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May
9
comment The composition of a nowhere-differentiable function with a differentiable function.
Probably not. That was the first thing that came to my mind but this problem is way before power series are introduced. I am starting to wonder whether this is a badly worded problem...
May
8
comment The composition of a nowhere-differentiable function with a differentiable function.
This was not clear to me either. I have copied the exercise as it appears on the book and I tried to assume it was nowhere differentiable.
May
8
comment The composition of a nowhere-differentiable function with a differentiable function.
I want to assume that $f$ is nowhere differentiable, otherwise the problem is pretty easy to handle.
Oct
6
comment Is this $\beta$-reduction well defined?
The question is whether it is possible to operate as it is.
Jul
4
comment A step in the proof of the Riemann Mapping Theorem
Are we considering the graph of $sin(1/x)$ as a subset of $\overline{\Bbb C}$?
Apr
19
comment Abelian group admitting a surjective homomorphism onto an infinite cyclic group
I just realised my stupid mistake! Just one more thing. How can I show that any element of $G$ is expressible as a product of two elements, the first of which is from $\ker(f)$ and the second from $im(g)$?
Apr
19
comment Studying the action of $GL(V)$ on the vector space $V$
Thank you very much! It is all clear now!
Apr
19
comment Abelian group admitting a surjective homomorphism onto an infinite cyclic group
I have tried to show that $\ker(g)$ is trivial, but I am failing to do so. My problem is that the $x$ you chose above might have finite order so you would have $nx=0$ with $n \neq 0$ thus the kernel will not be trivial. I cannot find a method to by-pass this problem.
Apr
16
comment Abelian group admitting a surjective homomorphism onto an infinite cyclic group
I am so very sorry for not understanding this immediately. Here is my point of confusion. I cannot see how $G = \ker(f) \times im(g) \implies G \cong \ker(f) \times \Bbb Z$, since $im(g) \subset G$ so you would get $G \cong \ker(f) \times im(g) \subset \ker(f) \times G$
Apr
12
comment Generalised eigenvalue is eigenvalue if it is in the field
Thank you very much for your help, Branimir!
Apr
12
comment Generalised eigenvalue is eigenvalue if it is in the field
I see my mistake! Thank you very much for pointing this out!
Mar
24
comment Imposing the topology of open rays in $\Bbb R$
Dear Brian, I would like to ask your permission to integrate your suggestions and hints into a complete answer in my original post. May I do so?
Mar
15
comment Imposing the topology of open rays in $\Bbb R$
Thank you very much for your most helpful comments!
Mar
3
comment Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]
Oh, OK! Thanks!
Feb
27
comment Mathematical preparation for postgraduate studies in Linguistics
@AveMaleficum I have further updated the post. Some useful links have been added, should you wish to consult them.
Feb
26
comment Mathematical preparation for postgraduate studies in Linguistics
Thank you! I will try to extend this as much as possible and provide links to other resources too. Did you check the one suggested in the update I made?
Feb
24
comment Definite Integral with a discontinuty
I don't think the problem is the discontinuity itself, but rather the fact that the integrand seems to blow up at $x = 0$.
Feb
22
comment Differential Equation : $f '' = f '$
You are indeed correct to point out that flaw. But that does not mean the proof is not rigorous. It only means that there is one more case to be checked. I will add it.
Feb
21
comment Differential Equation : $f '' = f '$
No harm done! (:
Feb
21
comment Differential Equation : $f '' = f '$
That's impossible! The first derivative will be some number but the second will be 0. How are they working?