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12:04 AM
I wonder for math majors, whether they can go straight to analysis or they still need calculus as a padding
 
analysis is very hard if you have never done proofs or developed intuition with computational stuff in calculus. VERY.
 
Right
 
In Germany there is no such thing like calculus
 
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?
 
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
 
12:08 AM
@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
 
@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
 
Vector than ever
 
@AkivaWeinberger Ahhhh. Ok the notation made no sense
Is it like $A_{ij}$?
 
Yeah, but computer science / programming -style
 
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
 
12:21 AM
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.
 
Guys, check this out. And this is not even close to being the complete list in my arsenal to increase T.
 
1
Q: Solids of Revolution around other functions.

TdonutRecently 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

jakI 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...

 
12:50 AM
@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
 
hi @TedShifrin how's it goign
 
Doing OK ... haven't seen you in a while. You doing OK?
 
yeeep. I've been on break.
Quite busy with work, but I think I was a little lazier than I ought to have been.
 
Bad, bad Antonios. :) You're taking after your algebra students? :P
 
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?
 
12:59 AM
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?
 
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.
 
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.
 
yeah very likely.
yeah, please.
Might be good to check it out for inspiration, anyways.
 
You can figure it out from here.
 
oh my. This is quite exciting LOL
i love reading course notes from classes and such.
 
1:03 AM
Glad to make you happy. :)
 
yeah thanks! how has it been the last few weeks?
 
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.
 
oh, any particular reason for that?
 
For which?
 
1:05 AM
anyone here good at combinatorics haha
 
Oh, my landlord decided he wanted to sell. Always a risk.
I can do the haha part, Drew.
 
mathb.in/23477 [Combinatorics]
@TedShifrin how are you
 
ahh, yeah that's no fun @TedShifrin.
Anyway, time for me to get back to this Serre reading :P
 
Talk later, Antonios.
 
have a nice evening. I'll be around more as of tomorrow, so no worries about that.
 
1:07 AM
Serre? group representations? number theory? alg geom?
 
Hello
 
@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.
 
@TedShifrin Same. I like analysis and continuousy things
And yes, you're right. You have to be careful to avoid double counting.
 
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.
 
right
 
1:11 AM
I'm about to take combinatorics and then I'll be closer to competent but let's see, could be fun
 
It is a cool question, though.
 
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
 
Not that I followed that.
 
not that I explained it very clearly :)
basically, the short story is induction on n /k don't seem to work :(
 
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.
 
1:15 AM
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
 
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.
 
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
 
Yeah, I think that's a good idea.
 
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.
 
Why say "a ring with more than one element and a unity" this is redundant
 
1:22 AM
because there are rings wiht one element
if you don't take a ring to be unital
 
Because there are rings with no unity.
$2\Bbb Z$ is (in some people's definition) an infinite ring with no unity.
 
there's also the trivial ring
R = {*}
 
The point is to understand why it's not redundant to have both requirements.
 
doesn't the latter imply the former
 
{*} is a ring with unity though
 
1:23 AM
oh
 
No, the $0$ ring has unity :)
 
0 = 1
 
is this basically the $1\neq 0$ thing yep ok
 
right.
 
ok, in my head i assumed the definition of unity to stipulate $1\neq 0$ :(
but this is not the case oc
 
1:25 AM
some authors take a ring to be unital by definition
 
Yup, @Drew. Artin's book does. My book does ... :)
 
do you know why its helpful to consider non-unital rings?
 
right. my text speaks plenty of rings with unity
 
Some people want to talk about $2\Bbb Z$ as a ring, in considering homomorphisms, etc.
 
does it make sense categorically though
like if you take 2Z to be a ring does that change Z being initial
 
1:27 AM
unital? No, why?
 
no, does that change Z being initial in the category of rings
 
Oh
Then you obviously don't require that a homomorphism takes $1_R$ to $1_S$.
 
@DrewBrady non-unital ring you mean a ring explicitly without unity (sometimes called a rng)?
 
