Mathematics - 2018-03-19 (page 1 of 3)

Secret

00:04

I wonder for math majors, whether they can go straight to analysis or they still need calculus as a padding

Ted Shifrin

analysis is very hard if you have never done proofs or developed intuition with computational stuff in calculus. VERY.

Secret

Right

Jakob Werner

In Germany there is no such thing like calculus

Corellian

So there's a 4x4 array of 1s and 0s and the problem says "A loop runs through all entries on row 2 and sets each $A[\text{row}][\text{column}]$ to equal $A[\text{row}+1][\text{column}]$. What is this supposed to even mean?

Jakob Werner

You do some very basic stuff in school, basically only differentiating and integrating polynomials and then in the first university semester you start with rigorous mathematics

Akiva Weinberger

00:08

@Corellian So row = 2? So you go to A[2][0] and set it equal to A[3][0]. Then you go to A[2][1] and set it equal to A[3][1]. Etc.

You're copying row 3 onto row 2, essentially

while forgetting the initial values of row 2

Krijn

@TedShifrin Real Analysis (here in the Netherlands, at least) is taught after only the one-dimensional part of calculus (the non-vector part, I mean)

Not coincidentally, I think, teaching students so much vector calculus is what bothers me most

Akiva Weinberger

Vector than ever

Corellian

@AkivaWeinberger Ahhhh. Ok the notation made no sense

Is it like $A_{ij}$?

Akiva Weinberger

Yeah, but computer science / programming -style

Corellian

Ok. Was going through the CS course on Brilliant and it was the 1st time encountering this. They didn't even name the array to begin with so it was confusing

But it seems obvious now knowing it just notates a given entry

Maneesh Narayanan

00:21

Do anyone know good reference book for the volume of solid revolving around any axis? I searched in Calculus books. I could find the theory of special cases for the above one. i.e, revolution around x-axis and y-axis only. I have a book. but it is not trustable. \textbf{Integral Calculus by Shanti Narayanan}. It has no ISBN. There is a disclaimer in the book that publisher was not responisble for the error.

please help me to find good reference.

The Testosterone Fanatic

Guys, check this out. And this is not even close to being the complete list in my arsenal to increase T.

Secret

1

Q: Solids of Revolution around other functions.

Recently I've been thinking about solids of revlution, and thought about an interesting experiment. Can you rotate functions around, for example, the line $f(x)=x$? And consequently, could you rotate a function around another entirely different function? To rotate solids around the line $f(x)=x$...

2

Q: Solid of revolution about a slanted line

I just thought about this idea and I decided to work on it. After taking on a general case, which proved to be too difficult, I tried a specific case. Something simple like the curve $y_1 = x^2$ rotating about the line $y = x$ Which is the same as rotating $y = \sqrt{x}$ about the x-axis. I k...

calculus solid-geometry

Ted Shifrin

00:50

@Krijn: If you were interested in geometry or applications instead of number theory and algebra, you'd be complaining that students don't know/understand enough multivariable calculus/analysis. I complain about mathematicians' ignorance about this all the time.

@Maneesh: One way to do it with lines is to rotate the plane to make the line horizontal. I had a student once who worked on the general case and wrote some stuff up, but I don't have it. I've never seen this in any book. In general (for curves more general than a line), you have to worry about whether it even makes sense.

heya @Antonios

Antonios-Alexandros Robotis

hi @TedShifrin how's it goign

Ted Shifrin

Doing OK ... haven't seen you in a while. You doing OK?

Antonios-Alexandros Robotis

yeeep. I've been on break.

Quite busy with work, but I think I was a little lazier than I ought to have been.

Ted Shifrin

Bad, bad Antonios. :) You're taking after your algebra students? :P

Antonios-Alexandros Robotis

loll not quite, I hope.

I think I needed a day or two off, though.

Makes things a lot more pleasant today.

Actually, I was going to send you an email. I was wondering if you could give me some tips on what would make a reasonable expository article/paper?

Ted Shifrin

00:59

Yes, I used cooking and dinner with friends and bridge as non-math things in grad school.

I'm not sure I'm an expert to answer that. To appear where?

