meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 3 months |
| seen | Apr 24 at 19:51 | |
| stats | profile views | 125 |
|
Mar 25 |
comment |
The special orthogonal group is a manifold $S_n(\mathbb{R})$ is the vector space of symetric matrices. For $f :X \rightarrow Y$ a differentiable function between $X$ and $Y$ two finite dimensionnal normed $\mathbb{R}$-vector spaces, we say $y \in Y$ is a regular value of $f$ if for any $x$ such that $f(x) = y$, $Df(x)$ is onto(or,equivalently, that $Df(x)$ has maximal rank). My remark is based on the following theorem : Let $f : U \subset \mathbb{R}^m \rightarrow \mathbb{R}^n$ be a function $\mathcal{C}^k$ on a domain U, and let $y$ be a regular value of $f$. Then $f^{-1}(\{y\})$ is a submanifold of $\mathbb{R}^m$ of dimension $m-n$ |
|
Mar 25 |
comment |
The special orthogonal group is a manifold Sorry I wrote without thinking. To make sure we are right, one can compute the tangent space at any point of $P \in O_n(\mathbb{R})$ : it is $P A_n(\mathbb{R})$ which is od dimension $\frac{n(n-1)}{2}$. |
|
Mar 25 |
revised |
The special orthogonal group is a manifold added 82 characters in body |
|
Mar 25 |
revised |
The special orthogonal group is a manifold added 82 characters in body |
|
Mar 25 |
answered | The special orthogonal group is a manifold |
|
Mar 22 |
answered | Uncountable sets and isolated points |
|
Mar 21 |
comment |
Prove that partial sums of $\sum_{n=1}^{\infty}{z^n}, z \in \mathbb{C}, |z|=1$ are bounded Don't forget the case $ z = 1$ :) Else, it's only the formula for a geometric progression ... |
|
Mar 21 |
comment |
What are the probabilities of getting a “Straight flush” in a poker game? very weak actually :) |
|
Mar 21 |
revised |
An Integral inequality added 73 characters in body |
|
Mar 21 |
answered | An Integral inequality |
|
Mar 20 |
answered | infinite limit of sequence $b_{n}\rightarrow \infty $ |
|
Mar 20 |
comment |
How to find the integral $\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$ Are you familiar with complex analysis ? |
|
Mar 18 |
answered | Homework question on whether two quotient spaces are homeomorphic |
|
Mar 17 |
comment |
Proving a family of orthogonal functions is complete over a certain interval I think you must had functions $\sqrt{\frac{2}{\pi}} \cos kx $ for your familie to be complete in the sense you mentionned. Then theory of Fourier series gives you that any $\mathcal{C}^{\infty}$ function which is orthogonal to every $u_k$ is zero. But since $\mathcal{C}^{\infty}$ is dense in $L^2$ , it's true for every function in $L^2$. |
|
Mar 16 |
awarded | Commentator |
|
Mar 16 |
comment |
$f$ an isometry from a hilbert space $H$ to itself such that $f(0)=0$ then $f$ linear. Since $f(0)=0$, $ ||f(x)|| = ||x||$. Then develop $||f(x) - f(y) ||^2 = ||x-y||^2$ |
|
Mar 16 |
comment |
$f$ an isometry from a hilbert space $H$ to itself such that $f(0)=0$ then $f$ linear. First thing to notice : $f$ conserves the inner product ... |
|
Mar 16 |
awarded | Critic |
|
Mar 13 |
answered | A question about analytic functions. |
|
Mar 12 |
comment |
Is the complement of a countable set in $\mathbb{R}$ dense? Application to convergence of probability distribution functions. Yes that's what i meant. I'm French, English is not my mother language so i translated directly an expression that is used in French :) |
