Functorial Weierstrass?
#1
This follows from the discussion held at: [note dump] iteration, actions, Bennett-like ops, (May 05, 2022), Tetration Forum.



(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.

  1. What do you mean with Weierstrass products there?
  2. Where \(h\) is defined?
  3. 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.
If I'm not totally mistaken all of those constructions \(\mathcal F(-,\mathbb C)\) are examples of sheaves over \(\mathbb C\), I'd need to check if the gluing axiom holds.

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 \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Reply
#2
Haha, I'm sorry Mphlee for confusing you. It's not as complicated as you're making it, lol. 

If you have a sequence of distinct points \(x_j \to 0\), if two functions \(f(x_j) = g(x_j)\) then \(f = g\). Weierstrass is basically used to construct a function \(F(x_j) = y_j\). If \(y_j = f(x_j)\), then \(F = f\). This can only be done if the sequence is summable though--at least, as far as I remember, you may be able to weaken it.


I'll take a moment to answer your confusions.

The set \(\mathcal{C}^1(U,\mathbb{C})\), is equivalent to the set of holomorphic functions. The confusion your having, is that complex differentiability at a point, does not mean holomorphic. But complex differentiability as a local construct, is equivalent to holomorphy. So if \(f\) is complex differentiable at \(z_0\), it does not mean it's holomorphic. But if \(f\) is complex differentiable for \(|z-z_0| < \delta\), then it's holomorphic.


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

Precisely that.

If you have an accumulating sequence \(x_k \to 0\), then \(f\) is completely determined by its behaviour on \(x_k\). This is a consequence of the identity theorem:

$$
H(x_k) = 0\,\,\,\text{implies}\,\,\, H = 0\\
$$

The trouble comes from using Weierstrass to actually construct \(f\) solely from its data at \(x_k\), that's where it gets a little tricky.

Weierstrass allows you to prescribe zeroes, yes! That's the main theorem, but it also allows you to interpolate an infinite sequence of points. For simplicity, I'll just explain how the basics of the argument would work. Weierstrass can assign zero at every point \(1/x_k\), because this sequence diverges. It won't look exactly like this, it's more difficult because we'd want to do this factorization on a simply connected domain.

Call this function:

$$
\begin{align}
W(z)\\
W(1/x_k) = 0\\
\end{align}
$$

Which will be entire. Now look at the function:

$$
P(z)= W(z)\sum_{k=0}^\infty \frac{1}{y_k}\frac{a_k }{z-\frac{1}{x_k}}
$$

Where: \(a_k = \lim_{z\to 1/x_k} \frac{z-\frac{1}{x_k}}{W(z)} = \frac{1}{W'(1/x_k)}\)

Then the function

$$
P(1/x_k) = 1/y_k
$$

If we call

$$
f(z) = 1/P(1/z)\\
$$

then \(f(x_k) = y_k\). This won't work for this example though, because \(f\) won't be holomorphic at \(0\), but there is a way to do this chain of arguments to make a similar \(f\) and be holomorphic at \(0\). There are multiple ways to reconstruct a holomorphic function from its values on accumulation points though, another one is just polynomial interpolation.

This prescription can be more difficult to do in practice, but this creates an interpolation algorithm.  I can't remember exactly how to do this for accumulation points near zero; it's a bit tricky; I think you have to use Blashke products or something. This isn't the correct way of doing it because it will not be holomorphic at \(0\), but there is a way to make it holomorphic at \(0\) I just can't remember off the top of my head.

It's named for someone else, not weierstrass, but it builds from weierstrass, I'll try to find it... When you start talking about advanced weierstrass stuff it gets very very highbrow...


Basically all I was saying, is we can equate a sheaf at zero to accumulation points at zero in \(\mathbb{C}^2\). And you can go one way and go back, and go the other way and also go back.

Sorry for derailing a bit, lol...
Reply
#3
Oh! Finally something I can fully grasp. This makes really lot of sense. What I take home from all of this is that complex functions are really well behaved gadgets. They possess a kind of elastic/rigid quality that reduces the degrees of freedom and stores a lot of global information of the objects in the local behavior, much like an hologram or a fractal, where every local  part contains the information on the whole (holos). Unlike real functions where is like we play with rabber, with some plastic material: even if we have infinitesimal data of all at all degrees (smoothness) we have still lot of freedom (eg. not every smooth is analytic).

What you say in the first paragraph seems incredible to me but also reinforces my previous statement about the rigidity of the holomorphic functions. Basically, you are saying that the global behavior of a function, i.e. its values at an uncountable number of points \(0\in U\subseteq \mathbb C\) is determined uniquely by its behavior at a countable subset of points*, because \({\rm im}x=\{x_n\,:\, n\in\mathbb N\}\) is at most countable....

*sure that set \({\rm im}x\) needs to "accumulate" around \(0\), i.e. for every open containing zero, the sequence must fall infinite times in it.

