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...
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...