Suppose $F_{p^k}$ is a finite field. If $F_{p^{nk}}$ is some extension field, then the primitive element theorem tells us that $F_{p^{nk}}=F_{p^k}(\alpha)$ for some $\alpha$, whose minimal polynomial is thus an irreducible polynomial of degree $n$ over $F_{p^k}$.
Is there an alternative to show...
As I read another question and R[x] had a counter example, which made me ask here whether the statement "no root implies irreducibility" hold in finite fields?
@PedroTamaroff No. I think there is one [that might include non-elementary functions], but I don't know it. (My memory is terrible, I'm happy that I can remember the closed form of $\sum \frac{z^n}{n!}$.)
Why is correlation in probability given by a dot product? I understand you are asking how much of one set of data is projecting onto another set but I don't see why projecting one set of data points onto another (as vectors) says anything about correlation of the data :(
For a given function $f$, the functional derivative of the functional $\mathcal{F}[\rho]=\int f(x,\rho(x))\,dx$ is well-known to be $\frac{\delta}{\delta \rho(x)}\mathcal{F}[\rho]=\frac{\partial f}{\partial \rho}.$Suppose I instead have a functional of the form $$\mathcal{G}[\rho]=\int \int^x g(x...
As far as I can tell, nobody has written down a proof of the Iwasawa decomposition for general Lie groups for decades. It's certainly not in any of the standard textbooks.
@bobby: I may end up posting a variation on it: Compute the functional derivative wrt to $\rho(x_0)$ of the iterated integral $$\int \int^{x_n}\cdots\int^{x_1} f(x_0,x_1,\cdots,x_n,\rho(x_0),\rho(x_1),\cdots,\rho(x_n))\,dx_0\,dx_1\,\cdots\,dx_n$$
Greiner says the chain rule holds verbatim, also he shows you how to work the integrals out by hand, it's definitely worth working through the 4 pages on functional integrals
I will save it and try when I finally sit down and write this shit up :)
Anyone have any intuition on why correlation of a set of random data points is given by a dot product of normalized vectors whose entries are the data points? Let alone how that relates to using Path integrals to test for correlation hehe
I understand you are asking how much of one set of data is projecting onto another set but I don't see why projecting one set of data points onto another (as vectors) says anything about correlation of the data :(
If $\zeta$ is an $n$-th root of unity, why is it that $\Bbb Z[\zeta]$ is only a PID for $n = 1,2,3,4,6$ (i.e. normal integers, Gaussian integers, Eisenstein integers)? Does it have something to do with the whether the $n$-gon tesselates?
@DanielFischer If $f(z)$ as an essential singularity at $c$, I am trying to find the kind of singularity of $e^{f(z)}$ at $c$. If $f$ had a pole, this would be an essential singularity.
@PedroTamaroff Incomplete. That $\exp$ is zero-free does not imply that a function $e^{f(z)}$ cannot have a continuous extension with a zero. It can't happen if $f$ is holomorphic (or meromorphic), but remember $\exp(-\lvert z\rvert^{-1})$.
@PedroTamaroff It suffices that it is bounded below. Then $f$ can have neither pole nor essential singularity at $c$.
But, we're in a slightly different situation, if $h(z) = e^{f(z)}$ in a punctured neighbourhood of $c$, then we still need to prove that it is impossible that $h$ can be continuously extended into $c$ by setting $h(c) = 0$.
@DanielFischer OK. I'll start over: suppose $f$ has an essential singularity at $c$. Then $e^f$ cannot have a removable singularity at $c$. Say $e^{f(z)}=h(z)$ with $h$ holomorphic every where. So yes, @DanielFischer, we still have to sort out that $h(c)\neq 0$.
@Semiclassical I was confused by how you said you had evaluated $\int_{0}^{q_{0}} \sqrt{\tan^2{q_0}-\tan^2q} \ dq$ by using the residue at infinity because the integrand doesn't have a Laurent expansion at infinity. But Daniel Fischer pointed out that you probably first made a substitution. Is that what you did?
