Talk:Mittag-Leffler Expansion for Cotangent Function

Logarithm of Product does not suffice. Neither does Logarithm of Infinite Product of Complex Functions. What you really need is Logarithmic Derivative of Infinite Product of Analytic Functions. (Note how the proof secretly uses logarithmtic derivatives, not derivatives of logarithms, because those aren't even continuous.) --barto (talk) (contribs) 16:49, 18 November 2017 (EST)

Voilà, hope you don't mind. It only remains to establish the locally uniform convergence, but this isn't hard, because it's even absolute, see Definition:Locally Uniform Absolute Convergence of Product: it reduces to series. --barto (talk) (contribs) 17:10, 18 November 2017 (EST)

Not at all. I'm just unsure about a few things (this is purely to add to my own knowledge, I'm not questioning anything you're saying, you're far more versed in this than me :p) Why doesn't the second one suffice? Had I known it existed, I probably would've used that instead. I would've thought that convergence of the sum directly follows from the convergence of the product which is given in the theorem, though on second thought that might require a bit more explanation. I'm confused by your comment in brackets, can you explain a bit more? Caliburn (talk) 17:27, 18 November 2017 (EST)

Sure :) In fact, Logarithm of Infinite Product of Complex Functions is not so useful, because it does not tell us anything about whether the $2\pi i k$ is constant. But okay, let's suppose we have a version of the theorem (cf the corollary) where it is true.

There's a second problem: A complex function rarely has a logarithm that is everywhere continuous, let alone differentiable. (For example, there is no logarithm of $z$ that is continuous on $\C\setminus0$.) That is, we can't just differentiate $\log f_n$, where $f_n$ is one of the factors.

What you can do to resolve this problem is to work locally, because it is a fact that a nonzero analytic function has locally an analytic logarithm. You can locally take analytic logarithms and differentiate, and that's exactly what is done in the proof of Logarithmic Derivative of Infinite Product of Analytic Functions. --barto (talk) (contribs) 17:58, 18 November 2017 (EST)

Ohhhhh. The fact that I was working over $\C \setminus \Z$ completely escaped my mind. In my head I was working over $\R$ where logs are obviously single valued, and the argument would've assumedly followed fine. Thanks! Caliburn (talk) 18:11, 18 November 2017 (EST)

Perhaps then we should create a real version as well, with your original proof. --barto (talk) (contribs) 18:15, 18 November 2017 (EST)

Oh yes, definitely, because it gives a second proof of the complex case by analytic continuation. --barto (talk) (contribs) 18:29, 18 November 2017 (EST)