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Let $A$ and $B$ be are non-empty compact sets. If $A\cap B=\varnothing$, then $d(A,B)>0$.
We're defining $d(A,B):=\inf\{\left|a-b\right|:a\in A\ \text{e}\ b\in B\}$.
I didn't use the fact that they're compact, so I feel something is missing. Is that right? Any feedbacks will be gratefully receive...
Is my logic correct?
$f:U(n)\rightarrow U(1)$ defined by $f(A)=\det A$ is a group
homomorphism so that the induced homomorphism $f^{*}: \pi_1(U(n))\rightarrow \pi_1(U(1))$ will be an isomorphism, right (I am not sure)? as $\pi_1(U(1))=\mathbb{Z}$ as $U(1)=S^1$
so $\pi_1(U(n))=\mathbb{Z}$.
Is there a closed form for this integral?
$$I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$$
I tried $u=e^x$ but it didn't simplify things much.
I think the closed form is a Bessel function, but I can't verify this directly.
I'd like to see the steps involved to reach the closed form.
In modular arithmetic, the integers coprime (relatively prime) to n from the set
{
0
,
1
,
…
,
n
−
1
}
{\displaystyle \{0,1,\dots ,n-1\}}
of n non-negative integers form a group under multiplication modulo n, called the multiplicative group of integers modulo n. Equivalently, the elements of this group can be thought of as the congruence classes, also known as residues modulo n, that are coprime to n.
Hence another name is the group of primitive residue classes modulo n.
In...
