Mathematics - 2014-07-06 (page 1 of 2)

blue

00:00

I need a nice name for groups which are inverse limits of locally finite systems of finite groups in which each group is covered by at most countably many other groups.

Pedro Tamaroff

LEL

What is a LFsystem? @blue

blue

locally finite refers to the partial ordering

hmm, thinking "pointy" group might do

or maybe "sandy"

or "particulate"

Pedro Tamaroff

Granulated?

@blue

blue

I like particulate best

Pedro Tamaroff

Well one always has "atomic".

Maybe that one is taken.

I think atomic sounds badass.

blue

00:09

someone on freenode told me it's unsolved how many subfields $\overline{\Bbb Q}$ has

Pedro Tamaroff

Cool.

Feel like solving it? =O

I have to go now.

blue

00:37

For any prime $p$, we can extend $v_p$ on $\Bbb Q$ to $\overline{\Bbb Q}$. Do so for each prime, including $p=\infty$. Call a sequence $\theta_1,\theta_2,\cdots\in\overline{\Bbb Q}$ locally ultimate (at $p$) if $\max |\theta_i-\theta_i'|_p\to\infty$ and $\min_{\theta_i\ne\theta_i'}|\theta_i-\theta_i'|\to0$. Whether or not a sequence is a locally ultimate at $p$ is independent of how we extended $v_p$.

Call a sequence $\theta_1,\theta_2,\cdots$ globally ultimate if every $\alpha\in\overline{\Bbb Q}$ is in cofinitely many $\Bbb Q(\theta_i)$. It would be cool if there is a relationship between locally and globally ultimate sequences. It would be like a "Galois approximation" theorem, similar to approximation theorems for adeles, or else a manifestation of Local-Global philosophy.

hmm, Krasner's lemma is probably useful here

blue

01:49

actually we probably want some better notion of "asymptotically dense"

perhaps "a sequence $S_i$ of subsets of a metric space is asymptotically dense if every open set nontrivially intersects cofinitely many $S_i$" is too strong

Ben

02:36

can reflecting a graph about the x-axis or y-axis change the total number of x and y intercepts?

blue

what do you think?

if you reflect an axis intercept across the other axis, does it still intercept the same axis?

robjohn

02:51

@N3buchadnezzar That is just $$\left[\frac12\log(1+x)^2\right]_0^1=\frac12\log(2)^2$$

@Chris'ssis that should be workable by considering the terms where $k+n=m$... let me see.

@Chris'ssis if I am correct, that should be $$2\sum_{m=1}^\infty(-1)^m\frac{H_{m-1}}{m}=\log(2)^2$$

which I computed the other day.

robjohn

03:12

These two being the same, I get the feeling this is not a coincidence...

Vibhav Pant

morning

me recalls a similar result @Chris'ssis has proved earlier

blue

08:01

did a bit of math, forgot everything I was doing in the water temple

Chris's sis

08:54

Greetings (gezzz, I was so tired yesterday)

Now it's much better. :-)

@robjohn hmmm, you mean that series is precisely $\log^2(2)$?

@N3buchadnezzar Numerically they don't match.

N3buchadnezzar

@Chris'ssis :p

Chris's sis

@N3buchadnezzar How is it going there? :-)

N3buchadnezzar

Fine, traveling to a friend in half an hour or so

Mats Granvik

09:31

Series reversion on a power series through repeated convolution, is that known?

(*Mathematica:*)
(*program start*)
(*coefficients (coeff) in power series can be changed*)
coeff = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
nn = Length[coeff];
A1 = Table[Table[If[n >= k, 1, 0], {k, 1, nn}], {n, 1, nn}];
MatrixForm[A1];
(*Iteratively downshift the matrix A1 one step and apply power series \
with coefficients of choice (coeff)*)
Monitor[Do[A2 = A1[[1 ;; nn - 1]];
A2 = Prepend[A2, ConstantArray[0, nn]];
A1 = Total[
Table[Table[
Table[(coeff)[[m]]*MatrixPower[A2, m - 1][[n, k]], {k, 1,

This particular program gives the Catalan numbers. But it can also be used on the power series for the Riemann zeta function.

Since they are matrix powers of the power series coefficients, does it mean that the inverse function can be expressed in terms of the function to be inversed? Or am I only dreaming?

