# Thread: Analytic function - estimation

1. ## Analytic function - estimation

Hi,

How can I show the following estimation:

$|c_n| \leq (\log||\phi||+||a||r)\frac{2}{r^n}$,

where

$c_n=\frac{1}{\pi i} \int_{|\lambda|=r}\log|\psi (\lambda)|\frac{d\lambda}{\lambda^{n+1}}$;
$\psi (\lambda)=\phi[exp(\lambda a)]$

I can show that

$\log|\psi (\lambda)| \leq \log||\phi||+|\lambda|\cdot||a||$

So that I guess that I should show

$\frac{1}{\pi i} \int_{|\lambda|=r}\frac{d\lambda}{\lambda^{n+1}} \leq \frac{2}{r^n}$

Probably I have to use the following estimation

$\|\int_c f(z)dz \| \leq ML$,

where

$|f(z)| \leq M$ and $L$ - lenght of the contour (in this problem $L=2\pi r$)

It is everything what I know and I've still problem with this estimation. From where we have $\frac{2}{r^n}$?

2. Originally Posted by Arczi1984
Hi,

How can I show the following estimation:

$|c_n| \leq (\log||\phi||+||a||r)\frac{2}{r^n}$,

where

$c_n=\frac{1}{\pi i} \int_{|\lambda|=r}\log|\psi (\lambda)|\frac{d\lambda}{\lambda^{n+1}}$;
$\psi (\lambda)=\phi[exp(\lambda a)]$

I can show that

$\log|\psi (\lambda)| \leq \log||\phi||+|\lambda|\cdot||a||$

So that I guess that I should show

$\frac{1}{\pi i} \int_{|\lambda|=r}\frac{d\lambda}{\lambda^{n+1}} \leq \frac{2}{r^n}$

Probably I have to use the following estimation

$\|\int_c f(z)dz \| \leq ML$,

where

$|f(z)| \leq M$ and $L$ - lenght of the contour (in this problem $L=2\pi r$)

It is everything what I know and I've still problem with this estimation. From where we have $\frac{2}{r^n}$?
Let $f(\lambda)=\frac{\log\left|\phi e^{\lambda a} \right|}{\lambda^{n+1}}$.

$|f(\lambda)| = \frac{\log\left|\phi e^{\lambda a} \right|}{r^{n+1}}$ since $|\lambda| = r$.

So $|f(\lambda)| \leq \frac{\log||\phi||+|\lambda|\cdot ||a||}{r^{n+1}} = \frac{\log||\phi||+||a||r}{r^{n+1}} = M$.

Thus $|c_n| \leq \frac{1}{\pi}ML = 2rM = (\log||\phi||+||a||r)\frac{2}{r^n}$.