Antonios-Alexandros Robotis

well, not really planning on submitting it anywhere except perhaps arxiv if I feel it's decent, but I'm meant to write a brief final "paper" summarizing the stuff I've read for my independent study.

This is due in - say - 2 months, but I figure that I may as well get something out of it I can show to committees come the fall.

Ted Shifrin

Probably too technical to be "expository."

If it's really masterful, something a serious student could learn from, you might check out the new AMS notes website (I put my diff geo text up there). I can give you the link.

Antonios-Alexandros Robotis

yeah very likely.

yeah, please.

Might be good to check it out for inspiration, anyways.

Ted Shifrin

You can figure it out from here.

Antonios-Alexandros Robotis

oh my. This is quite exciting LOL

i love reading course notes from classes and such.

Ted Shifrin

01:03

Glad to make you happy. :)

Antonios-Alexandros Robotis

yeah thanks! how has it been the last few weeks?

Ted Shifrin

Well, I have to move from my rented condo to one in the next building. So I'm dreading that. Especially with my degenerating neck. But I'll survive, I suspect.

Antonios-Alexandros Robotis

oh, any particular reason for that?

Ted Shifrin

For which?

Antonios-Alexandros Robotis

the move

Drew Brady

01:05

anyone here good at combinatorics haha

Ted Shifrin

Oh, my landlord decided he wanted to sell. Always a risk.

I can do the haha part, Drew.

Drew Brady

mathb.in/23477 [Combinatorics]

@TedShifrin how are you

Antonios-Alexandros Robotis

ahh, yeah that's no fun @TedShifrin.

Anyway, time for me to get back to this Serre reading :P

Ted Shifrin

Talk later, Antonios.

Antonios-Alexandros Robotis

have a nice evening. I'll be around more as of tomorrow, so no worries about that.

Drew Brady

01:07

Serre? group representations? number theory? alg geom?

Daminark

Hello

Ted Shifrin

@Drew: I'm not a good finite math person. Offhand, you'd think that with certain original strings you'd get double-counting if you don't keep track carefully.

Hi Demonark. Maybe Demonark knows how to do it, Drew.

Drew Brady

@TedShifrin Same. I like analysis and continuousy things

And yes, you're right. You have to be careful to avoid double counting.

Ted Shifrin

So, if you have a $0$ in a string and you put a $0$ either before or after it, it's the same. So such things need to be counted only once. But putting a $1$ before or after it will give different net strings.

Drew Brady

right

Daminark

01:11

I'm about to take combinatorics and then I'll be closer to competent but let's see, could be fun

Ted Shifrin

It is a cool question, though.

Drew Brady

Also, because someone may ask, induction doesn't seem to work.

for example you may assume the formula holds to go from n-k -> n-1, and then you'll try to argue why the formula holds to make the step from n-1 -> n and unfortunately, the overlaps cause a problem. (sums don't workout)

in other words, going from n-k -> n is not the same as going from n-k -> n-1 and then for each of the n-1 strings, going from n-1 -> n. Because for each extension of length n-1, you get as many unique things (purportedly) as the formula predicts, but between the set of extensions from each n-1 extension, you will almost always get many overlaps

Ted Shifrin

Not that I followed that.

Drew Brady

not that I explained it very clearly :)

basically, the short story is induction on n /k don't seem to work :(

Ted Shifrin

So it seems like you could choose where to put the additional digits. Then you just have to keep track of the double-counting I've referenced.

Drew Brady

01:15

yeah. the keeping track of the double counting is the tricky part

it is clear that the formula holds for the string of all 0s though.

because you can interpret the sum ranging over i = # of 1s to insert. and there are exactly n choose i different places to insert a 1

Ted Shifrin

Aha. So given a random sequence, if you put a new digit between two that are already there, you have $1$ option if those two are the same and $2$ options if they are different.

If they're different, there's no danger of double-counting with putting a digit somewhere else.

Drew Brady

this is also true. something i was thinking of is somehow introducing the "number of transitions"

= number of times you go from 1 -> 0

and vice versa

Ted Shifrin

Yeah, I think that's a good idea.

