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Jun
25
comment Is the Cauchy-Schwarz inequality ever used in Physics?
That's the perfect example! See, for instance, Sakurai “Modern Quantum Mechanics”, 1.4 .
Apr
15
comment How to estimate the axis of symmetry for an even function with error?
You are welcome! If the large error is on the $y_i$'s, I've added a second (maybe more practical) method that could work.
Apr
15
revised How to estimate the axis of symmetry for an even function with error?
added 671 characters in body
Apr
15
answered How to estimate the axis of symmetry for an even function with error?
Apr
8
awarded  Enlightened
Apr
7
awarded  Nice Answer
Mar
5
answered Show that $A =\left(\begin{smallmatrix}41&12\\12&34\end{smallmatrix}\right)$ is symmetric positive definite
Mar
4
comment Show that $A =\left(\begin{smallmatrix}41&12\\12&34\end{smallmatrix}\right)$ is symmetric positive definite
Hint for alternative method: is there any relationship beetween $A$'s eigenvalues and quantities like, for example $\det A$?
Jan
31
accepted Differential of the inversion of Lie group
Jan
22
awarded  Peer Pressure
Jan
22
comment Given $S \subset \Bbb{R}$, show $\textbf{int}(S)+\textbf{ext}(S)+\partial S =\Bbb{R}$
Hi @Kaytlyn, consider that some authors define $\partial S$ as the complement of $\text {int} S+\text {ext} S$ (for example Spivak in his book "Calculus on manifolds"). What's your definition of $\partial S$?
Jan
20
comment Does every differentiable ruled surfaces possess a global ruled parametrization?
Hi @Evgeny, thanks for the suggestion. If I make some progress, I will post. Is there anything I can do to improve my question?
Jan
18
awarded  Yearling
Jan
9
asked Does every differentiable ruled surfaces possess a global ruled parametrization?
Dec
9
awarded  Caucus
Nov
27
asked Surjective $\gamma \colon I \to M^1$, $\gamma (t_1)=\gamma (t_2)$ can be extended to a periodic parametrization of $M^1$
Sep
30
awarded  Explainer
Aug
30
comment Alternative proof: Matrix $A$ is similar to $B$ iff $\lambda I - A$ is equivalent to $\lambda I - B$
I'm sorry, I've never heard this terminology. I assume it means $\exists P,Q\in GL(n,\mathbb K)$ such that $\forall \lambda \in \mathbb K$, $\lambda I - A=P^{-1}(\lambda I -B)Q$, right?
Aug
30
comment Alternative proof: Matrix $A$ is similar to $B$ iff $\lambda I - A$ is equivalent to $\lambda I - B$
I think that you should state the theorem more precisely: what values is $\lambda$ free to take? Since in this form it is obviously false: take $\lambda =0$.
Aug
30
comment Alternative proof: Matrix $A$ is similar to $B$ iff $\lambda I - A$ is equivalent to $\lambda I - B$
Matrix equivalence: en.wikipedia.org/wiki/Matrix_equivalence