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I have this question that I took a shot at but I am not very familiar with Hermite or Laguerre, this my first time running across these type of polynomials and need some help please.

(a) The first four Hermite polynomials are $1, 2t,-2+4t^2,$ and $-12t+8t^3$. These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis of $\mathbb{P}_3$.

(b) The first four Laguerre polynomials are $1, 1-t, 2-4t + t^2,$ and $6-18t + 9t^2- t^3$. Show that these four Laguerre polynomials form a basis of $\mathbb{P}_3$.


(a) The first four Hermite polynomials will be shown to form a basis of $\mathbb{P}_3$ by showing that they are linearly independent and that the number of polynomials equals the dimension of $\mathbb{P}_3$. Consider the following linear combination of the four Hermite polynomials:

$x(1) + y(2t) + z(-2+4t^2) + w(-12t + 8t^3) = at^3 + bt^2 + ct + d$

The first four Hermite polynomials will be shown to be linearly independent by showing that the only linear combination of them that produces the zero polynomial ($0t^3+0t^2+0t+0$) is the trivial combination of zero times each polynomial.

That is all I could come about thus far with it. Can anyone refine or correct this if this is the wrong approach to the problem.

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2 Answers 2

Forget all this Hermite and Laguerre stuff. The fact is that any family of polynomials with all different degrees is linearly free. Hence any family of polynomials with degrees $0$, $1$, $\ldots$, $n$ is a basis of the vector space of polynomials of degree at most $n$ (the space you denote by $\mathbb{P}_n$).

A relatively more sophisticated way of saying the same thing is that any triangular matrix with no zero on its diagonal is invertible.

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The only other thing to note, that Hermite $H_i$ and Laguerre $L_i$ polynomials are orthogonal. In a sense that, $\int H_i(x) H_j(x) w(x) dx = 0$ when $j \neq i$, where $w(x) = e^{x^2}$, $\int L_i(x) L_j(x) v(x) dx = 0$ when $j \neq i$, where $v(x) = e^{-x}.$

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