Room for evinda and LutzL

evinda

17:33

Hello @LutzL !!!

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 $f'''(x)$?

LutzL

19:45

One-sided formulas of minimal length are always worse than symmetric formulas. I'd guess that your second error term is wrong, I'd expect that because of the symmetry the error is O(h^2).

f(x+h)-f(x-h) = 2f'(x)h+1/3·f'''(x)h^3+1/60·f^{(5)}(x)h^5+...

f(x+2h)-f(x-2h) = 4f'(x)h+8/3·f'''(x)h^3+8/15·f^{(5)}(x)h^5+...
so in the difference
[f(x+2h)-f(x-2h)]-2[f(x+h)-f(x-h)] = 2·f'''(x)·h^3 + 1/2·f^{(5)}(x)h^5+...

with more careful computation it should be possible to fold the remainder into the 5th derivative, so I'd expect at most a factor of 17/30 for h²·||f^{(5)}||_{\infty}

evinda

A ok... I will redo the calculations to make the expansion till the fifth derivative.
@LutzL Do we have to do so also for the first quotient difference?

@LutzL Could you explain me what you mean with "One-sided formulas of minimal length" ?

LutzL

20:01

The pure one-sided, uncentered powers of the difference operator.

evinda

If we have a difference quotient that is symmetric do we expect that the error is O(h^2) and if it is not symmetric that it is O(h) ? @LutzL

Also does this imply: $$\left| \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}-f'''(x) \right| \leq \frac{22}{4} h ||f^{(4)}||_{\infty}$$
that the difference quotient is an approximation of f''' or do we have to say that it tends to 0 while h->0 ? @LutzL

LutzL

by the extended mean value theorem

[ f(x+2h)-2f(x+h)+2f(x-h)-f(x-2h) - 2h^3·f'''(x) ] / [h^5]
=
[ 8f'''(x+2s)-2f'''(x+s)-2f'''(x-s) + 8f'''(x-2s) - 12 f'''(x) ] / [ 60 s^2]
=
[ [f'''(x+2s)-2f'''(x+s)+f'''(x)] + [f'''(x)-2f'''(x-s)+f'''(x-2s)] + 7[f'''(x+2s)-2f'''(x) +f'''(x-2s)] ] / [ 60 s^2]
=
[ f^{(5)}(x+s+t_1) + 28 f^{(5)}(x+2t_2) + f^{(5)}(x-s+t_3) ] / [ 60 s^2]

with t_{123} \in [-s,s] and s\in [0,h]

last equation line should be

=
[ f^{(5)}(x+s+t_1) + 28 f^{(5)}(x+2t_2) + f^{(5)}(x-s+t_3) ] / [ 60 ]

Both formulas are approximations for f'''(x) (and f'''(x-h) etc.), the first with error O(h), the second with error O(h²).

evinda

@LutzL Did you use this? mathworld.wolfram.com/ExtendedMean-ValueTheorem.html

LutzL

Yes, with, in that notation, f(a)=0=g(a), f the difference minus the differential, g(h)=h^5, a=0, b=h

evinda

20:19

Instead of using the extended mean value theorem can't we just use Taylor expansion?
For the first difference quotient if we would expand till the fifth term, we wouldn't get something that could help us to find the right bound, right? @LutzL

LutzL

20:38

Yes, the problem is always to assemble the remainder term with the smallest possible coefficient. Taylor for each term separately is too rough for that.

Of course to just get the error order with any coefficient, Taylor is sufficient.

evinda

So having done Taylor for both can we say that:

We notice that the order of accuracy of $\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}$ is O(h) and the order of accuracy of $ \frac{f(x+2h)-2f(x+h)+2f(x-h)-f(x-2h)}{2h^3}$ is $O(h^2)$ and thus the second difference quotient is a better approximation.

@LutzL Or could we explain it better?

evinda