- #71
Pi-is-3
- 49
- 13
@fresh_42 Is my Q15 proof wrong?
Sorry, no, and, yes, it was correct. It just slipped under the many posts about the Brownian motion and I hadn't seen it.Pi-is-3 said:@fresh_42 Is my Q15 proof wrong?
How can this be of degree ##4##, if ##2-\alpha^2=\beta^2\,##?nuuskur said:Oh, I think I understand now, partly..somewhat..hopefully..here we go again
Put [itex]\alpha := \sqrt{1-\sqrt{3}}[/itex] and [itex]\beta := \sqrt{1+\sqrt{3}}[/itex].
By Eisenstein the polynomial is irreducible, thus a minimal polynomial of [itex]\beta[/itex], hence
[tex]
[\mathbb Q(\beta),\mathbb Q] = 4
[/tex]
(a basis is given by [itex]1,\beta,\beta ^2,\beta ^3[/itex]). Now to find [itex][\mathbb Q(\alpha,\beta), \mathbb Q(\beta)][/itex]. Note that [itex]\alpha[/itex] is algebraic over [itex]\mathbb Q\subset \mathbb Q(\beta)[/itex] so by relevant hint in #68
[tex]
[\mathbb Q(\alpha,\beta),\mathbb Q(\beta)] = [\mathbb Q(\beta)(\alpha),\mathbb Q(\beta)] = 4,
[/tex]
so the field extension [itex]\mathbb Q(\alpha,\beta) /\mathbb Q[/itex] is of degree [itex]16[/itex] by the tower lemma. Not sure how to proceed at the moment. Does one observe how the automorphisms behave on the roots?
No, it just means that the minimal polynomial of ##\alpha## over ##\mathbb{Q}(\beta)## is not the same as over ##\mathbb{Q}##. You have a basis ##\{\,1,\beta,\beta^2,\beta^3\,\}## of ##\mathbb{Q}(\beta)## over ##\mathbb{Q}## and ##\alpha^2 \in \mathbb{Q}(\beta)##. So the only question is, whether ##\{\,1,\alpha\,\}## is linear independent over ##\mathbb{Q}(\beta)##, which is a simple linear equation.nuuskur said:Mhh, the hint in #68 is inaccurate, then. I need to check the basics.. again. These side comments are too esoteric.
nuuskur said:I understand my mistake: I missed the [itex]\alpha \in K[/itex] requirement. We don't have [itex]\alpha\in\mathbb Q(\beta)[/itex], hence argument that followed is gibberish.
Put [itex]\alpha := \sqrt{1-\sqrt{3}}[/itex] and [itex]\beta := \sqrt{1+\sqrt{3}}[/itex]. The polynomial [itex]x^4-2x^2 -2[/itex] is irreducible by Eisenstein, hence minimal polynomial of [itex]\beta[/itex], which yields [itex][\mathbb Q(\beta),\mathbb Q] = 4[/itex] with a basis of [itex]\{1,\beta,\beta ^2,\beta ^3\}[/itex]. Observe that [itex]\alpha^2 = 2-\beta ^2\in\mathbb Q(\beta)[/itex] so it suffices to have [itex]\{1,\alpha\}[/itex] to generate [itex]\mathbb Q(\alpha,\beta) / \mathbb Q(\beta)[/itex] and they are linearly independent over [itex]\mathbb Q(\beta)[/itex]. Thus
[tex]
[\mathbb Q(\alpha,\beta),\mathbb Q] = 8.
[/tex]
Per definition, the Galois group is the group of automorphisms on [itex]\mathbb Q(\alpha,\beta)[/itex] that pointwise fix [itex]\mathbb Q[/itex]. If [itex]\tau[/itex] is such an automorphism and [itex]\alpha[/itex] is a root, then [itex]\tau (\alpha)[/itex] is also a root. An automorphism is therefore determined by how it acts on the roots. This has to be (isomorphic to) a subgroup of [itex]S_4[/itex], one of order [itex]8[/itex] to be exact.
One also notices that the polynomial is even so the roots come in pairs: [itex]\alpha, -\alpha[/itex] and [itex]\beta,-\beta[/itex].
The resolvent cubic is reducible and there are two complex roots, so it must be the dihedral group.Math_QED said:I'm really looking for the exact isomorphism structure of the Galois group. There are only 5 groups of order 8 (up to isomorphism). Which one is it?
Why couldn't we have e.g. a parabola, with equal values at the extremes, but a maximum in the middle? Straight lines alone isn't sufficient. Of course it is in the end, but the constant rates have to be shown, either by analytic methods (easier) or by geometric theorems.JosephFG said:Yellow takes away from both red and blue, and the diagram is all straight lines, so, the rate at which yellow replaces blue and at which yellow replaces red are both consistent and constant.
Well, as already mentioned, your idea is correct. Here we simply can calculate the areas and do not need to consider the rate of change:JosephFG said:I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.
Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:
Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.
If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.
nuuskur said:I'm not even sure if any of the above makes sense or if I'm heading the right way.The representatives of the elements in the quotient ring are of the the form
[tex]
a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1},
[/tex]
where [itex]a_j\in \mathbb F_q[/itex], so more compactly, they can be written as n-tuples [itex](a_0,a_1,\ldots, a_{n-1})[/itex]
If [itex]e^1,\ldots, e^n[/itex] is a basis for [itex]\mathbb F_q^n[/itex] then some selections of the basis elements span cyclic subspaces, for instace the entire space itself is of course cyclic. Denote the basis elements as follows [itex]e^j := (e^j_1,\ldots,e^j_n)[/itex].
