Mathematics - 2013-10-18 (page 4 of 4)

Charlie

15:06

@skullpatrol in what sense?

skullpatrol

@Charlie Try it.

Charlie

@skullpatrol ?

skullpatrol

@Charlie To find out "in what sense" nothing happened.

Charlie

@skullpatrol ah, I.saw :P

skullpatrol

@Charlie :D

Charlie

15:11

@skullpatrol hehehe :D

@skullpatrol whats new, Skullucky?

skullpatrol

Chillin like a villain
how 'bout you Chuck?

Charlie

@skullpatrol chillin' too, reading a few things

skullpatrol

icic

Charlie

15:27

@skullpatrol I'll eat, Skull, brb :)

skullpatrol

@Charlie later pal

Quinn Culver

is it clear that the hilbert cube under the operation of componentwise addition mod 1 is a compact group?

i.e. is the operation of addition mod 1 continuous?

@amWhy, @Chris'ssis, @DanielFischer, or @robjohn, any idea?

Daniel Fischer

15:43

If you add modulo $1$, you never get $1$ as the result of an addition, so $\alpha([0,1]\times[0,1]) = [0,1)$ is not compact, where $\alpha(x,y) = (x+y)\bmod 1$. Thus addition modulo $1$ is not continuous.

robjohn

@QuinnCulver Are you talking about the Hilbert cube or the torus-like $S^\infty$?

Quinn Culver

I'm talking about $[0,1]^{\mathbb{N}}$

robjohn

@QuinnCulver Then Daniel Fischer's concern arises

Quinn Culver

hmmm, lemme think...

@DanielFischer Okay, I see, the image of the compact $[0,1] \times [0,1]$ isn't compact

Mats Granvik

tetration rocks!

Daniel Fischer

15:46

quintation pebbles!

Quinn Culver

@DanielFischer, @robjohn, but is it locally compact?

Daniel Fischer

@QuinnCulver The space is compact, but it's not a topological group (under componentwise addition modulo $1$).

robjohn

@DanielFischer sextation dusts 'em all

Charlie

@skull @skull @skull

Daniel Fischer

Troo dat, @robjohn.

skullpatrol

15:49

@Charlie @Charlie @Charlie

Charlie

@skullpatrol :?

skullpatrol

@Charlie :/

Quinn Culver

@DanielFischer Ya, I guess you're right

@DanielFischer Thanks a lot brah

Daniel Fischer

De nada.

Charlie

16:09

@DanielFischer você fala português, hablas español or both?

Valtteri

Good evening

Charlie

@Valtteri good evening

Valtteri

How do you do

Charlie

@Valtteri very fine, and you?

Valtteri

One does good, thank you. What is on the agenda for this Friday night?

Charlie

16:12

@Valtteri read, study

@Valtteri you?

Valtteri

Ah, just thinking. Many thing...Zeta function, prime counting function, profits of container ship companies and growth of cities. None of these things leads to anywhere but...

You might say I am bored

Charlie

@Valtteri sounds nice to me

Valtteri

May I ask why math is so fun? I had a meeting of a project about the container ships some days ago and the three of us had such fun. Gratuitous use of Greek letters gives off much laughter. hard to imagine some humanists deriving such amusement from a project.

Charlie

@Valtteri I dont know. I have fun with things in math that I don't understand. When in.the process to understand, I have fun. Naturally curious, this learning proccess is amusing . Even if you dislike something, if you get it, it is very pleasant. Math is a toy, we are like children.

Valtteri

True.

But does math have some goal? Deriving bounds, modelling the world?

It sometimes feels like all you ever get is either trivial or circular. You don't move forwards, but rather run to stay where you are.

Charlie

16:32

@Valtteri yes, of course. Everything has. But I believe that discover hiw math can be useful.for pratical life, is not our concern. Each carreer is focused on something, so I don't get why people say "when will I use this in life?" Well I dot, you may not realize, but you will end up usin.g. Math is wonderful

Bitrex

People who ask "when will I use this in life" don't "get it"

Valtteri

I don't disagree.

