@BalarkaSen Actually, just scratching around with pen and paper leads me to believe that the problem can be solved, on the domain where it actually makes sense (that is, for $y \geq 11/4$ where $y = \times^2+\times+3$). There's some unexpected magic going on here, which probably means there's a simpler solution.
Why would adopting the given convention make every complex number $(x,y)$ have a unique form $x+iy$? I mean, I don’t see what would be different if we didn’t adopt the convention? Would $\alpha\cdot(x,y)$ then just be undefined?
No. My objection was that you had failed to prove that it didn't matter which $y$ you chose to put into $g(y)$ if you had $x$ and you had $x = y^2+y+3$.
Not totally obvious I'd say, but it is, actually! If $z$ is a complex number, $e^z = \sum_{k = 0}^\infty z^k/k!$, so it's "absolutified" series is $\sum_{k= 0}^\infty |z^k/k!| = \sum_{k = 0}^\infty |z|^k/k!$, but that's just $e^{|z|}$!
the problem is my book defines $e^{z=x+iy}=e^x\cdot e^{iy}$. And then they show that $e^{iy}=\cos y+i\sin y$, assuming "reasonable convergence", like wtf.
I feel confused, even though I know what they're doing. I don't like what they're doing. anyhow, #endrant
o, and they define $e^{iy}=\cos y+i\sin y$
yea, alright it's ok. I was just annoyed for a sec
(I should have written they show that $\sum_{j=0}^\infty\frac{(iy)^j}{j!}=\cos y+i\sin y$, assuming reasonable convergence)
Hi, if $v_1$ is the only eigenvector of $A$ with eigenvalue $\lambda$, but the multiplicity of $\lambda$ is say $m=2$, then why does $(A-\lambda I)v_2 = v_1$ have a solution?
$v_2$ should then be the generalized eigenvector
I know that $(A-\lambda I)^2 = O_\text{mat}$. Does this help me somehow?
How would you show that $(A- \lambda I)v \neq 0$? When $v$ is orthogonal to $v_1$ then it's easy, since $v_1 \in N(A- \lambda I)$ it must be $v \notin N(A- \lambda I)$.
I'll try something. Since $v_1 \neq 0$ we need to have $v_2 \notin N(A-2I)$ to begin with. It may not exist but if it does, then it must not be in the nullspace. Now $v_1$ is in the nullspace $N(A-2I)$. So we want to show that the linear map $(A-2I): N(A-2I)^\perp \to N(A-2I)$ is surjective so that for every $v_1 \in N(A-2I)$ there exists $v_2 \in N(A-2I)^\perp$ such that $(A-2I)v_2=v_1$. By some theorem this happens if the map is injective.
A map is injective if its kernel is trivial. But we know that $N\left((A-2I)|_{N(A-2I)^\perp}\right)=\{0\}$ since only $x \in N(A-2I)$ are mapped to zero which we excluded with the complement.
Hi there, Any suggestions for a good resource(s) for function norms? I've just found out vector norms and matrix norms. No information is found in Calculus textbooks I've read.
I guess the connection here is that, in the quaternions, if $w$ is complex (no $j$ or $k$ component) then $wj=j\bar w$.
So $-w\bar z=w(jj)\bar z=(wj)(zj)$
and if $w=w_1+w_2i$, then the latter is $(w_1j+w_2k)(z_1j+z_2k)$
and then the thing I just said about pure imaginary quaternions happens
and the cross product is in the $i$ direction, with two explanations: $w\bar z$ is obviously complex, and the cross product is perpendicular to $w_1j+w_2k$ and $z_1j+z_2k$.
(The real part of $qr$ is the negative of the dot product, sorry)
Can someone give me a pointer how one would go about computing an expectation like $E[ \min { i | \Sum_{k=0}^i Z_k >= \gamma ]$ (where $Z_k$ iid Bernoulli)? Basically the expectation after how many steps the sum of iid random variables exceeds a threshold. I'm thinking this has something to do with "random walks", but am missing the specifics.
Problem: If $\{f_n\}$ is a sequence of measurable functions on the measurable set $E$ that is Cauchy in measure on $E$, then there is a measurable function $f$ such that $f_n \to f$ in measure. Attempt: I have already been able to show that there is a subsequence $\{f_{n_j}\}$ such that $f(x) := \lim_{j \to \infty} f_{n_j}(x)$ exists for each $x \in E$. I don't know how to use this to show that $f_n \to f$ in measure, though...
I was thinking of using the fact that $$\{x \in E : |f_n(x)-f(x)| \ge \eta\} \subseteq \{x \in E : |f_n(x) - f_{n_j}(x)| \ge \eta \} \cup \{x \in E : |f_{n_j}(x) - f(x) | \ge \eta \}, $$ but that would require showing that $f_{n_j} \to f$ in measure...
why do you expect anything natural in this case? I mean, anything natural would make use of the topology on $\Bbb R$, but if you don't include that information in the definition on the set, why should there be a topology that cares for the topology on $\Bbb R$
the natural topology for a purely set-theoretic object is the discrete toplogy
If I have a tower of field extensions $\Bbb Q\subseteq K_1\subseteq \cdots\subseteq K_n$ with $|K_i:K_{i-1}|=2$ for all $i$ (so that $|K_n:\Bbb Q|=2^n$) is there a simple explanation of why the minimal polynomial of $\alpha$ has only terms of even degree where $\alpha$ is a primite element of $K_n$?
