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evinda
evinda
17:35
@LutzL So can't it be $\delta_{h,-,2} f(x) := \left( \delta_{h,-}+\frac{h}{2} \delta_{h,-}^2 \right) ?
 
2 hours later…
LutzL
LutzL
19:14
Sorry, no, the last part of my last remark was off, probably was too late in the night. Up to the discussed sign the original formula is correct, as was aqain confirmed by the first part of my last remark.
evinda
evinda
@LutzL So you mean that it is right with - ?
LutzL
LutzL
Without the -? Where 3f(x) in the end has positive sign.
evinda
evinda
So it is equal to $\frac{-4f(x-h)+3f(x)+f(x-2h)}{2h}$ ? @LutzL
@LutzL And I have also an other question.
@LutzL $$\delta_{h, \text{right}}f(x)=(\delta_{h,+}^2-h \delta_{h,+}^3) f(x)=\frac{1}{h^2} (2f(x)-5f(x+h)+4f(x+2h)-f(x+3h))$$

Does it hold that $$|\delta_{h, \text{right}}f(x)-f''(x)| \leq \frac{11}{12} h^2 ||f^{(4)}||_{\infty}$$ ?

Because I find an other constant.
@LutzL I have found that $$\delta_{h, \text{right}}f(x)=f''(x)+\frac{h^2}{24} (64 f^{(4)}(\xi_2)-5 f^{(4)}(\xi_1)-81 f^{(4)}(\xi_3))$$
LutzL
LutzL
19:53
There it seems you will need to use an interpolation formula to get the remainder term estimate with only one point for the 4th derivative. This should be an actual question in the forum.
evinda
evinda
@LutzL Can I show what I tried?
@LutzL Is it possible that the error of a Taylor expansion is negative?
LutzL
LutzL
20:19
Mathematic formulas are very hard to read here and the problem is big enough for a proper question. Perhaps there is a trick that makes this very easy, I don't know it.

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