Charlie

@Bitrex yes,yes

Bitrex

Why listen to beautiful music? You're never going to "use it" in life, really

Valtteri

Of course the world would be better if a plumber integrated for 2 hours before reaching for the wrench.

Bitrex

16:36

@Valtteri The plumber is coming in a few hours to work on my bathroom - I hope he doesn't!

Charlie

@Bitrex it's what I say. I really feel sometimes, due to."society pressure", as.if I am being a useless.human for studying math, and things that cannot "help" humanity so directly. I just try to keep doing it because I simply like it,.

Valtteri

@Bitrex Of course you do. He comes in and takes some paper and pencil. "I will now need to concentrate, these integrals are tricky" and then he does math. For, you see, all physical phenomena are fundamentally mathematical, so by discovering the correct math, the plumber can do all the work in the mathematical realm, thus not sweating all over your floor.

Bitrex

@Valtteri LOL!

Valtteri

@Bitrex You cannot prove this doesn't work

Bitrex

@Valtteri Whether the toilet works or not at the end of the day is the proof.

Valtteri

16:45

Possibly. But the lavatory is mathematical. So all its functions are mathematical, thus any solution to its problems must be mathematical. The way the plumber does the math in his head, effects the world around him, thus the toilet is fixed.

Charlie

@Valtteri you're forcing the argument

Or mocking

Valtteri

Can't disagree with your assessment :)

Bitrex

@Charlie I work in a "practical" field, but I want to live in a world where talented people have the freedom to pursue art, music, philosophy, mathematics, for their own intrinsic beauty. Some people I know don't agree - they see people in these professions as somehow not contributing their fair share to society. I think that's BS.

Charlie

@Bitrex I totally agree.

Bitrex

I wouldn't mind working in pure mathematics myself, but unfortunately as they say "I ain't got the skills to pay the bills" :)

Valtteri

16:50

Some British dude made a study of collapse of empires. The last state before dissolving seems always to be "decadence", represented by huge emphasis of sports and cooking. Footballers and formula drivers and also chefs get paid huge sums and we watch them on TV and follow what they do in the media. Thus, the Western world is about to end.

Charlie

@Valtteri following this line, we should live like cavemen...

Bitrex

@Valtteri At least for my part, in the United States, people have been saying that the American way of life is decadent and about to fail since approximately the founding of the nation

However, here's a good book if you're interested: amazon.com/Why-America-Failed-Imperial-Decline/dp/1118061810

Valtteri

Why? If you can show that what you do benefits society, instead of some pointless satisfaction of instincts, surely it cannot contribute to the collapse of society.

One lives in Finland. And one is very nearly broke, so no book purchases for me. But thanks anyway

Bitrex

@Valtteri Ever watch any baseball?

Baseball is a mathematician's game.

Valtteri

Finnish baseball, much simpler that way

Alec Teal

16:54

Hey guys

@skullpatrol you here?

Bitrex

@Valtteri I've got to try some blood pancakes someday.

skullpatrol

@AlecTeal yes

Valtteri

Interesting non sequitur there

Alec Teal

@skullpatrol I need help with defining a step function, searching is not very helpful

[I want some answers checked]

Valtteri

Floor[x] is a step function

Alec Teal

16:57

@Valtteri that's not true

Well it might be....

Arguably not a finite number of partitions or values

Valtteri

and you would win that argument

if not a step, then a staircase

Alec Teal

Grr, BRB, sorry to say "help me later"

[brb as in ~hour]

Bitrex

@Valtteri I guess the mention of Finland triggered it

Valtteri

Yes, I understand. Tell me more about your fixation with blood as a food item.

skullpatrol

@AlecTeal here

Valtteri

17:04

Ah, that one. I remember seeing it on a course on PDE

@Bitrex And after that, could you tell me how would you define a city?

Alec Teal

[redacted]

Nevermind!

Valtteri

I need to go now. bye all

skullpatrol

later pal

Nimza

Hello!

skullpatrol

Hi

Nimza

17:15

@skullpatrol hi, how are you?

skullpatrol

@Nimza Fine thank you. How are you?

