Question

The first four laguerre polynomials are 1, 1 - t, 2 - 4t t2, and 6 - l 8t 9t2 - t 3 . show that these polynomials form a basis of lp'3?

Answer 1

The polynomials 1, 1-t, $2-4t+t^2$ and $6-18t+9t^2-t^3$ are linear independent and span $P_3$. So the first four Laguerre polynomials forms a basis of $P_3$.

We have the first four Laguerre polynomials. Here $P_3$ is the space of all polynomials of degree at most 3.

i.e., $P_3= \{ a+bx+cx^2+dx^3| a,b,c,d \in R\}$

So the dimension of $P_3$ is 4. Also there are 4 Laguerre polynomials.

Now to show that the first four Laguerre polynomials forms a basis, it remains to show that they are linear independent.

Consider the polynomials 1, 1-t, $2-4t+t^2$ and $6-18t+9t^2-t^3$. Write them into a matrix with the co-efficient of 1, t, $t^2$ and $t^3$ in each row.

$\left[\begin{array}{cccc}1&1&2&6\\0&-1&-4&-18\\0&0&1&9\\0&0&0&-1\end{array}\right]$

This matrix is already in its echelon form with all its pivots occupied. The pivot of first row is 1. The pivot of 2nd row is -1. The pivot of 3rd row is 1 and the pivot of 4th row is -1.

Hence the polynomials are linearly independent since the columns of the above matrix are linearly independent.

In conclusion, the first four Laguerre polynomials 1, 1-t, $2-4t+t^2$ and $6-18t+9t^2-t^3$  forms a basis for $P_3$.

Learn more about basis at brainly.com/question/13258990

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