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My lecture note says for $f\in C^2 _c $ the $|f(z+h)-(f(z)+f'(z)h+\frac{1}{2}f''(z)h^2)|\le \omega(h)h^2$ where $\omega$ is the modulus of continuity of $f''$.

I have been thinking this for an hour and I cannot get the answer without assuming third derivative. Could anyone help to show why this is true?
Thanks!

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The mean value form of the Taylor series is $$f(z+h) = f(z) + f'(z) h + \frac{1}{2} f''(\xi) h^2,$$ for some $\xi$ between $z$ and $z+h$. Then use $|f''(\xi) - f''(z)| \le \omega(|\xi-z|) \le \omega(h)$.

More detail: substituting the mean value form for $f(z+h)$ gives

\begin{align} &\left|f(z+h) - f(z) - f'(z) h - \frac{1}{2} f''(z) h^2\right|\\ &\le \frac{1}{2} |f''(\xi) - f''(z)| h^2\\ &\le \frac{1}{2} \omega(h) h^2. \end{align}

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    Thanks for the reply! I am a little confused here, do you mean that if the error bound$w(h)h^2$ works for $|f(z+h)-(f(z)+f'(z)h)|$ it also works for $|f(z+h)-(f(z)+f'(z)h+\frac{1}{2}f''(z)h^2)|$? – 2017-02-02
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    @Sherry See my edit – 2017-02-02
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    Oh, I see it now. Thank you so much! – 2017-02-02