I'd like to find the formula for the 3rd order numerical derivative in order to further implement a Python function for a time series (Python only have a function from scipy which takes an unknown variable). In order to demonstrate and calculate the general formula I just took the 2 main formulas : $$f'(x) = \frac{f(x+h) - f(x-h)}{2h}$$ $$f''(x) = \frac{f(x+h) - 2f(x) + f(x+h)}{h^2}$$
then I made the following hypothesis that : $$ f'''(x) = \frac{f''(x+h) - f''(x-h)}{2h}$$ where, $$ f''(x+h) = \frac{f(x+h+h)-2f(x+h)+f(x+h-h)}{2h}$$ $$ = \frac{f(x+2h)-2f(x+h)+f(x)}{2h}$$ and same calculation for the other, $$ f''(x-h) = \frac{f(x-2h)-2f(x-h) + f(x)}{2h}$$
which gives me at the end : $$f'''(x) = \frac{f(x+2h) + 2h(f(x-h) - f(x+h))-f(x-2h)}{4h^2}$$
But I just can't find anywhere if my calculation is right or not,
- Is my result good ?
- What about the calculation method ?
I'm doubting as I'm not studying at school anymore I don't have a teacher to correct me.
If you want to approximate $f'''(x)$ in terms of $f$ evaluated at $x-2h, x-h, x+h,x+2h$, i.e., $$f'''(x)\approx af(x-2h)+bf(x-h)+cf(x+h)+df(x+2h), $$ you better make sure that the result for $f'''(0)$ is zero for $f(x)=1$, $f(x)=x$, or $f(x)=x^2$, and is $=6$ for $f(x)=x^3$. This leads us to the conditions $$ a+b+c+d=0$$ $$ 2a+b-c-2d=0$$ $$ 4a+b+c+4d=0$$ and $$-8a-b+c+8d=\frac 6{h^3}. $$ The first three quickly lead to $d=-a$, $c=-b$, $b=-2a$, so that the last becomes $-8a+2a+2a-8a=\frac6{h^3}$ and ultimately $$\tag1 f'''(x)\approx \frac{-f(x-2h)+2f(x-h)-2f(x+h)+f(x+2h)}{2h^3}$$ The interesting question is whether it would be preferable to use other points of evaluation; four points seems adequate because we want to impose four conditions; symmetric with respect to $x$ seems adequate as well; however, evenly spaced points $x-3h,x-h,x+h,x+3h$ look attractive as well. By the same method as above, we can determine the best coefficients and find $$\tag2 f'''(x)\approx\frac{-f(x-3h)+3f(x-h)-3f(x+h)+f(x+3h)}{8h^3}$$ Which of these fomulae is better? Incidentally, both give us $f'''(0)=0$ for $f(x)=x^4$ (or any even function). For $f(x)=x^5$, we obtain $$ f'''(x)\approx 30h^2$$ from $(1)$ and $$ f'''(x)\approx 60h^2$$ from $(2)$. In this respect, it seems that $(1)$ is better, but take into account that by the wider spread of points, $(2)$ effectively uses a $\frac32$ times bigger $h$, and $(3/2)^2>\frac{60}{30}$.