This builds upon the INVERT transform in the oeis, communicated to me via Gary W Adamson.

The program here is a generalization of the INVERT transform.

0

Q: Algorithm for reversion of power series?

Is this an algorithm for power series reversion? As input I give the alternating reciprocals of the factorial numbers: {1/0!, -1/1!, 1/2!, -1/3!, 1/4!, -1/5!, 1/6!, -1/7!, 1/8!, -1/9!, 1/10!, -1/11!} Like this: (*Mathematica program start*) (*coefficients (coeff) in power series can be chan...

mathworld.wolfram.com/SeriesReversion.html

It is much easier to understand when you see the spreadsheet calculation of this.

Anyone?

There is no subtraction involved.

No alternating series. Only downshifting, multiplication, addition, and matrix multiplication.

robjohn

10:13

@Chris'ssis Do you get something else? I also tried it in Mathematica and it agrees.

Chris's sis

@robjohn you mean that $$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{k+n} \frac{\log(k+n)}{k n}=\log^2(2)$$?

robjohn

@Chris'ssis yes. Do you get something else?

Chris's sis

@robjohn NSum[(-1)^(k + n) Log[k + n]/ (k n), {k, 1, 10000}, {n, 1, 10000}] = 0.203667

NSum[(-1)^(k + n) Log[k + n]/ (k n), {k, 1, 10000}, {n, 1, 100000}] = 0.203947

NSum[(-1)^(k + n) Log[k + n]/ (k n), {k, 1, 1000000}, {n, 1, 1000000}] = 0.204296

NSum[(-1)^(k + n) Log[k + n]/ (k n), {k, 1, 10000000}, {n, 1,
10000000}] = 0.204305

@robjohn maybe my Mathematica version deceives me? What do you get for these values?

robjohn

@Chris'ssis You know what... never mind. I left out the log :-(

Chris's sis

@robjohn hehe :-)

robjohn

10:21

This should be $$2\sum_{k=1}^\infty(-1)^m\frac{H_{m-1}\log(m)}{m}$$

which I did not yet compute :-)

Chris's sis

@robjohn Yeap. :-)

robjohn

Or did I? I need to look back at some things...

Chris's sis

10:49

@robjohn did you?

robjohn

11:03

@Chris'ssis I'm not sure...

robjohn

11:17

That sum is $0.20430570664961148161$

Chris's sis

11:33

@robjohn The inverse symbolic calculator says "Wow, really found nothing.".

Shisui

What is the inverse symbolic calculator?
Also, good afternoon!

Chris's sis

@Shisui Hi - en.wikipedia.org/wiki/Inverse_Symbolic_Calculator

Shisui

What's the use of that calculator?
I don't get it ...

Chris's sis

@Shisui "A user will input a number and the Calculator will use an algorithm to search for and calculate closed-form expressions or suitable functions that have roots near this number."

Shisui

@Chris'ssis That's actually pretty cool!

Chris's sis

11:44

@Shisui Most of the time, it cannot help me (but this is not the case of the others).

c c

12:08

how do you interpret the convention $0.\infty=0$?

blue

@cc you don't, because it's not a convention and to the extent it's meaningful at all it's not true ($0\cdot\infty$ is an indeterminate form, not $=0$)

c c

(this convention is used here en.wikipedia.org/wiki/Lebesgue_integration)

> The convention 0 × ∞ = 0 must be used

I think it's that the nullity of the scalar, is more important than the vector

blue

that a user on wikipedia decided to set that interpretation to make an equality easier to present does not make it a "convention"; that word shouldn't be used like that

@cc it's because integrating 0 over a region of infinite measure is still 0

don't pretend like the stuff on the left-hand side doesn't mean anything; it's where the stuff on the right-hand side comes from in the first place, hence it serves as a guide for why we're interpreting $0\cdot\infty$ as $0$ on the right

c c

@blue it's not just wikipedia, I was reading math.u-psud.fr/~jflegall/IPPA2.pdf actually

@blue ok

mesel

@cc Is it Spanish ?

Daniel Fischer

12:22

@blue It is an established convention in the scope of integration. Outside that scope, it is different.

c c

12:52

and can we form a non-null integral over a region of measure 0?

robjohn

14:17

@Chris'ssis Yes, I had checked that out. That doesn't mean much :-)

Chris's sis

@robjohn The question is crazy difficult and crazy awesome at the same time. :-) (some research is needed there)

robjohn

@Chris'ssis I am guessing that you don't have a closed form for this yet.

