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.

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# Homework Help: Can I do this?

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