I looked up a problem in a book and struggled to follow the answer to it, I annotated the specific parts with (a), (b),…:
Question:
Find Polynomials $\{H_n\}_{n=0}^5$ with $H_n\in P_n$ that is polynomials of degree $n$ in the inner product
$$
(f,g)=\int_{-\infty}^\infty f(x)g(x)e^{-x^2}dx
$$
and remember (and use) the relation:
$$
\int_{-\infty}^\infty x^n e^{-x^2}dx=\frac{n-1}{2}\int_{-\infty}^\infty x^{n-2} e^{-x^2}dx
$$
The polynomial should be normalized so that the leading coefficient is $2^n$.
Answer:
Since the interval $(-\infty, \infty)$ is symmetric (I understand this) and the weight $w(x)=e^{-x^2}$ is even (this too), we know that $H_{2k}(x)$ is even and $H_{2k+1}(x)$ is odd. I do not understand this (a). Why does this hold? This seems to be the most important point.
It goes on to state that therefore (why? (b)) $H_{0}(x)=2^0$ and $H_{1}(x)=2^1 x$. Next we put $H_{2}(x)=2^2(x^2+a)$ (again why? (c)) and determine $a$ by imposing
$$
0=(H_2,H_0)=2^2\int_{-\infty}^\infty(x^2+a)e^{-x^2}dx=2^2(1/2 +a)\int_{-\infty}^\infty e^{-x^2}dx
$$
so we get $a=-1/2$ and $H_{2}(x)=(4x^2+2)$. Why do we compare $(H_2,H_0)$ at this point and say not $(H_2,H_1)$?(d)
I the next step one puts $H_{3}(x)=2^3(x^3+ax)$ (again why?(e)) and by imposing
$$
0=(H_3,H_1)=…
$$
we get again a. Why do we use $(H_3,H_1)$ here and not $(H_3,H_0)$ now for example?
Then again $H_{4}(x)=2^4(x^4+ax^2+b)$ is imposed through conditions $(H_4,H_2)$ and $(H_4,H_0)$. Why do we choose these two again?
I would appreciate your help very much, I really tried my best to get to the bottom and understanding all of it.
Best Answer
Let's skip symmetry and come back to it.
We are required to have $H_0 \in P_0$, the polynomials of degree $0$, so $H_0$ is a constant. We are told that its leading (only!) coefficient is $2^0$. So automatically, we have $$ H_0 = 2^0 = 1 \text{.} $$
$H_1 \in P_1$, so is linear, with leading coefficient $2^1$, so $$ H_1 = 2^1x + a = 2x+a \text{.} $$ We're supposed to be finding the Hermite polynomials, which are an orthogonal set of polynomials. You are given that the Hermite polynomials are a basis under the given inner product and are given their normalization, so we should have $$ (H_0, H_1) = 0 \text{.} $$ If $H_1$ is an odd function, meaning $a = 0$, this is automatic. So suppose that $H_1$ is not odd, that $a \neq 0$. Then \begin{align*} (H_0, H_1) &= \int_{-\infty}^\infty H_0 H_1 \mathrm{e}^{-x^2} \,\mathrm{d}x \\ &= \int_{-\infty}^\infty 1 (2x+a) \mathrm{e}^{-x^2} \,\mathrm{d}x \\ &= \int_{-\infty}^\infty 2x \mathrm{e}^{-x^2} \,\mathrm{d}x + \int_{-\infty}^\infty a \mathrm{e}^{-x^2} \,\mathrm{d}x \\ &= 0 + \int_{-\infty}^\infty a \mathrm{e}^{-x^2} \,\mathrm{d}x \\ &= a \sqrt{\pi} \text{,} \end{align*} where the integral with $2x$ in the integrand vanishes by odd symmetry. For orthogonality to hold, we must have $a = 0$.
Notice what we have said: because of odd symmetry in $H_1$, we obtain orthogonality. Without odd symmetry, we do not obtain an orthogonal polynomial relative to the given inner product.
How about $H_2 \in P_2$? Well, starting with $$ H_2 = 2^2 x^2 + b x + c \text{,} $$ we use the given reduction integral to compute \begin{align*} (H_0, H_2) &= (2+c)\sqrt{\pi} \\ (H_1, H_2) &= b\sqrt{\pi} \text{.} \end{align*} The computation of the latter uses odd symmetry to leave only the $b$ term and orthogonality forces $b = 0$. The former and orthogonality forces $c = -2$. Therefore, $$ H_2 = 2^2 x^2 - 2 $$ (correcting a sign error in the work you show). Notice that if we had been leveraging symmetry, we would have known $b = 0$ and ignored $(H_2, H_1)$.
By now, we have given sufficient evidence that your thought that the symmetry is important is correct, since symmetry explains the reasons for the work addressed in your subsequent questions. So let's try to get a handle on the symmetry.
By symmetry, any odd degree term in $p(x) \in P_n$ vanishes in the integral $$ \int_{-\infty}^\infty p(x) \mathrm{e}^{-x^2} \,\mathrm{d}x \text{,} $$ so only the even degree terms survive. Using the reduction integral, any $p(x) \in P_n$ with $n \geq 2$ is equivalent to some $q(x) \in P_{n-2}$. And iterating the reduction, we can find an equivalent polynomial in $P_0$ or $P_1$, depending on whether the degree or $p(x)$ is even or odd.
Let's pretend we don't know that odd degrees reduce to $P_1$ and even degrees reduce to $P_0$. Instead, we know that any polynomial reduces to an element of $P_1$, but accept that the leading coefficient might be $0$, so we may actually have an element of $P_0$. (This is the sort of immediate conclusion we can obtain from the reduction integral -- any term with degree ${}\geq 2$ can have its degree reduced by $2$, so we end up with an equivalent polynomial having only linear and constant terms, either or both of which could be zero.)
So, if $p(x)$ has positive even degree, we apply the reduction repeatedly until we find a $\hat{p}(x) = a x + b \in P_1$ We have shown $$ (p(x), H_0) = (\hat{p}(x), H_0) = (ax+b,1) \text{.} $$ We know that the odd degree term vanishes in this inner product, so really, we have $(b,1)$. For $p(x)$ to be one of the $H_n$, it is orthogonal to $H_0$, so that inner product is zero, so $b = 0$. Also, $$ (p(x), H_1) = (ax+b, 2x) = ((ax+b)\cdot 2x,1) \text{.} $$ The $2bx$ term contributes zero to the integral, so we find $$ (ax+b, 2x) = (2ax^2,1) \text{.} $$ Then reduction and orthogonality force $a = 0$. Unravelling what we have done, we discover that all the odd degree terms in $p(x)$ have coefficient zero. Therefore, if $p(x)$ is an $H_n$ with $n$ even and greater than $1$, then $p(x)$ has nonzero constant term and zero odd degree coefficients.
A similar argument for $p(x)$ an $H_n$ with odd degree greater than $1$ yields that $p(x)$ is an odd polynomial, so has zero coefficients for terms of even degree. These two arguments together are summarized by the symmetry comments in your recited solution.