@AlessandroCodenotti That's not true. Consider $\Bbb Q \subset \Bbb Q(\sqrt{5}) \subset \Bbb Q(\zeta_5)$. The minimal polynomial of $\zeta_5$ is $x^4+x^3+x^2+x+1$
@BAYMAX not sure what you mean by that. But you can write curl using the Levi-Civita symbol as $$\text{curl }\vec{A} = \sum_{i,j,k=1}^3 \epsilon_{ijk}\,\hat{x}_i \frac{\partial}{\partial x_j} A_k$$
the counterexample can be generalized: Suppse $K/ \Bbb Q$ is Galois such that $\operatorname{Gal}(K/\Bbb Q)$ is cyclic of order $4$, then the minimal polynomial of a primitive element can't have only terms of even degree
or in index form $(\text{curl }\vec{A})_i = \sum_{j,k} \epsilon_{ijk} \partial_{j} A_k$ where I've used the subscript form of $\partial/\partial x_j$ b/c I can't be arsed
Does this argument make sense? I have a tower $$\Bbb Q\subseteq \Bbb Q(i)\subseteq \Bbb Q(i,\sqrt{2})\subseteq \Bbb Q(i,\sqrt{2},\sqrt{3})\subseteq \Bbb Q(i,\sqrt{2},\sqrt{3},\sqrt{5})=K$$ and I know that degree $2$ extensions are Galois. The two automorphism of $K$ fixing the third extension are the identity and $\sqrt{5}\mapsto-\sqrt{5}$ so those are also in $\operatorname{Gal}(K/\Bbb Q)$.
By constructing the tower in a different order I get that $\sqrt{2}\mapsto -\sqrt{2}$ is also an authomorphism of $K$ and so on for the other roots. Composing those I get $16$ authomorphisms, hence the extension is Galois
Guys, is there a reason why my book defines the holomorphic property only for continuously differentiable functions, and not just simply differentiably functions?
you have to be a bit careful with extending automorphisms, but yeah one can make this work. But even computing the degree of $\Bbb Q(i,\sqrt{2},\sqrt{3},\sqrt{5})$ to be $16$ is a bit annoying if you do it without fancier machinery like discriminants
This whole mess is justified because I want to calculate the minimal polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}+i$ but I want to do so by showing that its roots must be $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}\pm i$ and using the expressions of the coefficents as symmetric functions of the roots
What do you mean with composite? Because I know that towers of Galois extensions don't necessarily mean that the last term of the tower is Galois on the first
@MatheinBoulomenos Because it's an exercise in my algebraic number theory pset and I'm looking for the least painful way to do it, I'd just throw it at a computer obviously if it were for me!
so if you have two fields $K_1$ and $K_2$ which lie inside a bigger field $F$, then the composite of $K_1$ and $K_2$ is the smallest subfield of $F$ that contains both $K_1$ and $K_2$
you can construct this e.g. by adjoining all elements of $K_1$ to $K_2$ or vice versa
if $k$ is a subfield of both $K_1$ and $K_2$ and $K_1=k(a_1, \dots, a_n)$ and $K_2=k(b_1, \dots, b_k)$, then $K_1 \cdot K_2$ (the composite) is $k(a_1, \dots, a_n, b_1, \dots, b_k)$
@ShaVuklia It's slightly harder to prove everything with the latter. Eg, if you want to prove Cauchy's theorem that $\oint_D f(z) dz = 0$ where $D \subset \Bbb C$ is a simply connected domain for a holomorphic function $f$, then you can directly invoke Gauss-Stokes theorem directly (because $f = u(x, y) + iv(x, y)$ is holomorphic implies the real and imaginary parts of $f'$ are curl-less vector fields - that's the content of the C-R equations).
But Gauss works only if $f$ is $C^1$.
If you don't assume it's $C^1$ you have to go through Goursat's lemma, which is proving Cauchy's theorem on a triangle using a Sierpinski subdivision trick.
@AlessandroCodenotti oh if it's for ANT, then you might actually use discriminants to simplify stuff like proving that $\sqrt{5}$ isn't already contained in $\Bbb Q(\sqrt{2},\sqrt{3})$
@MatheinBoulomenos Oh, definitely, showing that the degree of the extension is $16$ isn't a problem there, I was more wary of my argument above about extending automorphisms
@Semiclassical I started with the expectations for specific gammas <0, <1, <2 and then generalized from there. The expectation was $p^\gamma \sum_{n=0}^\infty \binom{n+\gamma-1}{n} (n+\gamma) (1-p)^n$ (where $\gamma$ is rounded up to the next integer), which reduces to just $\gamma/p$.