This means that, if we see from another direction, that... functions defined only on a finite/countable set, sure that set must be special, uniquely extends to an uncountable set of complex numbers.. THIS is magic.

Let \(x:\mathbb N\to \mathbb U\) and \(f,g\in{\mathcal H}(U,\mathbb C)\), where \(U\in {\mathfrak U}(0)\). Then your statement is literally the following

Quote:Injectivity of precomposition if \(x\in {\mathcal O}(1/n^{1+\delta})\subseteq {\rm Hom}(\mathbb N,U)\)
$$x^*:{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C)$$
is injective. This translates as: \(x^*f=x^*g\), i.e. \(\forall n.\, f(x_n)=g(x_n)\), implies \(f=g\)

I read this as: holomorphic functions are so rigid, that if they agree infinitely time inside a small region, then they agree on the entirety of that small region.

Corollary 0 The first shocking consequence is that the cardinality of \({\mathcal H}(U,\mathbb C)\) is equal or smaller then the cardinality of \(\mathbb C\).
$$|{\mathcal H}(U,\mathbb C)|\leq |{\rm Hom}(\mathbb N,\mathbb C)|=|\mathbb C|^{|\mathbb N|}=(2^{\aleph_0})^{\aleph_0}={\mathfrak c}$$
This translates as "there are not many holomorphic functions" and as we can associate to every holomorphic function \(f\) on \(U\) an unique complex number \(i(f)\in\mathbb C\).

This is so good that it seems suspicious. There should be some technicality that you omitted stating the conditions probably.

Corollary 1 Note that as an easy corollary we have that there exists (at least one) surjection
$${\mathcal H}(U,\mathbb C)\leftarrow {\rm Hom}(\mathbb N,\mathbb C):e$$ going in the opposite way sending each \(y:\mathbb N\to \mathbb C\) to an holomorphic \(e(y):U\to \mathbb C \) such that
$$e(x^*(f))=f$$
Question
We have injectivity of \(x^*\) for \(x\) well behaved, it decays in a good manner at zero. But is \(x^*\) also surjective? In other words, do every \(y\in {\rm Hom}(\mathbb N,\mathbb C)\) come as a restriction at \(x\) of some holomorphic function \(f\in {\mathcal H}(U,\mathbb C)\)?

Yes, you say. You define it to be \(y\mapsto P\mapsto {\iota\circ P \circ \iota^{-1}}\). Call this construction $${\mathfrak e}_x(y)$$
Quote:Surjectivity of \(x^*\) if \(x\in {\mathcal O}(1/n^{1+\delta})\subseteq {\rm Hom}(\mathbb N,U)\)
$$x^*:{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C)$$
is injective. That reads as: for every \(y\in {\rm Hom}(\mathbb N,\mathbb C)\) exists a \(f\in {\mathcal H}(U,\mathbb C)\) such that \(x^*f=y\).

Construction Let \(\iota(z)=1/z\), \(W_{x}\) the holomorphic function constructed by Weirstrass factorization theorem, it satisfies \(x^*W_{x}=W_x\circ x=0\). Define \(P_{x,y}(z)=W_x(z)\sum_{k=0}^\infty y_k\frac{a_{x,k}}{z-x_k}\) the function you defined in your post.
$${\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C):{\mathfrak e}_x\quad{\rm defined\,\, as}\quad {\mathfrak e}_x(y)=\iota \circ P_{\iota\circ x,\iota\circ y}\circ \iota$$
This construction satisfies  $$x^*{\mathfrak e}_x(y)={\mathfrak e}_x(y)\circ x=y$$
Corollary 3 \({\mathfrak e}_x\) must be injective. This means there is a bijection (\((x^*)^{-1}={\mathfrak e}_x\)).
$$x^*:{\mathcal H}(U,\mathbb C)\simeq {\rm Hom}(\mathbb N,\mathbb C):{\mathfrak e}_x$$
If this holds exactly as I stated, then it is like an upgraded version of the recursion theorem. The analogy is a bit blurry but \(x^*\) is like evaluation at \(0\) and \({\mathfrak e}_x\) is like recursion (of what?).

ENIGMA This is all cool and exciting... but then why haven't we solved the existence and uniqueness problem for tetration and for complex iteration in general?
I mean... just define \(y_k=1/m[k]n \) or \(w_k=1/m[i]k\). We find a function \(P\) that satisfies \(P(1/k^{1+\delta})=1/y_k\) and an \(f\) is an holomorphic interpolation with \(f|_{\mathbb N}(k^{1+\delta})=m[k]n\).
At this point we have only two possibilities. Since extensions are interpolations, every holomorphic extension must be an holomorphic interpolation. But it must be unique and obtained by this interpolation procedure.
So if the interpolation obtained by Weierstrass is an extension good; if it isn't we have proved that there is not an holomorphic extension.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Reply
#4
(05/17/2022, 08:38 PM)MphLee Wrote: Oh! Finally something I can fully grasp. This makes really lot of sense. What I take home from all of this is that complex functions are really well behaved gadgets. They possess a kind of elastic/rigid quality that reduces the degrees of freedom and stores a lot of global information of the objects in the local behavior, much like an hologram or a fractal, where every local  part contains the information on the whole (holos). Unlike real functions where is like we play with rabber, with some plastic material: even if we have infinitesimal data of all at all degrees (smoothness) we have still lot of freedom (eg. not every smooth is analytic).

