How Can We Show the Bound for $\delta_{h,-,2} f(x)$?

[evinda]

Hello! (Wave)

We define $\delta_{h,-,2} f(x) :=- \left( \delta_{h,-}+\frac{h}{2} \delta_{h,-}^2 \right) f(x)=\frac{1}{2h} \left( -f(x-2h)+4f(x-h)-3f(x)\right)$.

Let $f \in C^3[a,b]$. Then:

$$| \delta_{h,-,2} f(x)- f'(x)|\leq h^2 ||f'''||_{\infty}$$

I have tried the following:

$$f(x-h)=f(x)-h f'(x)+\frac{h^2}{2} f''(x)-\frac{h^3}{6} f'''(\xi_1), \xi_1 \in (x-h,x) \\ f(x-2h)=f(x)-2hf'(x)+2h^2 f''(x)-\frac{4}{3} h^3 f'''(\xi_2), \xi_2 \in (x-2h,x)$$

So $\delta_{h,-,2} f(x)=-f'(x)+\frac{2}{3} h^2 f'''(\xi_2)-\frac{1}{3} h^3 f'''(\xi_1)$.

But how can we say something about $| \delta_{h,-,2} f(x)- f'(x)|$ now that we have $\delta_{h,-,2} f(x)=-f'(x)+ \dots$ instead of $\delta_{h,-,2} f(x)=f'(x)+ \dots$ ? (Thinking)

 

[jvicens]

Hello! Based on the formula provided, we can see that $\delta_{h,-,2} f(x)$ is actually an approximation of $f'(x)$, with an error term of $h^2 ||f'''||_{\infty}$. This means that as $h$ gets smaller, the error term also decreases, and the approximation becomes closer to the true value of $f'(x)$. So even though the formula may seem to be calculating $-\delta_{h,-,2} f(x)$ instead of $\delta_{h,-,2} f(x)$, the error term ensures that it is still a valid approximation of $f'(x)$.

In other words, the negative sign in front of $f'(x)$ does not affect the accuracy of the approximation, as long as the error term is small enough. So we can still say that $|\delta_{h,-,2} f(x)- f'(x)|\leq h^2 ||f'''||_{\infty}$, regardless of the negative sign. I hope this helps clarify your doubts. Let me know if you have any other questions.