not bad to consider the variant $$\int _0^1\int _0^1 \int_0^1 \frac{\log ^2(1-x y z)}{x y z(1-x y z)} \ dx \ dy \ dz$$

Math Man

19:07

@Chris'ssistheartist, how do you come up with all these great questions?

Chris's sis the artist

@MathMan Thanks for that. They simply cross my mind with no effort, just a matter of some experience.

Math Man

@Chris'ssistheartist, is $log^2$ the base 2 logarithm?

Chris's sis the artist

@MathMan It's squared log (natural logarithm in base $e$).

Math Man

Oh like $(ln(1-xyz))^2$

Chris's sis the artist

Yeap.

Agawa001

19:12

@Chris'ssistheartist get used to use $ln$ instead of $log$, because programmers tend to interpret "log" as decimal logarithm

Chris's sis the artist

@Agawa001 Also in our country we use $\ln$, but after a while I used the more well-accepted notation, that is $\log$.

Agawa001

@Chris'ssistheartist yes log is widely accepted and considered

Chris's sis the artist

Yeap

Math Man

Can we do this? $$\int _0^1 \frac{ \int _0^1 \frac{\int_0^1 \frac{\log ^2(1-x y z)}{x(1-x y z)} \ dx}{y} \ dy}{z} \ dz$$

@Chris'ssistheartist

Chris's sis the artist

@MathMan It looks nice.

Math Man

19:21

@Chris'ssistheartist that's as far as I can go. Any hints on what to consider next?

Chris's sis the artist

@MathMan there is a problem with that one as far as I can see. You don't treat the whole thing as a triple integral, no, but you have separate single integrals to calculate.

Math Man

Well I tried to pull the $y$ and $z$ out to simplify it.

Chris's sis the artist

@MathMan Go this way and see what happens.

Math Man

Where do I go next? I don't see a way to use IBP, u-sub, or any of the other basic strategies yet?

@Chris'ssistheartist

Chris's sis the artist

@MathMan Before going there, do you see a way to do this one $$\int _0^1\int _0^1 \int_0^1 \frac{\log ^2(1-x y z)}{x y z(1-x y z)} \ dx \ dy \ dz$$?

Yours it looks a bit more advanced at first sight.

Math Man

19:33

Can we do some sort of strange u-sub where $u=xyz$

Chris's sis the artist

@MathMan For the integral you showed me I might like to start first doing the version in 2 variables.

Math Man

$du=xy+xz+yz$? No good?

Well my integral was just a version of yours.

Chris's sis the artist

@MathMan Sure, but it's a version of yours.

Math Man

So then $\int_0^1 \frac{\log ^2(1-x y z)}{x(1-x y z)} \ dx$.

(inner integral)

Chris's sis the artist

Let me work on it. The idea to make it nicely work by IBP would be to fit the integration limits such that the variable outer is the same as the variable in the upper integration limit (say).

Ah, yeah, that was funny.

Because of the small letters I thought in the first fraction iz $z$ instead o $x$.

@MathMan ^^^

$$\int _0^1 \frac{ \int _0^1 \frac{\int_0^1 \frac{\log ^2(1-x y z)}{x(1-x y z)} \ dx}{y} \ dy}{z} \ dz$$

Look on the top fraction, you are inclined to believe in the denominator you have $z$. When increasing you note that in fact it's $x$.

Math Man

19:43

@Chris'ssistheartist True. I just copy pasted yours though and added some fractions below.

Chris's sis the artist

@MathMan Yeah, they are the same. But anyone can be easily deceived by the looking of the integral as posed.

Agawa001

@MathMan the principle of your question is close to that of sudoku puzzle, i think noone down the earth would come up with exact mathematical form

Chris's sis the artist

Tell me if you also see a $z$ instead of $x$ without increasing the font. Curious to know.

It looks like $$ \frac{\log ^2(1-x y z)}{z(1-x y z)}$$ on top.

Math Man

Without increasing the font I can't tell. It could be easily seen as either.

Agawa001

@MathMan do you accept heuristics ? lower,higher boundaries?

i think so, after i read last part

Math Man

19:51

Sure. That would be great! $2*n!$ is one such lower bound for odd $n$.