What you say in the first paragraph seems incredible to me but also reinforces my previous statement about the rigidity of the holomorphic functions. Basically, you are saying that the global behavior of a function, i.e. its values at an uncountable number of points \(0\in U\subseteq \mathbb C\) is determined uniquely by its behavior at a countable subset of points*, because \({\rm im}x=\{x_n\,:\, n\in\mathbb N\}\) is at most countable....

*sure that set \({\rm im}x\) needs to "accumulate" around \(0\), i.e. for every open containing zero, the sequence must fall infinite times in it.

This means that, if we see from another direction, that... functions defined only on a finite/countable set, sure that set must be special, uniquely extends to an uncountable set of complex numbers.. THIS is magic.

Let \(x:\mathbb N\to \mathbb U\) and \(f,g\in{\mathcal H}(U,\mathbb C)\), where \(U\in {\mathfrak U}(0)\). Then your statement is literally the following

Quote:Injectivity of precomposition if \(x\in {\mathcal O}(1/n^{1+\delta})\subseteq {\rm Hom}(\mathbb N,U)\)
$$x^*:{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C)$$
is injective. This translates as: \(x^*f=x^*g\), i.e. \(\forall n.\, f(x_n)=g(x_n)\), implies \(f=g\)

I read this as: holomorphic functions are so rigid, that if they agree infinitely time inside a small region, then they agree on the entirety of that small region.

Corollary 0 The first shocking consequence is that the cardinality of \({\mathcal H}(U,\mathbb C)\) is equal or smaller then the cardinality of \(\mathbb C\).
$$|{\mathcal H}(U,\mathbb C)|\leq |{\rm Hom}(\mathbb N,\mathbb C)|=|\mathbb C|^{|\mathbb N|}=(2^{\aleph_0})^{\aleph_0}={\mathfrak c}$$
This translates as "there are not many holomorphic functions" and as we can associate to every holomorphic function \(f\) on \(U\) an unique complex number \(i(f)\in\mathbb C\).

This is so good that it seems suspicious. There should be some technicality that you omitted stating the conditions probably.

Corollary 1 Note that as an easy corollary we have that there exists (at least one) surjection
$${\mathcal H}(U,\mathbb C)\leftarrow {\rm Hom}(\mathbb N,\mathbb C):e$$ going in the opposite way sending each \(y:\mathbb N\to \mathbb C\) to an holomorphic \(e(y):U\to \mathbb C \) such that
$$e(x^*(f))=f$$
Question
We have injectivity of \(x^*\) for \(x\) well behaved, it decays in a good manner at zero. But is \(x^*\) also surjective? In other words, do every \(y\in {\rm Hom}(\mathbb N,\mathbb C)\) come as a restriction at \(x\) of some holomorphic function \(f\in {\mathcal H}(U,\mathbb C)\)?

Yes, you say. You define it to be \(y\mapsto P\mapsto {\iota\circ P \circ \iota^{-1}}\). Call this construction $${\mathfrak e}_x(y)$$
Quote:Surjectivity of \(x^*\) if \(x\in {\mathcal O}(1/n^{1+\delta})\subseteq {\rm Hom}(\mathbb N,U)\)
$$x^*:{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C)$$
is injective. That reads as: for every \(y\in {\rm Hom}(\mathbb N,\mathbb C)\) exists a \(f\in {\mathcal H}(U,\mathbb C)\) such that \(x^*f=y\).

Construction Let \(\iota(z)=1/z\), \(W_{x}\) the holomorphic function constructed by Weirstrass factorization theorem, it satisfies \(x^*W_{x}=W_x\circ x=0\). Define \(P_{x,y}(z)=W_x(z)\sum_{k=0}^\infty y_k\frac{a_{x,k}}{z-x_k}\) the function you defined in your post.
$${\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C):{\mathfrak e}_x\quad{\rm defined\,\, as}\quad {\mathfrak e}_x(y)=\iota \circ P_{\iota\circ x,\iota\circ y}\circ \iota$$
This construction satisfies  $$x^*{\mathfrak e}_x(y)={\mathfrak e}_x(y)\circ x=y$$
Corollary 3 \({\mathfrak e}_x\) must be injective. This means there is a bijection (\((x^*)^{-1}={\mathfrak e}_x\)).
$$x^*:{\mathcal H}(U,\mathbb C)\simeq {\rm Hom}(\mathbb N,\mathbb C):{\mathfrak e}_x$$
If this holds exactly as I stated, then it is like an upgraded version of the recursion theorem. The analogy is a bit blurry but \(x^*\) is like evaluation at \(0\) and \({\mathfrak e}_x\) is like recursion (of what?).

