Question about Taylor's Theorem

Question

I am reading a proof from my professor (proof of convergence of Secant Method in Numerical Analysis), and she makes use of Taylor series. We have a function $f \in C^2(a, b)$, and a point $z \in (a, b)$. She writes $$f(z+h) = f(z) + f'(z)h + \dfrac {f''(z)}{2}h^2 + O(h^3).$$

However, I don't understand how we get this result.

I thought that Taylor's Theorem said: Let $f \in C^{k+1}(a, b)$. Then for every $z \in (a, b)$ and for every $h$ satsifying $z+h \in (a, b)$, we have $$f(z+h) = f(z) + f'(z)h + \cdots + \dfrac {f^{(k)}(z)}{k!}h^k + \dfrac {f^{(k+1)}(\xi_h)}{(k+1)!} h^{k+1}$$ for some $\xi_h$ between $z$ and $z+h$.

Thus, if $f \in C^2(a, b)$, I could see us writing $$f(z+h) = f(z) + f'(z)h + \dfrac {f''(\xi_h)}{2}h^2$$ and (I guess assuming the second derivative is bounded) we could write $$f(z+h) = f(z) + f'(z)h + O(h^2)$$

However, I don't understand how to get what my professor wrote.

Thanks for the help.

Answer 1

Why did you stop at the square term? In your use of Taylor's theorem, we can say $$ f(z+h)=f(z)+f'(z)h + \frac{f''(z)}{2!}h^2 + \frac{f'''(\xi_h)}{3!}h^3. $$ The last term is $O(h^3)$ since obviously $h^3$ is, and for the other term in the product, $z\leq \xi_h\leq z+h$ so as $h\to 0$, we see that $\frac{f'''(\xi_h)}{3!}\to \frac{f'''(z)}{3!}=O(1).$

Answer 2

You an also write $$ f(z+h)=f(z)+f'(z)h+\frac{f''(z)h^2}2+\frac{h^2}2(f''(\zeta_h)-f(z)) $$ and by $f\in C^2$ you get that the last term is $o(h^2)$, which might still be sufficient for your purposes. To get $O(h^3)$ from this approach, you need that $f''$ is Lipschitz, that is, $f\in C^{2,1}([a,b])$.