Mathematics - 2016-12-17 (page 3 of 7)

Balarka Sen

12:01

@DHMO How are you so sure about this?

I mean it may happen that a lot of elements of the form a + b become equal for a bunch of different a's and b's, which will cut down the cardinality of A + B

Like {2, 3} + {2, 3} = {1, 4} in Z/5.

DHMO

Because (informally) you can fix an element in B, say b, and |A+b| must equal |A|

Balarka Sen

ah. Good call

My previous example is false anyway it was supposed to be {0, 1, 4}

ok, gotta run

mick

Any ideas ?? math.stackexchange.com/questions/2057948/…

Secret

Ok, here's my procedure in plain english:
1. I have a magma $(M,\circ)$ of n elements in the underlying set $M$.
2. I then impose a (partial) ordering onto $M$ so it becomes the set $S$
3. By taking the cartesian product of S with itself, and equating all ordered pairs into product terms of the magma, I will get the cayley table except the starting row and column that act as labels in the cayley table
4. I then tried to pick out all n rows of the cayley table, and similarly for columns
5. I then collect all these rows into a collection $\mathcal{F}$ and columns into a collection $\mathcal{G}$.

Illustrative example:

DHMO

@Secret step 2: how does imposing ordering change the set at all?

Secret

12:07

Because $M$ is unordered

DHMO

No, $M=S$

Imposing an ordering to {0,1,2} does not make it a different set

Secret

But aren't all cayley tables are equivalent under interchange of two rows and two columns correspond to the same label?

that is you can draw your cayley table begin with a,b,c, or b,a,c or c,b,a etc.?

DHMO

yes

You can say you obtain a new structure $(M,\circ,\preceq)$

Secret

Thus step 2 is trying to pick one of the many a,b,c arrangement to begin with, thus fixing the row and column ordering in the cayley table (since that will not be changed for the procedure to work)

DHMO

@Secret continue, that's just a minor flaw

Secret

12:13

step 3 generates the cayley table defined abstractly via a cartesian product of M with itself, with the ordering of the cartesian product given by the product/coordinate ordering analogous to matrices and arrays (or more simply the ordering of $\mathbb{R}^2$)

By this stage, it should be possible to label entries in this set like what is done in matrices

DHMO

could you give an example?

Secret

For example, suppose my magma is the group $\mathbb{Z}/3$ The cayley table of $\mathbb{Z}/3$ is given as follows:

Null

@DHMO antisymmetrical is the other extreme to symmetrical right? But it's not to be mixed up with not symmetrical.

DHMO

@Null yes

Sirmimer

Question regarding continued fractions: Why is [1,0,1,1,2,1,1,4,1,1,6,1,...,1,2n,1,...] = [2,1,2,1,1,4,1,1,6,1,...,1,2n,1,...]. Any pointers how I could prove/explain this?

DHMO

12:21

What is the minimum length of a equilateral triangle to house 8 unit circles? Wikipedia gives 9.293 but I don't know what the closed form is

Astyx

Hi again

DHMO

@Sirmimer that looks familiar

so $1+\dfrac{1}{0+\frac{1}{x}} = 1+x$?

yep, that's why

Sirmimer

DH any pointers how to evaluate a continued fraction expansion against/with another?

DHMO*

DHMO

let x = [1,1,2,1,1,4,1,1,6,...]

then LHS = 1+1/(0+1/(x)) and RHS = 1+x

@Astyx hi

3 mins ago, by DHMO

What is the minimum length of a equilateral triangle to house 8 unit circles? Wikipedia gives 9.293 but I don't know what the closed form is

Can you solve this?

Astyx

I'll try

DHMO

12:25

thanks

Sirmimer

Thank you DHMO, i'll have a closer look at it. Btw any idea why i cannot see the $...$ expression but instead it reads as "$...$" ?

mick

0

Q: Triangle analogue of matrix ??

We could modify the square matrix into a equilateral number triangle and define Sum and product in an analogue way. Here is a small example for triangles size 2. A B C X Y Z = A+X B+Y C+Z A B C * X Y Z = BX + AZ BY CX What has been done with those ? Or variants of it ...