Drew Brady

anyways, since the formula holds for 0^{n-k}, another thing I was thinking of is to show that by flipping the ith bit to a 0 you don't change the # of unique extensions

because if you can show this, we're done. (but this is also not very obvious)

its also not that intuitive.

Corellian

Why say "a ring with more than one element and a unity" this is redundant

Drew Brady

01:22

because there are rings wiht one element

if you don't take a ring to be unital

Ted Shifrin

Because there are rings with no unity.

$2\Bbb Z$ is (in some people's definition) an infinite ring with no unity.

Drew Brady

there's also the trivial ring

R = {*}

Ted Shifrin

The point is to understand why it's not redundant to have both requirements.

Corellian

doesn't the latter imply the former

Drew Brady

{*} is a ring with unity though

Corellian

01:23

oh

Ted Shifrin

No, the $0$ ring has unity :)

Drew Brady

0 = 1

Corellian

is this basically the $1\neq 0$ thing yep ok

Drew Brady

right.

Corellian

ok, in my head i assumed the definition of unity to stipulate $1\neq 0$ :(

but this is not the case oc

Drew Brady

01:25

some authors take a ring to be unital by definition

Ted Shifrin

Yup, @Drew. Artin's book does. My book does ... :)

Drew Brady

do you know why its helpful to consider non-unital rings?

Corellian

right. my text speaks plenty of rings with unity

Ted Shifrin

Some people want to talk about $2\Bbb Z$ as a ring, in considering homomorphisms, etc.

Drew Brady

does it make sense categorically though

like if you take 2Z to be a ring does that change Z being initial

Ted Shifrin

01:27

unital? No, why?

Drew Brady

no, does that change Z being initial in the category of rings

Ted Shifrin

Oh

Then you obviously don't require that a homomorphism takes $1_R$ to $1_S$.

Corellian

@DrewBrady non-unital ring you mean a ring explicitly without unity (sometimes called a rng)?

Ted Shifrin

Yes @Corellian

Drew Brady

I think it would make that no longer true. because the initial homomorphism given is usual n |-> n \cdot 1_R

but if there is no 1_R in the ring,then.....

anyways. I think we've answer Corellian's question

Corellian

01:29

thanks drew and ted

Drew Brady

no worries and sorry for my silly aside on categories lol

yeah lemme think more about this transitions idea

Daminark

@Drew for what it's worth, $0$ will then become a zero object (both initial and terminal)

Drew Brady

yeah I figured that out after being less stupid. Ted reminded me that the category is literally different (i.e., the morphisms change)

Ted Shifrin

Ted never says anything categorical.

Drew Brady

ha. you may not have intended to said anything categorical :)

`abstract nonsense'

Daminark

01:33

Schlag intensifies

Corellian

the (compound) word category-theoretically sounds funny

Ted Shifrin

Anyhow, that's a cool combo question, Drew. Let me know how you solve it.

Only if you have a perverted sense of humor, @Corellian.

Drew Brady

Ted its actually not a problem from a text or anything, it is something I came across working on a different problem.

Ted Shifrin

Sorta like how I discovered "stars and stripes" unknowingly working on my thesis.

Drew Brady

hah stars and stripes / balls and urns / stars and bars are now pretty standard I feel

Ted Shifrin

01:37

They probably were then, too, but I never took probability or combinatorics.

Just saying ...

I'm going to do dinner. See y'all later.

Drew Brady

Yeah I never took combinatorics or graph theory. I probably should

see ya!

Secret

For me, I prefer the term "non unital rings" for rngs

and only use rings when we have identity

Maneesh Narayanan

@TedShifrin actually. the reference book. Integral Calculus by Santi Narayanan is not a credible source. It has no ISBN number. That is why I was asking for an alternative reference.

Secret

Ted already said that no books he knew talks about solid of revolutions about arbitrary curves

Maneesh Narayanan

yes.

Secret

01:48

as for axes, you can always rotate the shape so that it aligns with the x or y axes

or use the suggestion in that MSE to integrate it directly

Maneesh Narayanan

mmm.

Secret

02:14

.i.sstatic.net/4qGia.png

I am not going to calculate the shaded areas of these

Corellian

@Secret what is the first one exactly?

