- #1

- 272

- 0

f is analytic and has all orders of derivatives.

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.

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.

Last edited: