Editing Polynomial interpolation (section)

=== Proof of Lagrange remainder ===
Set the error term as <math display="inline"> R_n(x) = f(x) - p_n(x) </math>, and define an auxiliary function:<math display="block"> Y(t) = R_n(t) - \frac{R_n(x)}{W(x)} W(t) \qquad\text{where}\qquad 
W(t) = \prod_{i=0}^n (t-x_i). </math>Thus:<math display="block"> Y^{(n+1)}(t) = R_n^{(n+1)}(t) - \frac{R_n(x)}{W(x)} \ (n+1)! </math>

But since <math>p_n(x)</math> is a polynomial of degree at most {{mvar|n}}, we have <math display="inline"> R_n^{(n+1)}(t) = f^{(n+1)}(t) </math>, and: 
<math display="block"> Y^{(n+1)}(t) = f^{(n+1)}(t) - \frac{R_n(x)}{W(x)} \ (n+1)! </math>

Now, since {{math|''x<sub>i</sub>''}} are roots of <math>R_n(t)</math> and <math>W(t)</math>, we have <math> Y(x)=Y(x_j)=0 </math>, which means {{mvar|Y}} has at least {{math|''n'' + 2}} roots. From [[Rolle's theorem]], <math>Y^\prime(t)</math> has at least {{math|''n'' + 1}} roots, and iteratively <math>Y^{(n+1)}(t)</math> has at least one root {{mvar|ξ}} in the interval {{mvar|I}}. Thus:
<math display="block"> Y^{(n+1)}(\xi) = f^{(n+1)}(\xi) - \frac{R_n(x)}{W(x)} \ (n+1)! = 0 </math>

and:
<math display="block"> R_n(x) = f(x) - p_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} \prod_{i=0}^n (x-x_i) .</math>

This parallels the reasoning behind the Lagrange remainder term in the [[Taylor's theorem|Taylor theorem]]; in fact, the Taylor remainder is a special case of interpolation error when all interpolation nodes {{math|''x<sub>i</sub>''}} are identical.<ref>{{cite web| url=http://www.math.okstate.edu/~binegar/4513-F98/4513-l16.pdf| title=Errors in Polynomial Interpolation}}</ref> Note that the error will be zero when <math>x = x_i</math> for any ''i''. Thus, the maximum error will occur at some point in the interval between two successive nodes.