Secret

Think the white circles are at fixed radius ratio wrt each other

i.e. they form a geometric series

A random thought that came to mind is whether there's a nice relation between the angle of the cone that circumscribe the circles with the radius ratio of the circles

Corellian

Hmm if the first white disk has half the radius whole disk then the blue area is 2/3 the whole area I think

Secret

yeah, that's a standard result. What's more interesting is the angle of the common tangents and how it depends on the radius ratio

Corellian

02:30

The second graphic looks like a cool logo :)

Secret

That's the big difference between geometry questions vs other maths questions: Even if they are too hard, at least they look pretty

One approach I had in deriving that relation is to consider the similar triangles formed by the diameters of each circle and the two common tangents. Since we knew the radius of all the circles (including the large blue one) in principle we have everything we need to derive the tan of the similar triangles and hence the angle substended by the common tangents

namely, we will have the following sequence (assuming the large circle has radius of 1)

$$\tan \theta = \sup (\frac{1-\frac{1}{r}}{\frac{1}{r}}, \frac{1-\frac{2}{r}-\frac{1}{r^2}}{\frac{1}{r^2}}, \frac{1-\frac{2}{r}-\frac{2}{r^2}-\frac{1}{r^3}}{\frac{1}{r^3}},\cdots...)$$

Isa

Hello, I have a probability question: why P(-.62<z<8.12)=1-P(z<-.62) ? What happen with the 8.12 number?

Secret

What is P, is stuff z > 8.12 somewhere near the tail of the pdf?

Isa

P means probability

mmh I don't know if it's near the tail of pdf

Secret

02:46

I think we need more context in order to figure out why P(x > 8.12) got cut off

Corellian

Standard normal distribution?

Anything beyond 8.12 is pretty negligible

Isa

yes it's the normal distribution

I've just saw the graph, so the range where information matters is -3 to 3 approximately?

Secret

yeah, for normal distribution, anything beyond 3 standard deviations is pretty negligible in most context

Corellian

Yeah. The area beneath the curve past 8 sigmas (i.e. standard deviations) is very small.

That's the basis of the so-called empirical rule (aka 3 sigma rule)

Isa

ok thanks!

Dattier

03:42

Hi,

@Niing r is a root of P(r) iff P(r)=0

Let $P_n(x)=\sum\limits_{k=0}^n \dfrac{x^k}{k!}$. Find $\forall x\in\mathbb R, \lim P_n(x)\times P_n(\frac{1}{x})\times x^n$.

Secret

$x^a-1$. If $a | b$ then $b=na$ for some integer $n$. Then $x^b-1 = (x^a)^n-1 = (x^a)^n-1^n = (x^a-1)(stuff)$ thus r is also a root of $x^b-1$

geometrically, I have no idea what that means

Dattier

reasoning with rotation

in the complex plane

Secret

yeah I know $x^a-1$ gives the ath roots of unity, but the $Q(x)=(x^{an}-x^{a(n-1)}+x^{a(n-2)}+\cdots...+1)$ does not seemed to have a clear geometric meaning

All we knew is that (ath roots of unity) Q(x) = (bth roots of unity)

Dattier

why are you introducing Q(x) ?

Secret

because that's the other part after your factorise $x^b-1$

i.e. what I wrote as "stuff" is actually Q(x)

Dattier

03:56

what's the problem ?

@Secret

Secret

$x^b-1 = (x^a-1)(x^{a(n-1)}-x^{a(n-2)}+x^{a(n-3)}+\cdots + 1)$ right?

Dattier

yes

Secret

$x^b-1$ and $x^a-1$ gave their corresponding roots of unity, but what is the geometric meaning of $(x^{a(n-1)}-x^{a(n-2)}+x^{a(n-3)}+\cdots + 1)$?

Dattier

$\dfrac{x^b-1}{x^a-1}$ a homographie

@Secret ok ?

