05/16/2022, 10:55 AM

This follows from the discussion held at: [note dump] iteration, actions, Bennett-like ops, (May 05, 2022), Tetration Forum.

Well, you massively overestimate me. It's hard for me to grasp all the implicit data in your comment. probably that's what most of the forum readers feel when reading my posts, like I'm leaving out lot of essential details because for me are trivialities.

I'll begin by reviewing my understanding of basics terminology/notations.

Observation (for my use of big-O) The Landau notation \(\mathcal O\) is basically a very confortable and well established abuse of notation to express how quickly a sequence \(x:\mathbb N\to X\), for \(X\) a metric space, where conververges or diverges. It is it's behavior at infinity. It workds by using some sequence \(s_n\) as a benchmark to measure the growth (to limit it from above) of the sequence \(x_n\). We write \(s_n={\mathcal O}(s_n)\) but formally we mean \(s_n\in{\mathcal O}(s_n)\). Because being \({\mathcal O}(s_n)\) is a property \({\mathcal O}(s_n)\) is the set of sequences that are eventually dominated by some scalar multiple of \(s_n\).

\({\mathcal O}(s_n)\subseteq X^\mathbb N \) Is thus a subset set of sequences, i.e. set functions from \(\mathbb N\) to a metric space \(X\), at least in the present thread*

*I know that the O-notation can be extended from sequences to functions, and from \(\to \infty\) to other limiting cases.

Definition (review) We call \({\rm Hom}(U,\mathbb C)\) the set of functions from \(U\subseteq\mathbb C \) to the set of complex numbers, this set contains as subsets all the following sets: let \(\mathbb C\) denote also a topological space.

Fundamentals (Cauchy/Goursat?) I'm not sure all the details on the definitions are correct. Anyways the core of complex analysis seems to be the the statement: to be \(\mathcal C^\omega\) (analytic) is equivalent to be \(\mathcal H\) (holomorphic).

$${\mathcal C}^{\omega}(U,\mathbb C)={\mathcal H}(U,\mathbb C)$$

Consequences If I'm not mistaken this should imply that also

$${\mathcal C}^{\infty}(U,\mathbb C)={\mathcal H}(U,\mathbb C)$$

Tbh the point that did stick to my mind was: with complex numbers if a function is \(1\)-differentiable then it is \(k\)-differentiable for every \(k\), i.e. it is smooth.

But also wikipedia seems to suggest that \({\mathcal C}^{1}(U,\mathbb C)\) is not exactly the same as \({\mathcal H}(U,\mathbb C)\) when it says "A function may be complex differentiable at a point but not holomorphic at this point.".

Ok, I'll start first by extracting what I understand from your statements. I'll fill up the fine details later, once the skeleton of the building is ready to support more weight.

Your statement (in my words) let \({\mathfrak U}(0)\) be the system of neighborhoods of \(0\) then we have a bijection

$${\mathcal O}(\frac{1}{n^{1+\delta}})\times {\mathcal O}(\frac{1}{n^{1+\delta}})\simeq \bigcup_{V\in {\mathfrak U}(0)}{\mathcal H}(V,\mathbb C)$$

It takes a pair of sequences \((x,y):\mathbb N\times \mathbb N\to \mathbb C\), where \(x\) is \({\mathcal O}(\frac{1}{n^{1+\delta}})\), and builds an unique holomorphic function \(f:V\to\mathbb C\);

Proof: It follows from some corollary of some "Weierstrass" black magic factorization... even if to me this sound as "if follows from Euler result".

Then, you add some details...

Your statement (in my words- 2nd attempt) there is a bijection between accumulation points and \(\bigcup_{V\in {\mathfrak U}(0)}{\mathcal H}(V,\mathbb C)\).

Then I need to understand how accumulation points relates to this.

Btw... looking at Weierstrass factorization on Wikipedia, it strikes me as a kind of generalization of the fundamental theorem of algebras. We can factor a polynomial using the information of its roots and associated multiplicity.

Example the set of polynomial functions \( k[x]\) is in bijection with the set of finite support, i.e. eventually zero, sequence of scalars. This is a bijection, if the field \(k\) is infinite, that sends each finite sequence \(a_n\) in the polynomial \(\sum a_nx^n\) and each polynomial to the sequence of its coefficients.

$${\rm coeff}:k[x]\simeq {\rm Hom}_{\rm fin\,supp}(\mathbb N,k):{\sum}$$

But, here where the Weierstrass Factorization enters, maybe: if \(k=\mathbb C\), I guess it's enough to be algebraically closed, we can factor each polynomial intro a product. We send each polynomial to its sequence, counting mutiplicity, of its roots \(\zeta_i\), and each sequence \(\zeta_i\) to the polynomial \(a\prod_i (x-\zeta_i)\)

$${\rm roots}:k[x]\simeq {\rm Hom}_{\rm fin\,supp}(\mathbb N,k):{\prod}$$

So are we talking about some way of associating to each holomorphic function a sequence of complex values, and viceversa, following the previous pattern?