Chris's sis the artist

Not sure yet what is the value of the triple integral. Let me calculate it ...

$$\int _0^1\int _0^1 \int_0^1 \frac{\log ^2(1-x y z)}{x y z(1-x y z)} \ dx \ dy \ dz$$

or in this form $$\large \int _0^1 \frac{ \int _0^1 \frac{\int_0^1 \Large \frac{\log ^2(1-x y z)}{x(1-x y z)} \ dx}{y} \ dy}{z} \ dz$$ (I used \Large to make it bigger - to me letters are clear now)

Agawa001

@MathMan who knows !

a 4 dimensional grid has just 4! solutions exactly

Math Man

@Agawa001 How did you get that???

@Chris'ssistheartist Yep can definitely see smallest $x$ clearly.

@Agawa001 do you mean an n by n by n by n grid?

Agawa001

@MathMan ?

4*4 grid of course

Math Man

@Agawa001 4*4 has no solutions.

Agawa001

20:00

it seems so

Math Man

@Agawa001 Give me an example solution. One of the 4! solutions.

Agawa001

no i just arranged the second row without paying attention to 3rd

now i cant fill lower lines

Math Man

Right

@Agawa001 Now the 8 queens problem can be done no matter which square you put the first queen into. Correct?

Agawa001

@MathMan the problem is totally irrelevant to n queens problem

Math Man

well if we can do the 8 queens problem 8 times on different sets of squares each time then we can number each set of queens and we will have completed the 8x8 grid.

Agawa001

20:09

@MathMan moreover, the solvability of n queens problem is indeed dependant to first queen emplacement

Chris's sis the artist

Almost 10 minutes here and I'm not done yet. Hope to finish it in 15 min.

Count me ...

On the last part ...

Math Man

@Agawa001 Ahh well that adds something important to my question!!! Then no solution exists for $n=8$ great!!!

Agawa001

@MathMan how ?

Chris's sis the artist

Give me one more minute!!!!!!!!!

Math Man

@Agawa001 For instance, give me a set of coordinates in which you can not place the first queen?

Chris's sis the artist

20:17

$$\Large{\text{MISSION ACCOMPLISHED}}$$

Math Man

@Chris'ssistheartist, I clock you in at 6 minutes and 31 seconds!!!

Chris's sis the artist

@MathMan It's a bit over 15 min I think ...

Now ...

awesome or what?

Anyway, let me add that to my book.

skull petrol

Indeed, an awesome display of speed.

Math Man

@Chris'ssistheartist great job!

Chris's sis the artist

@MathMan @skullpetrol Thank you

Math Man

20:24

@Agawa001, I found this link math.utah.edu/~alfeld/queens/all.gif now so it looks like no queen can be placed at (4,1). This means that in solving my problem for whatever number we place at (4,1) we will get stuck trying to place it somewhere else on the grid. Do you see the connection now?

Each number is it's own greatest enemy.

skull petrol

Some of your thinking skills are definitely highly developed. @Chris'ssistheartist

2

Chris's sis the artist

@skullpetrol Thanks. I don't wanna be modest, in general I'm not, but the truth is, as I often say it, I don't have any mathematical gift, all is done by extremely hard work. Perhaps some with gifts would have reached this perfomance much faster than me. It's great that an usual person can do such things, that's the most important lesson.

Agawa001

@MathMan yes that was my point, there is some squares where if you put a queen on, you wouldnt reach an optimal solution

i think the solution depends on n*n/4 first steps

Jeff

Linear algebra textbook example says to find the dimension of the subspace $W={(d, c-d, c)}$. The answer says "you can see that $W$ is spanned by the set $S={(0,1,1),(1,-1,0)}$ with no further explanation. How can I see that?

And how did they come up with those two vectors?

Math Man

@Agawa001 so is the n queens problem unsolvable given some specific starting squares for any even n?

If so, can you give me the proof!!!

Agawa001

20:30

@MathMan hmmm i cant extrapolate

Jeff

@Jeff Well, actually I can see it pretty easily. How did they come up with those two vecs?

Math Man

It would make a great partial answer to my question!

Sanic Hodgeheg

@Jeff you can rewrite that as $d(1,-1,0) + c(0,1,1)$

Agawa001

@MathMan again .... how is that relevant ?

