There is also some subtlety in the convergence of the BCH formula. It does not generally converge on a global scale. A quick glance at this paper should hopefully convince you of that.
I believe BCH always converges in connected, simply connected groups, though that does not apply in this case.
Sure. Alternatively, isomorphisms preserve cardinality, so as ## \mathbb Z[x]/(x,2) \cong \mathbb Z_2 ## and the right-hand-side has two elements, you get the same answer.
If you take the isomorphism route, I think it would be better to define the map \phi: \mathbb Z[x] \to \mathbb Z_2 by p(x) \mapsto [p(0)] . Namely, each element is mapped to its constant term mod 2. Your map is certainly surjective, so just check that the kernel is I, which isn't too bad...
If I recall correctly, the following result holds:
Let R be a ring and a,b \in R . If \bar b is the equivalence class of b in R/(a), then
R/(a,b) = [R/(a)]/(\bar b).
This essentially just says that if you are careful about the book keeping, you can quotient by (2,x) by first quotienting by...
It is certainly not simply connected, but I think you mean just connected in this case. Simple connectivity is unrelated to surjectivity of the exponential.
Rereading your question, I now understand what you are saying. Have you been doing as suggested? Use long division to break up the rational function, then use vela's comment about how x^0 = 1 and you'll get the answer your originally posted.
The 1 in front comes from using long division to isolate the 1/(1-x) term. I'm not sure what you're asking in your second question though. If you want to change the index of a summation, you can do it entirely artificially. Namely, if you want \sum_{n=1}^\infty x^n to look like a sum where the...
Homework Statement
Let R be an arbitrary ring, B and B' be left R-modules, and i: B' \to B be an R-module morphism. Show that if the induced map i^*: \operatorname{Hom}_R(B,M) \to \operatorname{Hom}(B',M) is surjective for every R-module M, then i: B' \to B is injective.
The...
I always stress to my colleagues that they take a serious look at this book: it is incredibly insightful, well structured, and just an all around fantastic read. While I often have to (sometimes physically) wrestle Dummit & Foote or Lang from the iron-clasp grip of their fingers, once they pick...
For multivariate calculus, do the same argument with partials. Though here you need not do that since your function is just a map F: \mathbb R \to \mathbb R .
Now the solution is not quite straightforward: you are going to need to check that the derivative of your F(x) is continuous. You...
If f \in C^k(\mathbb R, \mathbb R) then define g(x) = \int_1^x f(t) gt . We would like to show that g \in C^{k+1} , and hence that the (k+1) derivatives of g exist and are continuous. Well, \frac{d^{k+1} g}{dx^{k+1}} = \frac{d^k f}{dx^k} and this is continuous by assumption that f \in C^k .
My apologies, perhaps it would have been better if I had chosen the parameterization of S^1 = \mathbb R / 2\pi \mathbb Z before writing that line. All I meant is that if we take \gamma: S^1 \to M and \iota: S^1 \to S^1 to be loops in their respective codomains, then they are implicitly...
I hope that this is a foolish question and that someone can make quick meat out of it. If \gamma: S^1 \to M is a loop on an arbitrary manifold M, our goal is to analyze the tangent vectors to \gamma when the loop is traversed in the opposite direction.
Let \iota: S^1 \to S^1 be the map...