The Galois Group $Gamma(mathbb{Q}(sqrt[n]{2}):mathbb{Q})$ is trivial if $n$ is odd.

The full question is: Given a group $F_n=mathbb{Q}(sqrt(n){2})$ then prove that:

  • (A) If $n$ is odd then $Gamma(F_n:mathbb{Q})$={$rm{id}$} and
  • (B) If $n$ is even then $Gamma(F_n:mathbb{Q})congmathbb{Z_2}$

I am confused as to how to find either, I know that $F_n:mathbb{Q}$ is a non-normal extension. We were given the hint:

"For any $tauinrm{Aut}_mathbb{Q}(F_n)$ we must have that $tau(sqrt(n){2})in F_nsubseteqmathbb{R}$ is again a real $n$-th root of 2"

fitting – (Non)LinearModelFit hangs on “trivial” linear fit (Fit succeeds)

A simplified version of my function and data:

    myfunc(x_, a_) = 8*a*Exp(-8/Sqrt(x))/(Sqrt(x)*Abs(WhittakerW(E*I/Sqrt(x), 1/2, 8/10*I*Sqrt(x)*a))^2)
    mydata = {{4.5, 195}, {2.9, 175}, {2.1, 95}}

myfunc also needs a global rescaling. If I want to fit both parameter a and the scale, NonLinearFitModel works great out of the box and quickly:

    NonlinearModelFit(mydata, scale*myfunc(x, a), {scale, a}, {x})

If, however, I only want to find the best scaling for an a value of my choice, this:

   NonlinearModelFit(mydata, scale*myfunc(x, 4), {scale}, {x})

takes forever in my computer (I aborted after a while, I don’t know whether it would succeed at some point or not). I am surprised: why the “complicated” fit works so easily, and this “trivial” linear one not?

LinearModelFit is supposedly the right tool here, but:

   LinearModelFit(mydata, {myfunc(x, 4)}, {x}, IncludeConstantBasis -> False)

hangs in the same way. There must be something wrong I am doing.
I tried using 4.0 instead of 4 for a, enclosing with an N() or putting all constants , adding WorkingPrecision -> MachinePrecision, but I get the same result.

NonlinearModelFit works adding Method -> "NMinimize", although it is slow, and raises a warning “You asked for a non-linear method, note that the model is linear”. I do not really know what NMinimize is doing here, I guess it applies the “right” simplification to myfunc, so the computation is easier?

Finally, after wasting one day, Fit just works:

   Fit(mydata, {myfunc(x, 4)}, {x})

This reinforces my impression that I just need to apply some magical simplification, that Fit always employs, that NonlinearModelFit uses only for the two-parameter fit, and that Method -> "NMinimize" will cause. Is LinearModelFit evaluating myfunc much more times than nedeed? I am clueless.

LinearModelFit has several niceties that I would rather not give up for what appears to be such a stupid issue. Furthermore, I am probably going to learn a new important Mathematica quirk from this. Thus:

Why the above code, using (Non)LinearModelFit for a simple scale*f(x) model, hangs, and how should I modify it to make it work?

group theory – Non trivial homomorphisms $mathbb{Z}/3mathbb{Z} rightarrow text{Aut}(mathbb{Z}/7mathbb{Z})$

I have to find non trivial homomorphisms $varphi$, $varphi’$: $mathbb{Z}/3mathbb{Z} rightarrow text{Aut}(mathbb{Z}/7mathbb{Z})$. We know that Aut $(mathbb{Z}/7mathbb{Z}) cong (mathbb{Z}/7mathbb{Z})^times cong mathbb{Z}/6mathbb{Z}$. So we have to look order of elements of $mathbb{Z}/6mathbb{Z}$ that divide $3$. These elements are $(2)_6$ and $(3)_6$. In corrections these homomorphisms are defined as: $Big(varphi((1)_3)Big)((i)_7) = (2i)_7$ and $varphi’$:$Big(varphi((1)_3)Big)((i)_7) = (4i)_7$. The thing that i don’t understand, is what $(i)_7$ represents. For me, if we generalize for example $varphi$, we got $varphi((r)_3) = (2^r)_7$. I read a lot of articles on ”research” of homomorphisms, and i feel confused right now and can’t get an intuition for this. Thanks in advance for help.

real analysis – Nontrivial signed measure on Lebesgue measurable sets being trivial on Borel sets

Let $mathfrak{L}(mathbb{R})$ be the collection of Lebesgue measurable sets and $mathfrak{B}(mathbb{R})$ be the Borel sets.

Question: Is there a nontrivial signed measure on $mathfrak{L}(mathbb{R})$ that is trivial on $mathfrak{B}(mathbb{R})$?

Obviously, any positive measure that is trivial on $mathfrak{B}(mathbb{R})$ is also trivial on $mathfrak{L}(mathbb{R})$, since any Lebesgue measurable set is a subset of a Borel set.