Well...If I was 95% lost, after this I'm 99% lost.

(05/15/2022, 02:37 AM)JmsNxn Wrote: In this sense, we can refer to a sequence \((x_n,y_n) \to 0\) of complex numbers \(x_n,y_n \in \mathbb{C}\) are in bijection with holomorphic functions \(f(z)\) for \(z\) in a neighborhood of zero. Though we have to have a certain decay condition \(x_n = \mathcal{O}(1/n^{1+\delta})\) for \(\delta > 0\) (this could probably relaxed though).

This essentially creates a bijection between accumulation points and sheafs at zero. This would create an equivalence class of sorts, but it fits perfectly in this list, and I think it would work great for this.

I realized I missed a few steps here...

Weierstrass products for:

$$

h(1/x_n) = 1/y_n\\

$$

Always exist, and they exist based (with scrutiny), on the behaviour at infinity. But this creates a complex identity theorem (with a bonus Weierstrass construction); which we can say:

If \((x_n, y_n) \to 0\) and \(f(x_n) = y_n\), then these objects are bijective to each other (The object of sequences, to holomorphic functions at zero). And all we're asking is an infinite occurrences of \(\bullet\) that converge....

Damn, Mphlee, I hope this makes sense....

Well, you massively overestimate me. It's hard for me to grasp all the implicit data in your comment. probably that's what most of the forum readers feel when reading my posts, like I'm leaving out lot of essential details because for me are trivialities.

- What do you mean with Weierstrass products there?

- Where \(h\) is defined?

- Is \(f\) meant to be the unique holomorphic function at zero that maps the first sequence to the second sequence? But then the two sequences must have precise requisites on the mutiplicity of occurrences of each value in the sequence: if \(x_i=x_j\) for some pair \(i,j\in\mathbb N\) then \(y_i=y_j\);

I'll begin by reviewing my understanding of basics terminology/notations.

Observation (for my use of big-O) The Landau notation \(\mathcal O\) is basically a very confortable and well established abuse of notation to express how quickly a sequence \(x:\mathbb N\to X\), for \(X\) a metric space, where conververges or diverges. It is it's behavior at infinity. It workds by using some sequence \(s_n\) as a benchmark to measure the growth (to limit it from above) of the sequence \(x_n\). We write \(s_n={\mathcal O}(s_n)\) but formally we mean \(s_n\in{\mathcal O}(s_n)\). Because being \({\mathcal O}(s_n)\) is a property \({\mathcal O}(s_n)\) is the set of sequences that are eventually dominated by some scalar multiple of \(s_n\).

\({\mathcal O}(s_n)\subseteq X^\mathbb N \) Is thus a subset set of sequences, i.e. set functions from \(\mathbb N\) to a metric space \(X\), at least in the present thread*

*I know that the O-notation can be extended from sequences to functions, and from \(\to \infty\) to other limiting cases.

Definition (review) We call \({\rm Hom}(U,\mathbb C)\) the set of functions from \(U\subseteq\mathbb C \) to the set of complex numbers, this set contains as subsets all the following sets: let \(\mathbb C\) denote also a topological space.

- \({\mathcal C}(U,\mathbb C)\) is the set of continuous complex-valued functions over the open set \(U\subseteq\mathbb C \);

- \({\mathcal C}^1(U,\mathbb C)\) is the set of differentiable complex-valued functions over the open(?) set \(U\subseteq\mathbb C \), i.e. differentiable at every point of \(U\);

- \({\mathcal C}^k(U,\mathbb C)\) is the set of \(k\)-times differentiable complex-valued functions over the open set \(U\subseteq\mathbb C \);

- \({\mathcal C}^{\infty}(U,\mathbb C)\) is the set of smooth complex-valued functions over the open set \(U\subseteq\mathbb C \), i.e. infinite times differentiable over \(U\);

- \({\mathcal C}^{\omega}(U,\mathbb C)\) is the set of analytic complex-valued functions over the open set \(U\subseteq\mathbb C \), i.e. functions \(f:U\to \mathbb C\) s.t. for every \(a\in U\) exits an open disk \(U_a\) such that \(f|_{U_a}(z)=\sum_{i=0}^\infty c_n(z-a)^i\);

- \({\mathcal H}(U,\mathbb C)\) is the set of holomorphic functions over the (open?) set \(U\subseteq\mathbb C \), i.e. functions \(f:U\to \mathbb C\) s.t. for every \(a\in U\) exits an (open??) neighbourhood \(U_a\) such that \(f|_{U_a}\in {\mathcal C}^1(U,\mathbb C)\);

- \({\mathcal H}(\mathbb C,\mathbb C)\) is the set of entire functions.

