While doing convolution we flip one of the functions say $f_2$ and then circular shift it by the amount $\tau$ and place it over $f1$ and then take the inner product of it with $f1$. We do this for different shifts $\tau$ to get $f3$ at diffe with differentrent points...Now lets go to the situation right before taking the (innerproduct/dot product)...different points (in domain) of $f1$ are aligned with different points of $f_2$ due to the flipping and also circular shift........
now consider a point $P$ in (0,1)...and let $f_1$ be $n_1(P)$ times differentiable at $P$ and let the flipped and circular shifted (by $\tau$) version of $f_2$ be $n_2(P)$ times differentiable at $P$...Now I define $n$ equal to minimum value of $n_1(P)+n_2(P)$ $\forall P \in (0,1)$.
@tb : if there are singularities ( finitely differentiable points) in both the functions being convolved then there will be singularities in the result of the convolution too
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers.
Suppose that $\{g_n\}_{n=0}^\infty$ is a sequence of functions holomorphic on an open domain $D$, and $$f(z)=\sum_{n=0}^\infty g_n(z)$$ holds on some subdomain $C$, while $f$ itself is holomorphic on all of $D$. If we define $$S_n(z)=\sum_{k=0}^n g_k(z) \qquad \text{and} \qquad s(z)= \big(S_0(z),S_1(z),S_2(z),\dots\big),$$ is there necessarily a Banach limit $\phi$ such that $\phi(s(z))=f(z)$ on all of $D$?
The motivation for my question is the 'spooky' phenomenon of certain analytic continuations matching the values you would 'expect' using naive algebraic manipulations of the series.
The first problem (Exercise 1) We already know $N + mM \subset M$. Now suppose we want to show the reverse inclusion. Suppose that $x \in M$ is not in the submodule $mM$ (otherwise there is nothing to show). Then in the quotient, $\bar{x} = \bar{\alpha_1}\bar{b_1} + \ldots \bar{\alpha_n}\bar{b_n}$
@anon How do you define a Banach limit of a possibly unbounded sequence? A silly example to look at would be to take the domain to be the entire complex plane $f(z) = 1/(1-z)$ and $g_n(z) = z^n$...
@tb: Crap, I forgot they have to be $\ell^\infty$ sequences. Is there a more general type of summability method that doesn't have any size requirements?
@t.b. I suppose we could define "loose" limits to be functionals satisfying the same properties as Banach limits but on all of $\mathbb{R}^\mathbb{N}$. Would such a definition cause problems?
And, all of them analysts. Analysis of several complex variables, Futile attempts and Banach space specialists, operator theory, Quantum Probability. sigh
@anon Well, I'd have to think a bit (and I have never seen anything of the kind). The first reflex would be to try to extend the construction of a Banach limit to $\mathbb{R}^{\mathbb{N}}$. But before going down this road: isn't the reason many such identities hold due to the identity theorem for holomorphic functions?
@BenjaminLim Never mind. : ) I have to read the CA room, I've been meaning to do that ever since I got up this morning but this chat has been a big help in procrastinating : )
How does the identity theorem relate to something like x=1+2+4+8+16+... => 2x=x-1 => x=-1? I mean, there's an obvious p-adic interpretation, but different series don't have such an interpretation.
@MattN Well I was only interested in it.......Honestly my degree in my university demands that one maintains 80% average in order to stay in it, if I were really the type interested in marks I would forget doing AC at all.
@SivaramAmbikasaran Also picking infinitely many elements from $(0,1)$, since we know that $(0,1)$ has more than countably many elements we can choose an arbitrary subset of size $\aleph_0$. This still does not require the axiom of choice since we only chose from one set. (@Kannappan, this is relevant to you as well regarding what you guys discussed in the CA room).
I guess I do my best to follow closely every conversation in any room I frequent. I prefer not to frequent many rooms as it would take too much of my time... :-)
@KannappanSampath We know that $|(0,1)|=2^{\aleph_0}>\aleph_0$. This means that there is a countable subset of $(0,1)$ is true even without the axiom of choice. So now we only need to choose one element from the nonempty set $\{A\subseteq (0,1)\mid A\text{ countable}\}$.
He was having a go at mathematicians, for not being physicists. His argument was that since mathematicians, after proving something, would always say "it's obvious"; mathematicians can therefore only prove obvious things.
@DavidWallace I happen to know that because I used an excerpt from that story in a talk on the Banach-Tarski paradox. shudder two things I wouldn't do anymore: quote Feynman and give a public talk on BT...
@DavidWallace I definitely agree with the assessment. The problem with those is that they appear in every other popular math talk/text you come across.
No, this isn't fake, it's just not personally identifiable. FB requires real names, so an alias like "Carter Living" is fake (when that's not really my name).
Yeah you seem to have strange fantasies....but not me....i never understood the first part hence i didn't go for the later ones....anyway i enjoyed the movie
"Where are you right now?" "I'm sitting in my office right now." "I doubt that." "Why would you doubt that?" "If you were we'd be having this conversation face to face." *Click*.
@Kanna : but beware this one seemed to be open problem to me...as i did not find it books...they just stopped at Jordan's and Dini's test without any comments on necessary conditions
@MattN Whenever someone wants to tag a question with (series) it will be tagged (sequences-and-series). The tag (series) does not exist but gets re-invented over and over again.
@anon Congratulations. You've began Armageddon. The first seal is broken! Huzzah! Six more seals and then we all have to dance the danse macabre or play chess with the grim ripper.
