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I do not understand the part that I have underlined in green. I thought that if you fit a polynomial $P_n$to given data for say $n+1$ distinct points you got an approximation of $f(x)$, where there is an error $f-P_n$.

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usainlightning
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    I think it just says that if the function you're interpolating is a n-degree polynomial, then the approximation $P_n$ with $n+1$ points is actually exactly $f$, on every point. This is because $f-P_n$ is a (at most) n-degree polynomial, that has the $n+1$ interpolation points as roots, but a n-degree not nul polynomial has at most $n$ roots. So $f-P_n \equiv 0$ and $P_n$ is exactly $f$. – yago Apr 19 '14 at 17:43
  • Why are the interpolation points roots? – usainlightning Apr 19 '14 at 18:06
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    Polynomial interpolation problem is for points $x_1,\ldots,x_{n+1}$ is to find a n-degree polynomial $P_n$ such that $\forall i \geq n+1,P_n(x_i)=f(x_i)$, ie $P_n$ and $f$ takes the same values on $x_1,\ldots,x_{n+1}$. On the other hand, if $f$ is a n-degree polynomial, then the difference $f-P_n$ is also a (at most n)-degree polynomial. But by defintion, $\forall x \geq n+1, (f-P_n)(x_i)=f(x_i)-P_n(x_i)=0$ because $P_n(x_i)=f(x_i)$. So $x_1,\ldots,x_{n+1}$ are $n+1$ roots of $P_n-f$, but if its not null, he can have at most n distinct roots. So it is null, and $P_n=f$ – yago Apr 19 '14 at 18:18

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In general fitting a polynomial to points is a pretty bad way to approximate a function unless the function is a polynomial. Usually the Taylor polynomials are much better. The only way the underlined part you mentioned makes sense is if $f$ is a polynomial, otherwise the norm $||f - p||$ will not be zero.

user2566092
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