Jeff

@sanic OK, did that (but I put $c$ first and wrote the vecs vertically)

Tobias Kildetoft

20:32

@Jeff Think of the subspace as being given by two parameters $s$ and $t$ (so $(s,t)$ corresponds to $(t,s-t,s)$)

Jeff

@sanic ok, i see that now. it comes out to the def. of $W$ right away

@tobias OK

Tobias Kildetoft

Now their basis is the standard one in terms of $s$ and $t$ (i.e. $(1,0)$ and $(0,1)$)

Jeff

@tobias OK, I'm with you.

Tobias Kildetoft

@Jeff And that is probably how they found them

Forever Mozart

:( research is getting TEDIOUS

Tobias Kildetoft

20:35

@ForeverMozart research in what?

Math Man

@Agawa001 If for a given n the board can not be solved from every initial placement of a queen, then my problem can not be solved for that particular n*n grid.

Jeff

@ForeverMozart I've hard research is always tedious.

Forever Mozart

strange connected spaces

Tobias Kildetoft

I generally find research to be fun

Forever Mozart

it is not easy

Agawa001

20:36

@MathMan thats not answering me, but i doubt that there is no solution for n=8

Tobias Kildetoft

@ForeverMozart Well, if it was easy, someone else would have already done it

Jeff

@TobiasKildetoft I'm not seeing the jump from the standard basis for $R^2$ to the two vecs in my question. Can you add a little bit?

Tobias Kildetoft

@Jeff Do you agree that the standard basis is a "natural" thing?

Jeff

@TobiasKildetoft The standard basis is natural for $R^2$. I agree

Math Man

@Agawa001 rules 1 and 2 of my puzzle: no repeats in any row or column. No repeats in any diagonal. That's the same rules for the n queens. So if the queens would conflict on a given grid so would the digits which must be placed in the same squares.

Tobias Kildetoft

20:38

@Jeff Ok, so once we have our parametrization of the subspace, it is natural to consider what we get from the standard basis

Balarka Sen

heya @ForeverMozart

Forever Mozart

hi it's been a while

Balarka Sen

indeed. what have you been upto?

Forever Mozart

same old stuff, teaching and trying to prove something cool

how about you

Jeff

@TobiasKildetoft Yes, but what do you mean by that? What's the next step?

Balarka Sen

20:41

what have you been trying to prove?

i'm fine. i am studying math.

Tobias Kildetoft

@Jeff That was the final step. Those basis vectors are what you get from the standard basis using this parametrization

Chris's sis the artist

@skullpetrol That integral can be nicely related to another very nice series I worked with in my research. It's a matter of top research.

Math Man

@Agawa001 Take this solution for example. $\begin{matrix} 1&2&3&4&5\\3&4&5&1&2\\5&1&2&3&4\\2&3&4&5&1\\4&5&1&2&3\end{matrix}$ If we replace a number with a queen (say 4) we have $\begin{matrix} -&-&-&Q&-\\-&Q&-&-&-\\-&-&-&-&Q\\-&-&Q&-&-\\Q&-&-&-&-\end{matrix}$ If we cannot make a valid queen puzzle starting on the 4th square of the top row then we cannot place all of the 4's into the puzzle.

Chris's sis the artist

The amazing thing is that I can finished that without using the series and so I have now 2 different ways of calculating the series.

I'm amazed how well things work.

skull petrol

Indeed @Chris'ssistheartist

Agawa001

20:44

@MathMan oh , so its not validity of solution, what you want is solvability by every starting position !!!!

Chris's sis the artist

@BalarkaSen thanks for writing on channel 'multivariable calculus', that pushed me to create 2 very nice integrals that I'm going to add to my book.

Probably I publish in some magazine the triple version.

Balarka Sen

sure. can you tell me how you did your 2 variable version? i'm interested.

Math Man

@Agawa001 Well if any given n can be solved it can be solved in at least n! ways.

Chris's sis the artist

@BalarkaSen Give you a hint? Use the generating function of the harmonic series (in a clever way). In another way, just use Taylor series and some clever manipulation of the resulting series.

Jeff

@TobiasKildetoft OK, wait. You're saying to map vectors in the natural basis $v_1=(1,0), v_2=(0,1)$ to the new space? IOW $(1,0) \to (0,1,1)$ and $(0,1) \to (1, -1, 0)$? Those are the two vecs the example provides.

@TobiasKildetoft I think I'm good-to-go. TY!