@AlessandroCodenotti the extension stuff works because of this lemma: suppose $K$ and $F$ are fields and $f:K \to F$ is a field homomorphism and $f$ is a polynomial that is irreducible over $K$ and $\alpha$ is a root of $f$ contained in some extension of $K$, then for any element root $\beta$ of $f$ inside $F$, there is a unique homomorphism $\sigma:K(\alpha) \to F$ which extends the given homomorphism such that $\sigma(\alpha)=\beta$
In general, this lemma can be horrible to work with when you have towers, because you don't just need irreducibility over, say $\Bbb Q$, but over extensions of that, but in this case, the polynomials are degree $2$ for which irreducibility is easy even over larger fields
yeah
I didn't want to give the field homomorphism $K \to F$ a name actually
@AlessandroCodenotti do you know about divisibility relations of discriminants in towers of extensions? (there's also a formula for composite fields). You can show e.g. that $\sqrt{3}$ is not contained in $\Bbb Q(\sqrt{2})$ because $\Bbb Q(\sqrt{3})$ and $\Bbb Q(\sqrt{2})$ have different discriminants. And the discriminant of $\Bbb Q(\sqrt{2},\sqrt{3})$ has as only prime factors $2$ and $3$, so $\Bbb Q(\sqrt{5})$ can't be a subfield of that since that has discriminant $5$
Actually in this example why can't I use my approach of starting with automorphisms of the biggest field rather than extending automorphisms of the intermediate ones? I don't think I need this lemma
@MatheinBoulomenos Do you have a reference for that kind of stuff? I'd like to read more about it, in our course we did very little about discriminants and then moved on to Dedekind's theorem and showing that $\mathcal{O}_K$ is a Dedekind ring
if $L$ is an extension of a number field $K$ and $\mathfrak{P}$ is a prime of $L$ lying over the prime $\mathfrak{p} \subset \mathcal{O}_K$, how can I argue that the extension $\Bbb F_{\mathfrak{P}}/\Bbb F_{\mathfrak{p}}$ is Galois?
Anti Symmetrization bracket - Suppose $\theta$ is an object which are dependent on $m_{1},m_{2},..,m_{f}$, In other words $\theta = [\theta_{m_{1},m_{2},..,m_{f}}]$ then $\theta_{[m_{1},m_{2},..,m_{f}]} = \frac{1}{f!} \sum_{\pi \in S_{f}} (Sgn \pi) \theta_{m_{\pi_{1}},m_{\pi_{2}},..,m_{\pi_{f}}}$
It's easy when $K = \Bbb Q$ because then $\Bbb F_{\mathfrak{P}}$ is the splitting field of $X^{p^{f(\mathfrak{P}/p)}} - X$ where $f(\mathfrak{P}/p)$ is the inertial degree of $\mathfrak{P}$ over $p$, but does a similar argument work when $K \neq \Bbb Q$?
@BAYMAX not sure what you mean by that. But you can write curl using the Levi-Civita symbol as $$\text{curl }\vec{A} = \sum_{i,j,k=1}^3 \epsilon_{ijk}\,\hat{x}_i \frac{\partial}{\partial x_j} A_k$$
Where $\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1$, $\epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1$, and zero otherwise
You can convert that into your permutation language without much work.
I know that there is one switching $\pm\sqrt{5}$ because I know it's an extension of degree two (let's assume this is known/proved) over the previous field so it's also Galois and I know its automorphisms fixing the previous field
But it also an extension of degree two over $\Bbb Q(\sqrt{2},\sqrt{5},i)$ and that gives me an automorphism switching $\pm\sqrt{3}$
I get $4$ automorphisms, each switching a root with its opposite in this way, and composing those together I get 12 more
But saying that there is an isomorphism that fixes $\Bbb Q(\sqrt{2},\sqrt{5},i)$ that switches $\pm \sqrt{3}$ is the same thing as extending the identity on that field in a certain way
So it's really the same argument
Except that if you allow yourself to extend stuff that is not the identity, you only need one tower
And I think all the computations you need to do for the extending are already necessary if you just want the degree
If I were a grader, I wouldn't accept that. If you do everything without machinery, then the computations become a lot more ugly once you have more than 2 square roots
@AlessandroCodenotti Kummer theory describes extensions (and their Galois groups) that are given by adjoining some square roots under suitable assumptions which are all satisfied here
@MatheinBoulomenos I generally don't subtract point for a correct answer, even if it uses less theory than was intended. If that is possible, then it was my fault for not making the question suitable
What is homomorphism thaat makes $C_2\times C_2$ homomorphic image of $Z\times Z$ here? First /i thought that (even, even) should map to 0 and (odd, odd) should map to 1, but then what to map (1,2) to?
@AkivaWeinberger $X$ $\sigma$-compact metric (maybe works with a uniform space), $\mathcal F\subset C(X)$ equibounded and equicontinuous, then any sequence in $\mathcal F$ subconverges locally uniformly
