f is analytic and has all orders of derivatives.(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to prove something here but doing this makes me uneasy... is there anything wrong with saying:

If [tex]|f(z)| \leq |z^{n}|[/tex] then,

[tex]|f'(z)| \leq |nz^{n-1}|[/tex] so

[tex]|f''(z)| \leq |n(n-1)z^{n-2}|[/tex]

.

.

.

[tex]|f^{n}(z)| \leq n!z^{0}[/tex]

[tex]|f^{n}(z)| \leq n![/tex]

[tex]|f^{n+1}(z)| \leq 0[/tex]

[tex]|f^{n+1}(z)| = 0[/tex]

I can find no support for taking the derivative of the inequality like this in my calc book.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Can I do this?

**Physics Forums | Science Articles, Homework Help, Discussion**