## Linear Independence of Lagrange Polynomials

Lagrange polynomials are polynomials that interpolate the values of a function at certain points. We can sum Lagrange polynomials to interpolate the values of a function at several different points and approximate the function over an interval. Suppose the function
$f(x)$
satisfies
$f(0)=a, \: f(1)=b, \: f(2)=c$

The first of these is satisfied by the Lagrange polynomial
$f|_1(x)=\frac{x-1)(x-2)}{(0-1)(0-2)} = \frac{x^2-3x+2}{2}a$

The second of these is satisfied by the Lagrange polynomial
$f|_2(x)=\frac{x-0)(x-2)}{(1-0)(1-2)} = (-x^2+2x)b$

The third of these is satisfied by the Lagrange polynomial
$f|_3(x)=\frac{x-0)(x-1)}{(2-0)(2-1)}c = \frac{x^2-x}{2}c$

Adding these gives the approximation over the interval
$[0,2]$
:
\begin{aligned} f(x) & \simeq f_1(x)a+f_2(x)b+f_3(x)c \\ &=\frac{x^2-3x+2}{2}a + (-x^2+2x)b +\frac{x^2-x}{2}c \\ &=x^2(a/2-b+c/2)+x(-3a/2+2b-c/2)+a \end{aligned}

This expression is only equal to zero if each coefficient is zero i.e.
$a/2-b+c/2=-3a/2+2b-c/2=a=0$

Since
$a=0$
,
$-b+c/2=2b-c/2=0$

The only possibility is that
$b=c=0$
.
Then these Lagrange polynomials are linearly independent.