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I am trying to use Newtons algorithm for polynomial interpolation. The original polynomial is $p(x) = 3x^2+4x+7$ and the Points from which I am trying to interpolate the polynomial are $p(1) = 14$, $p(6) = 139$ and $p(7) = 182$.

Now as far as I know the formula for the interpolated polynomial should be $r(x) = a_0+a_1(x-x_0)+a_2(x-x_0)(x-x_1)$.

To find $a_0$ I calculate $y_0 = 14 = a_0$. Then, $y_1 = 139 = a_0 + a_1 (x_1-x_0)=14+a_1(6-1)$ so by solving for $a_1$ the result is $a_1 = 25$. At last, $y_2 = 182 = a_0 + a_1(x_1-x_0)+a_2(x_2-x_0)(x_2-x_1)=14+25(6-1)+a_2(7-1)(7-6)$ and solving for $a_2$ results in $a_2=\frac{43}{6}$.

By inserting the found values into the formula I get $r(x)=14+25(x-1)+\frac{43}{6}(x-1)(x-6)=\frac{43}{6}x^2-\frac{151}{6}x+32$.

This polynomial doesn't go though the last point though: $r(7)=\frac{43}{6}\cdot49-\frac{151}{6}\cdot7+32=207 \ne 182=p(7)$.

Am I doing something wrong or does this usually happen with this algorithm?

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    $\begingroup$ You must have made a calculation mistake since the interpolation of degree $n$ of a polynomial of degree $n$ should always be the original polynomial $\endgroup$ – Maximilian Janisch Aug 26 at 14:37
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At your calculation of $a_2$ you substituted the value of $x_1$ for $x$ instead of $x_2$, which should have caused the error.

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At last, $y_2 = 182 = a_0 + a_1(x_1-x_0)+a_2(x_2-x_0)(x_2-x_1)=14+25(6-1)+a_2(7-1)(7-6)$ and solving for $a_2$ results in $a_2=\frac{43}{6}$.

This should have been: $y_2 = 182 = a_0 + a_1(x_2-x_0)+a_2(x_2-x_0)(x_2-x_1)=14+25(7-1)+a_2(7-1)(7-6)$ and solving for $a_2$ results in $a_2=3$.

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Note that any interpolating polynomial should be exact in the given points. This is not the case here because you made a calculation mistake for $a_2$.

Note that (using notation for divided differences) we have $a_2=[14,139,182]=\frac{182-139}6-\frac{139-14}{30}=3$.

Indeed you get back the original polynomial.

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