Prove any polynomial of degree n that is orthogonal to ${1, x, ..., x^{n-1}}$ is a constant multiple of a Legendre Polynomial.

Question

The Legendre Polynomials are defined by $L_n(x) = \frac{d^n}{dx^{n}} (x^2 - 1)^n$. The inner product in this case is defined on $[-1, 1]$ as follows: $\langle f(x), g(x)\rangle = \int_{-1}^{1} f(x)g(x)dx$.

I'll denote the arbitrary polynomial of degree n by $P_n(x)$. Since it is orthogonal to $\{1, x, ..., x^{n-1}\}$, then it is orthogonal to $g_{n-1}(x) = a_0 + a_1 x + ... + a_{n-1}x^{n-1}$. Therefore $\langle P_n(x), g_{n-1}(x)\rangle = \int_{-1}^{1} P_n(x)g_{n-1}(x)dx = 0. $

If I expand $g_{n-1}(x)$, I have $n$ finite integrals, whose sum is 0. But how do I show that this implies $P_n(x) = k L_n(x)$ for some constant $k$? Am I taking the right approach?

Answer 1

The Legendre polynomials $L_0(x), \ldots, L_n(x)$. form a basis for the vector space of polynomials of degree $\leq n$. Hence any polynomial $p(x)$ of degree $n$ can be written uniquely as a sum $p(x) = \sum_{k=0}^n a_k L_k(x)$. By your argument, $p(x)$ is orthogonal to each $L_i(x)$ for $i < n$, so $$0 = \int_0^1 L_i(x) p(x) \, dx = \int_0^1 L_i(x) \sum_{k=0}^na_kL_k(x)\, dx = a_i \int_0^1 L_i(x) L_i(x) \, dx$$ which gives $a_i = 0$. So $p(x) = a_n L_n(x)$.

Edit: To see that the Legendre polynomials form a basis of the vector space of polynomials, note that $L_n(x)$ has degree $n$. In fact any sequence of polynomials $p_n(x)$ so that $p_n(x)$ has degree $n$ will form a basis. The reason is that the transition matrix from the standard basis $x^n$ to the $p_n(x)$ is lower triangular:

$$ \left(\begin{matrix} p_0(x)\\ p_1(x) \\ \vdots \\ p_n(x) \end{matrix}\right)= \left(\begin{matrix} a_{11} & 0 & \ldots & 0\\ a_{21} & a_{22} & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} &\ldots & a_{nn} \end{matrix}\right)\left(\begin{matrix} 1\\ x \\ \vdots \\ x^n \end{matrix}\right) $$

The diagonal entries are nonzero by definition since $p_k(x)$ has degree $k$, so the matrix has determinant $a_{11} a_{22} \cdots a_{nn} \neq 0$ and is nonsingular.

Answer 2

A simple answer.

Suppose there exsit a n-degree polynomial $P_n(x)=\sum_{i=0}^na_ix^{n-i}$ satisfy

$P_n(x)-cL_n(x)≠0$ for any c in R, which $L_n(x)$ is n-degree Legendre polynomial. We note $a_0^{L_n}≠0$ as the coefficient of $x^n$ in $L_n(x)$.

and $\int_{-1}^{1}P_n(x)x^mdx=0$, where $m=0,1,2,...,n-1$.

For $a_0≠0$, there must be a non-zero real number $c=a_0/a_0^{L_n}$, thus $Q(x)=P_n(x)-cL_n(x)≠0$ is a n-1 degree polynomial also satisfy

$\int_{-1}^{1}Q(x)x^mdx=0$, where $m=0,1,2,...,n-1$.

Because Q(x) can be expressed in $1,x,x^2,...,x^{n-1}$, So

$\int_{-1}^{1}Q^2(x)dx=0$.

But Q(x) is a continuous function！Q(x) must be zero everywhere in [-1,1], leading to contradiction!

So $P_n(x)-cL_n(x)=0$ for some c.

Q.E.D