Mariano Suárez-Alvarez
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15h
comment Galois group isomorphic to $\mathbb Z$
In other words, there is no such extension.
16h
comment How do i prove $\Gamma (m) \Gamma (1-m)= \frac{\pi}{\sin(m\pi)}$
@TaylorTed, it is somewhat impossible to do this without knowing complex analysis, really!
16h
comment How do i prove $\Gamma (m) \Gamma (1-m)= \frac{\pi}{\sin(m\pi)}$
Isn't this proved in any textbook treating $\zeta$, $\Gamma$ and so on?
1d
comment Why is the statement “all vector space have a basis” is equivalent to the axiom of choice?
The equivalence is a rather technical point, which is pretty much impossible to explain "briefly" to someone without the proper background. You are asking a rather difficult thing!
1d
comment Why is the statement “all vector space have a basis” is equivalent to the axiom of choice?
A good plan before asking this would be to study some formal set theory and get some background on the axiom of choice...
1d
comment Why do we need to specify the domain of an unbounded operator?
You end with «...then its domain is trivial», with weird sense of the word trivial!
1d
comment Why do we need to specify the domain of an unbounded operator?
@user1952009, that has nothing To do with anything.
1d
comment Why do we need to specify the domain of an unbounded operator?
I simply cannot tell what relation there is between that and the OP's question.
1d
comment Why do we need to specify the domain of an unbounded operator?
That is absurd. Any linear map from a subspace of a Banach space $X$ to another Banach space $Y$, bounded or not, can be extended to a linear map $X\to Y$ defined on the whole of $X$. If the original one was not bounded, neither is the new one.
1d
comment Why do we need to specify the domain of an unbounded operator?
@user1952009, that is not true. An unbounded operator may well be defined "everywhere".
1d
comment Why do we need to specify the domain of an unbounded operator?
No idea. You always have to say what the domain is.
1d
comment Why do we need to specify the domain of an unbounded operator?
Whenever you define a function, you have to make explicit its domain.
2d
answered nilpotent endomorphism on finitely generated modules over a domain
2d
comment $e^{(A+B)} = e^Ae^Be^{[A,B]}$ for non commuting A and B?
Please edit the question to fix it: as it stands, it makes no sense.
2d
comment Is torus w. disc removed homotopic to klein bottle w. disc removed?
@Anubhav.K, the OP is not removing two discs, he is removing one $2$-disc...
2d
comment Is torus w. disc removed homotopic to klein bottle w. disc removed?
See en.wikipedia.org/wiki/Wedge_sum; it should be explained in any sensible textbook on algbraic topology!
2d
comment Is torus w. disc removed homotopic to klein bottle w. disc removed?
@ajotatxe, I cannot tell what the connection is between that and the question!
2d
answered Is torus w. disc removed homotopic to klein bottle w. disc removed?
Feb
9
comment Can the real part of an entire function be bounded above by a polynomial?
What is the absolute value of $\exp if(z)$?
Feb
9
comment Proving $\mathbb{R}/\sim$ is homeomorphic to unit circle
The image of [0,1] by the quotient map $R\to R/\sim$ is the whole thing, and the map is continuous.