I have to show that the numerical differentation formula $$ f^\left(IV\right) (x) = \frac{1}{h^4} \left(f(x-2h)-4f(x-h)+6f(x)-4f(x+h)+f(x+2h)\right) + R_h(f) $$
converges.
How do I do that? Do I have to show something related to the truncation error $R_h(f) $ ?
Use the Taylor expansion of $f$ at $x$, $$f(x+kh)=f(x)+f'(x)(kh)+\frac{f''(x)}{2}(kh)^2+\frac{f'''(x)}{6}(kh)^3+\frac{f^{iv}(x)}{24}(kh)^4+o(h^4)$$ with $k=\pm1$ and $k=\pm2$. Now $$\Delta_1:=f(x+h)+f(x-h)=2f(x)+2\frac{f''(x)}{2}(h)^2+2\frac{f^{iv}(x)}{24}(h)^4+o(h^4).$$ Similarly $$\Delta_2:=f(x+2h)+f(x-2h)=2f(x)+2\frac{f''(x)}{2}(2h)^2+2\frac{f^{iv}(x)}{24}(2h)^4+o(h^4).$$ Then $$f(x-2h)-4f(x-h)+6f(x)-4f(x+h)+f(x+2h)=\Delta_2-4\Delta_1+6f(x)\\ =2f(x)+4f''(x)h^2+\frac{32f^{iv}(x)}{24}h^4 -8f(x)-4f''(x)h^2-\frac{8f^{iv}(x)}{24}h^4+6f(x)+o(h^4)$$ which simplifies to $$\frac{(32-8)f^{iv}(x)}{24}h^4+o(h^4)=f^{iv}(x)+o(h^4).$$