ENIGMA This is all cool and exciting... but then why haven't we solved the existence and uniqueness problem for tetration and for complex iteration in general?
I mean... just define \(y_k=1/m[k]n \) or \(w_k=1/m[i]k\). We find a function \(P\) that satisfies \(P(1/k^{1+\delta})=1/y_k\) and an \(f\) is an holomorphic interpolation with \(f|_{\mathbb N}(k^{1+\delta})=m[k]n\).
At this point we have only two possibilities. Since extensions are interpolations, every holomorphic extension must be an holomorphic interpolation. But it must be unique and obtained by this interpolation procedure.
So if the interpolation obtained by Weierstrass is an extension good; if it isn't we have proved that there is not an holomorphic extension.

Hey, Mphlee

I can clarify why you think this is too good to be true. Everything you are saying upto that point is exactly true. But when you get into interpolating functions things get difficult. You also kind of missed one of the points I was making.

Let's say we take \(x_k = \frac{1}{k}\), and lets try to interpolate as \(x_k \to 0\) where each value equals \(y_k = \frac{1}{e\uparrow \uparrow k}\). Well, you're forgetting that these are point pairs. So, as per the rigidity of holomorphic functions, the function \(1/e \uparrow \uparrow z\) CANNOT BE HOLOMORPHIC as \(|z| \to \infty\).

So, if we were to make an interpolation, we'd have some weird function \(f(z)\), such that \(f(1/k) = 1/e\uparrow \uparrow k\), but it will also be holomorphic at zero. That instantly disallows it from being tetration. This doesn't help too much with interpolating, as it will not satisfy the functional equation properly. This is largely because we are interpolating as points go to infinity, and there's no unique way to do that (there's an essential singularity at \(z=\infty\)).

The only way you have uniqueness is if... \((x_k,y_k) \to 0\) and \(f(z) \) is holomorphic about zero, and \(f(x_k) = y_k\). THATS where you have a uniqueness criterion.

On the flip side, if we take points which go to infinity... we have no such uniqueness.  As an example, consider \(g(z) = A(z)\sin(\pi z)\). You can make \(A\) whatever you want, and this function will still satisfy \(g(k) = 0\). This is because this function will not be holomorphic at \(z = \infty\), there's an essential singularity there.

So this idea extends really only for \(|z| < \delta\), and accumulation points at zero. We have to have holomorphy at the accumulated point. So trying to cheat the math and interpolating tetration's data points won't work unfortunately, though it'd be nice if it did.

This actually touches on the fractional calculus approach. Yes we are interpolating \(f(k) = \alpha \uparrow \uparrow k\), but we are also restricting that \(|f(z)| \le e^{\rho \Re(z) + \tau |\Im(z)|}\) for \(\rho \in \mathbb{R}^+\) and \(\tau < \pi /2\). It turns out, this produces a uniqueness condition at infinity. Where, if you have a function \(g(k) = 0\) and satisfies this bound, then \(g=0\). Notice, that the \(\sin(\pi z)\) function I wrote above, doesn't satisfy these bounds, so it doesn't matter to us.

All in all, what I wrote is only useful for local scenarios. It will not help us at interpolating sequences to infinity, while requiring they satisfy a functional equation. I mean, that's kind of obvious. If \((x_k,y_k) \to \infty\) and \(h(x_k) = y_k\), then the function \(b(z) = h(z) + W(z)\) is another solution to the same interpolation.

This only doesn't happen in a local scenario (or if you add some extraneous constraint (like the bounds for the fractional calculus)). 

So just keep in mind, when I was saying \(f(x_k) = y_k\) as \((x_k,y_k) \to 0\) completely determines a holomorphic function, I am also assuming that \(f\) is holomorphic at \(z=0\). And that's a crucial requirement. I think this is what you were forgetting for a moment. So unfortunately, this doesn't really help in the sense of "interpolating" tetration. lol.

Quote:We have injectivity of  for  well behaved, it decays in a good manner at zero. But is  also surjective? In other words, do every  come as a restriction at  of some holomorphic function ?

I'll try to find a reference, because I don't remember exactly. But, I'm pretty sure the answer to this is yes, if you restrict "what kind of" accumulation points. As in, as long as the accumulation points accumulate fast enough then yes. I'll try to find the answer for this, I can't remember off the top of my head.
Reply




Users browsing this thread: 1 Guest(s)