1,849 reputation
21036
bio website anhhuy.ch
location Switzerland
age 22
visits member for 3 years, 11 months
seen 10 hours ago

My name is Anh Huy Truong and I am currently studying mathematics at the ETH Zürich.


Feb
16
comment How can I find a non-negative interpolation function?
Do you have any idea what the constant you use to subtract depends on in order for the resulting function to being non-negative?
Feb
16
comment How can I find a non-negative interpolation function?
I was thinking about using the logarithm method after all by simply adding a constant to my values (e.g. $y_i + 1$) and then taking the logarithm. However, when I then take the exponential of the resulting interpolating polynomial, is there a formula to find the constant I have to subtract in order to get back? In the examples I tried out I could not find any regularity and most importantly the constant $-1$ does not work. :-(
Feb
16
comment How can I find a non-negative interpolation function?
I think the dirty solution will be best suited for my problem. I have values $y_i = 0$ which rules out taking the logarithm (or is there a solution for that issue?). However, if I am using Hermite interpolation (i.e. I want to interpolate derivatives as well), what values do I take for the derivatives? Simply the square roots?
Dec
9
comment Showing that a matrix is positive (semi-)definite
I'm sorry I didn't refer to the paper earlier. I asked for an alternative proof because the proof provided was not very intuitive to me, i.e. I would have never been able to come up with it. Since you came up independently with basically the same proof: Is there any logical reason for introducing that matrix and decomposing the matrix $D$ as in your answer? Is it just experience or is there a deeper reason? I have never seen this kind of decomposition yet which is why it was rather odd for me.
Dec
9
comment Showing that a matrix is positive (semi-)definite
@DavidSpeyer: I put it for completeness. Sorry if it confused anyone.
Dec
9
comment Is the matrix corresponding to an equivalence relation positive semidefinite?
My question actually arose from this. I had a graph and the matrix $A$ had entries $1$ if two vertices belonged to the same connected component and $0$ otherwise. However, I did not know about the eigenvalues of a clique (or maybe I just forgot).
Dec
9
comment Is the matrix corresponding to an equivalence relation positive semidefinite?
Sorry, I misread. That makes a lot more sense now.
Dec
7
comment Showing that a matrix is positive (semi-)definite
@Casteels: Was there a comment that is deleted now?
Dec
5
comment Showing that a matrix is positive (semi-)definite
@Casteels: I'm sorry I completely forgot. The diagonal entries are 1.
Nov
22
comment Planning a mockup maths class for high school related to river reactivation
(or just of river regulation, sorry, I missed the 5 minute edit time span)
Nov
22
comment Planning a mockup maths class for high school related to river reactivation
@WillieWong: It is the inversion of canalisation of a river.
Aug
21
comment Proving Sylow's first theorem
Which implication are you talking about?
Aug
21
comment Proving Sylow's first theorem
Yes, $S_i$ is not necessarily a subgroup. That is the point of the equation $\operatorname{Stab}_G(S_i) = S_i$: As the stabilizer IS a subgroup, it implies that $S_i$ is too.
Aug
21
comment Proving Sylow's first theorem
Aren't order and cardinality of sets the same thing?
Aug
21
comment Proving Sylow's first theorem
Yes, I meant that $p$ cannot divide the cardinality of the orbit $S_i$ (I never said order, did I?). Also, I fixed a mistake. The way I understand the proof is that we have an orbit such that $p$ doesn't divide its cardinality and then find $\operatorname{Stab}_G(S_i) = S_i$ (but I don't know why this equation holds). This equation implies that the stabilizer and $S_i$ have the same cardinality and as the stabilizer is a subgroup (of cardinality $p^n$), we are done.
Aug
21
comment Proving Sylow's first theorem
$S_i$ only represents the orbit $GS_i$.
Jul
25
comment Differentiating terms involving evaluation operator and Wronski matrix
Why does $D\Phi^T(z) = \frac{\partial}{\partial y} \Phi^{T}y|_{y=z}$? Is this because $\Phi^T(z)$ does not depend on the time variable?
Jul
3
comment Construction of field extensions
I see. The use of $X$ as both a variable and a polynomial caused the confusion.
Jul
3
comment Construction of field extensions
Just after submitting this question, it came to my mind that in the definition $a:=X+(P)$, the $X$ is probably not meant as a variable, but as the polynomial $X \in K[X]$, right?
Jun
26
comment Application of the Chinese Remainder Theorem
Could you edit it into your original answer if it's not too much trouble? I'd appreciate that a lot. I'm not sure if I'm supposed to know it but it can't be any harm.