Suppose [itex]C := \left\langle e^{j_1},\ldots, e^{j_k}\right\rangle,\ k\leq n[/itex] is a cyclic subspace. Now, considering the basis elements as polynomials we can consider the ideal generated by them (if [itex]R[/itex] is a commutative ring, then [itex]r_1R + \ldots + r_kR \subseteq R[/itex] is an ideal) and map [itex]C[/itex] to that ideal.
The subset of cyclic subspaces is a sublattice, because it's closed w.r.t [itex]\cap, +[/itex]. Is there an isomorphism (of lattices) between this sublattice and ideals of the quotient ring?
JosephFG said:I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.
Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:
Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.
If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.
Physics lover said:My solution to 11.
View attachment 246120
By the way i wanted to know whether can we solve the same integral but indefinite.If we can please provide a solution.
I agree but for the constant C which is not needed since we have here the definite integral ##\int_0^{\pi /2} \, dx = \pi /2##.Physics lover said:My solution to 11.
View attachment 246120
By the way i wanted to know whether can we solve the same integral but indefinite.If we can please provide a solution.
TGVF said:I agree but for the constant C which is not needed since we have here the definite integral ##\int_0^{\pi /2} \, dx = \pi /2##.
No idea for the same integral I but indefinite form,.
TGVF said:Equation can be rewritten as: ...
nuuskur said:I think #86 is not sufficient. We still need the other direction: image of a cyclic subspace is an ideal or else there are potential well-definedness problems of the supposed isomorphism of lattices. Right now, we have
[tex]
I \text{ ideal} \implies T^{-1}(I) \text{ cyclic}
[/tex]
We'd also need
[tex]
C\text{ cyclic} \implies T(C) \text{ ideal}
[/tex]
Put
[tex]
T : \mathbb F_q^n \to \mathbb F_q[x] / (x^n-1) =:R,\ (a_0,\ldots, a_{n-1}) \mapsto a_0 + \sum _{k=1}^{n-1} a_kx^{k}
[/tex]
Let [itex]C \subseteq \mathbb F_q^n[/itex] be cyclic. We show [itex]T(C)[/itex] is an ideal. As [itex]C[/itex] is a subspace and [itex]T[/itex] is compatible with addition, [itex]T(C)[/itex] is a subgroup. Take [itex]r_0 + \sum_{k=1}^{n-1}r_kx^k\in R[/itex] and [itex]a_0 + \sum_{k=1}^{n-1}\in T(C)[/itex]. We must show their product is in [itex]T(C)[/itex].
[tex]
\begin{align*}
&\left (r_0 + \sum_{k=1}^{n-1}r_kx^k\right ) \left (a_0 + \sum_{k=1}^{n-1}a_kx^k\right ) \\
=&r_0\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1}\right ) \\
+&r_1\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1} \right )x \\
+&r_2\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1}\right ) x^2 \\
&\vdots \\
+&r_{n-1} \left ( a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1} \right )x^{n-1}.
\end{align*}
[/tex]
As [itex]C[/itex] is a subspace, it is closed w.r.t multiplying by [itex]r_k[/itex]. We also saw in #86 that multiplying by [itex]x^k[/itex] shifts the coefficients, but [itex]C[/itex] is cyclic, thus closed w.r.t shifting. All of the additives are in [itex]T(C)[/itex], therefore [itex]T(C)[/itex] is an ideal.
Wow. That is the simplest proof for that question!fresh_42 said:My proof uses ##u(1-u)=\frac{1}{4}-(u-\frac{1}{2})^2< \frac{1}{4}## since squares are positive.
##\leq## as u=1/2 is possible - squares can be zero. I didn't go into detail with that part as it is very easy to show and a well-known result, too.fresh_42 said:My proof uses ##u(1-u)=\frac{1}{4}-(u-\frac{1}{2})^2< \frac{1}{4}## since squares are positive.
I think that's just a typo.mfb said:##\leq## as u=1/2 is possible - squares can be zero. I didn't go into detail with that part as it is very easy to show and a well-known result, too.
Hey, I was lazy: "<" is on the keyboard ##"\leq"## is not. These details are trivial and left to the reader ;-)mfb said:##\leq## as u=1/2 is possible - squares can be zero. I didn't go into detail with that part as it is very easy to show and a well-known result, too.
That reminds me of a statistics course lecture. The lector said something like "..and since ##f## is injective, it is invertible..". Similarly, in algebra, saying something is a subgroup of ##G## is to be read as "is an isomorphic copy of a subgroup of ##G##".fresh_42 said:Hey, I was lazy: "<" is on the keyboard ##"\leq"## is not. These details are trivial and left to the reader ;-)
or omitting "almost everywhere" in measure theory.nuuskur said:That reminds me of a statistics course lecture. The lector said something like "..and since ##f## is injective, it is invertible..". Similarly, in algebra, saying something is a subgroup of ##G## is to be read as "is an isomorphic copy of a subgroup of ##G##".