19,439 reputation
12238
bio website math.ntnu.no/~hanche
location Trondheim, Norway
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visits member for 2 years, 9 months
seen 4 hours ago

15h
comment Theorem of the convergence of the series of fourier!
Okay, I did that, then noticed that the question was closed as a duplicate. I wasn't too sure if the software would accept the edit under these circumstances, but it did. I won't edit it further, though.
15h
revised Theorem of the convergence of the series of fourier!
Expand a bit more
19h
comment Beautiful little number theory prob
Maybe try to find solutions to $u+v=uv$ first?
1d
answered Proving commutativity
1d
comment Theorem of the convergence of the series of fourier!
I'll give that answer a +1 for avoidance of the complex exponential function. If you do know the complex exponential, however, that is easier to understand.
1d
revised Theorem of the convergence of the series of fourier!
A small bit more
1d
revised Theorem of the convergence of the series of fourier!
Backslashify trig functions
1d
comment Theorem of the convergence of the series of fourier!
No wonder you couldn't do it on your own. It is quite hard if you don't know the trick (see my answer).
1d
answered Theorem of the convergence of the series of fourier!
1d
comment What does a domain being maximal mean?
Yes, since any attempt to expand the domain takes you across the branch cut, where the function jumps to a different value. (This is possible to avoid, but then you have to leave the complex plane behind and work with Riemann surfaces. But that is a topic for a more advanced course, perhaps.)
1d
comment Is $C(\Omega)$ a C*-algebra if $\Omega$ is not locally compact, nor compact?
That is correct. And that is all you need compactness for.
1d
comment What does a domain being maximal mean?
If there is no analytic extension of the function to a larger domain than the given one, the given domain is maximal (for that function).
1d
comment Is $C(\Omega)$ a C*-algebra if $\Omega$ is not locally compact, nor compact?
For one thing, if $C(\Omega)$ has unbounded members, what will you use for the norm on your C*-algebra? But the space of bounded continuous functions on $\Omega$ is always a C*-algebra. The proof is trivial.
1d
comment Use contour integration to calculate real integrals
Along the left side, you have $z=-\epsilon+iu$, and so $dz=-idu$. But more importantly, $1/z=(-\epsilon-iu)/(u^2+\epsilon^2)$. You have left out the factor $-\epsilon-iu$, and made a similar omission along the right side.
1d
answered Use contour integration to calculate real integrals
1d
comment Use contour integration to calculate real integrals
I added some backslashes to make your trigonometric functions look right. What is $\gamma^+$? You only define $\gamma$. I am also a bit confused by your rectangle. Are its vertices at $\pm\epsilon$ and $\pm\epsilon-i$? Why the long detour along the negative imaginary axis?
1d
revised Use contour integration to calculate real integrals
Backslashify trig functions
2d
comment Tensor product of R-algebras
I don't understand the question. What do you mean by advantage?
2d
comment How to prove $l^p$ is separable?
This is a very standard result, with a very standard proof. Essentially, take series with rational entries, only finitely many of which are non-zero.
2d
comment Question about Compact metric space
Any subset of a metric space is a metric space, by restricting the metric of the larger metric space – in this case, $\mathbb{R}^{n+1}$ – to the subset. Of course, the implicit assumed metric on $\mathbb{R}^{n+1}$ must be the Euclidean metric: $d(x,y)=\sqrt{\sum_{j=0}^n|x_j-y_j|^2}$.