Yes @Corellian
 
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
 
1:29 AM
thanks drew and ted
 
no worries and sorry for my silly aside on categories lol
yeah lemme think more about this transitions idea
 
@Drew for what it's worth, $0$ will then become a zero object (both initial and terminal)
 
yeah I figured that out after being less stupid. Ted reminded me that the category is literally different (i.e., the morphisms change)
 
Ted never says anything categorical.
 
ha. you may not have intended to said anything categorical :)
`abstract nonsense'
 
1:33 AM
Schlag intensifies
 
the (compound) word category-theoretically sounds funny
 
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.
 
Ted its actually not a problem from a text or anything, it is something I came across working on a different problem.
 
Sorta like how I discovered "stars and stripes" unknowingly working on my thesis.
 
hah stars and stripes / balls and urns / stars and bars are now pretty standard I feel
 
1:37 AM
They probably were then, too, but I never took probability or combinatorics.
Just saying ...
I'm going to do dinner. See y'all later.
 
Yeah I never took combinatorics or graph theory. I probably should
see ya!
 
For me, I prefer the term "non unital rings" for rngs
and only use rings when we have identity
 
@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.
 
Ted already said that no books he knew talks about solid of revolutions about arbitrary curves
 
1:48 AM
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
 
2:14 AM
I am not going to calculate the shaded areas of these
 
@Secret what is the first one exactly?
 
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
 
Hmm if the first white disk has half the radius whole disk then the blue area is 2/3 the whole area I think
 
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
 
2:30 AM
The second graphic looks like a cool logo :)
 
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?
 
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
 
2:46 AM
I think we need more context in order to figure out why P(x > 8.12) got cut off
 
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?
 
yeah, for normal distribution, anything beyond 3 standard deviations is pretty negligible in most context
 
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!
 
3:42 AM
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 $\forall x\in\mathbb R, \lim P_n(x)\times P_n(\frac{1}{x})\times x^n$.
 
$x^a-1$. 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
 
reasoning with rotation
in the complex plane
 
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)
 
why are you introducing Q(x) ?
 
because that's the other part after your factorise $x^b-1$
i.e. what I wrote as "stuff" is actually Q(x)
 
3:56 AM
what's the problem ?
@Secret
 
$x^b-1 = (x^a-1)(x^{a(n-1)}-x^{a(n-2)}+x^{a(n-3)}+\cdots + 1)$ right?
 
yes
 
$x^b-1$ and $x^a-1$ gave their corresponding roots of unity, but what is the geometric meaning of $(x^{a(n-1)}-x^{a(n-2)}+x^{a(n-3)}+\cdots + 1)$?
 
$\dfrac{x^b-1}{x^a-1}$ a homographie
@Secret ok ?
 
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
 
4:05 AM
in the complex plan, that is an application which keep cocyclicity
(a,b,c,d) is cocyclic if are in the same cercle
 
right
 
That's true, if you think $x^d-1$ like a rotation (but in the "real life"it's not a rotation)
@Secret : ok ?
 
yes
 
what's the geometric meaning of $z^2$ in complex plan ?
only on unit cercle
 
4:20 AM
$z^2-1=(z-1)(z+1)$ hmm...
 
without the translation z-1 (is more easy)
 
$z^2=zz$?
hmm...
 
what effect on the angles ?
 
are they preserved, i.e. conformal?
 
sorry no
give pi/3 and pi/5
this doubles the angles
 
4:25 AM
yup
 
what's the meaning "yup" ?
 
(ok, yes, I see, right,...)
 
ah, thanks
 
I don't know how to say that in french
 
"ok" or "je vois"
 
4:27 AM
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)
 
yes
but the norm changing (if is not unit)
 
yeah, that's what I mean by "it scales the circle"
 
now, we know : how to build with the rule and the compass?
 
build what with rule and compass?
 
in french "la régle et le compas"
maybe the lignes and the compass
the lines and the cercles
 
4:33 AM
Hi, could anyone help me reason about a sieve?
 
depends on what sieve
 
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
 
by the way all numbers of the form $2^{n-1}+1$ don't get eliminated
that should be $2^n$*
not n-1
 
@DanielCastle : can you formalize your question ?
 