DHMO

@Sirmimer i'm not seeing the difference

mick

A weird question 😊

Sirmimer

DHMO you see the "$" signs as well and not proper expressions/symbols?

mick

12:27

Anyone seen those triangle analogues of matrices ?

DHMO

I see the expressions not the "$" signs because I have LaTeX enabled

see room description

Sirmimer

Latex, gotcha. Thank you

Astyx

"Unit circle" means the radius is 1 here right ? @DHMO

DHMO

yes

mick

0

Q: Triangle analogue of matrix ??

We could modify the square matrix into a equilateral number triangle and define Sum and product in an analogue way. Here is a small example for triangles size 2. A B C $+$ X Y Z = A+X B+Y C+Z A B C * X Y Z = BX + AZ BY CX What has been done with those ? Or...

abstract-algebra soft-question linear-transformations

Null

12:31

what prevents us from seeing 1+2-3+4 as (1+2)-(3+4)? Addition before subtraction?

Secret

@DHMO example here

basically, the procedure is multiplication by elements in the magma on the left or right maps rows to rows and columns to columns

Astyx

@Null convention

DHMO

@Null 1+2-3+4 is interpreted as (1)+(2)+(-3)+(4)

by convention

Null

ah right right...

DHMO

addition/subtraction, whichever comes first

Null

12:33

then it makes indeed sense to say, there is really only addition and multiplication

Astyx

Yes

DHMO

which raises the question what the hell is 6/2(1+2)

Null

$6\cdot\frac{1}{2}\cdot(1+2)$

but i see how one could arive at something else

Sophie

I think it is better to interpret a/bc as (a/b)c

Astyx

Hi @Sophie

Sophie

12:36

good morning

Astyx

Any news on that $\theta(n^2)$ ?

Sophie

I gave up

DHMO

@Secret so $S=\{0,1,2\}$, $R_n=\{x \circ n,x\in S\}$

Astyx

Too bad

Sophie

hopefully Terence Tao will write a blog post dumbing it down for everyone else

DHMO

12:36

@Astyx are you trying my question?

Astyx

Yep

Sophie

what question?

Secret

@DHMO Yes

DHMO

@Secret what's the question?

Astyx

I guess the way to minimize the length is not to put the circles as close to each other as they can

Null

12:38

@Sophie if you think about it, the otherway around would make $\frac{a}{bc}$ really strange to express

DHMO

@Astyx so what's your length?

Astyx

@DHMO Don't have one yet

Is this an open problem like the last one you asked me to try ?

DHMO

no this isn't

Secret

@DHMO Initially I thought I can define $R_n= \{\ y\lvert \exists n \in S,\forall (x,n) \in S \times S, x\circ n=y\}$ and then some users said I have used the set builder notation wrongly, and I am not sure what went wrong with it

DHMO

wikipedia gave 9.293

I just don't know what the closed form is

Secret

12:40

I thought I am replacing fixed n, of the order pairs (x,n) in this set with the result $x \circ n$

thus I should be getting the exact same set as what you showed earlier?

DHMO

@Secret well, since $n$ is defined earlier, you shouldn't have $\exists n \in S$

also it shouldn't be $\forall$

it looks very similar to a for-loop to me

Secret

Ah I see. Yeah... most of my ideas are in the programmer persepctive and I am trying to write them in set builder notation in order to check with the maths guys whether they makes sense

Astyx

@Secret $\{y\mid \exists n\in S, \forall x\in S, x\circ n = y\}$

DHMO

@Secret @Astyx $R_n = \{y|\exists x \in S: x \circ n = y\}$, but still not as elegant as mine

Astyx

Yeah might be what DHMO said

I'm not sure what the set is supposed to be

DHMO

12:45

@Secret so what's the question

Secret

@Astyx That is the set of the row of the cayley table labelled n. So generalising it to all $n \in S$ you have all n rows of the cayley table, which is then packed into the collection $\mathcal{F}$

Astyx

Why is there an $n$ in indice of $R$ ?

