# Legendre polynomials n=1 calculation by hand

Considering the Legendre polynomials: $$P_n(x) = \sum_{m=0}^{n}a_{n,m}{x^m}$$ I know that: $$P_0=1$$ and $$P_1=x$$.

However given $$P_0$$ if I want to find $$P_1$$ by hand: $$\langle P_1 | P_0 \rangle = 0 = \int_{-1}^{1}(a_0+a_1x)dx = 2a_0 = 0 \longrightarrow a_0=0$$ Imposing normalization condition: $$\langle P_1 | P_1 \rangle = 1 = \int_{-1}^{1}a_1^2x^2 dx = \frac{2}{3}a_1^2 = 1 \longrightarrow a_1= \sqrt{ \frac{3}{2}}$$ So from my computations $$P_1= \sqrt{ \frac{3}{2}} x$$.

Where am I wrong?

• The normalization condition for Legendre polynomials $P_n(z)$ is $\int_{-1}^1 P_n(x) P_m(x) dx = \frac{2}{2n+1} \delta_{n,m}$. i.e $P_n(z)$ are not orthonormalized, they are only orthogonal. – achille hui Nov 18 '20 at 9:08

$$P_1' = x$$ is not normalized since
$$\langle P_1' | P_1' \rangle = \int_{-1}^{1} x^2 dx = \dfrac{2}{3}$$
$$P_1 = \sqrt{\dfrac{3}{2}}x$$ is normalized.
$$\langle P_1 | P_1 \rangle = 1$$