@DanielFischer OK, so $\infty$ cannot be removable since $f$ is not constant. It suffices I see it is not essential, and then it will be a pole, so $f$ is a polynomial.
@DanielFischer Because say we could choose $z_i\to 0$ with $f(1/z_i)\to 1$, and this would give with $w_i=f(1/z_i)$ that $f^{-1}(w_i)=1/z_i$, so that $f$ would not be holomorphic at $1$.
@PedroTamaroff $f^{-1}$ would not be holomorphic at $1$. But maybe it's nicer to say that $f(\mathbb{D})$ is a nonempty open set that doesn't intersect $f(\mathbb{C}\setminus \overline{\mathbb{D}})$.
Hence the latter isn't dense, and $\infty$ not essential by Casorati-Weierstraß.
@Semiclassical A couple of weeks ago on chat you said you evaluated $\int_{0}^{q_{0}} \sqrt{\tan^2{q_0}-\tan^2q} \ dq$ by using the residue at infinity. I wanted to ask you at the time, but I didn't. chat.stackexchange.com/transcript/36?m=18463371#18463371
Let $P_n(z)=1+z/2!+\cdots+z^n/(n+1)!$. I have to show that given $R>0$, I can take $n_0$ such that $P_n(z)$ has no real zeros of magnitude $<R$ for $n>n_0$.
What comes up instead is just a problem of integrating a rational function of sines/cosines over one period, which is handled in the usual way by residues
@PedroTamaroff For this specific $f$, since we know that $f(x) > 0$ for all $x\in\mathbb{R}$, generally, it's a compact set not containing $0$, so there is a $\delta > 0$ with $\lvert f(z)\rvert \geqslant \delta$ on $[-R,R]$.
In my last question ( the eureka theorem ) I asked a question about the proof in a comment ... barely noticeable I guess ... If someone would answer that would be appreciated !
P.S. @DanielF: I suspect that that OP who posted the question about holomorphic had a teacher whose definition is just complex differentiable. It's still ok to have the theorem holomorphic on a region $\Omega$ $\iff$ analytic on $\Omega$.
@Pedro: One of my advisees just found out he can't graduate (again) because his computer science class won't accept his programs if he skips class ... Guess what.
Just like their latest improvement to the OS screwed up my TeX fonts in typeset documents (not in Reader, though), their latest improvement to the OS for the iPad and iPhone has messed up Safari big time.
I was just carping about attendance policies ... since @Pedro rarely attends :D
But for my advisee, it's too late, @Mike. I'm telling him to investigate completing a CS course somewhere else and transferring it in. I think he can still graduate that way.
@DanielFischer Sure. But say, here in this sheet I have $2z^5+7z-1$. Rouche says that if $f,g$ are holomorphic on some domain $D$, $C$ is some countour on $D$ and $|f(z)|>|g(z)|$ on $C$; then $f$ and $f+g$ have the same number of zeros on $C$. Now consider $|z|<1$. Then I can take $f(z)=7z-1$ and $g(z)=2z^5$.
not necessarily, @Teddy. You need to make the simplices smaller, so that they are contained in evenly-covered neighborhoods. That was the essence of my hint yesterday.
@RandomVariable forgot to mention, that integral arose from my answer to a classical mechanics question, where i'd indicated how to correctly write the action variable as that integral
@Studentmath They've been advertising forever. But not on Mathematics, at least so far they haven't. Except the non-commercial community ads in the side-bar, they always had those.
Suppose $F_{p^k}$ is a finite field. If $F_{p^{nk}}$ is some extension field, then the primitive element theorem tells us that $F_{p^{nk}}=F_{p^k}(\alpha)$ for some $\alpha$, whose minimal polynomial is thus an irreducible polynomial of degree $n$ over $F_{p^k}$.
Is there an alternative to show...