DHMO

the $n$-th row

Secret

the indices label the row of the cayley table. So $R_a$, $R_b$, $R_1$ etc. are different rows

thus the indices served as a labelling of the rows in the set $T$ that is the cayley table, defined as, $T=\{\ x\lvert \forall (a,b) \in S \times S, a\circ b=x\}$

For step 6, I tried to define some kind of left "magma" action, analogous to a semigroup action as follows:

Define: $\cdot : S \times \mathcal{F} \rightarrow \mathcal{F}$ such that for $x \in S$ and $R_y=\{a,b,c,d,...\} \in \mathcal{F}$, $x\cdot R_y=\{x\circ a,x\circ b,x\circ c, x \circ d ...\}$

however one thing is that since the collection $\mathcal{F}$ is unordered but $S$ is , is it legal to take a cartesian product of that?

btw

10 mins ago, by Astyx

I'm not sure what the set is supposed to be

is a star mistake, I cannot undo the star

Astyx

Can't you ?

Just click on it again no ?

Secret

12:55

notice say: It is too late to undo this operation

Astyx

Meh

I guess it'll have to stay there then

No big issue

Secret

sorry about that...

Astyx

Haha don't worry I don't really care

Sophie

it's surprisingly hard to obtain a lot of terms of the continued fraction of $2^{1/3}$

paste.debian.net/plain/902789

mercio

um

Astyx

12:57

Lots of 1s

mercio

sophie the last time i looked at your question the terms were wrong

Secret

Hmm... I think I might have answered one of my question: So I am not mapping within the same object, I guess that's why Tobias Kindetoft said it is not an endomorphism

Sophie

@mercio I'm going to fix them now

mercio

but this list looks correct

Sophie

I wrote some code but for some reason it's going really slow

Astyx

13:01

What is your code ?

mercio

well the deeper you go the harder it gets

Sophie

hastebin.com/zesewopozu.cpp

DHMO

@Sophie I hope 1.259921049894873164767210607278228350570251 is infinitely precise

Sophie

it is precise enough

mercio

excuse me but why do you have floating points

DHMO

13:03

@Sophie no it is not

Sophie

because I need to use the floor operation

mercio

it is absolutely not precise enough

DHMO

you have way more terms than digits

I reckon that it is precise to the 20th terms

mercio

if you need floating points then you're automatically doing things the wrong way

DHMO

+1

Sophie

13:03

I only need p_f + t_1 * q_f + t_2 * r_f to floor to the right number

that is, some rational number + some rational * 2^(1/3) + some rational * 2^(2/3) can't be more than 1/2 out

so if these rationals get really big I might have a problem, but I don't think they get big enough

DHMO

that 2^(1/3) and that 2^(2/3) is not precise enough

have you checked the values of your rational numbers?

themselves getting big is not the problem; their denominator or their numerator getting big is the problem

and with 100 terms (which I reckon you obtained), the rationals should be at least 2^100

Sophie

no, if the rational gets big I have a problem, but if the denominator and the nominator gets big but their ratio is small I'm okay

DHMO

you can't distinguish between (10^100)/(10^101) and (10^100 + 1)/(10^101)

that's my point

Sophie

@DHMO I don't need to

DHMO

@Sophie how are you expressing the rationals?

Sophie

13:07

I'm using the boost multiprecision lib

DHMO

if the precision of the rationals exceed your precision of 2^(1/3) then you have a problem

Sophie

nope. Unless p_f + t_1 * q_f + t_2 * r_f floors to the wrong thing then I'm okay, and the precision lost when using the floating points is not enough to do that unless p,q, or r are very big

you can find the continued fraction of $\sqrt{61}$ only by knowing it is about 7.81 and rationalizing, which is what I'm doing here

DHMO

@Sophie try it on 7.81 and show us the result

Null

is there something like a bounded sum? To express that the sum diverges, but only because it adds at the end +1 and -1 (or similar)

DHMO

@Null their limsup and liminf exist but differ

mercio

13:13

how many terms did you compute ?

