Daniel Fischer

less tired

@infinity hm?

Possibly.

@infinity Somewhat surprised. It's been snowing most of the day, and after a while it stopped melting, so we got about four inches of snow today. That's somewhat rare in the last years.

@infinity ho

well ... tired

hi

In terms of Laurent expansion, a removable singularity means $a_n = 0$ for all $n < 0$, a pole means there is a $k < 0$ with $a_k \neq 0$ and $a_n = 0$ for all $n < k$, an essential singularity means $a_n \neq 0$ for infinitely many $n < 0$.

A pole is an isolated singularity that becomes removable after multiplying with $(z - z_0)^k$ for a suitable $k \in \mathbb{N}\setminus \{0\}$.

That's a simple pole. Add higer powers for the general case.

And isolated, in the Casorati-Weierstraß context. Some people call non-isolated singularities essential too.

Neither removable nor a pole.

ho?

@infinity No, that was $\lvert a_n\rvert$. $M$ is a bound for $\lvert f(z)\rvert$ on the circle $\lvert z\rvert = R$.

Note that the value of $\lvert f(z_0)\rvert$ is irrelevant, all that matters is that it's nonzero and independent of $r$ (and $R$).

Now choose appropriate $R$ and $r$, and recall the bound for $M$. (But you misplaced the logarithm, it's $\log \frac{M}{\lvert f(z_0)\rvert}$.)

Look at the last few lines before the section about the Poisson-Jensen formula.

(Note, some care has to be applied because Jensen's formula assumes $f(0) \neq 0$.)

On the other hand for a function with the given growth bound, what (roughly) does Jensen's formula say about the number (if the function doesn't vanish identically)?

Wrong constant. We have a disc, which brings in $\pi$, and we only know zeros in the first quadrant a priori, so $\pi/4$. But that isn't important, at least $c\cdot r^2$ for a $c > 0$ is what matters.

For large $r$, roughly how many zeros with $\lvert z\rvert < r$ are immediately implied by the assumptions on $f$?

So ask Jensen how many zeros a non-constant $f$ with the given growth condition can have in $\lvert z\rvert < r$, and look how many your $f$ must have. Conclude $f = 0$.

Or Jensen's formula?

Have you ever encountered Hadamard's theory of entire functions of finite order?

And the task is to find that both together are very restrictive conditions.

And it's given that $f$ has a certain growth.

And if you pick the minimising $r$, $\lvert a_n\rvert \leqslant e^n/n^n$. But that doesn't really help. What you need is a relation between the growth and the distribution of zeros.

Note that you can completely ignore the 2018, just consider $e^{-2018}\cdot f$ instead of $f$. That is just there because that question was given three years (two-and-a-half ...) ago, and some people find it cute to have the year number in a problem.

@infinity Should be $\lvert re^{it}\rvert + 2018$ in the exponent, the bound you're given is $e^{\lvert z\rvert + 2018}$.

Either Hadamard's theory, or Jensen's formula works.

Vaguely. But it requires more than just the growth, you also need to have a relationship between the growth and the distribution of zeros.

No, because it would be more correct to write $M(r)$ rather than just $M$, and $M(r)$ grows, typically much faster than $r^n$.

Supposing $M$ means what it ought to mean.

It's $r^n$ rather than $r^{n+1}$, don't forget that the length of the circle is $2\pi\cdot r$.

Not quite.

Well, now, after I fixed the forgotten $n!$.

That's Cauchy's integral formula (for the $n$-th derivative).

Or, we can estimate it for all $z$ with $\lvert z - z_0\rvert < \rho$, where $0 < \rho < r$, simultaneously.

We can specifically pick z = z_0$.

Now throw in the standard estimate (aka ML-estimate), and you get Cauchy's inequality for a point ($z = z_0$).

Well, we have the integral formula, $$f^{(n)}(z) = \frac{n!}{2\pi i}\int_{\lvert \zeta - z_0\rvert = r} \frac{f(\zeta)}{(\zeta - z)^{n+1}}\,d\zeta$$ if $\lvert z-z_0\rvert < r$.

That is the case of a single point mentioned above.

$a_n$ is the $n$-th coefficient of the Taylor series of $f$ about $0$, i.e. $a_n = \frac{f^{(n)}(0)}{n!}$.

For the specific problem at hand, if you have it available, use Hadamard's theory of entire functions of finite order.

The inequalities you obtain from Cauchy's integral formula (for derivatives). Bounds for the modulus (of the function or one of its derivatives) at a point or on some set in terms of a bound for the modulus of the function on some contour (typically a circle) surrounding the point/set in question.

C.R. equations are hardly ever used. This type of problem usually uses Liouville's theorem, Cauchy inequalities or related techniques (e.g. Phragmén-Lindelöf).

Yeah, I think Casorati appears only that one time.

Probably Casorati-Weierstraß.

If we don't interpret "speak" literally.

aye