Mathematics

t.b.

12:00 AM

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

Victor

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

robjohn

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

@Victor The 4D case of what?

Victor

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

t.b.

Most likely Hamilton or Grassmann in some form.

robjohn

@Victor What tb said :-)

Victor

12:09 AM

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

t.b.

archive.org/details/dielinealeausde00grasgoog

for example

Victor

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

J. M.

@JonasTeuwen This and this.

t.b.

@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

Victor

@tb - Thanks to you and robjohn

t.b.

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

robjohn

@tb Thanks for the assistance :-)

J. M.

@tb Good morning to you too!

robjohn

@JM good morning!

J. M.

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

robjohn

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

J. M.

12:18 AM

Good evening, @rob.

t.b.

Paul Garret's just great: "One might object to certain very-classical treatments which make Gamma appear as a singular thing. While I agree that it is of singular importance in applications within and without mathematics, the means of establishing its asymptotics are not."

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

robjohn

@tb where was that?

t.b.

@robjohn your Stirling answer

robjohn

@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.

J. M.

12:34 AM

At least he took some time to explain... :)

t.b.

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

robjohn

@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.

t.b.

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

robjohn

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.

t.b.

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

robjohn

@tb :-)

J. M.

1:14 AM

I need to step out too. See you guys later!

t.b.

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.

Zhen Lin

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

t.b.

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

Zhen Lin

Yes, now it does. Thanks!

t.b.

2:29 AM

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

t.b.

2:45 AM

I hate it when people go for cheap partial answers

Srivatsan

2:56 AM

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

t.b.

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

Srivatsan

Done, thanks.

t.b.

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

Srivatsan

Right.

t.b.

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

Srivatsan

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.

t.b.

@Srivatsan I count two answers: Bill gave both.

What especially annoys me about this question is that the OP asked at least 50 times a similar question and still he gets upvotes and still he doesn't see the answer...

Srivatsan

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.

Woah!

t.b.

@Srivatsan now that was quick :)

Srivatsan

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?

Zhen Lin

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

Srivatsan

3:17 AM

I didn't emphasise it but $A$ is not square.

Zhen Lin

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

Srivatsan

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

t.b.

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 $\geq 0$.

Srivatsan

Yes, eigenvalues are nonnegative. Then?

t.b.

their product is non-negative.

Zhen Lin

3:22 AM

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

Srivatsan

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.

Sivaraman

@Srivatsan How are you doing?

t.b.

3:43 AM

@Srivatsan yeah, I saw that :)

Incognito

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.

t.b.

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

Incognito

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

t.b.

(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)

Incognito

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

t.b.

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 + ...

Incognito

That... is interesting. Hrmm.

t.b.

@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.

Incognito

That makes a lot of sense.

The patterns in this series just continues to boggle my mind.

t.b.

You're not the only one :)

Incognito

4:45 AM

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

t.b.

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

Incognito

Hahaha! Thanks!