I was doing some basic analysis of the Dedekind eta function and some Dirichlet series and the following equality just fell out:
\sum_{k=1}^\infty\frac{\mu (k)-\varphi (k)}{k}\log \left( 1-\frac{1}{\phi^k} \right) = \prod_{k=1}^\infty \left( 1-\frac{1}{\phi^k} \right)^{2\pi i\frac{\mu...
Suppose we are given two functions:
f:\mathbb R \times \mathbb C \rightarrow\mathbb C
g:\mathbb R \times \mathbb C \rightarrow\mathbb C
and the equation relating the Stieltjes Integrals
\int_a^\infty f(x,z)d\sigma(x)=\int_a^\infty g(x,z)d\rho(x)
where a is some real number, the...
I was taking a break from studying from my real analysis, electrodynamics, and nuclear physics exams this week, and, being a math-phile, I decided to play around with the gamma-function for some reason. Anyway, I used the common product expansion of the multiplicative inverse, and I arrived at a...
My question is exactly what is stated in the title: Does Cayley's theorem imply that all groups are countable?
I don't see how a well defined transitive action of an uncountable group on itself. How could you possibly find a set of permutations sending a single element to every other element in...
I was just brushing up on some Algebra for the past couple of days. I realize that the lattice isomorphism theorem deals with the collection of subgroups of a group containing a normal subgroup of G. Now, in general, if N is a normal subgroup of G, all of the subgroups of larger order than N do...
Homework Statement
Let \left[a,b\right] be a closed bounded interval, f : [a,b] \rightarrow \textbf{R} be bounded, and let g : [a,b] \rightarrow \textbf{R} be continuous with g\left(a\right)=g\left(b\right)=0. Let f_{n} be a uniformly bounded sequence of functions on \left[a,b\right]. Prove...
I have just started diving into tensor analysis. To be honest, I didn't know whether to post this question in the vector analysis forum or this one. I have looked at a few books on the subject and scoured the internet, but I can't seem to find anything that answers this question. Or, maybe I...