I am reading Howie's Complex Analysis. There I see this remark:
The observation that $c$ and $d$ are inverse points is the key to showing that every circle can be represented as $\{z:|z-c|=k|z-d|\}$. Suppose that $\Sigma$ is a circle with centre $a$ and radius $R$. Let $c = a + t$, where $0 < t
Ugg, I'm struggling with understand the tricks used in holder duals, they basically substitute in a second variable and maximize over it. Then the dual becomes the auxiliary space the second variable lives in?
Venus is not Earth’s closest neighbor Calculations and simulations confirm that on average, Mercury is the nearest planet to Earth—and to every other planet in the solar system.
they're like vector spaces but your scalars don't form a field, just a ring
they come up like everywhere
@TobiasKildetoft yes
@AkivaWeinberger if k is a field and G is a group then k[G]-modules are k-representations of G (so study of modules subsume (modular) representation theory)
if M is a manifold then (locally free) $C^\infty(M)$-modules are vector bundles
@AkivaWeinberger structure theorem for finitely generated abelian groups say that they are all $\Bbb Z^r \oplus \bigoplus_{i=1}^t \Bbb Z/p_i^{n_i} \Bbb Z$
the generalization is structure theorem for finitely generated modules over principal ideal domains (Z being a PID)
they're all $R^r \oplus \bigoplus_{i=1}^t R/p_i^{n_i} R$
and this provides a "straightforward" proof of Jordan Normal Form theorem
because a $k$-vector space with a linear transformation to itself is just a $k[X]$-module, and $k[X]$ is a PID
yeah okay, so this is true in general... if you have a ring hom $R \to S$ then $S$-modules are $R$-modules, and the hom need not be injective or surjective, so "bigger" doesn't matter
@LeakyNun So a quick check, what goes wrong here. I take the polynomial ring in infinitely many variables, I kill the first variable and identify the result with the original ring by shifting all indices by one. Why is this not a surjective map that is not a bijection?
which provides an isomorphism of rings, but not of modules, since we would have to twist the action on the target, which would make it no longer an endomorphism.
Numerical integration does not exactly work with "infinity". One may choose a finite symmetric interval of integration and take the limit afterwards (which is the traditional way of doing these integrals), but Mathematica is not equipped to solve this integral completely symbolically
Although I conjecture that it is theoretically possible
Regardless, I wish to find a constant $\lambda_F$ defined implicitly by
I am reading Howie's Complex Analysis. There I see this remark:
The observation that $c$ and $d$ are inverse points is the key to showing that every circle can be represented as $\{z:|z-c|=k|z-d|\}$. Suppose that $\Sigma$ is a circle with centre $a$ and radius $R$. Let $c = a + t$, where $0 < t 