Fundamentals (Cauchy/Goursat?) I'm not sure all the details on the definitions are correct. Anyways the core of complex analysis seems to be the the statement: to be \(\mathcal C^\omega\) (analytic) is equivalent to be \(\mathcal H\) (holomorphic).

$${\mathcal C}^{\omega}(U,\mathbb C)={\mathcal H}(U,\mathbb C)$$

Consequences If I'm not mistaken this should imply that also

$${\mathcal C}^{\infty}(U,\mathbb C)={\mathcal H}(U,\mathbb C)$$

Tbh the point that did stick to my mind was: with complex numbers if a function is \(1\)-differentiable then it is \(k\)-differentiable for every \(k\), i.e. it is smooth.

But also wikipedia seems to suggest that \({\mathcal C}^{1}(U,\mathbb C)\) is not exactly the same as \({\mathcal H}(U,\mathbb C)\) when it says "A function may be complex differentiable at a point but not holomorphic at this point.".

Ok, I'll start first by extracting what I understand from your statements. I'll fill up the fine details later, once the skeleton of the building is ready to support more weight.

Your statement (in my words) let \({\mathfrak U}(0)\) be the system of neighborhoods of \(0\) then we have a bijection

$${\mathcal O}(\frac{1}{n^{1+\delta}})\times {\mathcal O}(\frac{1}{n^{1+\delta}})\simeq \bigcup_{V\in {\mathfrak U}(0)}{\mathcal H}(V,\mathbb C)$$

It takes a pair of sequences \((x,y):\mathbb N\times \mathbb N\to \mathbb C\), where \(x\) is \({\mathcal O}(\frac{1}{n^{1+\delta}})\), and builds an unique holomorphic function \(f:V\to\mathbb C\);

Proof: It follows from some corollary of some "Weierstrass" black magic factorization... even if to me this sound as "if follows from Euler result".

Then, you add some details...

Quote:Accumulation points \(\mathcal{A}\), given as \((x_n,y_n)\), maps exactly to sheafs at zero...

Remember when I say Weierstrass, we need to add that:

$$

\sum_{n=0}^\infty |x_n|<\infty\\

$$

So just keep that result, and you can make this identification with \(\bullet\) (infinite point wise operations, as sequences) across \(\mathbb{C}\) to make holomorphy at \(0\).

Your statement (in my words- 2nd attempt) there is a bijection between accumulation points and \(\bigcup_{V\in {\mathfrak U}(0)}{\mathcal H}(V,\mathbb C)\).

Then I need to understand how accumulation points relates to this.

Btw... looking at Weierstrass factorization on Wikipedia, it strikes me as a kind of generalization of the fundamental theorem of algebras. We can factor a polynomial using the information of its roots and associated multiplicity.

Example the set of polynomial functions \( k[x]\) is in bijection with the set of finite support, i.e. eventually zero, sequence of scalars. This is a bijection, if the field \(k\) is infinite, that sends each finite sequence \(a_n\) in the polynomial \(\sum a_nx^n\) and each polynomial to the sequence of its coefficients.

$${\rm coeff}:k[x]\simeq {\rm Hom}_{\rm fin\,supp}(\mathbb N,k):{\sum}$$

But, here where the Weierstrass Factorization enters, maybe: if \(k=\mathbb C\), I guess it's enough to be algebraically closed, we can factor each polynomial intro a product. We send each polynomial to its sequence, counting mutiplicity, of its roots \(\zeta_i\), and each sequence \(\zeta_i\) to the polynomial \(a\prod_i (x-\zeta_i)\)

$${\rm roots}:k[x]\simeq {\rm Hom}_{\rm fin\,supp}(\mathbb N,k):{\prod}$$

So are we talking about some way of associating to each holomorphic function a sequence of complex values, and viceversa, following the previous pattern?

Quote:Also, when you talk about Weierstrass about zero, you don't talk about Weierstrass' factorization theorem. Where as the latter factorization theorem is designed for \(\mathbb{C}\), the former is used for simply connected domains. But it is still ultimately, Weierstrass.

Well...If I was 95% lost, after this I'm 99% lost.

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)