I have a problem where I have to estimate the second derivative of a function at a point using given data. I have $f(1.2)$, $f(1.3)$, and $f(1.4)$. ($h=0.1$) I need to find $f''(1.3)$, so I use the Three-point endpoint formula (which has an error term of $\frac{h^2}{3}f^{(3)}(\xi_0)$) to find $f'(1.2)$ and $f'(1.4)$, and then use these two values with the three-point midpoint formula (which has an error of $\frac{h^2}{6}f^{(3)}(\xi_i)$) to find $f''(1.3)$.
So how to I compute the overall error with the operations I've done (endpoint formula twice, midpoint formula once)? Is it additive? Multiplicative?
I am confused by your approach which can have significant errors due to the effect of roundoff error, truncation, etc.
Why wouldn't you use one of the three point formulas for the second derivative directly?
For example:
$$\tag 1 \displaystyle f''(x_0) = \frac{1}{h^2}\left[f(x_0 - h) -2 f(x_0) +f(x_0 + h)\right]-\frac{h^2}{12} f^{(4)}(\xi),$$
for some $\xi$, where $x_0 - h \lt \xi \lt x_0 + h$.
You are given $f(1.2)$, $f(1.3), f(1.4) ~~\text{and}~~ h=0.1.$
From $(1)$, we can write:
$$\displaystyle f''(1.3) = 100\left[f(1.2) -2 f(1.3) +f(1.4)\right]-\frac{1}{1200} f^{(4)}(\zeta)$$
Are you being asked to do it by finding first derivatives and then to use the second derivative approach?