8

This is from here

$$\int^1_{-1}f_n(x)P_n(x)dx = 2(-1)^n\frac{a_n}{2^n}\int^1_0(x^2-1)^ndx=2(-1)^n\frac{a_n}{2^n}.I_n$$........(6)

I don't understand as in shouldnt it be like this, $$\int^1_{-1}f_n(x)P_n(x)dx = (-1)^n\frac{a_n}{2^n}\int^1_{-1}(x^2-1)^ndx=0$$ as they should cancel out even if the integral is non-zero.

Edited: Lastly, how does $\int^1_{-1}f_n(x)P_n(x)dx$ shows orthogonality?

newbie125
  • 529
  • Here $f_n$ is an arbitrary polynomial of degree $\le n$. If $f_n$ is of lower degree then the coefficient $a_n$ is $0$. In particular $\int P_m(x) P_n(x), dx = 0$ for any $m<n$. By symmetry this is also true when $m>n$, i.e. for $m\ne n$. – Erick Wong Nov 08 '16 at 03:42

3 Answers3

18

Here is a much simpler proof from Special Functions and Their Applications by N. N. Lebedev.

We begin with Legendre's differential equation \begin{equation} [(1-x^{2})P^{\prime}_{n}(x)]^{\prime} +n(n+1)P_{n}(x) = 0,\quad n \in \mathbb{Z}_{0}^{+} \label{eq:lp1} \tag{1} \end{equation}

The first step is to multiple equation \eqref{eq:lp1} by $P_{m}(x)$ and subtract it from equation \eqref{eq:lp1} written for $m$ and multiplied by $P_{n}(x)$.

\begin{equation} [(1-x^{2})P^{\prime}_{m}(x)]^{\prime}P_{n}(x) \,-\, [(1-x^{2})P^{\prime}_{n}(x)]^{\prime}P_{m}(x) + [m(m+1)-n(n+1)]P_{m}(x)P_{n}(x) = 0 \end{equation}

Rearrangement yields \begin{equation} \{(1-x^{2})[P^{\prime}_{m}(x)P_{n}(x)-P^{\prime}_{n}(x)P_{m}(x)]\}^{\prime} + (m-n)(m+n+1)P_{m}(x)P_{n}(x) = 0 \label{eq:lp2} \tag{2} \end{equation}

Integrating equation \eqref{eq:lp2} from -1 to 1, the first term goes to 0 and we are left with \begin{equation} (m-n)(m+n+1) \int\limits_{-1}^{1} P_{m}(x)P_{n}(x) dx = 0 \end{equation} or \begin{equation} \int\limits_{-1}^{1} P_{m}(x)P_{n}(x) dx = 0, \quad m \ne n \end{equation}

poweierstrass
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  • 2
    In eqn (2) how did you get the $(1-x^2)$ factor out of the derivative? –  Feb 07 '19 at 18:02
1

This is like showing that the eigenvectors of a self adjoint operator corresponding to different eigenvalues are orthogonal.

Say we have ( Sturm-Liouville) the operator

$$\phi\overset{S}{ \mapsto }(\rho \phi')'$$

and we want to show that the operator is symmetric. We calculate

$$\langle S\phi, \psi \rangle =\int_a^b(\rho \phi')' \cdot \psi = \rho \phi' \psi |_a^b - \int_a^b \rho \phi'\psi' \\ \langle S \psi , \phi\rangle=\int_a^b(\rho \psi')' \cdot \phi = \rho \psi' \phi |_a^b - \int_a^b \rho \psi'\phi'$$

Now assume moreover that $\rho(a) = \rho(b) = 0$. Then in both RHS the first term is $0$, and we see that $\langle S\phi, \psi \rangle = \langle S \psi , \phi\rangle$, and the rest is standard

orangeskid
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-2

By parity, for any even integrable function $f$, we have $$ \int_{-a}^af(x)dx=2\int_{0}^af(x)dx $$ giving here $$ \int^1_{-1}(x^2-1)^ndx=2\int^1_{0}(x^2-1)^ndx. $$

Olivier Oloa
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