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...
The so called product integral was developed in 1887 by Volterra according to http://en.wikipedia.org/wiki/Product_integral
And here is the page for multiplicative calculus, just in case anyone is interested:
http://en.wikipedia.org/wiki/Multiplicative_calculus
Yea I think so too. This is a little bit off topic, but, it was also my reason for looking at the properties of the gamma function. Have you seen any theory on extending a product over a continuous interval, as is done with the sum to create the integral?
I have tried to develop an approach...
Actually, it follows immediately if you change the LHS of the last line to the equivalent
-(log(gamma(1+s))-log(gamma(1))/s
since log(gamma(1))=log(1)=0 and the digamma function ψ is defined to be the logarithmic derivative of the gamma function, so that under the limit as s ->0 this is...
I didn't even think to look at digamma, but it seems that as usual, you are indeed correct. It follows straight from a series representation for it. Basically, the last line in my analysis is almost exactly this
ψ(x+1)=-γ+∑(1/k-1/(x+k))
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...
Note that under union with 0 this is in fact a group, if x is in G and y is in R and |x-y|<epsilon then x is equivalent to y if you consider y to be an element of G. In other words, once you set the precision, to consider such a y as an element of G also, the part of y that is smaller than...
What I'm saying for the real numbers, is that every time you try to fix some labeling of them, there exist some elements in the real line not in this labeling. How then can a transitive action, which just means that the action has a single orbit containing all permutations of the group, be defined?
I understand that it may be infinite. But i just don't see how it is possible to have a group acting on itself transitively if it is uncountable. If it's uncountable it can't be indexed by any subset of the natural numbers, so a set of permutations which permutes every element to every other...
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...
If n is the order of G, Aut(G) is isomorphic to the symmetric group on n letters. It's order is n!. A quotient of G by some normal subgroup is less than or equal to the order of G. Say, the order G/Z(G) is m. Then m<=n. How can this quotient possibly be of order n! unless n is less than or equal...
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...