Mathematics - 2017-06-26 (page 1 of 6)

Semiclassical

12:00 AM

upshot, though, is that you're right that there should be an entire family of such generalizations

by taking $T(x)=x+k$ instead of $x+1$.

Adeek

okay awesome will add you

Semiclassical

So $g(x+k) = f(x)+k\implies g(x)=f(x-k)+k$

Daminark

Awesome

Semiclassical

So yeah, there should be an infinite number of such conjectures.

Adeek

@Daminark added you

Semiclassical

12:01 AM

In fact, $g(mx+b)=mf(x)+b$ should work as well.

or any $T(x)$, really.

Daminark

Fantastic

Semiclassical

Though not every one would be equally interesting.

Dodsy

hm

Semiclassical

For instance, $T(x)=x$ is obviously trivial---that's just $g(x)=f(x)$.

And $T(x)=0$ would similarly be trivial. (Doing some number of iterations of Collatz and then replacing the output by 0 would be the same as setting the input to zero and then iterating another map. A valid property, but a really boring one.)

I don't see a useful lesson from this as far as proof goes, though. It's cute but doesn't tells us anything about $f$ itself.

As it is it's just symbol-juggling and that usually doesn't help much.

Dodsy

okay here's the deal though

if we write it like $\frac{n+x}{2}$ $\frac{kn}{2}$

first of all

k has to be odd

and so does x.

interesting.

I really did accidentally stumble upon the best in that form.

Semiclassical

12:11 AM

Here's a fun equivalent version of Collatz which Wikipedia gives.

Dodsy

?

they give one?

Semiclassical

They give a few, but not of the kind you came up with.

Dodsy

oic

Semiclassical

Suppose we have a string consisting of the letter $a$ written $n$ times.

And we have the following 'production rules a->bc, b->a, c->aaa.

What we'll do at each step is delete the first letter of the string and append the rule to the end.

So aaa -> aabc -> abcbc-> bcbcbc->cbcbca->... and so on.

evidently this is equivalent to Collatz.

Dodsy

ooo

Semiclassical

12:16 AM

It's the "tag system" version discussed on the Wikipedia page

I suspect the point is that, in going through that, eventually you'll clear out all instances of the letters b,c and be left with another string of just a's.

Dodsy

nice.

Semiclassical

And evidently the number of a's in this output string is the same as what the Collatz map would give you based on the number of a's in the input string.

So it's equivalent, though with a lot of steps in between.

Seems more a novelty than anything else, though.

Dodsy

hm

I think mine is a bit of a novelty

like "hey theres actually an infinitude of collatz conjectures"

Semiclassical

Yeah, alas.

Dodsy

though I haven't found another yet.

have you?

Semiclassical

12:20 AM

Well, one should be able to proceed as such.

Suppose we take $T(x)=x+2$, so that we want $g(x+2)=f(x)+2\implies g(x)=f(x-2)+2$

So that'd be $g(x)=(x-2)/2+2=x/2+1$ if $x$ even and $g(x)=(3(x-2)+1)/2+2=(3x-1)/2$ if $x$ odd.

Dodsy

hm, right.

Semiclassical

So that one should also be equivalent to Collatz, but shifted up by 2 now.

Dodsy

that's interesting.

Fawad

12:47 AM

Hi

What is adj(adj(A)) for invertible matrix A ?

Semiclassical

well, $\text{adj}(A)=A^{-1}\det{A}$

so $\text{adj}(\text{adj}(A))=(A^{-1}\det A)^{-1}\det((A^{-1}\det A)^{-1})$

Balarka Sen

Off-topic: I hate computing adjugates

Semiclassical

which...ew.

Balarka Sen

for a given numerical matrix

Semiclassical

But that should come out as $A\cdot (\det A)^{-1}\cdot (\det A)\cdot (\det A)^{?}$

Where I'm not sure of the last exponent which comes from pulling a scalar out of the determinant.

I guess it'll depend on the matrix size?

Balarka Sen

12:52 AM

@Semiclassical I think there's no -1 inside the det

Semiclassical

woops, you're right

Can't edit now, though. drat

Bottom line, I think, is that adj(adj(A))=cA where c is some exponent of det(A).

