Respuesta :
The polynomials 1, 1-t, [tex]2-4t+t^2[/tex] and [tex]6-18t+9t^2-t^3[/tex] are linear independent and span [tex]P_3[/tex]. So the first four Laguerre polynomials forms a basis of [tex]P_3[/tex].
We have the first four Laguerre polynomials. Here [tex]P_3[/tex] is the space of all polynomials of degree at most 3.
i.e., [tex]P_3= \{ a+bx+cx^2+dx^3| a,b,c,d \in R\}[/tex]
So the dimension of [tex]P_3[/tex] 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, [tex]2-4t+t^2[/tex] and [tex]6-18t+9t^2-t^3[/tex]. Write them into a matrix with the co-efficient of 1, t, [tex]t^2[/tex] and [tex]t^3[/tex] in each row.
[tex]\left[\begin{array}{cccc}1&1&2&6\\0&-1&-4&-18\\0&0&1&9\\0&0&0&-1\end{array}\right][/tex]
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, [tex]2-4t+t^2[/tex] and [tex]6-18t+9t^2-t^3[/tex] forms a basis for [tex]P_3[/tex].
Learn more about basis at https://brainly.com/question/13258990
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