Chris's sis

:-)

Balarka Sen

I think it doesn't have any closed form.

@Chris'ssis Why would I serially upvote your questions? =P

Chris's sis

@BalarkaSen ;)

Balarka Sen

14:27

@Chris'ssis I might as well be wrong, so I am not betting on anything. My base of intuition is that $\zeta'(s)$ usually doesn't have closed forms algebraically dependent of $\zeta(s)$, for $s \in \Bbb Q$

And this

4 hours ago, by robjohn

This should be $$2\sum_{k=1}^\infty(-1)^m\frac{H_{m-1}\log(m)}{m}$$

Looks mighty like some derivative of twisted zeta

@Chris'ssis This sum is related to this question of yours, right?

Chris's sis

@BalarkaSen Yeah.

Balarka Sen

@Chris'ssis ... I have seen some similar sums evaluated somewhere using Ramanujan's Master theorem.

I can't recall.

Chris's sis

@BalarkaSen I didn't see Cleo giving the closed form ...

Balarka Sen

@Chris'ssis Hahaha

Chris's sis

@BalarkaSen :-) Sometimes I fell myself as if I were a bit evil. :D

Balarka Sen

14:35

@Chris'ssis Evil? About what?

Chris's sis

@BalarkaSen about asking these "scaring people" - questions. :-)

Balarka Sen

@Chris'ssis Well, those questions surely scare the hell out of me.

Chris's sis

:-)))

Balarka Sen

But I guess notion of beauty differs from person to person.

Chris's sis

True.

eXtremiity

14:55

So does Math Stack Exchange use MathJax to render its TeX?

Balarka Sen

@robjohn Do you know of anything like "the longest answer in MSE"?

Just an arbitrary thought, as the shortest one is one with a 'W'.

$\Bbb S$?

ah.

Chris's sis

This is a personal research topic.

Balarka Sen

@Chris'ssis I was talking about the notation.

You mean this?

Chris's sis

@BalarkaSen Exactly!

Balarka Sen

ok. thanks.

Fun generalization. How does it arise?

Kölbig's paper doesn't seem to be available online. Oh well.

Chris's sis

15:19

Kölbig's paper? There should be a link there on mathworld.

Balarka Sen

There is no link.

robjohn

15:33

@BalarkaSen I have a script to find the longest answer of mine, perhaps I can modify it...

Bananarama

15:45

Hello gang!

Balarka Sen

@robjohn Really? That's cool!

robjohn

16:35

@BalarkaSen this is the longest post on math.

3

Studentmath

Anyone has good idea on some seminar paper subject in graph theory?

Balarka Sen

@robjohn Heh. Starred.

Pedro Tamaroff

17:06

Hello peeps

Bananarama

@PedroTamaroff hello

Balarka Sen

@PedroTamaroff Hey.

You got some fun group theory problem?

Shisui

17:23

Anybody got any fun integrals to solve?

(Preferably ones you'd encounter at the beginning of an undergrad degree)

Balarka Sen

@Shisui $$\int \frac{dt}{t}$$

Shisui

@BalarkaSen $$ \int \dfrac{\text{d}t}{t} = \log|t| + \mathcal{C} $$

Balarka Sen

@Shisui Prove it =P

Shisui

@BalarkaSen Does the proof have anything to do with differential equations?

Balarka Sen

@Shisui The cleverest way to do it probably is to sub $t = \exp(x)$ in the integral.

Here's a 'cheaters' way to do it : think of $t$ as a function of two variables and look at how the differential behave : if $t = uv$ then $$\int \frac{dt}{t} = \int \frac{vdu + udv}{uv} = \int \frac{du}{u} + \int \frac{dv}{v}$$ So you can "think" of the resulting function $F$ (without the arbitrary constant) to be satisfying $F(uv) = F(u) + F(v)$. You can show with some difficulty that $F$ is indeed inverse of $\exp$. For example, use the limit definition of $e$ and show that $F(e) = 1$.

=P

Shisui

17:33

@BalarkaSen How did you get the numerator?

Balarka Sen

@Shisui $d(uv) = udv + vdu$

Shisui

@BalarkaSen =S

nullgeppetto

I have started a bounty, but unfortunately I am a poor man, so I can offer only 50 reps! I really need this answer, so if you'd like, take a look, and/or help drawing more attention to it (e.g. via upvoting, if you think it worths it)!

