Given the Probabilists' Hermite differential equation: $$U''-xU'+\lambda U=0\tag{1}$$
A book question asks me to:
Find the first $4$ polynomial solutions (for $\lambda = 0,1,2,3$), each normalised such that the highest power of $x$ has a coefficient of unity.
So I substituted $$U=\sum_{n=0}^{\lambda}C_nx^n\tag{2}$$ into $(1)$ and equated coefficients of $x^n$ to get the recursion relation:
$$C_{n+2}=\frac{(n-\lambda)}{(n+1)(n+2)}C_n$$
So I am looking to find the values of the coefficients $C_0,C_1,C_2$ for $$U=\sum_{n=0}^{3}C_nx^n=C_0+C_1x+C_2x^2+x^3$$
where we are given that $\color{blue}{C_3}=1$.
Using the recursion relation I find that for $(n=0, \lambda=2)$: $$C_2=-C_0$$
and for $(n=1, \lambda=3)$: $$\color{blue}{C_3}=-\frac13C_1\implies C_1=-3$$
This is as far as I can get to in this question.
The book answer simply states:
Using the recurrence relations, and setting the coefficient of the
highest power to $1$, $U_0=1, \,U_1=x,\, U_2=x^2-1\,$ and $U_3=x^3-3x$
I know from this page on Wikipedia that the book answer is right, but the answer is not very helpful to me as I have no idea why $U_0=1$. Does this mean that $C_0=1$, if so how did the author deduce this from the recursion relation? I showed that $C_1$ is equal to $-3$. So why is $U_1\ne-3x$? The same misunderstanding follows for $U_2$ and $U_3$. In fact I don't understand why $U_3$ doesn't contain an $x^2$ term and a constant term. Also, where is the $x$ term in $U_2$ and the constant term in $U_1$?
I have only just started reading about Hermite polynomials so my understanding is very weak (apologies).
Is there any chance someone could explain as simply as possible how to obtain $U_0,U_1,U_2,U_3$?
Thank you.
Best Answer
You should add a second index to the recurrence formula in order to avoid confusion. $$C_{n+2}^{(\lambda)}=\frac{(n-\lambda)}{(n+1)(n+2)}C_n^{(\lambda)}$$ With this, it is an immediate consequence that in even indexed polynomial $U_0, U_2,\dots$ the odd coefficients $C_{1}^{(\lambda)}, C_{3}^{(\lambda)}$ are zero and analogous for the odd indexed polynomials. So you have the trivial cases
$$U_0(x)=1$$ $$U_1(x)=x$$
For $U_2(x) = C_{0}^{(2)} + x^2$ you get $$1=C_{2}^{(2)} = \frac{(0-2)}{(0+1)(0+2)}C_{0}^{(2)}$$ i.e. $\color{red}{C_{0}^{(2)}}=-1$ and $U_2(x)= x^2-1$.
For $U_3(x) = C_{1}^{(2)}x + x^3$ you get $$1=\color{red}{C_{3}^{(3)}} = \frac{(1-3)}{(1+1)(1+2)}C_{1}^{(3)}=-\frac{1}{3}C_{1}^{(3)}$$
i.e. $\color{red}{C_{1}^{(3)}}=-3$ and $U_2(x)= x^3-3x$.