Real Analysis – Midpoint Approximation Error Bound for $f(x) = x$

Question

To my understanding the error bound is given by
\begin{equation*}
E(\Delta x) = \frac{b-a}{24}M(\Delta x)^2 = \frac{(b-a)^3}{24n^2}M\text{.}
\end{equation*}
where $M$ is chosen by the following rule $|f''(x)| \le M$, so in the case $f(x) = x$, since the second order derivative is $0$, I can choose M arbitrarily close to $0$ therefore the error can be arbitrarily small?
I know that this is wrong, so how does one find the error bound estimate in this case ?

Answer 1

Using your notation, the actual error formula for the midpoint method is $$ E(\Delta x)= \frac{b-a}{24} (\Delta x)^2f''(\xi), \quad \xi \in (a,b). $$

So, since in this case $f''(\xi) = 0$, so is the error.