Balarka Sen

@Shisui You don't know how to prove that?

It's basic property of differentiation, we call it product rule.

Shisui

@BalarkaSen Nope. What are you differentiating $t=uv$ with respect to?

Balarka Sen

17:35

@Shisui Not really differentiating. It's the "differential". The operator.

Pedro Tamaroff

Suppose that $G$ is an extension of $H$ by $K$. If $H,K$ are solvable, so is $G$.

@BalarkaSen

Shisui

@BalarkaSen That's not been explained to me before. Could you explain it to me?

Balarka Sen

@PedroTamaroff Noted.

@Shisui What to explain? The differential, or the product rule?

Shisui

@BalarkaSen It might be worth noting that I have heard of the product rule before, but never in the context of differential operators.

@BalarkaSen The differential.

Balarka Sen

@Shisui Let $f(x)$ be some arbitrary function.

Consider the change $\Delta y = f(x + \Delta x) - f(x)$

And let $\Delta x \to 0$

Shisui

17:41

@BalarkaSen Oops

Balarka Sen

@Shisui ?

Shisui

@BalarkaSen My brain's not working today U_U

Balarka Sen

You realize that $\Delta y = f'(x) \Delta x + \epsilon$, right?

Shisui

Nope

The epsilon looks foreign

I get the rest

Balarka Sen

$(x + \Delta x)^2 - x^2 = 2x \Delta x + (\Delta x)^2$.

$(\Delta x)^2$ is your epsilon here/

Note also that $\epsilon/\Delta x \to 0$

So they are of the same asymptotic order.

Shisui

17:44

I don't understand how $\Delta y = f'(x) \Delta x + \epsilon$ links to $(x+\Delta x)^2 - x^2 = 2x\Delta x + (\Delta x)^2$

Balarka Sen

What's not to understand? Be precise.

Shisui

Would it be possible for you to explain how you went from one to the next in words as opposed to solely mathematical notation?

Balarka Sen

I am not sure what you mean by that.

Heya @TedShifrin

I am teaching someone about the differential operator here.

Ted Shifrin

Heya @Balarka

Shisui

Okay, we started off with $\Delta y = f'(x) \Delta x + \epsilon$

Balarka Sen

17:48

90% of the people gets confuzzled by Lebesgue notation, @TedShifrin

Shisui

I know that $ f'(x) $ is the derivative of the function (i.e. it's gradient)
I also know that $ \displaystyle \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = f'(x) $
Where does the epsilon come into play?

Ted Shifrin

What is Lebesgue notation?

Balarka Sen

@TedShifrin Treating $dy/dx$ as a fraction.

Ted Shifrin

That's not Lebesgue. It's Leibniz.

Balarka Sen

Oh, right.

I forgot.

Ted Shifrin

17:50

Uh huh.

Balarka Sen

I am not good with remembering names.

Ted Shifrin

Wait 'til you're my age!

Balarka Sen

@Shisui Epsilon doesn't come to play till you look at $\Delta y$, boy.

Shisui

Ok. $\Delta y$ is a change in $y$.
It can be approximated by $f'(x) \Delta x$

Ted Shifrin

Funny hearing a boy call someone a boy ... :D

Balarka Sen

17:52

Look at $y = x^2$, @Shisui

@Shisui Exactly. Approximated, not equalized.

Shisui

Alright! $y=x^2 \implies \Delta y \approx 2x \Delta x$

Balarka Sen

@Shisui $\Delta$?

Yes.

$\Delta y = 2x\Delta x + (\Delta x)^2$, to be precise.

Shisui

Where does the $$\Delta x)^2$ come from?

Balarka Sen

Subtract $x^2$ from $(x + \Delta x)^2$, boy.

All the math you've known you've forgotten?

(nudge-nudge-wink-wink @TedShifrin)

Shisui

$(x+\Delta x)^2 - x^2 = x^2 + 2x\Delta x + (\Delta x)^2 - x^2 = 2x \Delta x + (\Delta x)^2$

Balarka Sen

17:57

@Shisui So that's the $(\Delta x)^2$.

Shisui

I'm not getting this at all.

Good grief, who stole my mind?