Secret

Homography

In projective geometry, a homography is an isomorphism of projective spaces, induced by an isomorphism of the vector spaces from which the projective spaces derive. It is a bijection that maps lines to lines, and thus a collineation. In general, some collineations are not homographies, but the fundamental theorem of projective geometry asserts that is not so in the case of real projective spaces of dimension at least two. Synonyms include projectivity, projective transformation, and projective collineation. Historically, homographies (and projective spaces) have been introduced to study perspective...

Ah, projective geometry is my worst of the geometries, that's why

ok

Dattier

04:05

in the complex plan, that is an application which keep cocyclicity

(a,b,c,d) is cocyclic if are in the same cercle

Secret

right

Dattier

That's true, if you think $x^d-1$ like a rotation (but in the "real life"it's not a rotation)

@Secret : ok ?

Secret

yes

Dattier

what's the geometric meaning of $z^2$ in complex plan ?

only on unit cercle

Secret

04:20

$z^2-1=(z-1)(z+1)$ hmm...

Dattier

without the translation z-1 (is more easy)

Secret

$z^2=zz$?

hmm...

Dattier

what effect on the angles ?

Secret

are they preserved, i.e. conformal?

Dattier

sorry no

give pi/3 and pi/5

this doubles the angles

Secret

04:25

yup

Dattier

what's the meaning "yup" ?

Secret

(ok, yes, I see, right,...)

Dattier

ah, thanks

Secret

I don't know how to say that in french

Dattier

"ok" or "je vois"

Secret

04:27

ok, so... if the domain is a circle at the origin, $z^n$ will scale the circle but otherwise won't change its shape (because any angle that get multiplied by n times will stay in the circle)

Dattier

yes

but the norm changing (if is not unit)

Secret

yeah, that's what I mean by "it scales the circle"

Dattier

now, we know : how to build with the rule and the compass?

Secret

build what with rule and compass?

Dattier

in french "la régle et le compas"

maybe the lignes and the compass

the lines and the cercles

Daniel Castle

04:33

Hi, could anyone help me reason about a sieve?

Secret

depends on what sieve

Daniel Castle

one where you start with n then add n+1, n+2 etc and move to the next number not eliminated

i was wondering how its counting function might behave

Dattier

@Secret : images.math.cnrs.fr/…

Daniel Castle

by the way all numbers of the form $2^{n-1}+1$ don't get eliminated

that should be $2^n$*

not n-1

Dattier

@DanielCastle : can you formalize your question ?

Daniel Castle

04:38

sure i posted about it a while ago over at math.stackexchange.com/questions/2491459/… and have just come back to it

Secret

@Dattier I cannot read French, but I think I kinda recognise that. solving problems related to circles by using compass and ruler.

Dattier

@Secret : yes

@Secret : now, we know : how build $z^n$ by using compass and ruler. ok ?

$n\in \mathbb N$

Secret

@Dattier think so

Dattier

@Secret the next question is, how do that with the least possible step?

@DanielCastle : sorry I don't know

Secret

So... instead of
elimination 1: 2,4,6,8,10,12,14,...
we have
elimination 1: 1,2,3,4,5,...?

Daniel Castle

04:45

ok no worries

oh @Secret was that for my question?

because the first elimination would be 1, 2, 4, 7, 11, 16, 22

you add n, then n+1, then n+2 so you end up with the formula $\frac{1}{2}(b+1)(b+2c-2)+1$ for the $b$th eliminated number starting from some $c$

Secret

hmm...

Daniel Castle

on one hand it seems as the gaps are getting larger between eliminations, there would be more left, but then they would go on to eliminate more so it seems to balance out

the data I have seems to show it decreasing fairly slowly to some limit, but I have no idea if that would be a positive constant or 0

Secret

So elimination 1: 1, 2, 4, 7, 11, 16, 22,...
Remaining: min(N-elim1) = 3
elimination 2: 3,4,6,9,13,18,...
Remaining: min(N-elim1-elim2)=5
...

Daniel Castle

well for elimination 2 you would start by adding n, so n=3 you get 3, 6, 10, 15, 21, 28...

rather than 1

but you would go to 5 yes

Secret

05:00

I am wondering, is there a nice closed form for N-elim1-elim2-elim3...?

