Can I do this?

  • Thread starter futurebird
  • Start date
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.
Last edited:
Taking the derivative of both sides of the inequality is not justified. For example, consider f(x) = sin(e^x) for x > 1. |f(x)| < x^2 but it is not that case that |f'(x)| = |e^x cos(e^x)| < 2x (for example, when x=ln(2*pi)). However, the fact that youre comparing an analytic function to a polynomial in z makes me think that Taylor's theorem may help you.
Last edited:
Thanks. IT just seemed way too easy.

How do they solve differential inequalities then, I wonder.

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