Balarka Sen

@Shisui tsk tsk

Shisui

Can we start from the beginning again?

I'm slow on the uptake.

Balarka Sen

@Shisui I think I need break. You just go an pick up a book on modern treatment on calculus.

Shisui

We have a function $y=x^2$. From this we can deduce that $\Delta y \approx f'(x) \Delta x \approx 2x \Delta x$.

Balarka Sen

17:59

What have you read so far?

Shisui

@BalarkaSen Absolutely nothing

Balarka Sen

@Shisui I hate $\approx$. It doesn't tell us anything. $\sim$ is much better.

At least it tells something.

Shisui

I'm just going on intuition here as I'm yet to start undergrad maths.

Balarka Sen

@Shisui Try with Courant if you want to do these stuffs.

I have heard a lot of praise about Spivak but never read it.

An awesome book is Piskunov.

Integration doesn't make sense unless you know about differentials.

Just knowing it's the opposite of $d/dx$ is not useful enough to do stuffs with it.

Sawarnik

@BalarkaSen What?

@BalarkaSen Me too :(

Balarka Sen

18:02

@Sawarnik Piskunov. Search it.

@Sawarnik Differentials.

Sawarnik

@BalarkaSen I know, its not very great.

Balarka Sen

@Sawarnik It is great. Have you even tried to read it?

Sawarnik

@BalarkaSen No, it makes sense without them.

Balarka Sen

@Sawarnik Superficially.

Sawarnik

@BalarkaSen Ok.

Balarka Sen

18:03

Integrals are objects of differential equations.

@Sawarnik Stop arguing.

Anyone will agree with me.

Unless you get Leibneiz notation very well you can't possibly do integration.

Shisui

$\text{d}(uv) = u\text{d}v + v\text{d}u$

Are $u$ and $v$ functions of another variable?

@BalarkaSen

Balarka Sen

@Shisui $u$ and $v$ are variables.

Well, independence doesn't matter.

Sawarnik

@BalarkaSen Because its confusing at best.

Balarka Sen

@Sawarnik It's confusing, yes. But it's an essentially step towards modern analysis.

Shisui

If you can't explain it simply, you don't understand it well enough.

What happened to using words to convey the meaning of and theory behind ideas.

-.-

Balarka Sen

18:10

@Shisui I can explain it simply, not my fault if you forgot how to compute $(a + b)^2$ =P

Shisui

I didn't. I don't understand where you got $(x+\Delta x)^2 - x^2$ from.

The $\epsilon$ too.

Balarka Sen

$\Delta y = f(x + \Delta x) - f(x) = (x + \Delta x)^2 - x^2$

Shisui

That's where you got it from.

Now I just feel plain silly.

Balarka Sen

Heh.

Sorry, but I think I have to go.

Read up Courant.

Bye.

Sawarnik

Bye.

I should go too.

Shisui

18:14

See ya.

I'm staying.

Sawarnik

Lots of irritating things to cram my brain... :(

Shisui

What do you mean?

Hey @cc

c c

hey Shi

Shisui

@cc Could you possibly explain why $\text{d}(uv) = u\text{d}v + v\text{d}u$ ?

I'm aware of the product rule of differentiation but I've only seen it in this form, where $u$ and $v$ are functions of $x$. $$ \dfrac{\text{d}}{\text{d}x} (uv) = u \dfrac{\text{d}v}{\text{d}x} + v \dfrac{\text{d}u}{\text{d}x} $$

Hey @r9m

Could you help me out?

c c

18:46

Product rule

In calculus, the product rule is a formula used to find the derivatives of products of two or more functions. It may be stated thus: :(f\cdot g)'=f'\cdot g+f\cdot g' \,\! . or in the Leibniz notation thus: :\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}. In the notation of differentials this can be written as follows: : d(uv)=u\,dv+v\,du. The derivative of the product of three functions is: :\dfrac{d}{dx}(u\cdot v \cdot w)=\dfrac{du}{dx} \cdot v \cdot w + u \cdot \dfrac{dv}{dx} \cdot w + u\cdot v\cdot \dfrac{dw}{dx}. Discovery Discovery of this rule is credi...

