12:00 AM
In robjohn's picture I just went from the top left picture to the rightmost one.

@robjohn - So who discover the 4D case and prove it first?

@Victor I have no idea. That depends on what you mean "the 4D case".
@Victor The 4D case of what?

@robjohn - The 4D case of the change of variable formula.

Most likely Hamilton or Grassmann in some form.

@Victor What tb said :-)

12:09 AM
@tb - They really prove the 4d or higher cases? Which book or article?

@tb - i don't speak German...

@JonasTeuwen This and this.

@Victor I didn't intend to say you should read this (it is very difficult to read and understand -- it took half a century till people started to grasp the ideas). You should look in a modern calculus text, such as Salas and Hille

@tb - Thanks to you and robjohn

12:16 AM
@JM Hi, J.M.!
Good morning!
@Victor And if you want to learn about the history of vector analysis look at Crowe

@tb Thanks for the assistance :-)

@tb Good morning to you too!

@JM good morning!

Crowe's work shows that "natural" notation doesn't always come out on the first try.

I should remember that end of the day is morning for you :-)

12:18 AM
Good evening, @rob.

@robjohn: there was somebody on a comments upvoting spree :/

@tb where was that?

@tb I will have to look.
@tb I see. A whole bunch of '1's and '3' on my last comment. I still haven't heard back from Didier.
@tb a long wait from Didier the last time meant that he was generating a three part reply about how my answer was lacking.

12:34 AM
At least he took some time to explain... :)

@robjohn let's put it this way: robjohn 3 : Didier 1 :)

@JM Oh, I am very appreciative of the fact that he explained (his?) downvote.
I feel that my answer is more complete because of the extra explanation, but I fear that it may not be as easy to read. I only left out details that I thought could be justified pretty easily and were not key to the computation. I appreciate the criticism from everyone.

@robjohn You're right, maybe some rearrangement might increase readability. Let me think about what one could do (I have some ideas already).

12:53 AM
@tb Thanks. I will be going out for a meeting this evening, so I will probably not get to read your suggestions until I get back.

@robjohn Then I'll think some more before I write them. At the moment they are just half-baked organizational ideas.

@tb :-)

1:14 AM
I need to step out too. See you guys later!

1:47 AM
@ZhenLin: I thought a bit about writing up an answer to your question, but I simply can't do better than what Bott and Tu do: see here. Working through their exposition + the Mayer-Vietoris argument should give you an answer.

Which page is that? Google doesn't want to show me...

Page 34/35 and the following ones. (maybe the link works now?)

Yes, now it does. Thanks!

2:29 AM
@Matt I forgot to state that $g$ has compact support in the "non-evil" version... :s sorry about that.

2:45 AM
I hate it when people go for cheap partial answers

2:56 AM
tb: I think this deserves more tags; can you look at it?

@Srivatsan If you add (field-theory) this should be enough

Done, thanks.

@Srivatsan I just saw that there's (quadratics) [whatever that means] seems to fit.

Right.

@Srivatsan: can you explain to me what the appeal of such a question is that five people deem it worthwhile to answer it?

3:05 AM
@tb It's easy =)
I hope you found five different perspectives though
I said that people answer it because the question is easy. I don't know whether five answers is necessary (of which three of them say the same thing, I think). I also don't know why one of the answers is voted significantly more than the rest; appeared first perhaps.

@Srivatsan I count two answers: Bill gave both.

I see. I think NS's solution is slightly different. It basically considers 4^n not as a power of 4, but as a perfect square (and hence cannot be -1 mod 3). That is a different take.

@Srivatsan now that was quick :)

tb: quick question. How easy is it prove that $A^*A$ has (real) positive determinant for any matrix $A$? In fact, is that claim even true?

Use the fact that $\det$ commutes with complex conjugation, matrix multiplication, and is invariant under transposition...

3:17 AM
I didn't emphasise it but $A$ is not square.

Well, in that case, just prove the same thing for an arbitrary positive-definite self-adjoint matrix.

Are you saying that if $B^* = B$, then $B$ has positive det?

I would argue that $A^\ast A$ is diagonalizable and $\langle A^\ast A x, x \rangle = \langle Ax,Ax \rangle \geq 0$, so all eigenvalues are $\geq 0$.

Yes, eigenvalues are nonnegative. Then?

their product is non-negative.

3:22 AM
If you're looking for a proof that the determinant is strictly positive, there isn't one. (Take $A = 0$.)

Ha! How do I forget that.
@ZhenLin Sure, sure. In fact if $A$ is a "wide" matrix (#rows < #columns), then $A^*A$ is always singular.
Thanks, tb. [I am kicking myself; I realised that eigenvalues are nonnegative but then didn't know what to do with that.]
@tb It might interest you to know that 5 became 6 recently.

@Srivatsan How are you doing?

3:43 AM
@Srivatsan yeah, I saw that :)

4:09 AM
I have a question about the feasibility of something I'm looking to do with Fibonacci numbers. Is it possible for me to have an identity that can sum all the numbers between the first number and an nth number, omitting numbers that are modulo some value (ie, even numbers)?
I've seen the identity where the sum of the first n Fibonacci numbers is the (n + 2)nd Fibonacci number minus 1.

@Incognito You can take the sum of all Fibonacci numbers minus the sum of all even Fibonacci numbers

@tb Any such thing for even valued numbers, rather than even index numbers?

(I hope I understood your question correctly: this would give the sum of the 1st the 3rd the 5th up to the nth Fibonacci number)

I'm looking to sum 1+1+3+5+13+21+55+89, where 2, 8, and 34 have been excluded.

4:25 AM
I see. I feared that I might have understood you. Then I don't know -- but you'd have to omit every third summand.
@Incognito I don't know if this helps: You could as well sum the even-valued terms, which looks like a little less work (1+1) + (3+5) + (13+21) + ... = 2 + 8 + 34 + ...

That... is interesting. Hrmm.

@Incognito And I guess you're almost done then (at least for n = 3k): you can just compute (F(3k+2) - 1)/2 (the sum of all Fibonacci numbers up to 3k divided by 2). If n = 3k+1 or n=3k+2 this needs only a very small adjustment.

That makes a lot of sense.
The patterns in this series just continues to boggle my mind.

You're not the only one :)

4:45 AM
Now, I wonder if I can somehow take binet's formula into that... but I'll save that for tomorrow.

Sure you can, but I'll leave that to you :)

Hahaha! Thanks!
Anyway, thanks for the help. It's bedtime for me now.

G'night!

5:32 AM
hi all

5:57 AM
hi $\forall$

1 hour later…
6:58 AM
hi, Can I found the end point of an arc? I have arc-length, start point and theta values

7:13 AM
@coure2011, that's been asked on the site recently

hi \forall

hi $\exists$

wassup @QED

just woke up

k where r u from?

7:23 AM
UK

I see
I am from India

From the last 3 days trying to figure out a simple thing but not able to find how to do that
Last time I studied/user mathematic is some 4-5 years back... looking that I forgot everything

@coure2011, well it is a circular arc?

yes

7:26 AM
This is the coordinates for a circle (centerx + radius*cos(angle), centery + radius*sin(angle))

that does it

so pick some center point and radius, then plot points for every angle between startangle and endangle

the circle is not on origin
also I dont know the center point

you have to first calculate the center point from things you do know

how?

7:30 AM
how what

how to find center point of circle

@coure2011 given what?

Start-Point (Ax, Ay), Radius of the circle, inside angle of the arc, and length of the arc using r x theta
I have to find end point

actually the centre must be determined first

@coure2011, see the comment here math.stackexchange.com/questions/96295/…

7:33 AM
how can I find the center

of ourse you can find position of end point relative to centre point anytime

ok lets say I have center point Cx, Cy. how to find the end point then?

k then u have starting point (Ax, Ay)=(r cos t, r sin t)
from this find t
u have inside angle of arc
let it be s
so the angle for end point will be t+s
and the point then becomes (r cos (t+s), t sin(t+s))
you are done

how to find t???

what? I have told everything

7:38 AM
thats what from where my problem started :(
I cannot find t using Ax = r cos t
using the programming language I am using

u know Ax and r right?

@coure2011: that is the geometric way to find the center, using straight-edge and compass.

I have to apply my solution in a computer program
I have to find path of an arc. From start point to end point with radius r

7:42 AM
Okay, so you are given a point on the circle? the radius of the circle?
ah

Let me summarize my problem again

given two points and a radius, there are two arcs of circles.

I have 2 points. P1, P2 and Radius of the circle on which these two points are. I also know the angle (lets say its 60 deg) between Points P1 and P2. Now.
I have to find Points , starting from P1 towards P2 after each 10 deg angle (so total 6 points)

|P2-P1|=2r sin(a/2)
so the angle is extra info (or the radius is)

7:46 AM

wow great pun ;)

how to calculate next point after P1, 10 deg away from P1
so no solution for this? ok another thing
Can I calculate 3rd point of a trianlge, knowing 2 points, all 3 side's lengths and all 3 angles inside them

$\frac{P_2+P_1}{2}\pm\left(\frac{P_2-P_1}{|P_2-P_1|}\right)^R \sqrt{r^2-\left|\frac{P_2-P_1}{2}\right|^2}$
where $(x,y)^R=(y,-x)$

can you please format it correctly? i am not able to read it

@coure2011 have you not gotten the MathJax bookmark?

7:56 AM
no

not getting it all

@coure2011 not getting all of what?

mathjax bookmark

@coure2011 you mean the stuff with the scroll bar?

8:01 AM
u provided answer in $\ frac { and so on I am not able to read it how to read it Go to the link I supplied above and install the bookmark. then come back here. There are at least two ways to install it given there. I clicked on 'This Bookmark' link that goes to pastebin.com/UYgNvcVU and that page is empty @coure2011 I just clicked on the link that you provided and I see the page. its showing me empty page why not try this link? 8:04 AM yes this is working and after clicking render link its showing me formatted equation now how to show the equation u typed here @coure2011 You need to drag that link to your bookmark bar the "render MathJax" link in chrome there is no bookmark bar you have to make it visible.... just a sec thx not working Oh man, what a nice hangover. 8:09 AM @coure2011 docking the bookmark bar? yes its visible, I am dragging the link to that bar but nothing happens okay, then let's try something else... can you create a new bookmark on the bar? yes but for other links not for this render link any other way I can see the formatted equation? 8:14 AM Create one and copy the text from the first page to the destination The text above the scroll bar `javascript:(function(){if(window.MathJax===undefined){var%20script%20=%20document.createElement("script");script.type%20=%20"text/javascript";script.src%20=%20"https://d3eoax9i5htok0.cloudfront.net/mathjax/latest/MathJax.js?config=TeX-AMS_HTML";var%20config%20=%20'MathJax.Hub.Config({'%20+%20'extensions:%20["tex2jax.js"],'%20+%20'tex2jax:%20{%20inlineMath:%20[["$","$"],["\\\\\\\\\\\$$","\\\\\\\\\\\$$"]],%20displayMath:%20[["$$","$$"],["\\\$","\\\$"]],%20processEscapes:%20true%20},'%20+%20'jax:%20["input/TeX","output/HTML-CSS"]'%20+%20'});'%20+%20'MathJax.Hub.Startup.onload();';if%20(wind that text :-) again not getting it what to do okay what do u mean by "Create one and copy the text from the first page to the destination" @robjohn: pastebin! @coure2011 We don't appreciate inet srtcuts here. on this page there is a scroll bar with text that looks like "javascript:(function(){if(window.MathJax" above it... 8:18 AM yes? copy that text (using a triple click or control-A or whatever you can to select it all) copied and paste it into the destination for the bookmark on the bookmark bar @AsafKaragila We tried pastebin, but coure2011 sees a blank page. @AsafKaragila Do you need to sign up to get stuff from pastebin? @robjohn Well. For some reason the Google! Spy does not approve pastebins I guess. @robjohn Not that I know of. thanks done 8:21 AM @AsafKaragila Ah so that is why they see nothing. Possible. @coure2011 now go click on that bookmark while viewing this chat page I recall Matt (the Swiss one) was working on a Chrome plugin for the ChatJax. yes got it. thanks for ur patience$\frac{P_2+P_1}{2}\pm\left(\frac{P_2-P_1}{|P_2-P_1|}\right)^R \sqrt{r^2-\left|\frac{P_2-P_1}{2}\right|^2}$8:22 AM hahaha @ equation does that look better? yes bravo! its looking very complex where$(x,y)^R=(y,-x)$8:24 AM its used to find the 3rd vertex of triangle? That finds the center of the circle. @coure2011 Complex because the square root is of a negative number? there are two as I mentioned before @AsafKaragila complicated, I think Speak of the Devil! I think the game is very higher than my knowledge/skill 8:25 AM Buenos. namaste I shld leave the project which project? finding arc path @Matt If I would say that you are cheesy (not now, of course...) would that make you full of holes? 8:27 AM @tb I don't think it needs to have compact support. thanks to all for ur support I have a question @AsafKaragila No. Why am I cheesy? @coure2011 Could you possible use real words? let T1, T2,.....Tn be numbers such that 8:29 AM @Matt You're not, but you should you do something cheesy... You'd be Swiss cheese, and thus full of holes! kiya matlub real words? I'm sorry, I don't speak Moronic. What did you say? @AsafKaragila This is so asaf : ) How long have you been working on this joke? thats Hindi btw @Matt Just came to me moments before you joined when I said that you were working on a Chrome ChatJax. 8:32 AM @AsafKaragila I've given up on making the button. It should have been a matter of 10 min and turned out to be too much pain. Now I made a bookmark and it works even better on Chrome than on FF. It's super fast compared to FF. Let T1,T2,....Tn be numbers such that T_k= k no. of digits in decimal expansion of an irrational number, starting from [k(k+1)/2 +1]th digit in the decimal expansion. e.g. for pi, T_1=0.3, T_2=0.14 and so on Question: Under what conditions will the set of all such T_k's will have a limit point of [0,1) ? @Matt Any bookmark? Or a ChatJax one? @AsafKaragila The bookmark that contains the MathJax code. Oh. Very good. I am hungover. I have to prepare the class for today or go to a seminar about countably-saturated real closed fields and whatnot. I'm going to go: need to have coffee and do some work. See you later folks. 8:35 AM I can't really do both. Why not? (<-- rhetorical question) No time. (<-- Hungover answer) ah a typo. The revised question is: Let T1,T2,....Tn be numbers such that T_k= k no. of digits in decimal expansion of an irrational number, starting from [k(k+1)/2 +1]th digit in the decimal expansion. e.g. for pi, T_1=0.1, T_2=0.41, T_3=0.592 and so on Question: Under what conditions will the set of all such T_k's will have a limit point of [0,1) ? 9:03 AM @Matt Hm. Looking at the time and the fact that I'm still home I guess that I'm not going to that seminar. @NikhilBellarykar, looking at the digits of an irrational number is probably going to produce near impossible to solve problems @NikhilBellarykar I'd guess that normality can be used to show convergence. Much like QED I doubt it can be solved by hand in the general case. oh if the number is normal, that means every point is a limit point? I don't know that. If the number is normal then the probability that you get almost all the rationals in$[0,1)$is$1$, I believe. I am going to shower and head out the university. 9:30 AM thanks @AsafKaragila and @QED 9:41 AM hmm got some idea 10:27 AM @AsafKaragila You replaced the seminar with piddling around on chat... No I haven't. I replaced it with a hangover and trying to focus long enough to write my notes for the class. 11:01 AM Yay, it's storm time here. @Matt It's sunny here. Poor you. Quite. I haven't shaken that hangover yet, the alcohol is slowly sipping into my brain as the caffeine rushes the blood through my veins. I wonder how that class will go. You shouldn't shake it, you should stir it. (<-- reference to "shake" vs. "shake off") I take my hangover shaken, not stirred. (<-- Reference to James Bond) 11:13 AM lol I'm stuck at HCF and LCM again, May I ask the Q? Sure. The HCF and LCM of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of two number is? what does author means by bold lines? other two , or simply he mean the factors of lcm What's HCF? highest common factor? @Matt yes , GCF , GCD ETC 11:22 AM You have gcd(x,y) = lcm(x,y) = 23 and 13 | 23 and 14 | 23? @Matt sorry , I have typo in Q : The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of two number is? That sounds better : ) I've not done this before but on Wikipedia you can find this:$lcm(a,b) = \frac{|a \cdot b|}{gcd(a,b)}$So,$a \cdot b = 13 \cdot 14 = 13 \cdot 2 \cdot 7$. @Matt I had gone the far previously, but the main problem is how do I find which is larger now? woo Hm, "the other two factors" seems to imply that it has only 3 factors... Got it a.b are co-primes here , so I should be making pairs right? 11:31 AM No,$a,b$coprime means that$gcd(a,b) = 1$but the question says that$gcd(a,b) = 23$Why not use \gcd and \lcm? Just because. That's not a reason, that is a lack of one. I have a typo up there.$a \cdot b = 23 \cdot 13 \cdot 7 \cdot 2$using the Wikipedia formula @Matt hmm , now what to do? , that's where I'm stuck. I just need hint 11:44 AM I don't see how this is going to work. You only have one factor of$23$. So I think I misunderstand the question.$23$is prime so if$gdc(a,b) = 23$both$a$and$b$have to have a factor of$23$. Help, anyone? @Matt at the end of the day you want$g$to be in$C_c\$, no?

@tb, I don't understand his response
He acknowledges that there is a gap, but then said "<gap> is not a problem"
He might be thinking oft hat group theory problem where it's like
forall ab, exists x, ax = b
and forall ab, exists y, ya = b
does anyone know the one I'm thinking of...
I've seen it in a couple different books, had to really stop and think each time - it's a tough one
at least it's got some trick to it
It's even been asked on this site, but I couldn't find it
@tb, what did you mean "don't get it"

12:06 PM
@tb Bleh. Of course. T_T

What's the problem
?
my internet went down for a moment

I'm too easily distracted. I should be memorising theorems and instead I piddle around on youtube.
I think I want bigger hands. This looks easier to play with stronger and bigger fingers...

12:24 PM

"But only hands small!" : D

I wonder which piece was actually being performed there?

12:45 PM
@ZhenLin Do you play the piano?

hi guys

Hi Nikhil.

listening to music?

Yes. : )

cool
Tchaikovsky is my favourite

12:55 PM
I don't know much Tchaikovsky : ) The nut cracker and th 4 seasons and that's about it I think.

well the swan lake ballet with various dances in it is awesome
not to mention the 1812 overture
nutcracker is also good
brb in an hour Matt
bye

Byee, Nikhil!

@Matt: I have the same problem, see...

Not knowing much Tchaikovsky? Or having small hands? Or being too easily distracted? Or all of them?