I am studying finite difference methods on my free time.
Finite Difference table
So based on the table, if I want to extrapolate the next value of my function, I have to use ... $$f(x+dx)=f(x)+f'(x)+f^2(x)+f^3(x)+...+f^n(x)$$ which will be ... $$f(6)=125+61+24+6+0=216=6^3$$
Taylor's expansion on Forward Difference Method
$$f(x+dx)=f(x)+dxf'(x)+dx^2\frac{f^2(x)}{2!}+dx^3\frac{f^3(x)}{3!}+...+dx^n\frac{f^n(x)}{n!}$$
Say my $dx=1$, so if I want to find $f(6)$, I have to divide the derivatives I found with their respective factorial before summing all of the results up, no? If so, ... $$f(6)=\frac{125}{0!}+\frac{61}{1!}+\frac{24}{2!}+\frac{6}{3!}+\frac{0}{4!}=199\ne216$$
Why are their inter/extrapolation between these two different? Which one should I use to perform inter/extrapolation? Also, is the benefit of finding higher derivatives is just to increase the accuracy of inter/extrapolation? Is there any other benefits of finding higher derivatives? or, did I misunderstood?
Apparently you want to use $x=5$. For $f(x)=x^3$, we have $f(5)=125$, $f‘(5)=75$, $f‘‘(5)=30$, $f‘‘‘(5)=6$, which does lead to $216$.