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  • 0 posts edited
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  • 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
accepted Is $\mathbb{R}\times\{0,1\}$ a manifold?
Apr
6
revised Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
edited body
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
accepted Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
Apr
6
revised Is $\mathbb{R}\times\{0,1\}$ a manifold?
added 1 character in body
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
6
answered Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
Apr
6
asked Is $\mathbb{R}\times\{0,1\}$ a manifold?
Apr
2
revised Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
added 12 characters in body
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..
Apr
2
asked Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$
Mar
28
revised Finding the inverse of $f(x,y)=(e^{x}\cos(y),e^{x}\sin(y))$ around a neighborhood.
Some additional thoughts
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
24
asked Finding the inverse of $f(x,y)=(e^{x}\cos(y),e^{x}\sin(y))$ around a neighborhood.
Mar
16
accepted Directional derivative of determinant at the identity is the trace of the matrix?
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!