Reputation
869
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
5 15
Newest
 Tumbleweed
Impact
~7k people reached

  • 0 posts edited
  • 0 helpful flags
  • 198 votes cast
Apr
6
comment Is $\mathbb{R}\times\{0,1\}$ a manifold?
Yep, literally just realized that. But thank you very much :)
Apr
6
comment Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
@LeGrandDODOM yeah, that is correct, it follows from the fact that the function must attain a maximum and this is the only point fitting that. Just poorly worded by me as English is not my native tongue. Will fix.
Apr
6
comment Is $\mathbb{R}\times\{0,1\}$ a manifold?
@ZevChonoles well this does make sense. Can you please explain the errors of my construction then?
Apr
6
comment Is $\mathbb{R}\times\{0,1\}$ a manifold?
@DietrichBurde As of right now the only definition I know of a manifold is the one I stated, and I have no definition for a manifold with boundary. Does that mean my solution to the problem is correct?
Apr
2
comment Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
@ArchisWelankar the answer to what?
Apr
2
comment Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
@Bubububu any insight on how you got to this conclusion? I'm unable to extrapolate even that unfortunately..
Mar
24
comment Finding the inverse of $f(x,y)=(e^{x}\cos(y),e^{x}\sin(y))$ around a neighborhood.
@IvoTerek unfortunately I don't really have any complex analysis knowledge
Mar
24
comment Finding the inverse of $f(x,y)=(e^{x}\cos(y),e^{x}\sin(y))$ around a neighborhood.
Sorry, I don't understand this one. Where does the $r$ come from? And am I even allowed to define a function $g(e^{x}sin(y))=(sin(y))$?
Mar
16
comment Directional derivative of determinant at the identity is the trace of the matrix?
I tried thinking of it like this but got confused within all the permutations. Actually seeing this makes it very simple. Thank you!
Jan
2
comment Derivative of the function $(x)!$.
the $n^{\rm{th}}$ derivative of $x^n$ is $n!$, which is a natural number, it's not a function in $x$
Dec
29
comment Any transformation that commutes with a transformation commuting with $S$ must be a polynomial in $S$
That makes sense.. Had troubles using the fact S is diagonalizable, but this makes it doable for me.
Dec
29
comment Any transformation that commutes with a transformation commuting with $S$ must be a polynomial in $S$
@orangeskid love to see a proof of the general case. As I'm still shakey on this specific case don't think I'll be able to prove it myself
Dec
29
comment Any transformation that commutes with a transformation commuting with $S$ must be a polynomial in $S$
@Omnomnomnom didn't know that. Thanks. English is not my native language
Dec
21
comment All the ternary n-words with an even sum of digits and a zero.
About your addition: This is exactly what I did originally, but what about even strings of lenght n-1 with zeroes? you can add 1 to then, so I also need $f(n) $ in the definition of f-bar, which puts me in a loop
Dec
21
comment All the ternary n-words with an even sum of digits and a zero.
got it specifically for n=1 with starting condition f(0)=1 because of the 2. And yeah.. f-bar is wrong. always confused by these...
Dec
5
comment Finding suitable basis for a free abelian finitely generated group.
It looks like I am taking the exact same course exactly a year later, as the question fits the one appearing on my homework, no one explained anything about Smith normal form to us, and I have no idea what do to.. (also your name fits the area :P). Let's hope the single answer here will help me get on my way...
Dec
5
comment Finding the function of the power series $\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}$
That's a lot simpler than my way...
Dec
5
comment Finding the function of the power series $\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}$
@r9m yep, you are right...
Nov
28
comment Proving differentiability on every interval equals proving differentiability on all of $\mathbb{R}$?
Well, that should have been pretty obvious to me looking back at it.. Thanks for clearing it all up anyway
Nov
28
comment Proving differentiability on every interval equals proving differentiability on all of $\mathbb{R}$?
@JohnMa don't really think I am.. The theorem in Wikipedia doesn't mention anything on a bound for any closed interval. Would it have any affect on all of $\mathbb{R}$ to have such a bound? en.wikipedia.org/wiki/Uniform_convergence#To_differentiability