Nimza

@skullpatrol me too, thanks)

Chris's sis

This is one of the candidates to the title of the problem of the year (to me) $$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n }\right) \log\left(1+\frac{1}{2n }\right) \log\left(1+\frac{1}{2n+1 }\right)$$

Bitrex

Yikes

Mats Granvik

17:59

Based on these limits: http://functions.wolfram.com/ElementaryFunctions/ProductLog/09/
Is it known how to calculate LambertW(x) by tetration for x=1,2,3,4,5,...?

Clear[nn, t, n, k, i];
nn = 85;
t[1, 1] = 1;
t[n_, k_] :=
t[n, k] = If[n >= k, Exp[-Sum[N[t[n - i, k]], {i, 1, k - 1}]], 1];
Table[t[nn, k]^-1, {k, 1 + 1, 12 + 1}]
Table[N[n/ProductLog[n]], {n, 1, 12}]

Mats Granvik

18:21

@skullpatrol

hi

recursively by tetration

{1.76322, 2.34575, 2.85739, 3.32732, 3.76868, 4.18881, \
4.59206, 4.98129, 5.36656, 5.72993, 6.07439, 6.44781}

compared to the buildt in Mathematica command

{1.76322, 2.34575, 2.85739, 3.32732, 3.76868, 4.18876, \
4.59214, 4.9819, 5.36028, 5.72893, 6.08911, 6.44186}

skullpatrol

@MatsGranvik hi pal :-)

Mats Granvik

@skullpatrol do you know how to do addition?

Alizter

is there a factor that makes a function of x real only in the bound [0,1] and the rest complex?

skullpatrol

@MatsGranvik sometimes

Mats Granvik

@skullpatrol what is 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

?

MLT

18:35

hello all

pls help me out with this

4

Q: Joint distribution by independent distributions

We have $N$ independent discrete finite random variables (RVs) $X_1,\dots,X_i,\dots,X_N$ where RV $X_i$ has $M_i$ finite number of elements. We are free to choose any distribution $f_i$ for RV $X_i$ $\forall i=\{1,\dots,N\}$. Then consider the product set $Y = X_1\times\dots\times X_i\times\dots\...

Mats Granvik

@skullpatrol eh bad joke, I was trying to introduce you to tetration if you did not know it.

skullpatrol

@MatsGranvik thanks for the intro :D

MeLight

Hello dearest geeks

Jack M

hi

would anyone have a hint to offer me on a group theory problem?

MeLight

probably not me

sorry, most definitely not me

Jack M

18:46

the problem is to prove that the conjugacy class of $a$ is $\{a\}$ if and only if $a$ is in the center of the group

it's the "only if" part that I'm struggling with, although I'd appreciate a hint more than a solution

having trouble with it in part because I don't have much of an intuitive feel for what conjugacy really is

I was thinking maybe I could prove $ab$ is always conjugate to $ba$, but I can't see any way of doing that

not even sure it's true

Kevin Driscoll

If only anon were here......

Jack M

wait, I just got it

that was way simpler than I thought

skullpatrol

....see, just the mention of his name solves problems :-)

Jack M

$$a^{-1}(ab)a=ba$$

Enjoys Math

Abstract algebra help needed: texpaste.com/n/aqm91tj8

It's a basic question, not involving math but prerequisites

Fernando Martin

19:06

@EnjoysMath: I don't think learning category theory before abstract algebra is a good idea

Enjoys Math

I know enough though

I don't need to understand all of Lang's Algebra to start this book

Fernando Martin

Just my 2c

Karl Kronenfeld

@EnjoysMath You need to already understand the connection between hom and tensor before finding it in a book strictly devoted to category theory.

Enjoys Math

Is that in Lang's algebra?

:D

\o| I have that, if it is in there

Karl Kronenfeld

I wouldn't know.

JGab

19:17

Hola! Is there any way to bring back your question 'to the top'?

Posted mine in the middle of the night, got burried quickly.. and now I worked a part of it but nobody is viewing it haha

Enjoys Math

Hit it from all angles: favorite it , edit it, and add a comment!

lulz

JGab

Haha

Enjoys Math

@Karl, what's your fav algebra book?

Karl Kronenfeld

@EnjoysMath I don't play favorites.

Enjoys Math

recommends?

What book did you learn from?