Daniel Castle

so for the min of that? it seems fairly chaotic until you start sorting it based on the number of times it gets eliminated

where they seem to smooth out into linear relations

although the nth number not hypothetically eliminated by any < k is $k=2^n +1$, with that including all numbers, not just those that haven't been eliminated already

so for example $2^3+1$ = 9 is not eliminated by 1,2,...,8 definitely

but 1,2,...,8 are not necessarily potential eliminators

Secret

1+0,1+1,(1+1)+2,((1+1)+2)+3,(((1+1)+2)+3)+4,...,(((1+1)+2)+3+...)+n,...
3+0,3+3,(3+3)+4,((3+3)+4)+5,(((3+3)+4)+5)+6,...,(((3+3)+4)+5+...)+(3+n),...
5+0,5+5,(5+5)+6,((5+5)+6)+7,(((5+5)+6)+7)+8,...,(((5+5)+6)+7+...)+(5+n),...
...

Daniel Castle

yes exactly

sorry i just came back

Secret

not seeing any patterns yet other than 0,1,2,3,4,5,6,7,8,9,... is being added countably many times and each time it shift one step to the left

Daniel Castle

05:17

here is the counting function, so the number not eliminated < n divided by n: i.sstatic.net/g87jN.png

Secret

it looks really jittery near the beginning

Daniel Castle

yeah i noticed that, strange

it smooths out though

it keeps going past 0.25

so i was wondering if it perhaps tended to zero?

Secret

05:31

elim1 = min(N)+0+{0,1,1,1,1...}+{0,0,2,2,2,2...}+{0,0,0,3,3,3,3,3,...}+...
elim2 = min(N-elim1)+{0,min(N-elim1)}+{0,0,min(N-elim1)}+...+0+{0,0,1,1,1,...}+{0,0,0,2,2,2,2...}+{0,0,0,0,3,3,3,3,3,...}+...
elim3 = min(N-elim1-elim2)
this is getting nowhere

Daniel Castle

ah that's a shame, I actually haven't considered the entire set like that however

thanks for the insight at least

Secret

elim1=vecsum
{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,...}
{0,0,0,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,...}
{0,0,0,0,4,5,6,7,8,9,10,11,12,13,14,15,16,17,...}
...

Daniel Castle

ooh

Secret

n2=min(N-elim1)
elim2=vecsum
{n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,...}
{0,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,...}
{0,0,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,...}
{0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,0,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,...}
{0,0,0,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,...}
.... hmmm....

btw typo on elim1:

elim1=vecsum
{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,0,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,...}
{0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,0,0,3,3,3,3,3,3,3,3,3,3,3,3,3,3,...}

Daniel Castle

wow, how are you generating these?

Secret

05:44

I just look at the sum "vertically"

I don't gen these

Daniel Castle

wow ok

Secret

well, one thing that is clear is that regardless of which eliminate number you are in, you always have the following sequence:

sequence = vecsum
{0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...}
{0,0,0,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,...}
{0,0,0,0,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,...}
...
=
{0,0,1,t2,t3,t4,t5,t6,....}

where tk are the triangle numbers = 1+2+3+..+k = k(k+1)/2

you also have this sequence:

Daniel Castle

hmm ok, the triangle numbers make sense because of the way you add 1 each time

Secret

sequence2 = vecsum
{n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,...}
{0,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,...}
{0,0,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,...}
{0,0,0,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,...}
...
=
{n,2n,3n,4n,5n,6n,...}

where n = min(N-elim1-elim2-elim3-...elim(n-1))

Daniel Castle

ah ok that makes sense

Secret

05:52

so elim n is given by {0,0,1,t2,t3,t4,t5,t6,....}+{n,2n,3n,4n,5n,6n,...}

now, if only we can get a nice formula for min(N-elim1-elim2-elim3-...)... let's see...

Daniel Castle

yeah that's the tricky part, like I said it seems chaotic so I doubt it but perhaps one not in closed form?

Secret

I think one thing that might give us more information is to compute the "counting function for the eliminates"

Above we have derived the formula for the nth eliminate, thus in theory we have turned it into a recursion relation, thus once we have n, we will see the distribution of integers being eliminated from n and see if it increases or decreases

that is, plot elim1,elim2,elim3,elim4,...N

and see if it rises very quickly

and then compare this with the counting function, which I suspect it plots N,N-elim1,N-elim1-elim2,...

