Lagrange interpolation formula

Let $(x_1,y_1),(x_2,y_2),\mathrm{\dots },(x_n,y_n)$ be $n$ points in the plane ($x_i\ne x_j$ for $i\ne j$). Then there exists a unique polynomial $p(x)$ of degree at most $n-1$ such that $y_i=p(x_i)$ for $i=1,\mathrm{\dots },n$.

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

$$p(x)=\frac{f(x)}{(x-x_1)f^{\prime }(x_1)}{y}_1+\frac{f(x)}{(x-x_2)f^{\prime }(x_2)}{y}_2+\mathrm{\cdots }+\frac{f(x)}{(x-x_n)f^{\prime }(x_n)}{y}_n$$

where $f(x)=(x-x_1)(x-x_2)\mathrm{\cdots }(x-x_n)$.

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

$p(x)$ $=y_1{\displaystyle \frac{(x-x_2)(x-x_3)\mathrm{\dots }(x-x_n)}{(x_1-x_2)(x_1-x_3)\mathrm{\dots }(x_1-x_n)}}+y_2{\displaystyle \frac{(x-x_1)(x-x_3)\mathrm{\dots }(x-x_n)}{(x_2-x_1)(x_2-x_3)\mathrm{\dots }(x_2-x_n)}}$

$+\mathrm{\dots }+y_n{\displaystyle \frac{(x-x_1)(x-x_2)\mathrm{\dots }(x-x_{n-1})}{(x_n-x_1)(x_n-x_2)\mathrm{\dots }(x_n-x_{n-1})}}$

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
Canonical name	LagrangeInterpolationFormula
Date of creation	2013-03-22 11:46:21
Last modified on	2013-03-22 11:46:21
Owner	drini (3)
Last modified by	drini (3)
Numerical id	16
Author	drini (3)
Entry type	Theorem
Classification	msc 65D05
Classification	msc 41A05
Synonym	Lagrange’s Interpolation formula
Related topic	SimpsonsRule
Related topic	LectureNotesOnPolynomialInterpolation
Defines	Lagrange polynomial