Karl Kronenfeld

19:21

@EnjoysMath Algebra by Isaacs was pretty cool, but I quickly "outgrew" it. It is an excellent introduction to the topics but don't expect much else from it.

The Chaz 2.0

I'm about to blow my brains out here...

Fernando Martin

That must be a troll

The Chaz 2.0

It's the razor...

I'll give him the benefit of the doubt for about two more back-and-forths

skullpatrol

slap him with a dictionary

The Chaz 2.0

Well he's already been to wikipedia, so I guess we're out of options... /s

Karl Kronenfeld

19:34

@EnjoysMath Btw I agree with @FernandoMartin about the fact that category theory should be learned on its own only after a thorough investigation of a field like commutative algebra. From my own experience it is certainly possible to grasp the notions of category theory without very much knowledge & understanding of a field that takes advantage of category theory.
However, I don't believe it is possible to actually internalize the most important aspects of category theory without understanding some of its applications. You build up understanding by abstracting properties of diverse examples…

Enjoys Math

K so commutative algebra first. That should help me understand the examples better

I have a digital copy of Eisenbud's Commutative Algebra: with a view toward Alg Geom

it also has prerequisites in Lang's algebra

should finish that up first then

Karl Kronenfeld

@EnjoysMath I don't have it, but one of its criticisms is that it is pretty dry.

Atiyah-Macdonald Commutative Algebra is a surprisingly thin book for how much material it covers. That said, I have actually never read the entire thing.

The Chaz 2.0

Re: averages livememe.com/4t1g7mp

Enjoys Math

@Karl, that this: bookos.org/book/967351/83d390

?

Karl Kronenfeld

@EnjoysMath Looks like it.

Page count is right, 128 pages.

Enjoys Math

19:41

Thanks for all your help! I gtg

Karl Kronenfeld

Bye

skullpatrol

Hi @HenningMakholm how are you?

Arkamis

@TheChaz2.0 Sometimes, people don't even try

The Chaz 2.0

he's trying, maybe. in another chat room

Arkamis

Like, stop and read, people!

skullpatrol

19:56

he got it :D

average: The average of a set of numbers is the sum of the numbers divided by the number of numbers.

Nimza

Hi

skullpatrol

Hi

Nimza

Help me please, if a function $f \in C^{\infty}(\mathbb R^2 \setminus 0)$ and is not continuous at zero (limits from different directions don't agree) does it mean that it's Fourier transform doesn't tend to zero at infinity?

Daniel Fischer

20:12

No, the Fourier transform of every $L^1$ function tends to $0$ at $\infty$.

So if your function is in $L^1$, its Fourier transform will be in $C_0$.

Nimza

@DanielFischer very great thanks

@DanielFischer actually my function is just a projection of a function from the Schwartz class: $\langle f(x), x/|x| \rangle x / |x|$

Daniel Fischer

I sort of remembered you had asked that question.

Nimza

heh))

Daniel Fischer

I don't know how fast the Fourier transform decays, unfortunately.

Fernando Martin

Is there a way for TeX to be auto-rendered instead of clicking the bookmarklet each time?

Nimza

20:17

@DanielFischer but that it tends to zero is also good for me, thank you. do you know where to get the proof? maybe I could accommodate it...

@DanielFischer ah, no, thanks, I got the proof

Daniel Fischer

@Nimza Riemann-Lebesgue lemma.

@Nimza But for your case, I'm sure one knows a better result.

Nimza

@DanielFischer oho! I didn't know that this result is so cool-named

JGab

@FernandoMartin What do you mean by TeX auto-rendering and bookmarklet? In chat or in math.stack

Fernando Martin

In chat

I have a little bookmarlket that renders TeX here, it's the one in the sidebar here ->

Jack M

what do you guys think of Lang's textbooks?

Henning Makholm

20:33

I'm baffled by this edit. It changes a bad question to a somewhat better but completely different one, and it's not even by the OP. Are we running out of question numbers such that we have to reuse old ones?

skullpatrol

52 mins ago, by skullpatrol

Hi @HenningMakholm how are you?

