Numerically differentiated values and their corresponding x-coordinates

Question

If we numerically differentiate a given time series data consisting of N points by finite forward difference method, we will have N-1 points corresponding to first derivative. If it is a second derivative, we will have N-2 points and so on.

Let us say for the first derivative

$$ \approx\frac{f(x+\Delta x)-f(x)}{\Delta x} $$

I have searched several books and webpages, but no one explicitly describes what is the corresponding x value for the numerically differentiated value if we wish to plot those N-1 values.

In most science and engineering applications, we will not have an exact formula for f(x). One would use a set of data points ($x_1$, $y_1$), ($x_2$, $y_2$), . . . , ($x_n$, $y_n$) available to describe the functional dependence y = f(x). Many users ignore the $x_1$ and use the remaining $x_i$ for plotting N-1 differentiated points.

Others say that the first differentiated value corresponding to the average of $\frac{x_1+x_2}{2}$ i.e., this belongs to the center of $x_1$ and $x_2$.

What is the mathematically and rigorously correct way of dealing with $N-1$ values for the first derivative and $N-2$ values for the second derivative when we have N x-values? If we wish to plot them, how should we modify the x-coordinates?

EDIT: The reason for interest in the x-coordinates is utilitarian. The reason is that in chemical analysis applications, the derivative is used to locate the inflection points of titraton curve or detect a hidden peak in an over lapped spectrum. In such cases the interest is not in the accuracy of the value of the derivative but its corresponding x-coordinate. For example, in a potentiometric titration curve, the end point of titration is located by the first derivative, the inflection point's x-coordinate is the required volume.

Thanks.

Answer 1

The OP asks for a "reputable source", I would think that Press and Teukolsky's Numerical Recipes [section 5.7 in The Book] qualifies as such. As they explain, if you approximate $f'(x)\approx h^{-1}[f(x+h)-f(x)]$ the truncation error (from higher order terms in the Taylor expansion) is of first order in the small increment $h$. You can improve this to a truncation error of second order by symmetrizing, $f'(x)\approx (2h)^{-1}[f(x+h)-f(x-h)]$.

This can be readily generalized to higher order derivatives, just by repeatedly differentiating each term. The second derivative becomes $$f''(x)=h^{-2}\left[f(x+h)+f(x-h)-2f(x)\right].$$

_{This is equivalent to $\frac{1}{4}h^{-2}[f(x+2h) + f(x-2h) - 2f(h)]$, as discussed at this MSE question.}

Answer 2

A variant of the argument in Carlo Beenakker's answer: if the $x_i$ are equispaced points with distance $h$ one from the next, then $$\frac{f(x_{i+1})-f(x_i)}{h} - f'(x_i) = O(h),$$

$$ \frac{f(x_{i+1})-f(x_i)}{h} - f'(\frac{x_{i}+x_{i+1}}{2}) = O(h^2) $$ (for a sufficiently regular $f$).

This suggests that the choice that minimizes the error is assigning to each derivative the $x$ coordinate of the midpoint of the grid segment it was calculated on. (But in the end it's a choice, there is no 'right' or 'wrong' here.)