Order of growth of a complex polynomial

Question

Question

I want to determine the order of growth of a complex polynomial $p(z)$.

My attempt

We pick $$p(z)=a_nz^n+\ldots +a_1z+a_0$$

then we know that

$$|p(z)| \le |a_nz^n|+\ldots +|a_1z|+|a_0|$$

since $e^x$ is increasing we have that

$$|p(z)|<e^{|p(z)|} \le e^{|a_nz^n|+\ldots +|a_1z|+|a_0|}=e^{|a_n||z^n|+\ldots +|a_1||z|+|a_0|} \le max\{|a_i|\}e^{max\{|a_i|\}|z|^{n}}$$

The thing is that for small values of $z$ for example $z=0$ the above bound doesn't work and my definition says that the above should hold for all $z \in \mathbb{C}$. I am almost sure that the order of growth should be $n$ but I don't know if this is correct or if there is a better proof to get which is the order of growth of a complex polynomial that you can provide.

Definitions.

Let $f$ be an entire function. If there exist a positive number $\rho$ and constants $A,B >0$ such that

$$|f(z)| \le A e^{B|z|^{\rho}}$$

for all $z \in \mathbb{C}$ then we say that $f$ has order of growth $\le \rho$. We define the order of growth of $f$ as

$$\rho_f=inf\{ \rho \}$$

Dou you want the growth order of $p(z)$ or of $e^{p(z)}$? – Hagen von Eitzen Apr 15 '16 at 16:21 — Hagen von Eitzen, Apr 15 '16 at 16:21
The order of growth of $p(z)$ – user162343 Apr 15 '16 at 16:22 — user162343, Apr 15 '16 at 16:22

score 8 · Answer 1 · edited Oct 02 '22 at 04:07

For all real $r$ we have $e^r\ge 1+r$, hence for positive $r$ we have $$e^r=(e^{r/n})^n\ge (1+r/n)^n\ge 1+\frac1{n^n}\cdot r^n $$ or in other words with $\rho=\frac1n$ and $r=(n\left|z\right|)^\rho$, we have $$1+|z|\le e^{n^\rho|z|^\rho} $$ and so for any a linear polynomial $p(z)=az+b$ we find that for $n\in\Bbb N$, $\rho=\frac 1n$, $B=n^\rho$, $A=\max\{|a|,|b|\} $ $$ |az+b|\le|a|\cdot|z|+|b|\le A(1+|z|)\le Ae^{B|z|^\rho}.$$ As this works for any $n\in\Bbb N$, we conclude that $az+b$ has order of growth $0$. The key observation is that even the slightest exponetial grows faster than linear. (It also grows faster than any polynomial, bit as we have the fundamantal theorem of algebra available, we can reach the result for arbitrary polynomials by a helpful result that is interesting in its own right:

Next note that the growth rate of the product of two functions is at most the max of their growth rates. Indeed, if $|f(z)|\le Ae^{B|z|^\rho}$ and $|g(z)|\le A'e^{B'|z|^{\rho'}}$ then $$|f(z)g(z)|\le AA'e^{B|z|^\rho+B'|z|^{\rho'}} $$ and as $$B|z|^\rho+B'|z|^{\rho'}\le (B+B')|z|^{\max\{\rho,\rho'\}} +\max\{B,B'\}$$ we find $$|f(z)g(z)|\le AA'e^{\max\{B,B'\}}\cdot e^{(B+B')|z|^{\max\{\rho,\rho'\}}}. $$

So finally, as every polynomial is the product of linear factors, we conclude that every polynomial has order of growth $0$.

Thanks let me check your answer :) just give me some time to get it :) — user162343, Apr 15 '16 at 17:02
I think you have a typo right? when you wrote $(e^{r/n})^{2}=e^{r}$ — user162343, Apr 15 '16 at 22:00
You state the inequality $1+|z|\leq e^{n^{\rho}|z|^{\rho}}$ I see don't clearly how do you conclude that, are you sure about the calculations? — jnaf, Jun 03 '17 at 02:54

score 0 · Answer 2 · answered May 08 '20 at 21:33

Since polynomials have no poles and are asymptotically determined by their leading term, the inequality $ |p(z)| \leq Ae^{B|z|^\rho} $ for $ \rho > 0 $ actually follows from the fact that $$ \frac{e^{x^\rho}}{x^N} \longrightarrow \infty \quad \text{ as }\quad x \longrightarrow \infty $$ for all $ N \geq 1 $. In other words, we only have to trap the end behavior of the polynomial growth. Using basic calculus, letting $ y = e^{x^\rho} / x^N $, we get that $ \log y = x^\rho - N \log x $ which diverges to $ +\infty $ as $ x $ grows large.

Order of growth of a complex polynomial

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