Jack M

@HenningMakholm that's an absolutely terrible edit

was it approved?

oh, it was

Charlie

@Nimza hahaha

Nimza

@Charlie :))))

JGab

@FernandoMartin I dont see any other way.. you could use a chrome applet

Fernando Martin

20:43

@JGab that's ok, I just wanted to know if you guys used a workaround

thanks!

Charlie

@HenningMakholm hii! Long time no see! how are you???

Alec Teal

At some point I must look at adding LaTeX to xchat

MeLight

Hey guys, im working on image 2d transforms (with transformation matrices), everything works well except for rotation - the rotation matrix somehow effects scaling also. The higher the rotation angle, the smaller the outcoming image becomes. Any idea where I could begin looking for the problem?

Fernando Martin

@AlecTeal: Is this chatroom accesible via xchat/irc?

JGab

@MeLight Your rotation matrices are probably not normalized

Alec Teal

20:48

No @FernandoMartin but it ought to be done. Maybe I'll make a LaTeX interpreter that pretty-prints (across lines) first, then irssi and xchat could benefit

JGab

@MeLight you can check this by looking at the product $\mathbf{RR}^T$ it will be equal to the identity times a scaler if there is a scaling

MeLight

@JGab probably - I didn't even know I was supposed to normalize them.

@Jgab so, I simply normalize the rotation matrix before applying it?

JGab

@MeLight well you should not have to if you used the standard definition, but that will indicate a problem

@MeLight are you using $[cos(x) -sin(x); sin(x) cos(x)]$

MeLight

@JGab could you please elaborate on that: $\mathbf{RR}^T$ - not really sure what it means

@JGab yes, thats what i use

JGab

@MeLight then there should not be any scaling effect, the determinant of the transformation matrix is 1...

MeLight

20:51

@JGab ok, will keep digging :/ thanks

Donkey_2009

$\mathbf{RR}^T$

MeLight

yah, I thought it should render as something else too

JGab

@MeLight by $\mathbf{RR}^T$ I meant the rotation matrix times it's transpose. For similarity transforms it should be the identity

90intuition

I'm trying to prove that $ℤ[\sqrt{7}]$ is a domain. Suppose $(a+b\sqrt{7})(c+d\sqrt{7})=0$. some calculation Then $ac=-7bd$ and $bc=-ad$. I don't know how I can show $a,b,c,d\in \{0\}$ now.

Henning Makholm

21:07

@Charlie Busy, mostly.

Charlie

@HenningMakholm ah, nice :)

Daniel Fischer

@90intuition Can you just say that $\mathbb{Z}[\sqrt{7}]$ is a subring of $\mathbb{R}$?

90intuition

Every subring is a domain ?

Henning Makholm

@FernandoMartin Conversely: Is there a web-based IRC client with the look and feel of the SE chat engine?

Daniel Fischer

@90intuition Every subring of an integral domain is an integral domain.

Fernando Martin

21:10

@HenningMakholm Not that I know of

90intuition

@DanielFischer Hm.. let's see if I can proof that.

Henning Makholm

@FernandoMartin Didn't think so, would be nice though.

90intuition

@DanielFischer Oh that is trivial, isn't it ?

@HenningMakholm That would be really nice !

Daniel Fischer

Sort of, yes.

90intuition

If $x,y$ with $xy=0$ in the subring, then $x,y$ in the ring and $xy$ is still zero. So $x=0$ or $y=0$.

Trying to proof that $(7,X^2-2)$ is an maximal or prime ideal of $ℤ[X]$.

Can I say $X^2-9 \in (7,X^2-2)$. And $X^2-9=(X-3)(X+3)$ But $X-3,X+3$ are not in the ideal. Therefore it is not a prime ideal.

Henning Makholm

21:20

@90intuition Looks right.

Alec Teal

@DanielFischer one quick question (I wrote it down waiting for either you or Pedro), can you have a function $f:[a,b]\rightarrow\mathbb{R}$ where $a\le b$ or must it be strictly smaller than? I ask because a single point matches my definition of step, and I want to be sure about this edge case.

Karl Kronenfeld

@90intuition Yes, you can actually get that factorization of $(X^2-2)$ in $(\mathbb Z/7\mathbb Z)[X]=\mathbb Z[x]/(7)$.