Tobias Kildetoft

20:46

@Jeff Right

Balarka Sen

@Chris'ssistheartist Hmm, let me see.

Math Man

@Agawa001 The part about the n queens is simply helping me to eliminate n values. It doesn't create solutions as it now stands.

Jeff

Anyone remember where to get a history of this room (like from when I asked my question until about now)?

Math Man

@Jeff at the top of the page click "full transcript"

Balarka Sen

@Chris'ssistheartist I see, you are using the Taylor series for $\log^2(1-t)/t(1-t)$ and then plugging it there with $t = xy$?

Chris's sis the artist

20:49

@BalarkaSen Not really. You can use -$\displaystyle \sum_{n=1}^{\infty} H_n x^n =\frac{\log(1-x)}{1-x}$, and then replace $x$ by $xy$

Jeff

@MathMan TY. Is there any way to save it? I've saved it as an .html (but not sure if that will work for long).

Balarka Sen

@Chris'ssistheartist Why not? Seems like you can arrive at a serious using that too.

@Chris'ssistheartist Hmm.

How do you tackle the squared-log, then?

Chris's sis the artist

@BalarkaSen You already dealt with a part. Now you can deal with the remaining part of the integrand (using series again).

Math Man

@Jeff not an easy way that I know of.

Balarka Sen

Oh, yeah, the remaining part is $\log(1-xy)/xy$, which you can just write as $\log(1-xy)/(1-xy) \cdot (1-xy)/xy$ and use that harmonic series again.

Jeff

20:55

@MathMan OK. I think I have the procedure I just learned down now anyway (I just typed it up!). Thanks.

Math Man

@Agawa001 did my explanation make sense?

Chris's sis the artist

@BalarkaSen Or simple Taylor series.

Balarka Sen

Fair enough. Cool. You gotta teach me some of these when I get to multivariable integration.

Chris's sis the artist

@BalarkaSen hehe, it would be easier to me to give you my book. :D

There I explain all very clear, even if you don't wanna learn, you learn it. :D

Balarka Sen

Well, the book has not been published yet. The best thing available is the to-be author. :P

Chris's sis the artist

20:58

:D

Balarka Sen

OK, I gotta get some work done.

Chris's sis the artist

Soon I'll publish an article with its sister, that is more advanced, but that I slay it very very bad ... :-)

(poor infinite series ...)

OK, this is some stuff a bit more advanced. Let's go back to an interesting integral, although very simple

What is the most brilliant way to calculate this one? $$\int_0^1 x^{n-1} \log(1-x) \ dx$$

No Taylor series, no double integrals, just some very simple and clever tricks.

(no pen and paper allowed - for some more fun :-))

Agawa001

21:26

@MathMan thats reminds me of how did i solve this, nice approach my twin :D

sory, internet cut-off, i think my isp is throttled

Math Man

@Agawa001 I made a room for our discussion here chat.stackexchange.com/rooms/info/30604/…

Chris's sis the artist

22:17

@skullpetrol it seems no one brings the calculation of the integral above at the artwork level.

@BalarkaSen calculating the integral above in 4, 5 ways is one of the things to do before going in the area of multiple integrals. It's a very nice challenge, to try to do that in a very simple way.

skull petrol

@Chris'ssistheartist not everybody is an artist :-)

Chris's sis the artist

@skullpetrol :D

@robjohn you're silent for days again.

Agawa001

@skullpetrol what happens to your old nick, mutant !

Chris's sis the artist

@skullpetrol robjohn is but he seems less interested in this stuff these days.

skull petrol

@Agawa001 what do you mean?

Perhaps he's busy @Chris'ssistheartist

Agawa001

22:25

@skullpetrol patrol -> petrol

Chris's sis the artist

@skullpetrol Yeap, I think so.

skull petrol

Skill got the patrol @Agawa001 :-)

Agawa001

@skullpetrol you werent skill patrol ?

skull petrol

Yes, I am both.

skull patrol = skill patrol + skull petrol

Agawa001

nice set of names, pal

skull petrol

22:33

thanks buddy :-)

Karim Mansour

hey

skull petrol

22:49

hey

morphic

hey

robjohn

@Chris'ssistheartist I've been talking, at least according to the logs... but I have been out of town today for a while.

Chris's sis the artist

@robjohn hehe, OK :-)

Transcript for

$Mathematics$ Mathematics