Daniel Castle

ah ok, so plotting the number of eliminations <n for some eliminate k?

Dattier

06:13

Let $P_n(x)=\sum\limits_{k=0}^n \dfrac{x^k}{k!}$. Find $\forall x\in\mathbb R, \lim\limits_{n \rightarrow \infty} P_n(x)\times P_n(\frac{1}{x})\times x^n$.

Secret

06:37

@DanielCastle yeah, I suspect it might reveal something, especially if the line is near horizontal

micsthepick

07:28

I am having some problems in attempting to create an expression in $z$ where $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, z=x+iy$$

any hints or guidance would be appreciated

Shobhit

@micsthepick use $Re(z)=x=\frac{z + \bar z}{2}$ and $Im(z)=y=\frac{z- \bar z}{2i}$

micsthepick

I thought of that, but I don't think that is the end answer that I should be looking for

is it possible to go from there to something nicer?

(technically it is not strictly an expression in $z$)

overexchange

07:56

For expression 2018, How to derive math expression like (((2 + 0) - 1) * 8) * 8 * 8 * 4 - 30 ?

micsthepick

@overexchange pardon?

overexchange

08:12

@micsthepick I need to build such expression

micsthepick

so you wan't to make something like sub(add(2, 0), 1)...?

overexchange

cs61a.org/assets/slides/01-Functions_full.pdf#page=124

Instead of add, sub use +, -

add(2,3) -> 2+ 3

micsthepick

so what specifically are you struggling with? I see that you can convert a simple expression

overexchange

Good question..

micsthepick

generally, the idea is to work from the innermost brackets outwards, just like how the order of operations works

overexchange

08:19

let me paste the expression tree

2000                + 10    + 8
(2*1000)            + 10    + 8
2*(2*500)           + 10    + 8
2*(2*(2*125))       + 10    + 8
2*(2*(2*(5*25)))    + 10    + 8
2*(2*(2*(5*(5*5)))) + 10    + 8
2*(2*(2*(5*(5*5)))) + (2*5) + 8
2*(2*(2*(5*(5*5)))) + (2*5) + (2*4)
2*(2*(2*(5*(5*5)))) + (2*5) + (2*(2*2))

this is what I thought of following the rule:

1. Evaluate the operator and then the operand subexpressions
2. Apply the function that is the value of the operator subexpression to the arguments that are the values of the operand subexpression
So they evaluate to values

which looks different from ` (((2 + 0) - 1) * 8) * 8 * 8 * 4 - 30 `

@micsthepick Did you get my question?

micsthepick

so, would mul(add(2, 3), add(4, 5)) -> mul((2+3), (4+5)) -> (2+3) * (4+5)?

Slereah

08:51

Where is Balarka when you need him

Balarka Sen

Over there

Slereah

Mister Balarka!

Consider this : I have two manifolds with boundaries

BUT

They are in fact fiber bundles

$\pi_i : E_i \to M_i$

Such that $M_i$ is a manifold with boundaries, and $\pi_i(\partial E_i) = \partial M_i$

Now consider the gluing of those two manifolds along the boundary

which also forms a fiber bundle trivially

Can we say something about its structure group?

I get the feeling that even if both bundles have the same structure group, the resulting bundle will not necessarily have this group

Probably depends on the gluing function

via the local trivialization on the glued region

Balarka Sen

Structure group of a fiber bundle doesn't make much sense tho

Slereah

what do you mean

Balarka Sen

I don't understand the question. What is "structure group of a fiber bundle"?

Slereah

08:57

The subgroup of $\text{Diff}$ that the transition functions of the trivializations are a part of

Balarka Sen

There is no such unique subgroup. All of $\text{Diff}(F)$ is a possible structure group, then.

Slereah

true

Hm

Balarka Sen

So I don't understand the question.

Slereah

How about this : if both bundles are tensor bundles, is the resulting bundle also the tensor bundle?

Balarka Sen

What is a tensor bundle if you're not talking about vector bundles?

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