Let $f:[a,b]\rightarrow R$ be a function that is not $C^{(n+1)}$ on $[a,b]$ but its $n$-th derivative is a Lipschitz function? How does the Lagrange's interpolating polynomial formula change? How does the error approximation change? How can I use it in order to find Simpson's and Newton's formulas of quadrature? I have searched a lot but I could not find a proper demonstration in any numerical analysis book. Please help!
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1$\begingroup$ Why should the Lagrange formula change at all? It does not use the function, just its values in the nodes. $\endgroup$– Federico PoloniCommented Apr 1, 2023 at 18:50
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$\begingroup$ Have you tried examining what happens in low-degree cases such as $n=1$ or $n=0$? $\endgroup$– Federico PoloniCommented Apr 1, 2023 at 18:54
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1$\begingroup$ @FedericoPoloni The usual remainder term is traditionally written with $f^{(n+1)}(\xi)$, which now is incorrect and should be replaced by $F$ where $F$ is some number between the $\sup$ and $\inf$ of the $n+1$-st derivative (existing a.e.) or just between $-L$ and $L$ where $L$ is the LIpschitz constant, if that is all that is known. The other changes are similar and the best way to see what should be done is not to search the literature, but to go over proofs and figure things out from the first principles. This advice applies to this situation and far beyond. Voting to close. $\endgroup$– fedjaCommented Apr 1, 2023 at 19:08
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$\begingroup$ @fedja I agree with you that the remainder formula changes, I was just pointing out that the formula for the interpolating polynomial doesn't. $\endgroup$– Federico PoloniCommented Apr 1, 2023 at 20:03
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$\begingroup$ @FedericoPoloni I have no doubts that this was what you had in mind, but it was rather clear that it was not what the OP meant by "Lagrange formula" :lol: $\endgroup$– fedjaCommented Apr 1, 2023 at 20:32
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