The picture contains the formula of the error which occurs when we interpolate a function $f(x)$ to a polynomial $P(x)$, my questions is: what is the difference between this error and the absolute error $|f(x)-P(x)|$? and what does $f\in C^{n}[a,b]$ refers to?
1 Answers
Presumably, it should be $f \in C^{\color{red}{n + 1}}[a, b]$, i.e. $f$ is $n+1$ times continuously differentiable on the closed interval $[a, b]$.
Then, for any $x \in [a, b]$, you have the formula $$ f(x) - P(x) = \frac{f^{(n+1)}(\xi(x))}{(n+1){!}} (x-x_0) \cdots (x-x_n) , $$ where $P$ is the polynomial of degree $n$ interpolating $f$ in the (distinct) nodes $x_0, \ldots, x_n \in [a, b]$, and $\xi : [a, b] \to [a, b]$ is some function...
If you want $|f(x) - P(x)|$, just wrap both sides in absolute value.
There is no difference between the "actual error" and the formula. The formula gives the actual, exact error. But you typically do not know the function $\xi$ to compute it that way. Therefore, this formula is usually employed do obtain an upper bound on that error by estimating $f^{(n+1)}$.
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OK! but what is the difference between the actual error and the error given in that formula? – soso sos Mar 13 '17 at 13:21
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@sososos: what is the "actual error"? – user66081 Mar 13 '17 at 13:24
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It is mentioned here: http://math.stackexchange.com/questions/1151001/lagrange-interpolating-polynomials-error-bound/2184699#2184699 – soso sos Mar 13 '17 at 13:28
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That other question wrote "error bound" when it could have just said "error". In other words, the formula is the actual errror. – David K Mar 13 '17 at 15:07