Balarka Sen

Yah

Well, no, did you mean det(A)^c * A?

Semiclassical

Right.

Balarka Sen

Ok, I agree

Semiclassical

some power of det(A), yeah. poor wording

Easiest way to check would be to take A=bI

Balarka Sen

12:54 AM

Should be (det(A))^{n-1} I think

Semiclassical

Sound right.

But, uh

I don't actually care enough to check :/

Balarka Sen

adj(A) = A^-1 * det(A) => adj(adj(A)) = (A^-1 * det(A))^-1 * det(A^-1 * det(A)) = A * det(A)^-1 * det(A^-1) * det(A)^n = A* det(A)^(n-2)

Semiclassical

works for me.

Balarka Sen

So it's $\text{adj} \text{adj}(A) = A \cdot \text{det}(A)^{n-2}$.

Just for the sake of being complete.

and having nothing better to do

Semiclassical

Right. Where $n$ is the matrix size.

Balarka Sen

12:59 AM

Mhm

Semiclassical

kinda neat that adj(adj(A))=A for the n=2 case.

Though inevitable, come to think of it.

in that case, adjugate flips signs on the off-diagonal and entries on the main diagonal.

Balarka Sen

Yeah so you get the same determinant, right?

Semiclassical

right.

Balarka Sen

which cancel etc

Semiclassical

yeah. not very exciting.

Balarka Sen

1:01 AM

i HATE computing numerical adjugates tho

mess up everytime

and i gotta do those for school

Semiclassical

ew

Balarka Sen

solving equations are slightly easier because the value of the k-th unknown variable is just (determinant of equation matrix with k-th column replaced with constant coefficients of the system)/(determinant of equation matrix)

I think

Semiclassical

yeah, cramers rule blah blah blah

hard to really be excited about any of that, though.

Balarka Sen

is cramer's rule that, or adj(A)/det(A) = A^-1, or both?

yeah meh

Semiclassical

Cramer's rule is that, yeah.

Balarka Sen

1:05 AM

Ah ok

Semiclassical

Jacobi's formula is somewhat neater: $d\,\det A = \text{tr}(\text{adj}(A)\,dA)$

plus the corollary that $\text{tr}\ln A=\ln\det A$

Balarka Sen

in what sense are we differentiating

Semiclassical

with respect to some parameter $t$ which $A$ depends on.

Balarka Sen

ah ok. so just termwise diff wrt t

Semiclassical

should really be d/dt on the left and dA/dt on the right

Balarka Sen

1:07 AM

considering A(t) as a function R --> R^mxn

Semiclassical

but meh. same difference.

Yeah.

Balarka Sen

got it

interesting

Semiclassical

As a special case you can think of that as differentiating with respect to a particular entry of A.

Balarka Sen

Sure, sure

Semiclassical

(I am blatantly stealing that from Wikipedia)

which gives $\dfrac{\partial}{\partial A}_{ij}\det A=\text{adj}^T(A)_{ij}$

Which is neat.

Balarka Sen

1:10 AM

Wait where did the trace term go?

Semiclassical

Good question. Hm.

I think the point is that when you differentiate A w/r/t A_{ij} what you get is just a matrix which is zero everywhere except the ij entry which is 1.

So when you then do matrix multiplication and trace you just pull out one entry.

Balarka Sen

I don't see why you should get 0 everywhere. Consider [t, t^2; t^3, t^4]

differentiate respect to A_11

which is t

Semiclassical

which I guess amounts to $dA=dt\, e_i e_j^T$.

oh.

No, I think that's meant in the sense that A can be understood as a function of its entries (all n^2 of them)

Balarka Sen

Oh, so you keep the rest constant

Just like that

Semiclassical

Right.

Liable to be confusing, though.

Balarka Sen

1:14 AM

So you're basically differentiating (x_1, x_2, x_3, ..., x_n) with respect to x_1 :P Not very exciting

Semiclassical

Yuuuup.

And, plus, adj^T is just a matrix of cofactors

Balarka Sen

Ya

Semiclassical

so it really just tells you that differentiating a determinant w/r/t one of its entries gives the corresponding cofactor.

