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Let's say we start with a polynomial like

$$ f(x) = a_{1} x^n + a_{2} x^{n-1} + \cdots + a_{n} $$

then we take $2n$ points over this function, and we try to find the polynomial interpolation, using those $2n$ points (and so we will try to find a polynomial of grade $2n-1$).

So now my question is, is this new polynomial equal to the first one?

My question comes from the statement "the polynomial interpolation will find the best polynomial that fits those points", but i can't understand how it will find a better polynomial then the original one, using $2n-1$ monomial (maybe it will assign the others coefficient, from $a_{n+1}$ to $a_{2n}$?)

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  • $\begingroup$ Please use MathJax to typeset math formulas. Also: “grade” should be “degree”. $\endgroup$ Jun 16, 2020 at 17:20
  • $\begingroup$ @ArturoMagidin sorry i'm new, i'll check it out now $\endgroup$ Jun 16, 2020 at 17:20
  • $\begingroup$ Welcome, and thank you. $\endgroup$ Jun 16, 2020 at 17:21
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    $\begingroup$ @ArturoMagidin better? $\endgroup$ Jun 16, 2020 at 17:23
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    $\begingroup$ @Berto99 Yes. If you pick $2n$ different points from $f(x)$ then there is only one $2n-1$ degree interpolation polynomial namely $f(x)$. $\endgroup$ Jun 16, 2020 at 17:27

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Given $k+1$ distinct points $x_0,x_1,\ldots,x_k$ and $k+1$ values $b_0,b_1,\ldots,b_k$, there is one, and only one, polynomial $p(x)$, either equal to $0$ or of degree at most $k$ such that $p(x_i) = b_i$ for $i=0,1,\ldots,k$.

The existence can be established using, for example, Lagrange interpolation.

For the uniqueness clause, if $p(x)$ and $q(x)$ are two polynomials, each either the zero polynomials or of degree at most $k$, with $p(x_i)=q(x_i)=b_i$ for each $i$, then $p-q$ is either the zero polynomial or has degree at most $k$ and has at least $k+1$ roots (namely, $x_0,x_1,\ldots,x_n$). Since a nonzero polynomial of degree at most $k$ has at most $k$ roots, it follows that $p-q$ is the zero polynomial, so $p=q$.

That means that if you start with a nonconstant polynomial of degree $n$, and then try to use interpolation to find a polynomial that agrees with the given polynomial at $2n$ points, then (since $2n-1\geq n$ for nonconstant polynomials) the polynomial you get out of the interpolation process is the one you started with, because that one works and has degree at most $2n-1$, and there is at most one that works and has degree at most $2n-1$. So it’s the one you already have.

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  • $\begingroup$ thank you again, really appreciated $\endgroup$ Jun 16, 2020 at 17:35

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