DHMO

@mercio 10000

Astyx

@DHMO $$2 + 2\sqrt{3} + {2\over3} \sqrt{33}$$

Not my work

www2.stetson.edu/~efriedma/cirintri

DHMO

@Astyx thanks

Astyx

jstor.org/stable/2324212?seq=1

My pleasure

It looks veyr interresting

Null

mmh, I saw a similar thing with overlapping circles. don't quite remember it tho

ah no

Astyx

13:16

mathworld.wolfram.com/TrianglePacking.html

mercio

it took me 5 seconds

Null

with overlapping small triangles

@Astyx yy

Secret

(Refined with maths chat guys help)
Def. 1. All elements in a Cayley table

Given a magma $(M,\circ)$ of $n$ elements, Construct an ordered set $S=\{s_1,s_2,\dots,s_n\}$ where $s_i \in M$ for all $i\in\{1,2,3,\dots , n\}=N$. Then the cayley table $T$ is given by $T=\{\ x\lvert \forall (a,b) \in S \times S, a\circ b=x\}$. The transpose is defined by $T^\textrm{T}=\{\ x\lvert \forall (b,a) \in S \times S, a\circ b=x\}$.

Def 2. Rows and columns and collections

A row $R_i$ is given by $R_i= \{a\circ i, a \in S\}$ where $i \in S$. Similarly, a column $C_i=\{i\circ b, b\in S\}$. Let $\mathcal{…

Astyx

That $n=6$ case for triangles in triangles is funky

mercio

sophie I don't think your list is correct

DHMO

13:18

me neither

Secret

@DHMO Question = does the above formal writing achieve what I said earlier on how the procedure works?

DHMO

but @Sophie insists that the precision is enough

mercio

oh wait

nevermind your text file doesn't have all 10000 terms

Sophie

where did you get 10000 terms from?

mercio

my computer

Sophie

13:19

how?

Null

@NaCl hi, all roger?

mercio

I typed a few things

Astyx

@Sophie @DHMO @mercio wolframalpha.com/input/…

NaCl

@Null I have a quite difficult programming task to solve, but besides that, yeah

mercio

I made a list of the entries that were larger than $100$

Secret

13:20

sorry typo, on the proof of lemma 2, need to fix
Actually the proof of lemma 2 become much harder as the rows and columns are not defined as subsets of $T$ and thus we cannot exploit the symmetry of rows and columns in $T$

mercio

in the first $10^5$ terms

Astyx

Less than 2 seconds for mathematica to compute the 100k first terms

trilolil

Hello everyone

What does this formula mean:

Astyx

And something around 30 seconds to compute the first 10 million terms

DHMO

@mercio Surprisingly @Sophie is right; the precision is enough for 1000 terms

trilolil

13:22

$\sum_{m,n} (D - R_{m,n})$ where D and R are both 4x4 matrices

Astyx

It's not that surprising considering the method she is using

DHMO

@Secret an ordered set is not a set.

trilolil

substract of each element of D the sum of all the elements of matrix R?

Sophie

@Astyx yes exactly, I'm doing all this rationalizing so I don't need to have stupidly high precision

run this to get the continued fraction of $\sqrt{61}$ hastebin.com/bilaqecaxi.cpp

Secret

https://en.wikipedia.org/wiki/Partially_ordered_set
https://en.wikipedia.org/wiki/Total_order

???

DHMO

13:24

@Secret wikipedia did not give a notation of an ordered set

see proofwiki.org/wiki/Definition:Partially_Ordered_Set

Astyx

Gotta go now

Good afternoon to all of you

Secret

I see ( changing "ordered set" to "poset" in the formalism)

trilolil

anyone?

DHMO

@Secret it is still not a set.

$S = \{s_1,s_2,\ldots,s_n\}$ is bad notation

the cayley table's set-builder notation is not good

Null

@DHMO bad to wrong, depending what you mean

DHMO

13:27

also it isn't a table, it is just another set

kayak

Hi guys

Sophie

@DHMO the $\sqrt{61}$ is easier to understand. The important part is that if you have $\lfloor a+b\sqrt{61}\rfloor=\lfloor a+ b\times 7.81\rfloor$ for rational numbers $a,b$ unless b is very big

Secret

@DHMO I am not sure how I can write the elements in S, K,L etc. without using too many a,b,c,d,e etc.