For the signed case, I have tried doing Jordan decomposition but it doesn’t seem work. It is hard (if ever possible) to show $(mu|_{mathfrak{B}(mathbb{R})})^+ = mu^+|_{mathfrak{B}(mathbb{R})}$ and $(mu|_{mathfrak{B}(mathbb{R})})^- = mu^-|_{mathfrak{B}(mathbb{R})}$.

Background: I am trying to prove (or disprove) that if $mu$ and $lambda$ are measures on $mathfrak{L}(mathbb{R})$, then $mu|_{mathfrak{B}(mathbb{R})} = lambda|_{mathfrak{B}(mathbb{R})}$ implies $mu = lambda$.

Is any periodic function trivial

Say I define

xi(s) = xi(1-s)

Since this function is just a lagged copy of itself, over some period, isnt anything which holds for a period true of all other periods?

ag.algebraic geometry – When is a twisted form coming from a torsor trivial?

Consider a sheaf of groups $G$, equipped with a left torsor $P$ and another left action $G$ on some $X$. Form the contracted product $P times^G X := (P times X)/sim$ where $sim$ is the antidiagonal quotient: $(g.p, x)sim (p, g.x)$.

Q1: When is $Ptimes^G X$ trivial? I.e., when do we have an isomorphism $P times^G X simeq X$?

Partial answer: $P times^G X simeq X$ over $(X/G)$ iff $P times (X/G)$ is a trivial torsor over the stack quotient $(X/G)$.

Proof: We can rewrite $P times^G X$ as a contracted product of two torsors $(P times (X/G))times^G_{(X/G)} X$. Then we contract with “$X^{-1}$” — the inverse to contracting with $X$ as a torsor over $(X/G)$ and we win. (as in B. Poonen’s Rational Points on Varieties, section

Am I allowed to do this? This argument probably shouldn’t have to appeal to algebraic stacks and may be somewhat dubious.

Q2: If I have one isomorphism $P times^G X simeq X$, can I choose another one that lies over $(X/G)$? Or at least is $G$-equivariant?

Q3: Is there a natural way to write the triviality of such a twisted form?

I first thought $P times^G X simeq X$ iff $P$ was trivial, which is clearly false for trivial actions on $X$. Then I was excited to have the pullback $* to BG$ represent triviality of the twisted form $P times^G X$ as well as the torsor $P$. Is there a natural representative of the sheaf of isomorphisms between $P times^G X$ and $X$?

These can all be sheaves, although I’m primarily interested in $G = GL_n, PGL_n, SL_n$, etc. acting on $X = mathbb{A}^n, mathbb{P}^n$ as appropriate. More ambitious is $G = text{Aut}(X)$ for even simple $X$. I’d be happy with answers in any level of generality.

Due Diligence Statement: I’m a novice in the area of “twisted forms” of varieties, so I apologize if the above is evident or obtuse. I checked all the “similar questions” listed here and couldn’t find an answer.

ordinary differential equations – $lambda$ values of $y”-(frac{1}{4}+frac{lambda}{x}).y=0$ such that there is a non trivial solution

What are the $lambda$ values such that $$; y”-(frac{1}{4}+frac{lambda}{x}).y=0 ;;; 0<x<infty$$ $$ y(0)=0 ; ; ; lim_{xrightarrowinfty}y(x)=0$$
have a non trivial solution.

Trying to use the Frobenius method, I found that $; (s^2+s-lambda).a_1=0;$ and the recurrence relation $; a_k=frac{-1}{k.(k-1)}.(frac{1}{4}.a_{k-2}+lambda.a_{k-1});$. These results that I got are correct ? How can I get the answer by these results ?

Complex conjugation inducing a trivial map on the fundamental group

Let $V$ be a smooth projective complex variety defined over the rationals such that $G=pi_1(V)$ is a non-abelian finite simple group. Can the map $Gto G$ induced by complex conjugation be trivial?

memory – Is it trivial to protect from double free just by LD_PRELOADing a custom malloc/calloc and free?

Can’t one just implement a malloc/calloc wrapper that adds the returned pointer address to a global hash table prior to returning, and then a free wrapper that checks for the presence of the pointer in the table prior to freeing (returning early if it isn’t present), and then LD_PRELOAD these malloc/calloc and free functions with a program like Firefox, in order to protect from double frees? Is there a reason why the standard malloc/calloc and free functions don’t use such a technique, or why there isn’t a secure variant that is suggested similarly to how strcpy_s is suggested in place of strcpy?

Generation of trivial cofibrations of the Bousfield localization

Assume $ mathfrak {M} $ is a category of appropriate celluar model left and $ S $ is a set of cofibrations $ mathfrak {M} $. What are the trivial cofibrations that generate $ L_S mathfrak {M} $? Are they $ J cup S $, or $ J $ is all about generating trivial cofibrations of $ mathfrak {M} $?