hello. I'm trying to find the limit as h goes to 0 of: [ 1 - cos (3h) ] / [ cos^2 (5h) - 1 ] .... I tried applying the identity sin^2 x + cos^2 x = 1 but I do not seem to be arriving anywhere. Could someone please give me a nudge in the right direction?
In this answer http://mathoverflow.net/a/65441/50868, why can we say that $X$ is diagonal, and why can we say that $\prod_{i=1}^{n}(1 + \lambda_{i}) \geq 1 + \prod_{i=1}^{n} \lambda_{i}$ ?
@SwapnilTripathi, try to prove that a polynomial $f$ is separable $\iff$ $f$ and $f'$ have distinct roots. That is, $f$ and $f'$ are relatively prime to each other.
'It suffices to prove that if X is positive definite and Hermitian, then det(I+X)≥(1+detX). We may conjugate X by a unitary matrix U and assume that X is diagonal.'
@KajHansen I am just making a wild guess as it is all that comes to my mind. f and f' are both polynomials over some field K. So their gcd is also a polynomial over K? And that makes f reducible.
@TedShifrin Could you tell me why in mathoverflow.net/a/65441/50868 we can say that $X$ is diagonal ? 't suffices to prove that if X is positive definite and Hermitian, then det(I+X)≥(1+detX). We may conjugate X by a unitary matrix U and assume that X is diagonal.'
@Studentmath: The number of people $Y$ who enter an elevator on the ground floor is a Poisson random variable with mean $10$. If there are $N$ floors above the ground floor and if each person is equally likely to get off at any one of those $N$ floors, independently of all the others, what is the expected number of stops that the elevator will make before discharging all its passengers?
I know. But what does that tell you @SwapnilTripathi? Look closer. If you are supposing for contradiction that the GCD is nontrivial, then what must $h$ be?
@evinda If the codomain is $\mathbb{C}$, the image is trivially contained in $\mathbb{C}$. The fact about the constants tells you that $\operatorname{im} \phi = \mathbb{C}$.
@KajHansen Thank you so much!! And thanks for clarifying this too, I hadn't thought of it. I read this somewhere as a remark below a corollary"f(x) is separable iff f'(x)\ne 0 over a field of characteristic 0". I didn't quite get where should f'(x) be non zero. On whole field? Guess not. If we consider g(x)=x^2-2. Then g'(0)=0 but g is irreducible.
@DanielFischer I see why the image is trivially contained in $\mathbb{C}$. How could we prove it, that if $\phi(z)=z \forall z \in \mathbb{C}$, that $\operatorname{im} \phi = \mathbb{C}$?
What is the maximum dimension of a vector space of $\mathcal{M}_n(\mathbb{R})$ containing only nilpotent matrices ? ($\mathcal{M}_n(\mathbb{R})$ : matrices $n\times n$ with coefficients in $\mathbb{R}$)
I don't really know how to solve this problem.There must be a way to give some good upper b...
@Ted if we look at one person, we can mark as success when he gets out, failure otherwise. We are looking for the first time of success, with $p=1/n$. So it's like Geometric distribution with $p=1/n$. I think.
Besides the fact that we are capped with $n$ trials
Oh. I searched many websites which stated f'(x)\ne 0 for irreducible polynomials and used this corollary. Now I know why. :) Thanks! I've been stuck on many problems since the past few days and no one interested in this topic came online on chat. So I have one last problem too. Give hints on this too. "Separable extension of separable extension is separable."
What is the maximum dimension of a vector space of $\mathcal{M}_n(\mathbb{R})$ containing only nilpotent matrices ? ($\mathcal{M}_n(\mathbb{R})$ : matrices $n\times n$ with coefficients in $\mathbb{R}$)
I don't really know how to solve this problem.There must be a way to give some good upper b...