DHMO

@kayak 안넝

kayak

I asked a question and I found it should be on 'Cross Validated'-site. How can I move my question?

DHMO

13:28

@Secret I mean, $(M,\le)$ is an ordered set, where $M$ is still the set

Sophie

and just like you can write $\frac{1}{a+b\sqrt{61}}$ as $x+y\sqrt{61}$ I'm doing the same with $\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}$

DHMO

imposing an order to a set does not make the set different

kayak

Or can I copy my question to another site without making a new question?

DHMO

@Sophie yes, thanks; that's a nice way to make continued fraction

kayak

I asked a question and I found it should be on 'Cross Validated'-site. How can I move my question?
Or can I copy my question to another site without making a new question?

ypercubeᵀᴹ

13:29

@kayak You can either delete and repost or flag your q to "moderator's attention", explain what you want and they could do it.

Secret

@DHMO So I can use $M$ directly as there's already some ordering specified thus allowing me to uniquely specify the cayley table I am trying to construct?

DHMO

@Secret yes

kayak

Oh thanks @ypercubeᵀᴹ

DHMO

@Secret you know, your "transpose" really does nothing

maybe you're thinking of matrices

Secret

That is the main question. As mentioend previosuly, I only need the entries in the cayley table, but with an ordering such that you can label them as if they are matrix entries. That is what I don't know how to do cause I need the rows of it to do that procedure

Yes, I am indeed treating the cayley table as if it is like a matrix, and I am trying to find a way to do that so that you can do an abstract algebra version of row operation on them

DHMO

13:32

@Secret the problem is that you can't impose your desired ordering to your $T$

Secret

I found the procedure works (as demonstrate in the $\mathbb{Z/3}$) example but I am not sure how to formalise it

DHMO

because there would be different entries whose value are equal

Sophie

@mercio can you paste me a list of those terms you found?

Secret

@DHMO ah, that's a problem, cause sets don't allow duplicate elements...

DHMO

yes, so you should be building a matrix instead of a set

mercio

13:34

the terms greater than $100$ in the first $10000$ entries

[534; 121; 186; 372; 186; 220; 255; 7451; 113; 151; 4941; 108; 354; 195;
304; 204; 201; 136; 127; 119; 124; 131; 101; 119; 328; 190; 101; 260; 157;
620; 101; 138; 12737; 117; 2897; 2078; 103; 105; 223; 662; 103; 495; 215;
135; 421; 1269; 118; 150; 227; 230; 105; 844; 160; 158; 179; 507; 138;
161; 111; 143; 193; 1013; 277; 190; 253; 186; 280; 145; 189; 194; 331;
134; 452; 307; 275; 105; 509; 339; 122; 208; 115; 106; 339; 117; 214;
6614; 113; 119; 192; 168; 540; 267; 135; 114; 932; 292; 314; 1118; 217;

Secret

Hmm, in that case, I need to rethink this..., especially on how matrix addition will be defined in such procedure...

but anyway, thanks for clarifying all that maths misconception I have made

trilolil

Hello everyone,

I don't know how to interprete this formula:
$\sum_{m,n} (D - R_{m,n})$ where D and R are both 4x4 matrices

Does this mean I have to substract of each element of the matrix D the sum of all the elements of matrix R?

Secret

I thought too much of a programmer more than a mathematician

DHMO

Each circle has radius 1

what the hell is the side length of the bottom-right triangle?

Sophie

I wonder how mathematica is doing it

they can't be calculating $2^{1/3}$ to that high precision, or they'd run out of memory

DHMO

13:38

@Sophie that's often a commercial secret

Sophie

maybe they're doing something similar to what I'm doing but more efficiently

mercio

they're way better than me lol

Secret

looks like you compute $(D-R_{m,n})$ for each n,m, and then add all these differences up

I guess R_m,n might be a 4x4 matrices of R_m,n copied everywhere

so there are mn of these R_m,n matrices need to be subtracted from D, and then add up the results

DHMO

A proper subset of high-schoolers believe that $\pi = \dfrac {22} 7$...

