MHB Comparing Difference Quotients for Approximating $f'''(x)$

[evinda]

Hello! (Wave)

I have to show that the following difference quotients are approximations of $f'''(x)$.

$$\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3} \\ \frac{f(x+2h)-2f(x+h)+2f(x-h)-f(x-2h)}{2h^3}$$

Which approximation is more accurate? Justify your answer.I found the Taylor expansion of $f(x+3h) , f(x+2h), f(x+h)$ and found that

$$\left| \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}-f'''(x) \right| \leq \frac{22}{4} h ||f^{(4)}||_{\infty}$$

Have we shown now that $\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}$ is an approximation of $f'''(x)$?
Or do we have to show that the above tends to $0$ ?

Similarly, I found that

$$\left| \frac{f(x+2h)-2f(x+h)+2f(x-h)-f(x-2h)}{2h^3}-f'''(x) \right| \leq \frac{3}{4} h ||f^{(4)}||_{\infty}$$

The second difference quotient is a better approximation because of the smaller constant, right ?

But how could we justify it formally? (Thinking)

 

[I like Serena]

The second difference quotient is a better approximation because of the smaller constant, right ?

But how could we justify it formally? (Thinking)

Hey evinda! (Smile)

I think that what you have is already sufficiently formal.

The only thing I can think of to improve it, is to set the expressions equal to $\frac{22}{4} h f^{(4)}(x+\theta h)$ respectively $\frac{3}{4} h f^{(4)}(x+\xi h)$, where $0\le\theta\le 1$ and $0\le\xi\le 1$.
I'd consider that sharper than giving an upper bound. (Thinking)

 

[evinda]

Hey evinda! (Smile)

I think that what you have is already sufficiently formal.

The only thing I can think of to improve it, is to set the expressions equal to $\frac{22}{4} h f^{(4)}(x+\theta h)$ respectively $\frac{3}{4} h f^{(4)}(x+\xi h)$, where $0\le\theta\le 1$ and $0\le\xi\le 1$.
I'd consider that sharper than giving an upper bound. (Thinking)

Nice... Thanks a lot! (Smile)