Lagrange interpolation formula

Let $(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{n},y_{n})$ be $n$ points in the plane ($x_{i}\neq x_{j}$ for $i\neq j$). Then there exists a unique polynomial $p(x)$ of degree at most $n-1$ such that $y_{i}=p(x_{i})$ for $i=1,\ldots,n$.

Such polynomial can be found using Lagrange’s interpolation formula:

 $p(x)=\frac{f(x)}{(x-x_{1})f^{\prime}(x_{1})}y_{1}+\frac{f(x)}{(x-x_{2})f^{% \prime}(x_{2})}y_{2}+\cdots+\frac{f(x)}{(x-x_{n})f^{\prime}(x_{n})}y_{n}$

where $f(x)=(x-x_{1})(x-x_{2})\cdots(x-x_{n})$.

To see this, notice that the above formula is the same as

 $\displaystyle p(x)$ $\displaystyle=y_{1}\frac{(x-x_{2})(x-x_{3})\dots(x-x_{n})}{(x_{1}-x_{2})(x_{1}% -x_{3})\dots(x_{1}-x_{n})}+y_{2}\frac{(x-x_{1})(x-x_{3})\dots(x-x_{n})}{(x_{2}% -x_{1})(x_{2}-x_{3})\dots(x_{2}-x_{n})}$ $\displaystyle\phantom{=}\qquad+\dots+y_{n}\frac{(x-x_{1})(x-x_{2})\dots(x-x_{n% -1})}{(x_{n}-x_{1})(x_{n}-x_{2})\dots(x_{n}-x_{n-1})}$

and that for all $x_{i}$, every numerator except one vanishes, and this numerator will be identical to the denominator, making the overall quotient equal to 1. Therefore, each $p(x_{i})$ equals $y_{i}$.

Title Lagrange interpolation formula LagrangeInterpolationFormula 2013-03-22 11:46:21 2013-03-22 11:46:21 drini (3) drini (3) 16 drini (3) Theorem msc 65D05 msc 41A05 Lagrange’s Interpolation formula SimpsonsRule LectureNotesOnPolynomialInterpolation Lagrange polynomial