Balarka Sen

Simple enough

Semiclassical

yeah.

Balarka Sen

1:15 AM

Hm, so how do I prove this identity. Cofactor expansion and induct? lol

Hey that can work

Semiclassical

dunno. I advertise, I don't derive.

Balarka Sen

Spoken like a true physicist

Semiclassical

yuuup

Wikipedia gives a proof of Jacobi's formula, but it's pretty boring.

not very hard, but not very interesting

Balarka Sen

yeh

i'll check it out later. i think induction does prove it

Semiclassical

probably

Balarka Sen

1:18 AM

the identity is more cute looking

Semiclassical

How'd your physics test come out? Sounded like you were pretty solid on it.

Balarka Sen

LOL

me too

even though I'm asian

@Semiclassical I did good on it. It wasn't a serious test, just a preparatory thing.

Semiclassical

mmkay.

What've you got next?

Balarka Sen

Kirchoff et al it seems

Kirchoff's law is pretty dank

Semiclassical

Yeah.

How are the teaching it? The usual (and more physically sensible) version is where each branch is labelled by a current.

Balarka Sen

1:23 AM

That's the first law isn't it? The vector sum of the currents is 0 at each branch

Semiclassical

Right.

Balarka Sen

the voltage law is about loops on the circuit diagram

total voltage = total emf

i think

Semiclassical

There's another way to formulate the current law which is neat.

Namely, you associate a (formal) current to each face of your circuit.

With some particular orientation as well, e.g. clockwise or counter-clockwise

If you've got a common edge between two faces, then you add the associated currents from the adjacent faces with appropriate directions.

What's nice about this is that the current rule is then satisfied automatically.

Balarka Sen

Ah yes sure

"Mesh law" something something

Semiclassical

Right.

What's especially cute is that this meshes well with how you might think of a circuit from the pov of cellular homology

the boundary of a 2-chain, after all, is a sum of 1-chains.

and the boundary of a 1-chain is a difference of 0-chains.

Balarka Sen

1:33 AM

Hm, yes, I see the connection

Semiclassical

I picked this up from a John Baez post here: math.ucr.edu/home/baez/week293.html

Though he doesn't seem to pose it in the way I've indicated. Hmm.

Balarka Sen

Wow fun perspective

Semiclassical

I suspect that the difference with his is that, for him, $\delta I=0$ is an imposed condition

But if $I$ is already the boundary of a 2-chain, that's automatic.

Balarka Sen

Kirchoff's current law and voltage law are just a cycle condition, $dI = 0$ and $dV = 0$.

I'm bookmarking this man. Thanks!

Semiclassical

oh, hey, he even has what I just said later down there: $I=\delta J$ where $J$ is a set of 'mesh currents'

@BalarkaSen Though, he seems to put I and V on different footings. Namely, I is a 1-chain and V is a 1-cochain.

Balarka Sen

1:38 AM

Yep.

Semiclassical

so it'd be $\delta I=0$ and $dV=0$.

Balarka Sen

Yeah I use the same differentials for both homology and cohomology

Semiclassical

And then $P=V(I)$ is automatic. Neat.

ahhh

Balarka Sen

I'll have to look at this more thoroughly. Thanks.

Semiclassical

np. I may have to do so as well

Quite overkill for any Kirchoff stuff you'll do, of course.

Dodsy

1:40 AM

I love doing long division in my head

:}

Semiclassical

lol

Balarka Sen

Heh, true

Semiclassical

the only thing better then mental long division of numbers is mental long division of polynomials /s

Dodsy

o

Fawad

$(tA)^{-1}=\dfrac 1t A^{-1}$ ?

Dodsy

1:41 AM

but I bet you can do like $\frac{45}{252}$ in your head pretty easily.

Semiclassical

main problem there is figuring out what 252/9 is

Fawad

t is scalar and A invertible matrix

Semiclassical

i can tell it factors (2+5+2=9 yay) but

a little harder to hold it in my head.

Dodsy

see what I do is just make $45=450$

then do the division

Semiclassical

heh, that works.

