A problem related to Legendre polynomial

Question

Let $P_n(x)$ be the Legendre polynomial of degree n such that $P_n(1)=1$, $n=1,2,\dotsc$ if $$\int_{-1}^1\left(\sum_{j=1}^{n}\sqrt {j(2j+1)}p_j(x)\right)^2=20$$

Then what is $n$?

1. 2
2. 3
3. 4
4. 5

I have no idea how to find or which formula is applicable here? Please help.

Thanks.

Answer 1

Let $h:[-1,1]\to\mathbb{R}$ given by $h(x)=\sum_{j=1}^{n}\sqrt {j(2j+1)}p_j(x)=\sum_{j=1}^{n}a_jp_j(x)$ for $a_j=\sqrt{j(2j+1)}$. Notice that you are calculating $\|h\|_2^2$ on $L^2[-1,1]$.

Since Legendre polynomials are orthogonal in this space, we have, by Pythagoras, $$\|h\|_2^2=\sum_{j=1}^{n}\|a_jp_j\|_2^2=\sum_{j=1}^{n}|a_j|^2\dfrac{2}{2j+1}=\sum_{j=1}^{n}j(2j+1)\dfrac{2}{2j+1}=2\sum_{j=1}^nj=n(n+1).$$

So $n=4$