Inequality: $$-\frac{22}{24}h^2 ||f^{(4)}||_{\infty}\leq 0$$

[mathmari]

Hey!

Does the following inequality hold?

$$|-\frac{5}{24}h^2f^{(4)}(x_1)+\frac{64}{24}h^2f^{(4)}(x_2)-\frac{81}{24}h^2f^{(4)}(x_3)| \\ \leq |-\frac{5}{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2|\max_x |f^{(4)}(x)| \\ = |-\frac{22}{24}h^2| ||f^{(4)}||_{\infty}= \frac{22}{24}h^2 ||f^{(4)}||_{\infty}$$

where $h>0$

 

[Opalg]

Hey!

Does the following inequality hold?

$$|-\frac{5}{24}h^2f^{(4)}(x_1)+\frac{64}{24}h^2f^{(4)}(x_2)-\frac{81}{24}h^2f^{(4)}(x_3)| \\ \leq |-\frac{5}{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2|\max_x |f^{(4)}(x)| \\ = |-\frac{22}{24}h^2| ||f^{(4)}||_{\infty}= \frac{22}{24}h^2 ||f^{(4)}||_{\infty}$$

where $h>0$

No, because $f^{(4)}(x_2)$ might have a different sign from $f^{(4)}(x_1)$ and $f^{(4)}(x_3)$. Unless you know that $f^{(4)}(x_1)$, $f^{(4)}(x_2)$ and $f^{(4)}(x_3)$ all have the same sign, the best you can say is that $$\left|-\frac{5}{24}h^2f^{(4)}(x_1)+\frac{64}{24}h^2f^{(4)}(x_2)-\frac{81}{24}h^2f^{(4)}(x_3)\right| \leqslant \left(\frac5{24} + \frac{64}{24} + \frac{81}{24}\right)h^2\|f^{(4)}\|_{\infty} = \frac{150}{24}h^2\|f^{(4)}\|_{\infty}.$$

 

[mathmari]

I used the Taylor expansion three times, once for $f(x+h)$, once for $f(x+2h)$ and once for $f(x+3h)$ and the $f(x_1), f(x_2), f(x_3)$ of the above expression are the remainders of each expansion. Do we know if they have all the same sign?

 

[mathmari]

I want to show that $$|\delta_{h,r}f(x)- f''(x)| \leq \frac{11}{12} h^2 ||f^{(4)}||_{\infty}$$ where $$\delta_{h,r}f(x)=\frac{1}{h^2} (2f(x)-5f(x+h)+4f(x+2h)-f(x+3h))$$

I applied the Taylor expanson at $f(x+h)$, $f(x+2h)$ and $f(x+3h)$ and $f(x_1), f(x_2), f(x_3)$ are the corresponding remainder of each expansion.

$$f(x+h)=f(x)+h f'(x)+\frac{h^2}{2} f''(x)+\frac{h^3}{6} f'''(x)+\frac{h^4}{24} f^{(4)}(x_1), x_1 \in (x,x+h)$$

$$f(x+2h)=f(x)+2hf'(x)+2h^2 f''(x)+\frac{4}{3} h^3 f'''(x)+\frac{16}{24}f^{(4)}(x_2), x_2 \in (x,x+2h)$$

$$f(x+3h)=f(x)+3hf'(x)+\frac{9}{2}h^2 f''(x)+\frac{27}{6}h^3 f'''(x)+\frac{81}{24} h^4 f^{(4)}(x_3), x_3 \in (x,x+3h)$$Substituting at $$\delta_{h,r}f(x)=\frac{1}{h^2} (2f(x)-5f(x+h)+4f(x+2h)-f(x+3h))$$ we get
$$\delta_{h,r}=f''(x)+\frac{h^2}{24}(64 f^{(4)}(x_2)-5f^{(4)}(x_1)-81 f^{(4)}(x_3))$$

Since $$|-\frac{5}{24}h^2f^{(4)}(x_1)+\frac{64}{24}h^2f^{(4)}(x_2)-\frac{81}{24}h^2f^{(4)}(x_3)| \\ \leq |-\frac{5}{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2|\max_x |f^{(4)}(x)| \\ = |-\frac{22}{24}h^2| ||f^{(4)}||_{\infty}= \frac{22}{24}h^2 ||f^{(4)}||_{\infty}$$
we have

$$|\delta_{h,r}f(x)- f''(x)| \leq \frac{11}{12} h^2 ||f^{(4)}||_{\infty}$$

($h>0$)

Is this correct? Is there somewhere a mistake?