Alec Teal

@DanielFischer also can I say "the subinterval of (a,b)\P" where P is a set of points $\in$ (a,b) to define partitions, I believe so. (P is a subset of (a,b))

That's it

I lied do you read $f|_s$ as "f over s" in your head? or restricted to?...

Daniel Fischer

@AlecTeal $[a,a] = \{a\}$, so you can have a function defined on that, no problem. $(a,a) = \varnothing$, so it is a subinterval.

@AlecTeal "Restricted to".

90intuition

@KarlKronenfeld So I get $ℤ/7ℤ[\sqrt{2}]$ ?

Daniel Fischer

21:26

Whether you call a function $f \colon [a,a] \to \mathbb{R}$ a step function or not would depend on whether your definition of step function contains "non-degenerate interval" or not.

90intuition

Oh wait you don't mean that.

Alec Teal

@DanielFischer I lied again, one more, is a partition defined as $(P_{k-1},P_k)$ or that with [] instead? I've seen [] used but where the step function is discontinuous it can take a different values. Unless subinterval means "the open interval"

I'm going with ( ) because that makes sense

90intuition

@KarlKronenfeld I see what you mean now.

Karl Kronenfeld

@90intuition Yeah, I would prefer to work in $F_p[X]$ over $\mathbb Z[X]$ when possible, just because the former is a PID.

Daniel Fischer

@AlecTeal Everybody can make their own definitions, but usually, one considers the open intervals.

Karl Kronenfeld

21:30

@90intuition In particular, in a PID if a nonzero ideal $I$ is contained properly in any proper ideal, then it is not prime.

This just makes your proof slicker.

Henning Makholm

@90intuition The problem with that is that $\mathbb F_7$ already contains a square root of 2, namely 3 (and 4). So when you try to adjoin an additional one, things go awry.

90intuition

What does PID mean ?

Henning Makholm

Principal ideal domain.

90intuition

I know what a principal ideal is, and what a domain is. But I didn't hear about a principal ideal domain.

Karl Kronenfeld

@90intuition $\mathbb Z$ and $k[X]$ for any field $k$ are the usual examples of PIDs.

Henning Makholm

21:33

@90intuition A principal ideal domain is an integral domain in which every ideal is principal.

Karl Kronenfeld

;)

90intuition

Aah I see now. Interesting.

I'm never sure if I can always do tricks like: $\mathbb{Q}[X,Y]/(X^2+3) = \mathbb{Q}[\sqrt{-3}][Y] = \mathbb{Q}[\sqrt{-3}][X]$. In my reader they do those tricks all the time, without giving any more details.

Karl Kronenfeld

@90intuition Read it as $R[X]/\mathfrak aR[X]$ where $\mathfrak a$ is an ideal of $R$. Then prove that the quotient is indeed $(R/\mathfrak a)[X]$.

@90intuition If you're stuck on any part of that message, let me know. I can give hints on the proof btw.

90intuition

I'll give it a try.

$$\varphi : R[X] → (R/aR)[X] : \sum_{i=0}^{n} a_i X^i ↦ \sum_{i=0}^{n} \bar{a_i} X^i$$

Karl Kronenfeld

Exactly.

90intuition

21:48

Suppose $f \in \ker \varphi$. Then $f(x)=0$. then $a_i \in aR$. Then $f \in aR[X]$.

Suppose $f \in aR[X]$. Then $a_i=aa_i'$. Then $f=\sum a_i X^i = a \sum a_i'X^i$ Then this maps to $0 ⋅ \sum a_i' X^i$. So $f \in \ker \varphi$.

Something like that ?

Karl Kronenfeld

Actually $\mathfrak a$ is not an element of $R$, it is an ideal. I will write $I$ instead of $\mathfrak a$ in the future.

But all you need is that the coefficients of $f$ are in $I$, so their images via the canonical projection $R\to R/I$ are 0.

90intuition

But what I find strange, is to write $\mathbb{Q}[X]/(X^2+3) ≅ \mathbb{Q}/(X^2+3) [X]$

Karl Kronenfeld

$(X^2+3)$ is not an ideal of $\mathbb Q$.