4:38 AM
sure i posted about it a while ago over at math.stackexchange.com/questions/2491459/… and have just come back to it
 
@Dattier I cannot read French, but I think I kinda recognise that. solving problems related to circles by using compass and ruler.
 
@Secret : yes
@Secret : now, we know : how build $z^n$ by using compass and ruler. ok ?
$n\in \mathbb N$
 
@Dattier think so
 
@Secret the next question is, how do that with the least possible step?
@DanielCastle : sorry I don't know
 
So... instead of
elimination 1: 2,4,6,8,10,12,14,...
we have
elimination 1: 1,2,3,4,5,...?
 
4:45 AM
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 $c$
 
hmm...
 
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
 
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
...
 
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
 
5:00 AM
I am wondering, is there a nice closed form for N-elim1-elim2-elim3...?
 
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
 
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),...
...
 
yes exactly
sorry i just came back
 
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
 
5:17 AM
here is the counting function, so the number not eliminated < n divided by n: i.stack.imgur.com/g87jN.png
 
it looks really jittery near the beginning
 
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?
 
5:31 AM
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
 
ah that's a shame, I actually haven't considered the entire set like that however
thanks for the insight at least
 
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,...}
...
 
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,...}
 
wow, how are you generating these?
 
5:44 AM
I just look at the sum "vertically"
I don't gen these
 
wow ok
 
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:
 
hmm ok, the triangle numbers make sense because of the way you add 1 each time
 
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))
 
ah ok that makes sense
 
5:52 AM
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...
 
yeah that's the tricky part, like I said it seems chaotic so I doubt it but perhaps one not in closed form?
 
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,...
 
ah ok, so plotting the number of eliminations <n for some eliminate k?
 
6:13 AM
Let $P_n(x)=\sum\limits_{k=0}^n \dfrac{x^k}{k!}$. Find $\forall x\in\mathbb R, \lim\limits_{n \rightarrow \infty} P_n(x)\times P_n(\frac{1}{x})\times x^n$.
 
6:37 AM
@DanielCastle yeah, I suspect it might reveal something, especially if the line is near horizontal
 
7:28 AM
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
 
@micsthepick use $Re(z)=x=\frac{z + \bar z}{2}$ and $Im(z)=y=\frac{z- \bar z}{2i}$
 
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 $z$)
 
7:56 AM
For expression 2018, How to derive math expression like (((2 + 0) - 1) * 8) * 8 * 8 * 4 - 30 ?
 
@overexchange pardon?
 
8:12 AM
@micsthepick I need to build such expression
 
so you wan't to make something like sub(add(2, 0), 1)...?
 
Instead of add, sub use +, -
add(2,3) -> 2+ 3
 
so what specifically are you struggling with? I see that you can convert a simple expression
 
Good question..
 
generally, the idea is to work from the innermost brackets outwards, just like how the order of operations works
 
8:19 AM
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?
 
so, would mul(add(2, 3), add(4, 5)) -> mul((2+3), (4+5)) -> (2+3) * (4+5)?
 
8:51 AM
Where is Balarka when you need him
 
Over there
 
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
 
Structure group of a fiber bundle doesn't make much sense tho
 
what do you mean
 
I don't understand the question. What is "structure group of a fiber bundle"?
 
8:57 AM
The subgroup of $\text{Diff}$ that the transition functions of the trivializations are a part of
 
There is no such unique subgroup. All of $\text{Diff}(F)$ is a possible structure group, then.
 
true
Hm
 
So I don't understand the question.
 
How about this : if both bundles are tensor bundles, is the resulting bundle also the tensor bundle?
 
What is a tensor bundle if you're not talking about vector bundles?
 

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