Sophie

@mercio am I going insane or do these get less common and also smaller?

@DHMO the empty set hopefully

DHMO

13:41

@Sophie unfortunately it is not the empty set

mercio

@sophie I really have no idea of how the list should look like, sorry

trilolil

@Secret A matrix of matrices?

mercio

I'm still stuck on computing the list for $10^5$ entries lol

Secret

either the stuff you just deleted (which looks more reasonable), or what I said above so that the sum is a matrix

I strongly suspect it is$\sum_{m,n} D_{m,n}-R_{m,n}$ because this is more commonly seen

trilolil

@Secret I don't think you are correct... This is the source: ti.com/lit/an/bpra065/bpra065.pdf p13 - (6)

@Secret indeed, but they did write something else...

Secret

13:45

the distortion measure?

trilolil

@Secret yes

Secret

(Note the brackets in that line are not actually balanced...)
you take the difference between the matrixes D (indices are ,j) and R (indices are k,l). There are mn of these differences, which are those that you need to sum up and then square to get the distortion measure

that is, for each n,m subtract each matrix R from the matrix $\Gamma(D)$, and then sum all these resulting differences up and square

trilolil

@Secret true, didn't notice the unbalanced brackets. Indeed this interpretation makes much more sense!

it is unclear due to the unbalanced brackets. If they added that one bracket before the m,n subscript it would have made much more sense. Right?

Secret

indeed

from that equation, I am guessing the distortion measure is the sum of sqaures of the difference of the transformation (distortion) introduced before and after the transformation from the domain to the range in that image processing

Secret

DHMO: I already forgot how to get the unique circle inscribed in the triangle, thus not sure how to work from the sides there. But it seems as n increases, the circles seemed to fill in more area of the larger and larger triangle

DHMO

13:56

maybe

Jonathan Krill

14:13

hey guys, anyone know if that statement ist true?:

Given $n$ integers of which none are divisible by $n$ you can always find two integers which sum is divisible by $n$.

DHMO

@JonathanKrill no, for example take n=3 and consider 1,4,7.

Balarka Sen

otoh what is true is that you can always find a subset of those integers which sums to a multiple of $n$. irrelevant though

Jonathan Krill

ah yeah this makes sense you need a subset

thank you

DHMO

@BalarkaSen how would you prove that?

mercio

inb4 empty subset

Balarka Sen

14:17

nonempty subset I meant. sorry about that

DHMO

I'm curious about the proof

mercio

it's the pigeonhole principle

applied to $x_1 ; x_1+x_2 ; x_1+x_2+x_3 ; \ldots$

Balarka Sen

yup

DHMO

@mercio oh, thanks

Balarka Sen

the point is they should give all the nonzero residue classes mod n

DHMO

14:23

@BalarkaSen what is the ratio of the radii of the outer circle and the inner circle?

Balarka Sen

idk man

DHMO

could you calculate it?

Balarka Sen

if you keep joining the radii of the 4 obvious circles which starts and ends at a circle tangential to the big circle, you'd get 3 lines sitting in a zig-zagged shape, of equal length. i'd guess the angles between two consecutive ones is 120 degrees by inspection

if you can prove that angle thing you can probably calculate with that construction

Null

@DHMO I see you like circles, so we put circles in a circle ;)

DHMO

can we just consider the six circles in one side of the reflectional symmetry?

I don't like to guess that the angles are 120 degrees

Balarka Sen

14:28

scratch of equal length. the first is 3r, second is 2r and the last is 3r. but yeah.

this is what I had in mind

Null

what does contain mean in this context? math.stackexchange.com/questions/2062262/…

does it mean, the highest power that is less or equal?

DHMO

@Null "contained in" means "which divides"

Balarka Sen

@DHMO Me neither. But I don't know a simple way to prove that.

