# Why are higher order derivatives linked to the higher orders of convergence?

My textbook defined the order of convergence by (original image link)

For an iterative process of the form $x_{n+1} = g(x_n)$, the order of convergence is first order when $|g'(x)| < 1$ but $|g'(\alpha)| \neq 0$. The order of convergence is second order when $|g'(x)| < 1$ and $|g'(\alpha)| = 0$ but $|g''(\alpha)| \neq 0$. Similar statements can also be made about higher orders of convergence.

Now, my first question is:
Is the order of convergence third if the absolute value of the derivative of $g$ of $x$ less than one and the absolute value of the derivative of $g$ of $x$ at alpha is equal to $0$, and the absolute value of the second derivative of $g$ of $x$ at alpha is equal to zero also, but the absolute value of the third derivative of $g$ of $x$ at alpha is not equal to zero?

And, my second question is:
Why does the inclusion of higher derivatives suggest that an iterative function converges more quickly?

If $x_n$ converges to $\alpha$, then $\alpha$ is a fixed point of $g$, that is $g(\alpha)=\alpha$.
Now suppose that $g^{(k)}(\alpha)=0$ for $k=1,\cdots,q-1$ but $g^{(q)}(\alpha)\ne0$.
Let $e_n = x_n - \alpha$. Then, by Taylor's theorem, $$e_{n+1}=\frac{g^{(q)}(\xi_n)}{q!}e^q_n$$
It is this sense that higher derivatives of $g$ at the fixed point are related to the order of convergence of the sequence.