So you can't apply that result I quoted.

90intuition

But you said that I should read it that way. No ?

Karl Kronenfeld

You should read $\mathbb Q[X,Y]/(X^2+3)$ as $(\mathbb Q[X])[Y]/(X^2+3)$. :)

90intuition

21:59

But what has that to do with $R[X]/(a)$ is $(R/(a)[X]$ ?

Karl Kronenfeld

Let $R=\mathbb Q[X]$ and $I=(X^2+3)$.

I am using $I$ instead of $\mathfrak a$ to represent that ideal.

90intuition

So I get $R[Y]/I = (R/I) [Y] = (\mathbb{Q}[X]/(X^2+3)) [Y]$

Aaah okay, I see

But that is another question. Can I always say because $X^2+3=0$ and this means $X=\sqrt {-3}$. Therefore this becomes $\mathbb{Q}[\sqrt{-3}][Y]$.

Because I can't take (negative) square roots in $\mathbb{Q}$. This seems like strange reasoning to me.

Because in my reader they give no more details. Or should I think of it as, you can proof this the exact same way as you have proven that $ℝ[X]/(X^2+1) ≅ ℂ$ ? And therefore they don't give detaisl.

Mats Granvik

Tetration of complex numbers:

Clear[nn, t, n, k, i];
nn = 85;
t[1, 1] = 1;
t[n_, k_] :=
t[n, k] = If[n >= k, Exp[-Sum[N[I*t[n - i, k]], {i, 1, k - 1}]], I];
Table[t[nn, k]*(k - 1)*I, {k, 1 + 1, 6 + 1}]
Table[N[ProductLog[n*I]], {n, 1, 6}]

By tetration:
{0.374699 + 0.576413 I, 0.683408 + 0.743386 I, 0.898668 + 0.826935 I,
1.06393 + 0.87901 I, 1.19448 + 0.915874 I, 1.31837 + 0.936584 I}

Mathematicas ProductLog:
{0.374699 + 0.576413 I, 0.683408 + 0.743386 I, 0.89871 + 0.826955 I,
1.06384 + 0.879803 I, 1.19805 + 0.917331 I, 1.31129 + 0.945888 I}

Correct that. Imaginary numbers, not any complex number.

The convergence is slow, and would need about 1000 iterations. But that is too much for Mathematicas recursion limit.

Instead I used 85 iterations.

Karl Kronenfeld

22:18

@90intuition It has to do with field extensions really. You use a $\sqrt{-3}$ from $\mathbb C$. For terminology, go here.

Mats Granvik

22:29

Now it seems possible to tetrate also complex numbers of form (1+I).

The limitation seems to be that the complex number has to be a integer multiple. Or not?

Bitrex

@MatsGranvik Cool question. I hope someone has an answer!

Mats Granvik

22:44

thanks. Just in order to try to generalize:

Clear[nn, t, n, k, i];
nn = 75;
t[1, 1] = 1;
t[n_, k_] :=
t[n, k] =
If[n >= k,
Exp[-Sum[N[(1/ZetaZero[1])*t[n - i, k]], {i, 1, k - 1}]], (1/
ZetaZero[1])];
Table[t[nn, k]*(k - 1)*(1/ZetaZero[1]), {k, 1 + 1, 6 + 1}]
Table[N[ProductLog[n*(1/ZetaZero[1])]], {n, 1, 6}]

By tetration of some complex number:
{0.00736674 - 0.0697969 I, 0.0235635 - 0.136089 I,
0.0466203 - 0.197008 I, 0.07436 - 0.251797 I, 0.104914 - 0.300521 I,
0.136894 - 0.343688 I}

Mathematicas ProductLog:

No generalization really. Still integer multiples of a complex number.

masfenix

Can someone help me understand this line: "Since $K \in L^2(\mathbb{R}^{2d})$ the modulus is in $L^1$. If anyone is confused, $K$ is the kernel used in the Hilbert Schmidt Integral Operator.

pourjour

23:08

what is the argument of 0

Kevin Driscoll

23:50

@pourjour Its not defined

pourjour

@KevinDriscoll ok thanks

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