Null

@DHMO ah thanks

Balarka Sen

maybe one would have to use symmetry in some way

DHMO

14:33

@BalarkaSen can we consider the equilateral hexagon on one side of the symmetry?

mercio

put $8$ circles along the perimeter of the large circle

Balarka Sen

ah yeah that's the thing, @DHMO

DHMO

I have $\alpha$, $\alpha$, $\alpha-60^\circ$, and $120^\circ$ and two angles left

mercio

then add the $2$ in the middle

and check if they touch at the other end ?

DHMO

@mercio easier said than done?

mercio

14:34

yes

DHMO

@BalarkaSen do you know any results for the angles of equilateral hexagons?

Balarka Sen

not off the top of my head. but note that the angle at the other end is easier to compute

you have the diameters of the three kissing circles making a equilateral triangle

so a 60 degree there

and the two other angles are equal by symmetry, so 2x = 360 - 60 = 300. aka x = 150.

DHMO

@BalarkaSen wait, what x?

Null

lol, funny profile incoming math.stackexchange.com/users/278967/schrodingerscat

Balarka Sen

the angle between the 2nd and the 3rd line I drew in my picture, from left to right

Sirmimer

14:38

Question regarding this post, 4): http://math.stackexchange.com/questions/1039741/define-sequence-and-convergence

Question 4) Why can you substitute L in on both sides of $L=\dfrac{L^2}{3}+\dfrac{18}{L}$

Ramanujan

@DHMO do I need to by heart all inverse trigonometry formulae?or is there something?

DHMO

@Ramanujan what do you mean?

Balarka Sen

once that angle is computed i think it should be easier to do stuff with the hexagon (because you know one angle of the hexagon is 150)

Ramanujan

DHMO

@BalarkaSen oh, but what about the other two angles

Ramanujan

14:43

I have a long list of formulae which are all different :(

DHMO

No, learn how to derive them

Balarka Sen

Unsure what you mean by the other two. There's only one, the angle between the 1st line and the 2nd.

DHMO

don't recite the formulas

@BalarkaSen oh, I mean the other one

how do you know the two angles are equal?

Balarka Sen

ah, alright

you mean where i said 2x = 300? by symmetry. the whole figure is symmetrical about an axis

mercio

yeah one angle is easy to determine

but for the others I don't think there is an easy way

DHMO

14:45

yes I know that angle is 150 but I'm asking about the other angle

formed by the other two lines

Balarka Sen

oh. no idea, still thinking

ok, i give up

Mary Star

Let $E/F$ be a finite extension and $K \leq K \leq E$, where $K/F$ is normal.
Let $a,b\in \in E$ be $F$-algebraic conjugent, i.e., they have the smae minimal polynomial over $F$, and for some $n\in \mathbb{N}$ let $a^n\in K$.
I want to show that then $b^n\in K$.

Since $K/F$ is normal, we have that all the irreducicle polynomials over $F$ that have one root in $K$ have all the roots in $K$.

Could you give e a hint how we could show that $b^n\in K$ ?

I have found in my notes the following theorem:

Secret

15:41

ok, the stuff I have been trying to do regarding cayley tables looks suspiciously similar to this

Algebra representation

In abstract algebra, a representation of an associative algebra is a module for that algebra. Here an associative algebra is a (not necessarily unital) ring. If the algebra is not unital, it may be made so in a standard way (see the adjoint functors page); there is no essential difference between modules for the resulting unital ring, in which the identity acts by the identity mapping, and representations of the algebra. == Examples == === Linear complex structure === One of the simplest non-trivial examples is a linear complex structure, which is a representation of the complex number...

hmm...

meanwhile rewriting all the formalism using matrices, which is determined by maths users to be more suitable in acheving my aim

Sophie

15:52

can I use latex on the programming stackexchange?

DHMO

I don't think so

you might want to convert them to images

codecogs.com/latex/eqneditor.php

@Sophie ^

Sophie

I use the texer on AOPS for that

artofproblemsolving.com/texer

hopefully they'll come up with better algorithms

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$Mathematics$ Mathematics