College Math Teaching

October 3, 2016

Lagrange Polynomials and Linear Algebra

Filed under: algebra, linear albegra — Tags: — collegemathteaching @ 9:24 pm

We are discussing abstract vector spaces in linear algebra class. So, I decided to do an application.

Let P_n denote the polynomials of degree n or less; the coefficients will be real numbers. Clearly P_n is n+1 dimensional and \{1, x, x^2, ...x^n \} constitutes a basis.

Now there are many reasons why we might want to find a degree n polynomial that takes on certain values for certain values of x . So, choose x_0, x_1, x_2, ..., x_{n-1} . So, let’s construct an alternate basis as follows: L_0 = \frac{(x-x_1)(x-x_2)(x-x_3)..(x-x_{n})}{(x_0 - x_1)(x_0-x-x_2)..(x_0 - x_{n})}, L_1 = \frac{(x-x_0)(x-x_2)(x-x_3)..(x-x_{n})}{(x_1 - x_0)(x_1-x-x_2)..(x_1 - x_{n})}, ...L_k = \frac{(x-x_0)(x-x_1)(x-x_2)..(x-x_{k-1})(x-x_{k+1})...(x-x_{n})}{(x_k - x_1)(x_k-x-x_2)..(x_k - x_{k-1})(x_k - x_{k+1})...(x_k - x_{n})}. ....L_{n} = \frac{(x-x_0)(x-x_1)(x-x_2)..(x-x_{n-1})}{(x_{n}- x_1)(x_{n}-x-x_2)..(x_{n} - x_{n})}

This is a blizzard of subscripts but the idea is pretty simple. Note that L_k(x_k) = 1 and L_k(x_j) = 0 if j \neq k .

But let’s look at a simple example: suppose we want to form a new basis for P_2 and we are interested in fixing x values of -1, 0, 1 .

So L_0 = \frac{(x)(x-1)}{(-1-0)(-1-1)} = \frac{(x)(x-1)}{2}, L_1 = \frac{(x+1)(x-1)}{(0+1)(0-1)} = -(x+1)(x-1),
L_2 = \frac{(x+1)x}{(1+1)(1-0)} = \frac{(x+1)(x)}{2} . Then we note that

L_0(-1) = 1, L_0(0) =0, L_0(1) =0, L_1(-1)=0, L_1(0) = 1, L_1(1) = 0, L_2(-1)=0, L_2(0) =0, L_2(1)=1

Now, we claim that the L_k are linearly independent. This is why:

Suppose a_0 L_0 + a_1 L_1 + ....a_n L_n =0 as a vector. We can now solve for the a_i Substitute x_i into the right hand side of the equation to get a_iL_i(x_i) = 0 (note: L_k(x_i) = 0 for i \neq k ). So L_0, L_1, ...L_n are n+1 linearly independent vectors in P_n and therefore constitute a basis.

Example: suppose we want to have a degree two polynomial p(x) where p(-1) =5, p(0) =3, p(1) = 17. . We use our new basis to obtain:

p(x) = 5L_0(x) + 3 L_1(x) + 17L_2(x) = \frac{5}{2}(x)(x-1)  -3(x+1)(x-1) + \frac{17}{2}x(x+1) . It is easy to check that p(-1) = 5, p(0) =3, p(1) = 17


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