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 $y_i$.