Question

Show that P1 (x) is orthogonal to [Pn (x)] ^2
on the interval (1,-1)

Answer 1

We need to prove that $P_1(x)$ is orthogonal to $(P_n(x))^2$ on the interval $(-1,1)$ $\Leftrightarrow$ $\int\limits_{-1}^{1}P_1(x)(P_n(x))^2dx=0$

Proof:

$P_1(x)=x$ is an odd function on $(-1,1)$ .

One of the properties of Legendre polynomials is

$P_n(-x)=(-1)^nP_n(x)$

Using this property, we have

$(P_n(-x))^2=((-1)^nP_n(x))^2=(-1)^{2n}(P_n(x))^2=(P_n(x))^2$

So, $(P_n(x))^2$ is an even function on $(-1,1)$

Therefore, $P_1(x)(P_n(x))^2$ is an odd function on $(-1,1)$ .

We can use the fact that

$f(x)$ is odd $\Rightarrow \int\limits_{-a}^{a}f(x)dx=0$

for function $P_1(x)(P_n(x))^2$ .

$P_1(x)(P_n(x))^2$ is odd $\Rightarrow \int\limits_{-1}^{1} P_1(x)(P_n(x))^2 dx=0$ .