Dodsy

1:43 AM

yeah

so that's 1.

and then I count the remainders

and then do it again

I really do enjoy doing it, I find it relaxing

Adeek

hi @arctictern

Semiclassical

right. that's basically the usual by-hand version of long division

Dodsy

o really

Balarka Sen

@Semiclassical I actually messed up an integral for time constraint while computing the electric field for an infinitely long straight wire. But it panned out fine when I tried it after coming back home :( I did an alternative on the exam anyway

Semiclassical

Sure.

Dodsy

1:44 AM

I barely remember learning division, to be honest.

It took me a while to get it again.

Semiclassical

i guess what i do is note that 45 and 252 have 9 as a common factor because they both add to 9.

Dodsy

hm.

that's an interesting thought.

where do you go after that.

Semiclassical

45/9 = 5 is automatic

Dodsy

also long division of polynomals I can not do in my head

at all

oh I see.

Semiclassical

i then mentally note that 252 is closest to 270, specifically 252=270-18, and then I can divide by 9 easily to get 252=9 * 30 - 9 * 2 = 9 * 28

Dodsy

1:45 AM

so you reduce it to $\frac{5}{28}$

Semiclassical

Right.

Dodsy

Brb, going to gf's house

Semiclassical

Of course, if you asked me for the actual decimal expansion...

Adeek

@Dodsy math is more important

Semiclassical

well, i'd get annoyed at you, and then I'd do long division by hand :P

mostly on the principle that dividing by something with a factor of 7 is always annoying.

Dodsy

1:54 AM

okay so now you've reduced it to 5/28

I think it's mostly estimation

I don't usually reduce it.

so if I had 45/252 I'd say okay that is 1

and then I have 198

so I guess 7

then I have to basically double check

so it still is a lengthy process

but once you get to .17

that's pretty good

especially if someone just asks you off hand what percentage is 45 of 252

you can say oh about 17%.

and then if you compute the next digit

you can even say "close to 18%"

Semiclassical

2:28 AM

@Dodsy The other way to get an estimate is to use some Taylor series shenanigans:

$$displaystyle \frac{5}{28} = \frac{5}{30-2}=\frac{5/30}{1-2/30}\approx \frac{5}{30}\left(1+\frac{2}{30}\right)$$

Daminark

@Adeek kek

Semiclassical

Nice thing at this point is that the fractions only involve multiples of 10 and 3.

So that's $1/6=0.166\ldots$

and $10/900=1/90=0.011\ldots$

which add together to give $0.177\ldots$

blarg, how'd I miss that displaystyle. that's just embarrassing.

anyways, $0.1\overline{7}$ agrees with $5/28 \approx 0.17857$ to within half a percent. Not bad.

But one can monkey around with the fraction in similar ways to get (possibly better) approximations.

heh, including the rather simple: 45/252 = 180/1004, which is nearly 180/1000=0.180

which blah blah blah leads to 9/50 - 9/12500 = 18/100 - 72/100000 = 0.18000 - 0.00072 = 0.17928.

or one can do long division and avoid such silliness :P

oh, piddle. should be 180/1008.

Adeek

2:57 AM

@Daminark I am getting stuck in a stupid calculation :S

it is so stupid and long

and it is boring

Daminark

Calculations are meh

Adeek

My prof is like that he wants every little detail to be checked

even if it is long and nuance

Typhon

hi

@Daminark not dankness.... Daminankness.

Adeek

3:21 AM

Is it true that finitely generated projective module is flat ?

Daminark

3:42 AM

shrug

@Typhon heh

Zee

@Daminark grothendieck wanna be

Daminark

His style is interesting for sure but I don't want to emulate any one person

Zee

@Daminark what's his style? Defining the problems away?

I don't think he really had a style or a philosophy contrary to what it appears, he was just god dang smart, that's why he never had any strong following, kinda the case with Wittgenstein as well

Semiclassical

Ehhhh, to say Wittgenstein didn't have a following seems a bit of a stretch. (He just didn't have a following of people who actually understood him.)

The logical positivists in Vienna were fans of his, for instance.

Daminark

3:58 AM

I mean he was very smart, just that he was able to see connections and then codified them

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