Editing Polynomial interpolation (section)

===First proof===
Consider the [[Lagrange polynomials|Lagrange basis functions]] <math>L_0(x),\ldots,L_n(x)</math> given by:
<math display="block">L_j(x)=\prod_{i\neq j}\frac{x-x_i}{x_j-x_i} 
= \frac{(x-x_0)\cdots(x-x_{j-1})(x-x_{j+1})\cdots(x-x_n)}
{(x_j-x_0)\cdots(x_j-x_{j-1})(x_j-x_{j+1})\cdots(x_j-x_n)}.</math>

Notice that <math>L_j(x)</math> is a polynomial of degree <math>n</math>, and we have <math>L_j(x_k)=0</math> for each <math>j\neq k</math>, while <math>L_k(x_k)=1</math>. It follows that the linear combination:
<math display="block">p(x) = \sum_{j=0}^n y_j L_j(x)</math>
has <math>p(x_k)=\sum_j y_j \,L_j(x_k) = y_k </math>, so <math>p(x)</math> is an interpolating polynomial of degree <math>n</math>.

To prove uniqueness, assume that there exists another interpolating polynomial <math>q(x)</math> of degree at most <math>n</math>, so that  <math>p(x_k)=q(x_k)</math> for all <math>k=0,\dotsc,n</math>. Then <math>p(x)-q(x)</math> is a polynomial of degree at most <math>n</math> which has <math>n+1</math> distinct zeros (the <math>x_k</math>). But a non-zero polynomial of degree at most <math>n</math> can have at most <math>n</math> zeros,{{efn|This follows from the [[Factor theorem]] for polynomial division.}} so <math>p(x)-q(x)</math> must be the zero polynomial, i.e. <math>p(x)=q(x)</math>.<ref name="Epperson 2013">{{Cite book |last=Epperson |first=James F. |title=An introduction to numerical methods and analysis |date=2013 |publisher=Wiley |isbn=978-1-118-36759-9 |edition=2nd |location=Hoboken, NJ}}</ref>