I think this is attributable to Leibniz

well $\text{d} f$ is the differential of $f$, $\dfrac{\text{d}f}{\text{d}x}$ is the derivative with variable x

Shisui

@cc Ok. So here's my burning question for the day. What's the difference between a differential of a function and the derivative of said function with respect to a variable?

c c

for simple functions of a real variable, $f'(a)$ is the derivate in a (a real value), $h \to f'(a).h$ is the differentiate function (a linear function)

for $\Bbb R^n$ the differentiate becomes $Df(a) : h \to grad f(a) .h$

@Shisui this proof of product rule using chain rule is fun: en.wikipedia.org/wiki/Product_rule#Chain_rule

Shisui

19:10

@cc Which one?

The direct consequence of the chain rule?

c c

this one en.wikipedia.org/wiki/… is a typical physician proof

Balarka Sen

19:27

@MikeMiller

Mike Miller

wat

oh well

Balarka Sen

@MikeMiller

Bob

@BalarkaSen what is the difference between small o and big o notation ??

Balarka Sen

If 1 --> N --> K is exact then 1 --> Aut_K(G) --> Aut_N(G) --> Aut_N(K) is exact

c c

Is it possible to have a function almost everywhere null, but with an integral not null?

Balarka Sen

19:40

@Bob f(x) = o(g(x)) if f(x)/g(x) tends to 0

as x tends to infinity i mean.

and f(x) = O(g(x)) if the limit tends to 1

@Bob you got the right kind of question to ask me.

=D

Bob

ya .. and thanks @BalarkaSen

c c

no, I think, f(x) = O(g(x)) is |f(x)/g(x)| <= M for x tends to infinity

Balarka Sen

@cc right. in general, i use ~ so there is mostly no difference for us number theorists.

Bob

i have a function with $O(1/n^2)$ what will be it $o()$

c c

but they are different in analysis

Balarka Sen

19:43

@Bob i can say o(n) =P

as well as o(n^100).

c c

sin(x)/x^2 is O(1/x^2) but not equivalent

Balarka Sen

@cc heh.

c c

@BalarkaSen else could you explain me Galois theory in simple and short terms?

Balarka Sen

pretty much any thing is O and o anything if you can make it up.

c c

:p, always been curious to learn it

Balarka Sen

19:46

@cc Depends if you are a geometer or an algebraist.

Either of them like different explanations.

c c

maybe more geometer, but not good

Balarka Sen

@cc Very well then I guess I'll go for algebraic one.

You know about field extensions?

c c

hmm no

Balarka Sen

ok.

take $\Bbb Q$

then take some equation $x^2 - 2 = 0$.

are the roots in $\Bbb Q$?

c c

no

Balarka Sen

19:49

@cc so adjoin the root. (which we call it $\sqrt{2}$). like have another field $\{a + b\sqrt{2} : a, b \in \Bbb Q\}$

exercise : verify it's a field.

c c

yes it is, multiplication, gives a $\sqrt{2}^2 = 2$

Balarka Sen

cool.

so that's what we call field extension.

having a field $\Bbb Q$ we get a bigger field $\Bbb Q(\sqrt{2})$ by adjunction of an algebraic elt.

@cc You know how to compute automorphism groups of groups?

Bob

@BalarkaSen means you are adding a member with $sqrt{2}times another entry$

c c

but this element must not be contained in the extended field?

Balarka Sen

@cc which element?

c c

19:52

$\sqrt 2$, for example does $\Bbb Q (3/2) (= \Bbb Q) $ make a sense?

Balarka Sen

@cc $\Bbb Q(3/2) = \Bbb Q$

c c

right

Balarka Sen

So I wouldn't call it a field extension/

c c

@BalarkaSen No

Balarka Sen

@cc do you know what's a field automorphism?

google it.

Bob

19:55

@BalarkaSen algebraic elt means ???

c c

A Morphism I remember it a little mathworld.wolfram.com/Morphism.html

Balarka Sen

@Bob something you get by finite composition of addition multiplication and inversion togather with taking roots.

in formal terms, an algebraic over $K$ is a root of a polynomial over $K[x]$.

c c

Ok an automorphism is a bijective morphism with same fields/sets at each 'end'

Balarka Sen

@cc yes, and it preserves operation.

c c

ok

Balarka Sen

19:59

f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b)

c c

it preserves equality and operations

Balarka Sen

right. so now look at the automorphisms of $\Bbb Q(\sqrt{2})$ which fixes elements of $\Bbb Q$ pointwise.

can you identify them?

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